The top of a swimming pool is at ground level. If the pool is 2.40 m deep, how far below ground level does the bottom of the pool appear to be located for the following conditions? (The index of refraction of water is 1.333.)

Answer:

5.994 kgm/s

Explanation:

Momentum: This can be defined as the product of the mass of a body to the velocity of that body. The S.I unit of momentum is kgm/s.

Mathematically, momentum can be expressed as,

M = mv.................................... Equation 1

Where M = momentum of the baseball, m = mass of the baseball, v = velocity of the baseball.

Given: m = 155 g = (155/1000) kg = 0.155 kg. v = 87 miles per hour = 87(1600/3600) m/s = 38.67 m/s.

Substituting into equation 1

M = 0.155(38.67)

M = 5.994 kgm/s.

Thus the momentum of the baseball = 5.994 kgm/s

The magnitude of the momentum of a baseball with a mass of 155 g thrown at 87 mph is approximately 6.03 kg·m/s after converting units and applying the momentum equation.

To calculate the magnitude of the momentum of a baseball thrown at a speed of 87 miles per hour, we use the formula p = mv, where p is the momentum, m is the mass in kilograms, and v is the velocity in meters per second. First, we need to convert the mass of the baseball from grams to kilograms (155 g to 0.155 kg) and the speed from miles per hour to meters per second (87 mph to 38.88 m/s, using the conversion factor 1 mile = 1600 meters and 1 hour = 3600 seconds).

Then, we substitute the values into the momentum equation to find the momentum: p = (0.155 kg)(38.88 m/s) = 6.0264 kg·m/s. Therefore, the magnitude of the momentum of the baseball is approximately 6.03 kg·m/s when rounded to two decimal places.

Explanation:

We know magnetic field due to a charge q moving with a velocity v [tex]=\dfrac{\mu_oq}{4\pi r^3}(v \times r)[/tex].

Case A:

r=0.5i

Putting value of r, q and v .

We get, [tex]B=-2.34\times 10^{-5} \ k.[/tex] [tex](\ j\times i=-k)[/tex]

Case B:

r=0.5 j

B=0 [tex]( j\times j=0)[/tex]

Case C:

We get, [tex]B=2.34\times 10^{-5} \ i\ (Since\ , j \times k=i)[/tex]

Case D:

We get , [tex]B=8.28\times 10^{-6} \ T \ i[/tex].

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Answer:

a) 11 m/s

b) 0.0564 s

Explanation:

Given:

m = 2100 kg

vi = 22 ..... m/s before collision

vf = 0 ......after collision to stop

Δs = 0.62   distance traveled after collision .. crumpling of truck

Part a

[tex]V_{avg} = \frac{vi- vf}{2}\\\\V_{avg} = \frac{22- 0}{2}\\\\ V_{avg} = 11 m/s[/tex]

Part b

[tex]vf = vi + a*t\\vf^2 = vi^2 + 2*a*s\\\\0 = 22 + a*t\\t = -22 /a\\\\0 = 484 +2*a*(0.62)\\a = - 390.3232 m/s^2\\\\t = -22/(-390.3232)\\\\t = 0.0564 s[/tex]

(a) The required average speed of the truck during collision is 11 m/s.

(b) The required time interval for the collision is 0.058 s.

(c)  The required magnitude of the average force exerted by the wall on the truck is  [tex]7.96 \times 10^{5} \;\rm N[/tex].

(d) The required ratio of the force on the truck and the gravitational force is 38.67 : 1.

(e) The required approximations are:

The section of analysis that deals with the motion of any object in one dimension are known as linear kinematics. The terms such as speed, velocity, and acceleration are the variables under kinematics.

Given data:

The mass of the truck is, m = 2100 kg.

The speed of the truck is, v = 22 m/s.

The distance crumpled by the truck is, d = 0.62 m.

(a)

Since the truck is going to stop finally (v' = 0) therefore the average speed is calculated as,

[tex]v_{av.}=\dfrac{v - v'}{2}\\\\ v_{av.}=\dfrac{22 - 0}{2}\\\\ v_{av.}=11 \;\rm m/s[/tex]

Thus, the required average speed of the truck during collision is 11 m/s.

