Answer:
The other student (59kg) moves right at 7.44 m/s
Explanation:
Given;
mass of the first student, m₁ = 77kg
mass of the second student, m₂ = 59kg
initial velocity of the first student, u₁ = 0
initial velocity of the second student, u₂ = 0
final velocity of the first student, v₁ = 5.7 m/s left
final velocity of the second student, v₂ = ? right
Apply the principle of conservation of linear momentum;
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(77 x 0) + (59 x 0) = (-77 x 5.7) + (59 x v₂)
0 = - 438.9 + 59v₂
59v₂ = 438.9
v₂ = 438.9 / 59
v₂ = 7.44 m/s to the right
Therefore, the other student (59kg) moves right at 7.44 m/s
Final answer:
The physics problem involves two students on skateboards applying a force to each other and moving in opposite directions. To find the velocity of the second student, the conservation of momentum is used, considering no external forces act on them.
Explanation:
The subject of this question is physics, specifically relating to the principle of conservation of momentum in a system of two students standing on skateboards. When the two students push off each other, they start moving in opposite directions due to this principle. Conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it. In this scenario, the system comprises the two students on skateboards.
To find how fast the second student moves, one would use the formula:
Total momentum before the shove = Total momentum after the shove
0 = (mass of student 1 \(\times\) velocity of student 1) + (mass of student 2 \(\times\) velocity of student 2)
We have the mass and velocity of student 1 (77 kg, moving left at 5.7 m/s), and the mass of student 2 (59 kg), but we need to calculate the velocity of student 2 after the push to ensure momentum is conserved.
A 10.0-g bullet is fired into a stationary block of wood having mass 5.00 kg. The bullet embeds 10 pts into the block and the speed of the block-and-bullet after the collision is 0.600 m/s. Find a) the original speed of the bullet, b) the mechanical energies of the block-bullet system before and after the collision, c) the percentage of mechanical energy lost to heat.
Answer:
A. The initial velocity of the bullet is [tex]= 300.6m/s[/tex]
B. Mechanical energy of the system before and after collision: 451.80 J, 0.9018 J
C. Percentage of K.E lost to heat is = 99.8 %
Explanation:
From conservation of linear momentum,
[tex](m_{1}v_{1} +m_{2}v_{2})= (m_{1}+m_{2})v[/tex]
let the mass of the block be m1 and velocity = v1
let the mass of the bullet be m2 and velocity = v2
Let the final velocity of the system be v.
A. Plugging our parameters into the equation, we have:
[tex][(5 \times 0) +(0.01\times v_{2})]= 5.01 \times 0.6[/tex]
[tex]v_{2}=\frac{3.006}{0.01}= 300.6m/s[/tex]
Hence, the initial velocity of the bullet is [tex]= 300.6m/s[/tex]
B. The mechanical energies of the system exist in form of kinetic energy.
I. Kinetic energy of the system before collision:
[tex]0.5 \times 5\times 0^{2} + 0.5 \times 0.01 \times 300.6^{2}= 451.80 J[/tex]
II. Kinetic energy after collision:
[tex]0.5\times 5.01 \times 0.6^{2}= 0.9018 J[/tex]
C. Change in Mechanical Energy = [tex]451.8 - 0.9018 J= 450.9J[/tex]
[tex]\frac{450.9}{451.8} \times 100 =99.8%[/tex]
Percentage of K.E lost to heat is = 99.8 %
3. A hydraulic lift is used to jack a 950-kg car 45 cm off the floor. The diameter of the output piston is 23 cm, and the input force is 375 N. a. What is the area of the input piston? (10pts) b. What is the work done in lifting the car 45 cm? (10pts) c. If the input piston moves 15 cm in each stroke, how high does the car move up for each stroke? (10pts) d. How many strokes are required to jack the car up 45 cm? (10pts) e. Show that the energy is conserved. (10pts)
Answer:
a. A input = 0.001669 m²
b. W= 4.193775 KJ
c. h output = 0.60357 cm
d. n = 74.55638 strokes
e. 4.1937 N = 4.1937 N
Explanation:
Hydraulic jacks lift loads using the force created by the pressure in the cylinder chamber by applying small effort. It works on Pascal's principles which explains that the pressure at a certain level and through a mass of fluid at rest is the same in all the directions.
Parameters given:
Diameter of the output piston, d = 0.23 m
Mass of the car, m= 950 kg
Force applied at the input piston, f = 375 N
Height, h = 0.45 m
(a) Finding the area of the input piston:
First, we use Pascal's principle to find the area
(f ÷ A output) ÷ (f ÷ A input)
Where A= area
g = 9.81
A output = πd² ÷ 4
(f ÷ (πd² ÷ 4)) = (f ÷ A input)
[(950 x 9.81) ÷ ((3.14 x 0.23²) ÷ 4) ] = 375 ÷ A input
9319.5 ÷ 0.0415 = 375 ÷ A input
A input = 0.001669 m²
(b) Finding the work done in lifting the car 45 cm
Work done, W = force, f x distance ( which in this case is height, h)
= (950 x 9.81) x 0.45
W= 4.193775 KJ
(c) Finding how the car move up for each stroke if the input piston moves 15 cm in each stroke.
W output = W input
F x h output = F x h input
= (950 x 9.81) x h output = 375 x 0.15
h output = 0.0060357 m
h output = 0.60357 cm
(d) Finding the number of strokes that are required to jack the car up 45 cm
n = h ÷ h output
n = 45 ÷ 0.60357 cm
n = 74.55638 strokes
(e) How the energy is conserved
W output = W input
F x h output = F x h input x n
(950 x 9.81) x 0.45 = 375 x 0.15 x 74.55638
4.1937 N = 4.1937 N
The area of the input piston will be 0.00167 square meters, the work done in lifting the car will be 4193.78 J, the height moved by car in each stroke will be 0.6035 cm, the number of strokes required to lift the car by 45 cm will be 74.57, and the energy conservation is justified.
Given information:
Mass of the car is [tex]m=950[/tex] kg
The lift of the car is [tex]h=45[/tex] cm.
The diameter of the output piston D is 23 cm.
The input force f is 375 N.
Let the output force be F and the input piston diameter be d.
Hydraulic lift follows Pascal's principle.
(a)
So, the area of the input piston will be calculated as,
[tex]\dfrac{F}{\frac{\pi}{4}D^2}=\dfrac{f}{\frac{\pi}{4}d^2}\\\dfrac{\pi}{4}d^2=\dfrac{f\frac{\pi}{4}D^2}{mg}\\\dfrac{\pi}{4}d^2=A=0.00167\rm\;m^2[/tex]
(b)
The work done in lifting the car will be,
[tex]W=mgh\\W=950\times 9.81\times0.45\\W=4193.78\rm\;J[/tex]
(c)
The input piston moves 15 cm in each stroke.
The height h' moved by car in each stroke will be calculated as,
[tex]mgh'=f\times 0.15\\h'=0.006035\rm\;m=0.6035\rm\;cm[/tex]
(d)
The number of strokes required to lift the car by 45 cm will be,
[tex]n=\dfrac{h}{h'}\\n=\dfrac{45}{0.6035}\\n=74.57[/tex]
(e)
The energy is conserved when output work is equal to input work.