(b)

Now, apply the first kinematic equation of motion as,

[tex]v' = v + at[/tex]

Here, a is the linear acceleration and t is the time interval for the collision.

Solving as,

[tex]0 = 22 + at\\\\ a = -22/t[/tex]

Now, apply the second kinematic equation as,

[tex]v'^{2}=u^{2}+2ad\\\\ 0^{2}=22^{2}+2 \times \dfrac{-22}{t} \times 0.62\\\\ \dfrac{27.28}{t}=484\\\\ t = 0.058 \;\rm s[/tex]

Thus, we can conclude that the required time interval for the collision is 0.058 s.

(c)

The expression for the magnitude of average force exerted by wall on truck is,

[tex]F_{av.} = \dfrac{mv}{t}[/tex]

Solving as,

[tex]F_{av.}=\dfrac{2100 \times 22}{0.058}\\\\ F_{av.}=7.96 \times 10^{5} \;\rm N[/tex]

Thus, the required magnitude of the average force exerted by the wall on truck is  [tex]7.96 \times 10^{5} \;\rm N[/tex].

(d)

The ratio of force of the truck and the gravitation force on the truck is,

[tex]= \dfrac{F_{av}}{mg}[/tex]

Here, g is the gravitational acceleration.

Solving as,

[tex]=\dfrac{7.96 \times 10^{5} \;\rm N}{2100 \times 9.8}\\\\ =38.67[/tex]

Thus, the required ratio of the force on the truck and the gravitational force is 38.67 : 1.

(e)

The approximations that were necessary to make these approximations include the negligence of the horizontal component of a force and the assumption of a constant force exerted by the wall.

This is because:

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Answer:

a) 73.48 m/s

b) 313.92 m

Explanation:

Data provided in the question:

ascending velocity = - 5 m/s      [ negative sign depicts upward movement]

Time taken by bag to hit the ground, t = 8 s

a) from the Newton's equation of motion

we have  

[tex]s=ut+\frac{1}{2}at^2[/tex]

where,  

u is the initial speed  

a is the acceleration = 9.81 m/s²   (since it is a case of free fall )

s is the distance

thus,

[tex]s=(-5)(8)+\frac{1}{2}(9.81)(8)^2[/tex]

s = - 40 + 313.92

s = 273.92 m

from

v = u + at

v is the final speed

v = -5 + (9.81)(8)

or

v = 73.48 m/s

b) Distance traveled by balloon  = Speed × Time

= 5 × 8

= 40 m

Therefore,

Altitude of the balloon

= Distance traveled by bag + Distance traveled by balloon

= 273.92 m + 40 m

= 313.92 m

The speed of the bag as it hits the ground is 5 m/s. The altitude of the balloon when the bag hits the ground is 40 meters.

To determine the speed of the bag as it hits the ground, we can use the equation for velocity:

v = u + at

Since the bag is dropped with an initial velocity of 5 m/s and there is no acceleration in the vertical direction, the final velocity of the bag as it hits the ground is also 5 m/s.

To determine the altitude of the balloon when the bag hits the ground, we can calculate the distance traveled by the bag using the equation for distance:

s = ut + (1/2)at^2

Plugging in the values, we get:

s = 5(8) + (1/2)(0)(8^2) = 40 meters

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Answer: Δβ (dB) = -13.1dB

Explanation:

The intensity of sound is inversely proportional to the square of the distance between them.

I ∝ 1/r²

I₁/I₂= r₂²/r₁² .....1

When the listener increases his distance from the source by a factor of 4.49.

Then,

r₂/r₁= 4.49

From equation 1

I₁/I₂ = (4.49)²

I₁/I₂ = 20.16

I₂/I₁ = 1/20.16

The change in sound intensity in dB can be given as

Δβ (dB) = 10 log(I₂/l₁) = 10log(1/20.6) = -13.1dB

The change in the sound intensity level in dB is -13.1 dB.