So, the energy conservation can be verified as,
[tex]W_i=W_o\\f\times0.15\times n=mgh\\375\times 0.15\times74.57=950\times 9.81\times0.45\\4194.28\approx4194.1[/tex]
The input and output works are approximately equal. The value isn't exactly equal because of the rounding-off.
So, energy conservation is justified.
Therefore, the area of the input piston will be 0.00167 square meters, the work done in lifting the car will be 4193.78 J, the height moved by car in each stroke will be 0.6035 cm, the number of strokes required to lift the car by 45 cm will be 74.57, and the energy conservation is justified.
For more details, refer to the link:
https://brainly.com/question/2590504
When a conducting plate swings through a magnetic field, which is true?
a. Energy is transferred from mechanical energy to thermal energy.
b. Energy is transferred from mechanical energy to the magnetic field.
c. Energy is transferred from the magnetic field to thermal energy.
d. Energy is transferred from mechanical energy to thermal energy and then back to mechanical energy.
e. Energy is transferred from the magnetic field to mechanical energy.
Answer:
Energy is transferred from mechanical energy to thermal energy and then back to mechanical energy ( D )
Explanation:
when conducting plate swings through a magnetic field, energy is transferred from mechanical energy to thermal energy and then back to mechanical energy and this is due to the induction of eddy currents during the process of that the conducting plate swings through the magnetic field.
An eddy current is induced in conductor when the magnetic field in which the conductor is located varies.
Answer:
Energy is transferred from mechanical energy to thermal energy. (A)
Explanation:
As the conducting plate is swinging, we know it posseses some form of mechanical energy (either K or U). Because the plate is conducting, charge is transfered, yielding energy transfer to thermal.
Power of sunlight on Earth The Sun emits about 3.9 * 1026 J of electromagnetic radiation each second. (a) Estimate the power that each square meter of the Sun’s surface radiates. (b) Estimate the power that 1 m2 of Earth’s surface receives. (c) What assumptions did you make in part (b)? The distance from Earth to the Sun is about 1.5 * 1011 m and the diameter of the Sun is about 1.4 * 109 m
Answer:
a)6.34 x [tex]10^{7}[/tex]W/m²
b)1.37 x [tex]10^{3[/tex] W/m²
c) see explanation.
Explanation:
a)The relation of intensity'I' of the radiation and area 'A' is given by:
I= P/A
where P= power of sunlight i.e 3.9 x [tex]10^{26}[/tex] J
and the area of the sun is given by,
A= 4π[tex]R_{sun}[/tex] => 4π[tex](\frac{1.4*10^{9} }{2} )^{2}[/tex]
A=6.15 x [tex]10^{18}[/tex]m²
[tex]I_{sun}[/tex] = 3.9 x [tex]10^{26}[/tex] / 6.15 x [tex]10^{18}[/tex] => 6.34 x [tex]10^{7}[/tex]W/m²
b) First determine the are of the sphere in order to determine intensity at the surface of the virtual space
A= 4π[tex]R[/tex]
Now R= 1.5 x [tex]10^{11[/tex]m
A= 4π x 1.5 x [tex]10^{11[/tex] =>2.83 x [tex]10^{23[/tex] m²
The power that each square meter of Earths surface receives
[tex]I_{earth[/tex] = 3.9 x [tex]10^{26}[/tex]/2.83 x [tex]10^{23[/tex] =>1.37 x [tex]10^{3[/tex] W/m²
c) in part (b), by assuming the shape of the wavefront of the light emitted by the Sun is a spherical shape so each point has the same distance from the source i.e sun on the wavefront.
The power per square meter reaching Earth's surface from the Sun can be estimated by considering the power output of the Sun and the distance between the Sun and Earth. The maximum power that reaches Earth's surface is approximately 1.30 kW/m². The assumptions made in this estimation include the perfect absorption of sunlight by the Earth's surface and the average radius of the Earth.
Explanation:The power per square meter reaching Earth's surface from the Sun can be estimated by considering the power output of the Sun and the distance between the Sun and Earth. The power output of the Sun is given as 4.00 × 10^26 W. To estimate the power reaching Earth's surface, we need to take into account that part of the Sun's radiation is absorbed and reflected by the atmosphere. The maximum power that reaches Earth's surface is approximately 1.30 kW/m².
To estimate the power received by 1 m² of Earth's surface, we can use the fact that the area of the Earth facing the Sun is πR², where R is the radius of the Earth. Assuming the Earth is a perfect sphere, we can take the average radius of the Earth as Re ≈ 6,371 km. Therefore, the power received by 1 m² of Earth's surface can be calculated as (1.30 kW/m²) × (πR²) / (4πRe²), where Re is the radius of the Earth.
In part (b), the assumptions made include the perfect absorption of sunlight by the Earth's surface, the average radius of the Earth, and the assumption that the Earth is a perfect sphere.
You throw a small toy ball that is covered with suction cups at the center of your glass patio door. When it hits it sticks to the glass, and because the door was not latched, it causes the door to swing open with an angular velocity of 0.19 rad/s. If the ball has a mass of 120 g and the patio door can be treated as a uniform box that is 2.3 m high, 1.0 m wide, and 0.030 m thick with a mass of 7.5 kg what speed did you throw the ball at
Answer:
Explanation:
moment of inertia of the door M = 1 /3 m ( l² + b² + d² )
= 1 / 3 x 7.5 x ( 2.3² + 1² + .03² )
= 1 / 3 x 7.5 x 6.2909
I = 15.7272.
moment of inertia of the door + small toy
15.7272 + .12 x .5²
= 15.7272 + .03
= 15.7572
Applying conservation of angular momentum law
mvr = M ω
m is mass of the toy thrown with velocity v at distance r from the axis , M is moment of inertia of door + toy and ω is angular velocity ith which door opens.
.12 x v x .5 = 15.7572 x .19
v = 49.89 m /s .
A ball of mass 1.00 kg is thrown at a door with velocity v1x = 12.0 m/s. The rectangular door has a mass of 30.0 kg and is 1.00 m wide. The ball strikes the door perpendicular to it, a horizontal distance 75.0 cm away from the hinges, and causes the door to swing open with an angular speed of 1.20 rad/s. What is the velocity v2x of the ball after striking the door? (The moment of inertia of a thin rectangular plate with mass M and dimensions a × b and the axis along the edge of length b is I = 1/3Ma2.)"
Answer:
The velocity of the ball after collision is [tex]v_{2x} = -4 m/s[/tex]
Explanation:
From the question we are told that
The mass of the ball is [tex]m_b = 1.00\ kg[/tex]
The velocity of the ball is [tex]v_{1x}= 12.0 \ m/s[/tex]
The mass of the rectangular door is [tex]m_d = 30 \ kg[/tex]
The width of the door is [tex]a = 1.00 \ m[/tex]
The distance of impact from the hinge is [tex]L = 75 \ cm = \frac{75}{100} = 0.75 \ m[/tex]
The angular speed of the door is [tex]w = 1.20 \ rad/s[/tex]
So the moment of inertia of the door is given from the question as
[tex]I = \frac{1}{3} M a^2[/tex]
substituting values
[tex]I = \frac{1}{3} * 30 * (1)[/tex]
[tex]I = 10 \ kg \cdot m^2[/tex]
According to the law of angular momentum conservation
[tex]L_i = L_f[/tex]
Where [tex]L_i[/tex] is the initial angular momentum of the system(the door and the ball) which is mathematically represented as
[tex]L_i = m_b * v_{1x} + Iw_i[/tex]
so [tex]w_i[/tex] is the initial angular speed of the door which is zero
So
[tex]L_i = m_b * v_{1x}[/tex]
[tex]L_f[/tex] is the final angular momentum of the system(the door and the ball) which is mathematically represented as
[tex]L_f = I w + m_b v_{2x} * L[/tex]
So
[tex]m_b * v_{1x} = I w + m_b v_{2x} * L[/tex]
Substituting values
[tex]1 * 12 * 0.75 = 10* 1.2 * v_{2x} * 0.75[/tex]
[tex]v_{2x} = -4 m/s[/tex]
The negative sign show a reversal in the balls direction
Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patterns on a screen a distance 4.80m away from the slits.
Part A
Which laser has its first maximum closer to the central maximum?
Part B
What is the distance \Delta _y__max-max_ between the first maxima (on the same side of the central maximum) of the two patterns?
Express your answer in meters.
Part C
What is the distance \Delta _y__max-max_ between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?
Express your answer in meters.
Answer:
A) first laser
B) 0.08m
C) 0.64m
Explanation:
To find the position of the maximum you use the following formula:
[tex]y=\frac{m\lambda D}{d}[/tex]
m: order of the maximum
λ: wavelength
D: distance to the screen = 4.80m
d: distance between slits
A) for the first laser you use:
[tex]y_1=\frac{(1)(d/20)(4.80m)}{d}=0.24m\\[/tex]
for the second laser:
[tex]y_2=\frac{(1)(d/15)(4.80m)}{d}=0.32m[/tex]
hence, the first maximum of the first laser is closer to the central maximum.
B) The difference between the first maximum:
[tex]\Delta y=y_2-y_1=0.32m-0.24m=0.08m=8cm[/tex]
hence, the distance between the first maximum is 0.08m
C) you calculate the second maximum of laser 1:
[tex]y_{m=2}=\frac{(2)(d/20)(4.80m)}{d}=0.48m[/tex]
and for the third minimum of laser 2:
[tex]y_{minimum}=\frac{(m+\frac{1}{2})(\lambda)(D)}{d}\\\\y_{m=3}=\frac{(3+\frac{1}{2})(d/15)(4.80m)}{d}=1.12m[/tex]
Finally, you take the difference:
[tex]1.12m-0.48m=0.64m[/tex]
hence, the distance is 0.64m
What phase difference between two otherwise identical traveling waves, moving in the same direction along a stretched string, will result in the combined wave having an amplitude 1.5 times that of the common amplitude of the two combining waves? Express your answer in (a) degrees, (b) radians, and (c) as a fraction of the wavelength.
Answer:
a) 82.8°
b)1.44 rad
c)0.23λ
Explanation:
Wave function form for each wave will be
-> wave 1:
[tex]y_{1}[/tex](x,t)=[tex]y_{m[/tex]sin( kx - ωt)
-> wave 2:
[tex]y_{2[/tex](x,t)=[tex]y_{m[/tex]sin( kx - ωt +φ)
The summation of above two functions is the resultant wave.
Y(x,t)=[tex]y_{m[/tex]sin( kx - ωt)+[tex]y_{m[/tex]sin( kx - ωt +φ)
Using the trigonometric addition formula for sin i.e
sin(A)+ sin(B) = 2 sin(A+B/2) cos(A-B/2)
Y(x,t)=2[tex]y_{m[/tex]cos(φ/2) sin(kx - ωt +φ)
When comparing to the general wave form, the amplitude is 2[tex]y_{m[/tex]cos(φ/2)m
Also, [tex]y_{m[/tex]cos(φ/2)= 1.5 [tex]y_{m[/tex]
φ/2=[tex]cos^{-1}[/tex](1.5/s)
(a) φ= 82.8°
(b)φ= 82.8π/180 =>1.44 rad
(c) one complete wave is 2π radians
Therefore, for two waves, φ/2π= 1.44/2π => 0.23 fraction of a complete wave from each other i.e they are separated by 0.23λ
The Franck-Hertz experiment involved shooting electrons into a low-density gas of mercury atoms and observing discrete amounts of kinetic energy loss by the electrons. Suppose instead a similar experiment is done with a very cold gas of atomic hydrogen, so that all of the hydrogen atoms are initially in the ground state. If the kinetic energy of an electron is 11.1 eV just before it collides with a hydrogen atom, how much kinetic energy will the electron have just after it collides with and excites the hydrogen atom? Kfinal = eV
Answer:
the final kinetic energy is 0.9eV
Explanation:
To find the kinetic energy of the electron just after the collision with hydrogen atoms you take into account that the energy of the electron in the hydrogen atoms are given by the expression:
[tex]E_n=\frac{-13.6eV}{n^2}[/tex]
you can assume that the shot electron excites the electron of the hydrogen atom to the first excited state, that is
[tex]E_{n_2-n_1}=-13.6eV[\frac{1}{n_2^2}-\frac{1}{n_1^2}]\\\\E_{2-1}=-13.6eV[\frac{1}{2^2}-\frac{1}{1}]=-10.2eV[/tex]
-10.2eV is the energy that the shot electron losses in the excitation of the electron of the hydrogen atom. Hence, the final kinetic energy of the shot electron after it has given -10.2eV of its energy is:
[tex]E_{k}=11.1eV-10.2eV=0.9eV[/tex]
The final kinetic energy the electron will have just after it collides with and excites the hydrogen atom is; E_k = 0.9 eV
The formula for the energy of the electron of hydrogen atom in its nth orbit is given by the formula;
E_n = -13.6eV/n²
Where n is the principal quantum number.
Thus;
For n = 1;
E_1 = -13.6/1²
E_1 = -13.6 eV
At n = 2;
E_2 = -13.6/2²
E_2 = -3.4 eV
Thus, the energy that the electron looses is;
E_2 - E_1 = -3.4 - (-13.6)
E_2 - E_1 = 10.2 eV
We are told that the kinetic energy of an electron is 11.1 eV just before it collides with a hydrogen atom.