The given parameters;

The relationship between intensity of sound and distance is calculated as follows;

[tex]I = \frac{k}{r^2} \\\\I_1r_1^2 = I_2r_2^2\\\\I_2 = \frac{I_1 r_1^2 }{r_2^2} \\\\I_2 = \frac{I_1 r_1^2}{(4.49r_1)^2} \\\\I_2 = \frac{I_1r_1^2}{20.16r_1^2} \\\\I_2 = \frac{I_1}{20.16} \\\\\frac{I_2}{I_1} = \frac{1}{20.16}[/tex]

The change in sound intensity in dB is calculated as follows;

[tex]\Delta \beta = 10 \ log[\frac{I_2}{I_1} ]\\\\\Delta \beta = 10 \times log [\frac{1}{20.16} ]\\\\\Delta \beta = -13.1 \ dB[/tex]

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Answer:

D

Explanation:

Static friction is a force that keeps an object at rest. It must be overcome to start moving the object. Once an object is in motion, it experiences kinetic friction.

When a bus suddenly stops, a passenger tends to keep moving forward due to inertia (Newton's first law). The force causing you to fall forward in this instance is the insufficient static friction between you and the floor of the bus, which isn't enough to counteract your forward momentum.

The force acting on you that causes you to fall forward when the bus comes to an immediate stop is B) the force due to static friction between you and the floor of the bus. This is because of Newton's first law, also known as the law of inertia, which states that an object at rest tends to stay at rest and an object in motion tends to stay in motion, unless acted upon by an external force.

While you're standing in the moving bus, both you and the bus are moving forward. When the bus suddenly stops, your body tends to keep moving forward due to inertia, causing you to fall forward. Here, the static friction between you and the floor of the bus isn't enough to counteract your momentum, leading to your forward fall.

It's important to note that other forces such as gravity and normal force are in effect too. Gravity pulls you downward, and the bus floor exerts an upward normal force to counterbalance it. These two forces are balanced and don't contribute to your forward movement. The force causing the unbalanced motion (falling forward) is the lack of sufficient friction to oppose your inertia.

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Geographically throughout this area of Mexico, Central America Caribbean is located the Cocos plate. This area is scientifically known as the Central American subduction zone.

In order for a volcano to form, there is usually a clash between the technical plates that generates the elevation of the ground and the connection with ducts that release the magma from the earth. If this entire area is a subduction area, it will also be a land stress release area where volcano lines will be formed, that is, it is a convergent plate boundary area

The relationship between the epicenters of the earthquake and the position of the volcano should be explained below.

When the Mexico area should be geographically located, so the Caribbean of central America should be located on the Cocos plate. So this area should be called as the subduction zone. At the time when the volcano should be created so there is normally clash that lies between the technical plats where it generated the ground elevation and the linked with the ducts due to this, it releases the magma from the earth.

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Answer:

Explanation:

Given

Charge on metal sphere [tex]Q=12\mu C[/tex]

no of electrons [tex]n=5.9\times 10^{13}[/tex]

Charge on each electron [tex]q=-1.6\times 10^{19}\ mu C[/tex]

Charge by Possessed by Electrons [tex]Q_2=-1.6\times 10^{19}\times 5.9\times 10^{13}[/tex]

[tex]Q_2=-9.44\mu C[/tex]

Net Charge on Sphere [tex]Q_{net}=Q+Q_2[/tex]

[tex]Q_{net}=12-9.44=2.56\mu C[/tex]

Answer

Organ Pipe A has both end open.

Organ Pipe B has one end open.

speed of sound, v = 343 m/s

The fundamental frequency = 270 Hz

wavelength of the pipe when both end open

  λ = 2 L

we know,

 [tex]\lambda = \dfrac{v}{f}[/tex]

now,

 [tex]L = \dfrac{v}{2f}[/tex]

inserting all the values

 [tex]L = \dfrac{343}{2\times 270}[/tex]

  L_A = 0.635 m

length of pipe A is equal to 0.635 m

b) since third harmonic of pipe B is equal to second harmonic of pipe A

     [tex]f_B = f_A[/tex]

     [tex]\dfrac{n_BV}{4L_B} = \dfrac{n_AV}{2L_A} [/tex]

     [tex]L_B= \dfrac{2n_BL_A}{4n_A}[/tex]      

     [tex]L_B= \dfrac{2\times 3 \times 0.635}{4\times 2}[/tex]

        L_B = 0.476 m

length of pipe B is equal to 0.476 m.

Answer:

[tex]1.67\times 10^{-27}kg[/tex]

Explanation:

We are given that mass of proton

[tex]1.672623\times 10^{-27}kg[/tex]

There are seven significant figures.

We have to round off.

If we round off to three  significant figures

The  thousandth place  of given mass of proton  is less than five therefore, digits on left side of  thousandth  place remains same and digits on right side of thousandth place and thousandth  replace by zero

Therefore, the mass of proton can be written as

[tex]1.67\times 10^{-27}kg[/tex]

Hence, the mass of proton is [tex]1.67\times 10^{-27}kg[/tex]

To round the mass of a proton to seven significant figures, the correct way is to round up the last significant figure if it is 5 or greater, and if it is less than 5, simply drop the remaining digits.

To round the mass of a proton, we look at the digit right after the desired number of significant figures, which in this case is the seventh significant figure. If this digit is 5 or greater, we round up the last significant figure. If it is less than 5, we simply drop the remaining digits.

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The speed in the second segment is 36 m/s, in the third segment is 144 m/s, and the volume flow rate through the pipe is 3.38 L/s.

A: Using the principle that the volume flow rate remains constant in an incompressible fluid, we can calculate the speed in the second segment to be 36 m/s and the speed in the third segment to be 144 m/s.

B: The volume flow rate through the pipe can be calculated by applying the equation Q = Av, where Q represents the flow rate, A is the cross-sectional area, and v is the velocity of the fluid. Given the diameter changes, the volume flow rate is 3.38 L/s.

C: The primary concept involved in this problem is the relationship between cross-sectional area and fluid velocity as the diameter changes along the pipe, affecting the flow rate.

Free energy is a thermodynamic potential that can be used to calculate the maximum reversible work that can be performed by a thermodynamic system at a constant temperature and pressure. It is fulfilled that if the energy change is less than zero it will mean that the relationship will proceed towards the product, while if the relationship is greater than zero the reaction will proceed towards the reactant. Therefore the correct option is D.

The free energy change of a reaction can determine the reaction direction

Answer:

D = 271.54 m

Explanation:

given,

1. car accelerates at 4.6 m/s² for 6.2 s

2. constant speed for 2.1 s

3. slows down at 3.3 m/s²

distance travel for case 1

using equation of motion

 [tex]d_1 = u t +\dfrac{1}{2}at^2[/tex]

 [tex]d_1 =\dfrac{1}{2}\times 4.6\times 6.2^2[/tex]

      d₁ = 88.41 m

case 2

constant speed for 2.1 s now, we have to find velocity

v = u  + at

v = 0 + 4.6 x 6.2

v = 28.52 m/s

distance travel in case 2

d₂ = v x t

d₂ = 28.52 x 2.1 = 59.89 m

for case 3

distance travel by the car

v² = u² + 2 a s

final velocity if the car is zero

0² = 28.52² + 2 x (-3.3) x d₃

6.6 d₃ = 813.39

 d₃ = 123.24 m

total distance travel by the car

D = d₁ + d₂ + d₃

D = 88.41 + 59.89 + 123.24

D = 271.54 m

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

[tex]v = \dfrac{2\pi R}{T}[/tex]

[tex]v = \dfrac{2\pi\times 8}{4.30}[/tex]

  v = 11.69 m/s

now, Force does the ring push on her at the top

[tex]- N - m g = \dfrac{-mv^2}{R}[/tex]

[tex] N + m g = \dfrac{mv^2}{R}[/tex]

[tex] N = \dfrac{mv^2}{R}- m g[/tex]

[tex] N = m(\dfrac{v^2}{R}- g)[/tex]

[tex] N = 58\times (\dfrac{11.69^2}{8}- 9.8)[/tex]