Thus, the final kinetic energy the electron will have just after it collides with and excites the hydrogen atom is;
E_k = 11.1 - 10.2
E_k = 0.9 eV
Read more at; https://brainly.com/question/13444136
A satellite and the International Space Station have the same mass and are going around Earth in concentric orbits. The distance of the satellite from Earth\'s center is twice that of the International Space Station\'s distance. What is the ratio of the centripetal force acting on the satellite compared to that acting on the International Space Station
Answer:
[tex]\dfrac{F_2}{F_1}=\dfrac{1}{4}[/tex]
Explanation:
G = Gravitational constant
r = Distance between Earth and object
M = Mass of Earth
m = Mass of object
Centripetal force on the space station
[tex]F_1=\dfrac{GMm}{r^2}[/tex]
Centripetal force on the satellite
[tex]F_2=\dfrac{GMm}{(2r)^2}\\\Rightarrow F_2=\dfrac{GMm}{4r^2}[/tex]
From the question the required ratio is
[tex]\dfrac{F_2}{F_1}=\dfrac{\dfrac{GMm}{4r^2}}{\dfrac{GMm}{r^2}}\\\Rightarrow \dfrac{F_2}{F_1}=\dfrac{1}{4}[/tex]
The ratio is [tex]\dfrac{F_2}{F_1}=\dfrac{1}{4}[/tex]
3.7.6. A satellite is in a circular orbit around the Earth. If it is at an altitude equal to twice the radius of the Earth, 2RE, how does its speed v relate to the Earth's radius RE, and the magnitude g of the acceleration due to gravity on the Earth's surface?a)b)c)d)e)
Answer:
Explanation:
For acceleration due to gravity
g = GM / R² ---------------------------- ( 1 )
At height 2R , radius of orbit is 3R .
If speed of satellite be v
m v² / 3R = GMm / (3R)² ( centripetal force = gravitational force )
v² = GM / 3R -------------------------------------- ( 2 )
Dividing (1) and (2)
v² / g = R / 3
v² = gR / 3
v = √( gR / 3 )
3. Jack and Zack are standing on a crate at rest on a frictionless horizontal surface. Jack has a mass of 70 kg, Zack has a mass of 45 kg, and the crate has a mass of 15 kg. In what follows, we will see that both jump of the crate. You may assume that they push themselves off with a speed of 4 m/s relative to the crate and in a direction that is essentially horizontal.a) what is the final speed of the crate if Jack and Zack jump simultaneously, in the same direction
Answer:
22.4 m/s
Explanation:
To find the speed of the crate you take into account the momentum conservation. The momentum before Jack and Zack jump must be equal to the momentum after. Then you have:
[tex]p_b=p_a\\\\[/tex]
Furthermore, you take into account that the momentum before, when crater, Jack and Zack are at rest, the momentum is zero. Hence:
[tex]p_b=(m_Z+m_J+m_c)v=0\\\\m_Z=45kg\\\\m_J=70kg\\\\m_c=15kg\\\\p_a=m_Zv_Z+m_Jv_J+m_cv_c[/tex]
but you know what are the velocities of Zack and Jack after they jump.
[tex]p_a=p_b=0\\\\-v_c=\frac{m_Zv_Z+m_Jv_J}{m_c}\\\\-v_c=\frac{(70kg)(4m/s)+(45kg)(4m/s)}{15kg}=-22.4\frac{m}{s}[/tex]
the minus sign means that the velocity of the crater is in an opposite direction of the spee of Jack and Zack.
hence, the speed of the crater is 22/4m/s
The final speed of the crate when Jack and Zack jump off simultaneously in the same direction is calculated to be approximately 3.54 m/s. The negative sign indicates opposite direction to the jump.
To solve for the final speed of the crate, we will use the principle of conservation of momentum. Since the system is initially at rest, the total initial momentum is zero.
Let's denote the mass of Jack as mJ, the mass of Zack as mZ, and the mass of the crate as mC. Also, let the velocity of Jack, Zack, and the crate relative to the ground be VJ, VZ, and VC respectively.
[tex]mJ = 70 kg[/tex]
[tex]mZ = 45 kg[/tex]
[tex]mC = 15 kg[/tex]
Since Jack and Zack both push off with a speed of 4 m/s relative to the crate:
VJ = VC + 4 m/s
VZ = VC + 4 m/s
Using conservation of momentum:
Initial momentum = Final momentum
[tex]0 = mJ(VC + 4) + mZ(VC + 4) + mCVC[/tex]
[tex]0 = 70(VC + 4) + 45(VC + 4) + 15VC[/tex]
Combine and solve for VC:
[tex]0 = 70VC + 280 + 45VC + 180 + 15VC[/tex]
[tex]0 = 130VC + 460[/tex]
[tex]-460 = 130VC[/tex]
[tex]VC = -460 / 130[/tex]
[tex]VC = -3.54 m/s[/tex]
The negative sign indicates the crate moves in the opposite direction to their jump. Thus, the final speed of the crate is 3.54 m/s.
Which of the following statements ARE correct?
Group of answer choices:
A) If net torque is zero, the object will be moving at constant velocity or remain static.
B) If the net torque is zero, the object will not rotate or rotate at constant angular velocity.
C) If net torque is zero, the angular momentum is conserved.
D) If net force is zero, the object will be rotate at constant angle velocity or not rotate at all.
Answer:
Option A, C and D are correct statements.
Explanation:
Torque is the twisting effect of a force which causes an object to acquire angular acceleration.
The direction of the torque depends on the direction of the force on the axis. The SI unit for torque is the Newton-meter.
The net torque is the sum of the individual torques.
If the net torque on a rotatable object is zero then:
1. It will be in rotational equilibrium and the angular velocity of the body remains constant.
2. It will be in rotational equilibrium and not able to acquire angular acceleration. Thus, it will not be able to acquire angular velocity.
3. Angular momentum will be conserved.
If the forces on an object are balanced, the net force is zero.
If there net force on an object is zero, its speed and direction of motion do not change, including if it is at rest, the object will not accelerate and the velocity will remain constant.
By considering the basic fundamentals of Torque, the statements (A), (C) and (D) are correct.
The given problem is based on the concept and fundamentals of Torque. Torque is the twisting effect of a force which causes an object to acquire angular acceleration.
The direction of the torque depends on the direction of the force on the axis. The SI unit for torque is the Newton-meter. The net torque is the sum of the individual torques.
If the net torque on a rotatable object is zero then:
It will be in rotational equilibrium and the angular velocity of the body remains constant. It will be in rotational equilibrium and not able to acquire angular acceleration. Thus, it will not be able to acquire angular velocity. Angular momentum will be conserved. If the forces on an object are balanced, the net force is zero. If there net force on an object is zero, its speed and direction of motion do not change, including if it is at rest, the object will not accelerate and the velocity will remain constant.Thus, we can conclude that by considering the basic fundamentals of Torque, the statements (A), (C) and (D) are correct.
Learn more about the fundamentals of Torque here:
https://brainly.com/question/19247046
An object is moving along a line with velocity v (t )equals A t squared space plus space B meters per second at time t seconds. Its initial position, at time t equals 0, is s (0 )equals C meters. If A equals 15, B equals 8 and C equals space 3, then Give the object's position function s (t ). Give the distance traveled by the object between t equals 1 and t equals 2 seconds and record your answer below.
Answer:
- S(t) = 5t³+8t+3
- 19.7m approximately
Explanation:
Given the velocity if an object to be
v(t) = At² + B where t is the time in seconds
Velocity is the change in displacement of a body with respect to time.