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

The force does the ring push on her at the top of the ride will be [tex]N=422.36\ Newton[/tex]

It is Given that

Time rotation  T= 4.30 s

Mass m= 58 kg

Now the Velocity will be calculated as

[tex]V=\dfrac{2\pi r}{T} =\dfrac{2\pi 8}{4.30} =11.69 \frac{m}{s}[/tex]

Now by balancing the forces

[tex]N=\dfrac{mv^2}{R} -mg[/tex]

[tex]N=m(\dfrac{v^2}{R} -g)[/tex]

[tex]N=58\times (\dfrac{11.69^2}{8} -9.8)[/tex]

[tex]N=422.36 \ Newton[/tex]

Thus the force does the ring push on her at the top of the ride will be [tex]N=422.36\ Newton[/tex]

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Answer:

[tex]V=-130000x+1080000[/tex]

Explanation:

Linear Dependence

Some variables are known or assumed to have linear dependence which means the graph of the ordered pairs (x,V) is a straight line.

If we know two points of the line, we can come up with the exact equation and therefore make predictions for other values of x

The linear depreciation gives us these points (2,820000) and (5,430000)

The general equation of the line is

[tex]V=mx+b[/tex]

Where V is the machine value and x is the  number of years after purchase. We need to find the values of m and b.

Replacing the first point

[tex]820000=m(2)+b[/tex]

[tex]2m+b=820000[/tex]

Replacing the second point

[tex]5m+b=430000[/tex]

Subtracting them  

[tex]-3m=390000[/tex]

[tex]m=-130000[/tex]

Replacing in any of the equations, say, the first one

[tex]2(-130000)+b=820000[/tex]

Solving for b

[tex]b=820000+260000[/tex]

[tex]b=1080000[/tex]

The formula for the machine value V is  

[tex]\boxed{V=-130000x+1080000}[/tex]

Formula for the machine value V in x year is v = -130000(x) + 1,080,000

What information do we have?

Cost of machine after 2 year = $820,000

Cost of machine after 5 year = $430,000

Liner depreciation equation

v = mx + b

After 2 year

820,000 = m(2) + b

820,000 = 2m + b .........   Eq1

After 5 year

430,000 = m(5) + b

430,000 = 5m + b .........   Eq2

Eq2 - Eq1

3m = -390,000

m = -130,000

From Eq

430,000 = 5m + b

430,000 = 5(-130,000) + b

b = 1,080,000

Liner equation of tha cost.

Amount of machine = mx + b

Amount of machine = -130000(x) + 1,080,000

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To solve this problem we will apply the laws of gaus that relate the Electric Flow as the charge of the object on the permittivity constant of free space. Mathematically this is

[tex]\phi = \frac{Q}{\epsilon_0}[/tex]

Rearranging to find the charge,

[tex]Q = \phi \epsilon_0[/tex]

Here

Q = Charge

[tex]\phi =[/tex] Electric Flux

[tex]\epsilon_0 =[/tex] Permittivity of free space

The total flux would be

[tex]\phi_T = \phi_1+\phi_2+\phi_3+\phi_4+...+\phi_{\infty}[/tex]

[tex]\phi = ( 1500+2200+4600-1800-3500-5400 ) N\cdot m^2 / C[/tex]

[tex]\phi = - 2400 N\cdot m^2 / C[/tex]

Replacing we have that,

[tex]Q = (-2400 N\cdot m^2/C)( 8.85*10^{-12} C^2 / N \cdot m^2)[/tex]

[tex]Q = -21240 * 10^{-12} C[/tex]

[tex]Q = - 21.24 nC[/tex]

Therefore the charge Q inside a rectangular box is -21.24nC

Final answer:

Using Gauss's Law, the charge Q inside the box can be calculated by summing the electric flux values across all six surfaces and then multiplying by the permittivity of free space, resulting in a charge of approximately -1.24 x 10^-8 C.

Explanation:

To determine the charge Q inside the rectangular box, we need to use Gauss's Law, which relates the electric flux through a closed surface to the charge enclosed within that surface. The net electric flux (Φnet) through a closed surface is equal to the charge inside (Q) divided by the permittivity of free space (ε0). Mathematically, this is expressed as Φnet = Q/ε0.