V(t) = dS(t)/dt
Making S(t) the subject of the formula
dS(t) = v(t)dt
Integrating both sides
∫dS(t) =∫v(t)dt
S(t) = ∫(At²+B)dt
S(t) = At³/3+Bt + C... 1
In its initial position s(0) = 0
t = 0, s = 0
S(0) = A(0)³/3+B(0)+C
S(0)= C
a) The object's position function s(t) if A = 15, B = 8 and C = 3 can be gotten by substituting this value into eqn 1
S(t) = 15t³/3+8t+3
S(t) = 5t³+8t+3
b) For the distance traveled by the object between t equals 1 and t equals 2 seconds
When t = 1s
S(1) = 5(1)³/3+8(1)+3
S(1) = 5/3+8+3
S(1) = 5+24+9/3
S(1) = 38/3 m
When t = 2secs
S(2) = 5(2)³/3+8(2)+3
S(2) = 40/3+16+3
S(2) = (40+48+9)/3
S(2) = 97/3
Distance travelled between this times will be 97/3-38/3
= 59/3
= 19.7m approximately
A beaker of water rests on an electronic balance that reads 875.0 g. A 2.75-cm-diameter solid copper ball attached to a string is submerged in the water, but does not touch the bottom. What are the tension in the string and? a. Find the tension in the string. (10pts) b. What is the new balance reading? (10pts)A beaker of water rests on an electronic balance that reads 875.0 g. A 2.75-cm-diameter solid copper ball attached to a string is submerged in the water, but does not touch the bottom. What are the tension in the string and? a. Find the tension in the string. (10pts) b. What is the new balance reading? (10pts)
Answer:
a
The tension in the string is [tex]T = 0.85 N[/tex]
b
The new balance reading is [tex]M_b = 885.86 g[/tex]
Explanation:
From the question we are told that
The mass of the beaker of water is [tex]m = 875 .0g[/tex]
The diameter of the copper ball is [tex]d = 2.75 cm = \frac{2.75}{100} = 0.0275 m[/tex]
There are two forces acting on the copper ball
The first is the Buoyant force of the water pushing it up which is mathematically represented as
[tex]F = \rho V g[/tex]
Where [tex]\rho[/tex] is the density of water which has value of [tex]\rho = 1000 kg/m^3[/tex]
g is the acceleration due to gravity [tex]g= 9.8 \ m/s^2[/tex]
[tex]V[/tex] is the volume of water displaced by the copper ball which is mathematically evaluated as
[tex]V = \frac{4}{3} \pi r^3[/tex]
The radius r is [tex]r = \frac{d}{3} = \frac{0.0275}{2} = 0.01375 m[/tex]
Substituting value
[tex]V = \frac{4}{3} * 3.142 * (0.01375)^3[/tex]
[tex]V = 1.08 9 *10^{-5 } m^3[/tex]
Substituting for F
[tex]F = 1000 * 1.089 *10^{-5} * 9.8[/tex]
[tex]F = 0.1067 N[/tex]
The second force is the weight of the copper ball which is mathematically represented as
[tex]W_c = mg[/tex]
Now m is the mass which can be mathematically evaluated as
[tex]m = \rho_c * V[/tex]
Where is the density of copper with value of [tex]\rho_c = 8960 kg /m^3[/tex]
So [tex]m = 8960 * 1.089 *10^{-5}[/tex]
[tex]m = 0.0976[/tex]
So the weight of copper is
[tex]W_c = 0.09756 * 9.8[/tex]
[tex]W_c = 0.956 N[/tex]
Now the tension the string would be mathematically evaluated as
[tex]T = W_c - F[/tex]
So [tex]T = 0.956 - 0.1067[/tex]
[tex]T = 0.85 N[/tex]
From this value we that the string is holding only 0.85 N of the copper weight thus (0.956 - 0.85 = 0.1065 N ) is being held by the balance
Now the mass equivalent of this weight is mathematically evaluated as
[tex]m_z = \frac{1.0645 }{9.8 }[/tex]
[tex]m_z = 0.01086 kg[/tex]
Converting to grams
[tex]m_z = 10.86 g[/tex]
So the new balance reading is
[tex]M_b = 875.0 +10.86[/tex]
[tex]M_b = 885.86 g[/tex]
Three long wires all lie in an xy plane parallel to the x axis. They intersect the y axis at the origin, y = d, and y = 2d, where d = 10 cm. The two outer wires each carry a current of 6.8 A in the positive x direction.
1. What is the magnitude of the force on a 2.8 m section of either of the outer wires if the current in the center wire is 3.3 A.
(a) in the positive x direction and (b) in the negative x direction?
Answer:
[tex]F = 2.55*10^{-4} \ N[/tex]
F = [tex]3.808 * 10^{-6} \ N[/tex]
Explanation:
The magnetic field due to the first wire can be written as:
[tex]B_1 = \frac{\mu _o I_1}{2 \pi (2 d)}[/tex]
[tex]B_1 = \frac{\mu _o I_1}{4 \pi d}[/tex]
The magnetic field due to the second wire is as follows:
[tex]B_2 = \frac{\mu _o I_2}{2 \pi d}[/tex]
The net magnetic field B = [tex]B_1 +B_2[/tex]
[tex]B = \frac{\mu _o I_1}{4 \pi d} +\frac{\mu _o I_2}{2 \pi d} \\ \\ B = \frac{ \mu_o }{2 \pi d}(\frac{I_1}{2}+ I_2)[/tex]
Also; the magnetic force on wire segment l = 2.8 m
[tex]F = I_1lB = \frac{ \mu_o }{2 \pi d}(\frac{I_1}{2}+ I_2)[/tex]
Replacing all the values ; we have :
[tex]F = \frac{ 4 \pi * 10^{-7}*6.8*2.8}{2 \pi * 10*10^{-2}}(\frac{6.8}{2}+ 3.3)[/tex]
[tex]F = 2.55*10^{-4} \ N[/tex]
b)
Magnetic field due to the first wire is :
[tex]B_1 = \frac{\mu _o I_1}{4 \pi d}[/tex]
[tex]B_1 = \frac{4.0*10^{-7}*6.8}{4 \pi *10*10^{-2}}[/tex]
[tex]B_1 = 6.8*10^{-6} \ T[/tex]
Magnetic field due to the second wire is :
[tex]B_2 = \frac{\mu _o I_2}{2 \pi d} \\ \\ B_2 = \frac{4*10^{-7}*3.3}{2 \pi *10*10^{-2}}[/tex]
[tex]B_2 = 6.6*10^{-6}T[/tex]
[tex]B_1>B_2[/tex]
Net magnetic field B = [tex]B_1 - B_2[/tex]
B = [tex](6.8*10^{-6} - 6.6*10^{-6})T[/tex]
B = [tex]2*10^{-7} \ T[/tex]
Magnetic force F = [tex]I_1lB[/tex]
F = [tex]6.6*2.8 *2*10^{-7}[/tex]
F = [tex]3.808 * 10^{-6} \ N[/tex]
Once you’ve identified the initial and final conditions, you’re ready to solve for the unknown quantity in your problem. Boyle’s Law expresses the pressure-volume relationship as P1V1=P2V2, so you will need to solve this equation for the unknown quantity and then plug in your known values to calculate the unknown. What pressure would it take to compress 250. L of helium gas initially at 1.00 atm into a 2.00 L tank at constant temperature? Express your answer with the appropriate units.
Answer:
125atm
Explanation:
Given:
initial pressure [tex]P_{1[/tex]= 1 atm
initial volume [tex]V_{1[/tex]= 250L
final volume [tex]V_{2[/tex]=2L
Required:
Final Pressure [tex]P_{2[/tex]=?
As Boyle’s Law expresses the pressure-volume relationship as
[tex]P_{1[/tex][tex]V_{1[/tex]= [tex]P_{2[/tex] [tex]V_{2[/tex]
(1)(250)= [tex]P_{2[/tex] (2)
[tex]P_{2[/tex] = 250/2
[tex]P_{2[/tex]=125atm
Therefore, it would take pressure of 125atm.
Using Boyle's Law, the pressure required to compress a gas of 250 L at 1 atm into a smaller volume of 2 L at constant temperature is calculated to be 125 atm.
Explanation:The student is attempting to solve this problem using Boyle's Law, which states that the pressure and volume of a gas have an inverse relationship when temperature is held constant. In this context, the formula for Boyle's Law is P1V1 = P2V2.
Here, the initial conditions are a volume V1 of 250 L and a pressure P1 of 1.00 atm. The final volume V2 is given as 2.00 L, and we are asked to solve for the final pressure P2.
Replace the given values into the Boyle's Law equation: (1.00 atm) * (250.L) = P2 * (2.00 L). From this, we can solve for P2 which sums to: P2 = ((1.00 atm) * (250.L)) / (2.00 L) = 125 atm. The pressure required to compress the helium gas into a 2L container would be 125 atm.
Learn more about Boyle's Law here:https://brainly.com/question/21184611
#SPJ11
The density of air under normal conditions is about 3 1.2 kg/ m . For a wind speed of 10 m/s, find (a) the actual power carried in the wind; (b) the maximum power (Betz limit) extractable by a wind turbine with a blade area of 10 m2; and (c) the actual power extracted by a wind turbine with a blade area of 10 m2 and a power coefficient of 0.46.
Betelgeuse is 100,000 times more luminous than our sun, which means that it releases an estimated 3.846 x 1031 W of luminous light. If an exoplanet existed with the same mass (5.972 x 10^24 kg) and twice the radius of earth (1.27 x 10^7 m) and half the distance (7.5 x 10^10 m) from Betelgeuse what would the intensity of the light at the surface of that "earth" look like? a) What is the intensity of Betelgeuse at the "earth’s" surface?
Answer:
5.4 × 10⁸ W/m²
Explanation:
Given that:
The Power (P) of Betelgeuse is estimated to release 3.846 × 10³¹ W
the mass of the exoplanet = 5.972 × 10²⁴ kg
radius of the earth = 1.27 × 10⁷ m
half the distance (i.e radius r ) = 7.5 × 10¹⁰ m
a) What is the intensity of Betelgeuse at the "earth’s" surface?
The Intensity of Betelgeuse can be determined by using the formula:
[tex]Intensity \ I = \frac{P}{4 \pi r^2}[/tex]
[tex]I = \frac{3.846*10^{31}}{4 \pi (7.5*10^{10})^2}[/tex]
I = 544097698.8 W/m²
I = 5.4 × 10⁸ W/m²
Answer:
1.97*10^14 W/m^2
Explanation:
To find the intensity of the light emitted by Betelgeuse you taken into account that the light from Betelgeuse expands spherically in space.
You use the following formula:
[tex]I=\frac{P}{A}=\frac{P}{4\pi r^2}[/tex] (1)
I: intensity of light
r: distance from Betelgeuse to the exoplanet = (1/2)*7.5*10^10m
P: power = 100,000*3.486*10^31 W
By replacing the values of the parameters in (1) you obtain:[tex]I=\frac{100000(3.486*10^{31}W}{4\pi (0.5*7.5*10^{10}m)^2}=1.97*10^{14}\frac{W}{m^2}[/tex]
One propulsion method that is currently being studied for interstellar travel uses a large sail. The intent is that sunlight striking the sail creates a force that pushes the ship away from the sun, much as wind propels a sailboat. Does such a design have any hope of working and, if so, should the surface facing the sun be shiny like a mirror or black, in order to produce the greatest force
Answer:
Yes, the design has a hope of working.
A dull face should yield more force.
Explanation:
According to the de broglie theory, and the wave-particle duality, waves can be assumed and made to behave as a matter and vice versa. Under these circumstances, the impinging wave can act as a particle (photon) colliding with a relatively stationary body, yielding some of its momentum to the relatively stationary body. With this, the momentum of the photons can be used to propel the ship by its solar sail.
Making the face dull should yield more force in the sense that the incident photon is not reflected and all of its momentum is yielded to the sail compared to when it is reflected (bouncing off) off the sail. This will be a case of inelestic collision between the sail and the photon.
A current flowing through you of more than 5 ma is considered dangerous. Why do we see warnings about high voltage, rather than high current? What voltages are dangerous even if the skin is dry? (Dry skin has a resistance of 100 kohms while wet skin can be much less.) An electroencephalogram (EEG) is a recording of differences in voltage observed on the scalp by attaching electrical contacts. If voltage differences that are observed are about 0.5 mv, and the resistance of the skin is about 10 kohms, what size currents are involved?
Answer:
V = 500 volts
I = 50 nA
Explanation:
We usually see 'high voltage' sign rather than 'high current' this is due to the fact that the amount of current depends upon the resistance of the body,
The relation of voltage, current and resistance is given by Ohm's law,
V = IR
I = V/R
As you can see, current and resistance are inversely related, the lower the resistance of the body, the higher will be the current flowing through the body.
What voltages are dangerous even if the skin is dry?
A dry human body has a resistance of approximately 100 kΩ and it gets reduced if the body is wet.
For a current of 5 mA, the corresponding voltage is,
V = 0.005*100,000
V = 500 volts
Therefore, for a dry human body, a voltage of 500 volts would be dangerous.
If voltage differences that are observed are about 0.5 mV, and the resistance of the skin is about 10 kΩ, what size currents are involved?
I = V/R
I = 0.0005/10,000
I = 50×10⁻⁹ A
I = 50 nA
Therefore, a current of 50 nA would be flowing through the EEG.
Inductors in parallel. Two inductors L1 = 1.31 H and L2 = 2.24 H are connected in parallel and separated by a large distance so that the magnetic field of one cannot affect the other. (a) Calculate the equivalent inductance. (Hint: Review the derivations for resistors in parallel and capacitors in parallel.Which is similar here?) (b) What is the generalization of (a) for N = 10 similar inductors L = 3.22 H in parallel?
Answer:
a) 0.83H
b) 3.22/N Henry
Explanation:
Given two inductors L1 = 1.31 H and L2 = 2.24 H connected in parallel, their equivalent inductance derivative is similar to that of resistance in parallel.