The net flux is the algebraic sum of the fluxes through each surface, so we have:

Φnet = (electric flux 1) + (electric flux 2) + (electric flux 3) + (electric flux 4) + (electric flux 5) + (electric flux 6)

Substituting the given values, we get:

Φnet = (+1500 N·m2/C) + (+2200 N·m2/C) + (+4600 N·m2/C) + (-1800 N·m2/C) + (-3500 N·m2/C) + (-5400 N·m2/C)

Summing these yields:

Φnet = -1400 N·m2/C

Assuming ε0 is the permittivity of free space (approximately 8.854 x 10-12 C2/N·m2), we can find Q:

Q = Φnet ε0

Q = (-1400 N·m2/C)(8.854 x 10-12 C2/N·m2)

Q ≈ -1.24 x 10-8 C

Thus, the charge Q inside the box is approximately -1.24 x 10-8 C.

Answer:

Explanation:

The Balmer series in a hydrogen atom relates the possible electron transitions down to the n = 2 position to the wavelength of the emission that scientists observe. In quantum physics, when electrons transition between different energy levels around the atom (described by the principal quantum number, n) they either release or absorb a photon. The Balmer series describes the transitions from higher energy levels to the second energy level and the wavelengths of the emitted photons. You can calculate this using the Rydberg formula.

Answer:

(a). The power delivered to the element is 187.68 mW

(b). The energy delivered to the element is 57.52 mJ.

Explanation:

Given that,

Charge [tex]q=5\sin4\pi t\ mC[/tex]

Voltage [tex]v=3\cos4\pi t\ V[/tex]

Time t = 0.3 sec

We need to calculate the current

Using formula of current

[tex]i(t)=\dfrac{dq}{dt}[/tex]

Put the value of charge

[tex]i(t)=\dfrac{d}{dt}(5\sin4\pi t)[/tex]

[tex]i(t)=5\times4\pi\cos4\pi t[/tex]

[tex]i(t)=20\pi\cos4\pi t[/tex]

(a).We need to calculate the power delivered to the element

Using formula of power

[tex]p(t)=v(t)\times i(t)[/tex]

Put the value into the formula

[tex]p(t)=3\cos4\pi t\times20\pi\cos4\pi t[/tex]

[tex]p(t)=60\pi\times10^{-3}\cos^2(4\pi t)[/tex]

[tex]p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi t}{2})[/tex]

Put the value of t

[tex]p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi\times0.3}{2})[/tex]

[tex]p(t)=30\pi\times10^{-3}(1+\cos8\pi \times0.3)[/tex]

[tex]p(t)=187.68\ mW[/tex]

(b). We need to calculate the energy delivered to the element between 0 and 0.6 s

Using formula of energy

[tex]E(t)=\int_{0}^{t}{p(t)dt}[/tex]

Put the value into the formula

[tex]E(t)=\int_{0}^{0.6}{30\pi\times10^{-3}(1+\cos8\pi \times t)}[/tex]

[tex]E(t)=30\pi\times10^{-3}\int_{0}^{0.6}{1+\cos8\pi \times t}[/tex]

[tex]E(t)=30\pi\times10^{-3}(t+\dfrac{\sin8\pi t}{8\pi})_{0}^{0.6}[/tex]

[tex]E(t)=30\pi\times10^{-3}(0.6+\dfrac{\sin8\pi\times0.6}{8\pi}-0-0)[/tex]

[tex]E(t)=57.52\ mJ[/tex]

Hence, (a). The power delivered to the element is 187.68 mW

(b). The energy delivered to the element is 57.52 mJ.

To solve the problem, the power at a certain moment (t=0.3s) is calculated by substitifying the values of voltage and charge into the power formula. The energy delivered to the element can be calculated by integrating the power function over the period from 0 to 0.6 s.

The subject of this question is related to electrical power and energy in a circuit, and it requires a knowledge of trigonometry. The instantaneous power in an electrical circuit is given by the product of charge (q) and voltage (v). Hence we can calculate the power at t = 0.3s by substituting the given values into the power formula.