Derivation:
If the voltage across an inductor
VL = IXL
I is the current
XL is the inductive reactance
XL = 2πfL
VL = I(2πfL)
L is the inductance.
From the formula, I = V/2πfL
Given two inductors in parallel, different current will flow through them.
I1 = V/2πfL1 (current in L1)
I2 = V/2πfL2 (current in L2)
Total current I = I1+I2
I = V/2πfL1 + V/2πfL2
I = V/2πf{1/L1+1/L2}
V/2πfL = V/2πf{1/L1+1/L2}
1/L = 1/L1+1/L2 (equivalent inductance in parallel)
Given L1 = 1.31 H and L2 = 2.24
1/L = 1/1.31 + 1/2.24
1/L = 0.763 + 0.446
1/L = 1.209
L = 1/1.209
L = 0.83H
The equivalent inductance is 0.83H
b) Given similar inductors L = 3.22H in parallel, the equivalent inductance will be:
1/L = 1/3.22+1/3.22+1/3.22+1/3.22+1/3.22
+1/3.22+1/3.22+1/3.22+1/3.22+1/3.22
1/L = 10/3.22 (since that all have the same denominator)
L = 3.22/10
If N = 10, the generalization of 10 similar inductors in parallel will be:
L = 3.22/N Henry
Problem 9) If you had light of wavelength 700nm when emitted, what would the wavelength be for an observer who is a red shift z = 0.05 from that source?
Answer:
The answer is 735 Nanometer
Explanation:
Solution to this problem
Given that:
Solution)
The Redshift is categorized by the difference related between wavelength(emitted) and wavelength (observer).
Now,
The wavelength can be computed as :
1+z=\lambda (observer)\div \lambda (emitted)
hence,
1 + 0.05 = λ(observer) / 700 nm
Gives
= 1.05*700 = 735 nm
Therefore, The wavelength from that source for an observer would be 735 nanometer
A what is a negatively charged Atoms
Answer:
Negatively charged or positively charged atom is generally termed as ANION/CATION. Short Explanation: If an atom loses electrons or gains protons, it will have a net positive charge and is called a Cation. If an atom gains electrons or loses protons, it will have a net negative charge and is called an Anion.
Explanation:
An infinitely long wire carrying a current of 4.5 A is bent at the origin and extends in both the y-direction and x-direction. (In other words, the wire makes a 90 degree angle with the x axis and y axis and has a bend at the origin) Find the magnetic field at the point x = 3 cm and y = 2 cm.
Answer:
6.444 × [tex]10^{-5}[/tex] T
Explanation:
Find the given attachments
The magnitude of the magnetic field at the point [tex](x = 3 \text{ cm}, y = 2 \text{ cm})[/tex] is approximately [tex]\( 1.08 \times 10^{-5} \text{ T} \)[/tex].
The magnitude of the magnetic field [tex]\( B \)[/tex] at a distance [tex]\( r \)[/tex] from a long straight wire carrying a current [tex]\( I \)[/tex] is given by:
[tex]\[ B = \frac{\mu_0 I}{2\pi r} \][/tex]
where [tex]\( \mu_0 \)[/tex] is the permeability of free space [tex]\( \mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} \)[/tex].
For the horizontal wire, the distance from the wire to the point is [tex]\( y = 2 \text{ cm} = 0.02 \text{ m} \)[/tex]. Thus, the magnetic field due to the horizontal wire is:
[tex]\[ B_x = \frac{\mu_0 (4.5 \text{ A})}{2\pi (0.02 \text{ m})} \][/tex]
Next, consider the vertical section of the wire (along the y-axis). Similarly, the magnetic field due to this section at a perpendicular distance [tex]\( r \)[/tex] is given by the same formula. Here, the distance from the wire to the point is [tex]\( x = 3 \text{ cm} = 0.03 \text{ m} \)[/tex].
The magnetic field due to the vertical wire is:
[tex]\[ B_y = \frac{\mu_0 (4.5 \text{ A})}{2\pi (0.03 \text{ m})} \][/tex]
The total magnetic field [tex]\( B \)[/tex] at the point is the vector sum of [tex]\( B_x \)[/tex] and [tex]\( B_y \)[/tex]. We can use Pythagoras' theorem to find the magnitude of the resultant magnetic field:
[tex]\[ B = \sqrt{B_x^2 + B_y^2} \][/tex]
Substituting the expressions for [tex]\( B_x \)[/tex] and [tex]\( B_y \)[/tex] we get:
[tex]\[ B = \sqrt{\left(\frac{\mu_0 (4.5 \text{ A})}{2\pi (0.02 \text{ m})}\right)^2 + \left(\frac{\mu_0 (4.5 \text{ A})}{2\pi (0.03 \text{ m})}\right)^2} \][/tex]
Plugging in the value of [tex]\( \mu_0 \)[/tex] and simplifying, we find:
[tex]\[ B = \sqrt{\left(\frac{4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} (4.5 \text{ A})}{2\pi (0.02 \text{ m})}\right)^2 + \left(\frac{4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} (4.5 \text{ A})}{2\pi (0.03 \text{ m})}\right)^2} \][/tex]
[tex]\[ B = \sqrt{\left(\frac{18 \times 10^{-7} \text{ T}\cdot\text{m}}{0.02 \text{ m}}\right)^2 + \left(\frac{18 \times 10^{-7} \text{ T}\cdot\text{m}}{0.03 \text{ m}}\right)^2} \][/tex]
[tex]\[ B = \sqrt{\left(9 \times 10^{-6} \text{ T}\right)^2 + \left(6 \times 10^{-6} \text{ T}\right)^2} \][/tex]
[tex]\[ B = \sqrt{81 \times 10^{-12} \text{ T}^2 + 36 \times 10^{-12} \text{ T}^2} \][/tex]
[tex]\[ B = \sqrt{117 \times 10^{-12} \text{ T}^2} \][/tex]
[tex]\[ B \approx 1.08 \times 10^{-5} \text{ T} \][/tex]
6) A pulsar is a rapidly rotating neutron star. The Crab nebula pulsar in the constellation Taurus has a period of 33.5 × 10−3 s, radius 10.0 km, and mass 2.8 × 1030 kg. The pulsar’s rotational period will increase over time due to the release of electromagnetic radiation, which doesn’t change its radius but reduces its rotational energy. (a) What is the angular momentum of the pulsar? (b) Suppose the angular velocity decreases at a rate of 10−14 rad/s2 . What is the torque on the pulsar?
Answer:
a) L = 2.10x10⁴⁰ kg*m²/s
b) τ = 1.12x10²⁴ N.m
Explanation:
a) The angular momentum (L) of the pulsar can be calculated using the following equation:
[tex] L = I \omega [/tex]
Where:
I: inertia momentum
ω: angular velocity
First we need to calculate ω and I. The angular velocity can be calculated as follows:
[tex] \omega = \frac{2 \pi}{T} [/tex]
Where:
T: is the period = 33.5x10⁻³ s
[tex] \omega = \frac{2 \pi}{T} = \frac{2 \pi}{33.5 \cdot 10^{-3} s} = 187.56 rad/s [/tex]
The inertia moment of the pulsar can be calculated using the following relation:
[tex] I = \frac{2}{5}mr^{2} [/tex]
Where:
m: is the mass of the pulsar = 2.8x10³⁰ kg
r: is the radius = 10.0 km
[tex] I = \frac{2}{5}mr^{2} = \frac{2}{5}2.8\cdot 10^{30} kg*(10\cdot 10^{3} m)^{2} = 1.12 \cdot 10^{38} kg*m^{2} [/tex]
Now, the angular momentum of the pulsar is:
[tex] L = I \omega = 1.12 \cdot 10^{38} kg*m^{2}*187.56 rad/s = 2.10 \cdot 10^{40} kg*m^{2}*s^{-1} [/tex]
b) If the angular velocity decreases at a rate of 10⁻¹⁴ rad/s², the torque of the pulsar is:
[tex] \tau = I*\alpha [/tex]
Where:
α: is the angular acceleration = 10⁻¹⁴ rad/s²
[tex]\tau = I*\alpha = 1.12 \cdot 10^{38} kg*m^{2} * 10^{-14} rad*s^{-2} = 1.12 \cdot 10^{24} N.m[/tex]
I hope it helps you!
Final answer:
To find the angular momentum of the pulsar, we can use the formula L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Explanation:
To find the angular momentum of the pulsar, we can use the formula L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
The moment of inertia for a solid sphere can be calculated using the formula I = (2/5)mr², where m is the mass and r is the radius.
Using the given radius and mass of the pulsar, we can calculate the moment of inertia.
Once we have the moment of inertia, we can use the given period of the pulsar to calculate the angular velocity.
To find the torque on the pulsar, we can use the formula τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
The angular acceleration can be calculated using the given rate at which the angular velocity is decreasing.
Using these formulas and the given values, we can find the angular momentum and torque of the pulsar.
A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 V. It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 V. If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?
Complete Question
A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 [tex]\Omega[/tex]. It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300[tex]\Omega[/tex] . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?
Answer:
a
The additional resistance is [tex]R_z = 4.4 \Omega[/tex]
b
The rate at which internal energy increase at the supply is [tex]Z_1 = 32 W[/tex]
c
The rate at which internal energy increase in the battery is [tex]Z_1 = 32 W[/tex]
d
The rate at which internal energy increase in the added series resistance is [tex]Z_3 = 70.4 W[/tex]
e
the increase rate of the chemically energy in the battery is [tex]C = 48 W[/tex]
Explanation:
From the question we are told that
The open circuit voltage is [tex]V = 40.0V[/tex]
The internal resistance is [tex]R = 2 \Omega[/tex]
The emf of each battery is [tex]e = 6.00 V[/tex]
The internal resistance of the battery is [tex]r = 0.300V[/tex]
The charging current is [tex]I = 4.00 \ A[/tex]
Let assume the the additional resistance to to added to the circuit is [tex]R_z[/tex]
So this implies that
The total resistance in the circuit is
[tex]R_T = R + 2r +R_z[/tex]
Substituting values
[tex]R_T = 2.6 +R_z[/tex]
And the difference in potential in the circuit is
[tex]E = V -2e[/tex]
=> [tex]E = 40 - (2 * 6)[/tex]
[tex]E = 28 V[/tex]
Now according to ohm's law
[tex]I = \frac{E}{R_T}[/tex]
Substituting values
[tex]4 = \frac{28}{R_z + 2.6}[/tex]
Making [tex]R_z[/tex] the subject of the formula
So [tex]R_z = \frac{28 - 10.4}{4}[/tex]
[tex]R_z = 4.4 \Omega[/tex]
The increase rate of internal energy at the supply is mathematically represented as
[tex]Z_1 = I^2 R[/tex]
Substituting values
[tex]Z_1 = 4^2 * 2[/tex]
[tex]Z_1 = 32 W[/tex]
The increase rate of internal energy at the batteries is mathematically represented as
[tex]Z_2 = I^2 r[/tex]
Substituting values
[tex]Z_2 = 4^2 * 2 * 0.3[/tex]
[tex]Z_2 = 9.6 \ W[/tex]
The increase rate of internal energy at the added series resistance is mathematically represented as
[tex]Z_3 = I^2 R_z[/tex]
Substituting values
[tex]Z_3 = 4^2 * 4.4[/tex]
[tex]Z_3 = 70.4 W[/tex]
Generally the increase rate of the chemically energy in the battery is mathematically represented as
[tex]C = 2 * e * I[/tex]
Substituting values
[tex]C = 2 * 6 * 4[/tex]
[tex]C = 48 W[/tex]
A wire with length 3.5 m and mass 0.03 kg is aligned horizontally and connected by conducting loops to two vertical frictionless metal rods. The rod is oriented in the East-West direction. What is the magnitude of the current that has to be run through the wire so that it hovers in place without sliding up or down? You may ignore magnetic forces caused by the vertical wires. Earth's magnetic field: 0.00005 Tesla or 0.5 gauss. Earth's gravity: 9.8 N/kg
Answer:
Current flowing in the wire is 1680 A
Explanation:
It is given length of wire l = 3.5 m
Mass of the wire m = 0.03 kg
Magnetic field B = 0.00005 Tesla
Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]
Mg force acting on the wire will be equal to Lorentz force acting on the wire.
Therefore [tex]mg=IBl[/tex]
[tex]0.03\times 9.8=I\times 0.00005\times 3.5[/tex]
[tex]I=1680A[/tex]
Therefore current flowing in the wire is 1680 A
A red laser from a physics lab is marked as producing 632.8 nm light. When light from this laser falls on two closely spaced slits, and interference pattern formed on a wall several meters away has bright fringes spaced 5.00 mm apart near the center of the pattern. When the laser is replaced by a small laser pointer, the fringes are 5.14 mm apart. What is the wavelength of the light produced by the laser pointer?
Given Information:
Wavelength of the red laser = λr = 632.8 nm
Distance between bright fringes due to red laser = yr = 5 mm
Distance between bright fringes due to laser pointer = yp = 5.14 mm
Required Information:
Wavelength of the laser pointer = λp = ?
Answer:
Wavelength of the laser pointer = λp = ?
Explanation:
The wavelength of the monochromatic light can be found using young's double slits formula,
y = Dλ/d
y/λ = D/d
Where
λ is the wavelength
y is the distance between bright fringes.
d is the double slit separation distance
D is the distance from the slits to the screen
For the red laser,
yr/λr = D/d
For the laser pointer,
yp/λp = D/d
Equating both equations yields,
yr/λr = yp/λp
Re-arrange for λp
λp = yp*λr/yr
λp = (5*632.8)/5.14
λp = 615.56 nm
Therefore, the wavelength of the small laser pointer is 615.56 nm.
Choose the appropriate word choices below that correctly complete the blanks in the following three sentences: At the beginning, the object is at the origin and moving to the left. In the first half of its motion, the object moves to the left with ________ velocity. Then, the motion of the object changes. In the second half of its motion, the object________ due to a ________ acceleration that is ________ in magnitude.
Answer:
Constant
Accelerates
Constant
Positive