Power, p = qv = (5 sin 4πt) * (3 cos 4πt) = 15 sin 4πt cos 4πt. At t = 0.3s, power p = 15 sin 4π * 0.3 cos 4π * 0.3 = 15 sin 1.2π cos 1.2π.

For part (b), the energy delivered to the element between 0 and 0.6 s can be obtained from the integral of the power function over the given time interval, which requires integration skills in calculus.

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Answer:

27000 Nm

Explanation:

The boom end at A is fixed and end at B is subjected to a 3kN force. Boom AB has length of 9m. The moment about A is the product of force 3kN (3000 N) at B and the moment arm of 9m

M = FL = 3000 * 9 = 27000 Nm

So the moment about A is 27000 Nm

Answer:

Explanation:

Given

Area of capacitor Plates [tex]A=L\times L[/tex]

distance between plates is d

capacitance C is given by

[tex]C=\frac{\epsilon A}{d}[/tex]

[tex]C=\frac{\epsilon \cdot L^2}{d}[/tex]

Provided V is Voltage

[tex]Charge(Q)=capacitance(C)\times Voltage(V)[/tex]

If L is doubled

Capacitance [tex]C'=\frac{\epsilon \cdot (2L)^2}{d}[/tex]

[tex]C'=4\times \frac{\epsilon \cdot L^2}{d}[/tex]

Electric field is given by

[tex]E=\frac{Q}{\epsilon _0A}[/tex]

[tex]E_i=\frac{Q}{\epsilon _0L^2}---1[/tex]

[tex]E_f=\frac{Q}{\epsilon _0(2L)^2}---2[/tex]

divide 1 and 2 we get

[tex]\frac{E_i}{E_f}=\frac{(2L)^2}{L^2}[/tex]

[tex]\frac{E_f}{E_i}=\frac}{1}{4}[/tex]

Answer:

Yes it is possible to charge balloon to several thousand of volts and the balloon will also store several Joules of energy.

Explanation:

By rubbing a balloon on one's hair or on different types of clothing, the balloon either gain or loss electron.

If the balloon gains electron it becomes negatively charged, it contains more electron and subsequently charged to several thousand volts.

Also, if the balloon losses electron, it becomes positively charged. In this case it contains more proton, which makes the balloon positively charged to several thousand volts.

However, amount of joules depends on volts produced in the balloon.

Volt = Joules/coulomb,

Joules = volts*coulomb

Because the charge of the particles (electron and proton) are small, amount of joules will always be small than volts.

So, it is possible to charge balloon to several thousand of volts and the balloon will store several Joules of energy

Answer:

Electric field strength=8.81×10⁶V/m

Explanation:

Given Data

voltage v= 74.0 mV

Membrane thickness d=8.40 nm

To find

Electric field strength E=?

Solution

Electric field strength =voltage/Membrane thickness

[tex]E=v/d\\E=\frac{74.0*10^{-3} }{8.40*10^{-9} }\\ E=8.81*10^{6}V/m[/tex]

Answer:

(a) 1.16 s

(b)0.861 Hz

Explanation:

(a) Period : The period of a simple harmonic motion is the time in seconds, required for a object undergoing oscillation to complete one cycle.

From the question,

If 1550 cycles is completed in (30×60) seconds,

1 cycle is completed in x seconds

x = 30×60/1550

x = 1.16 s

Hence the period is 1.16 seconds.

(b) Frequency : This can be defined as the number of cycles that is completed in one seconds, by an oscillating body. The S.I unit of frequency is Hertz (Hz).

Mathematically, Frequency is given as

F = 1/T ........................... Equation 1

Where F = frequency, T = period.

Given: T = 1.16 s.

Substitute into equation 1

F = 1/1.16

F = 0.862 Hz

Hence thee frequency = 0.862 Hz

Answer

given,

near point = 18 cm

far point = 40 cm

a) The lens should form an upright, virtual image at far point from the distant object.

therefore, f = q  = -40 cm = -0.4 m

where f is the focal length.

the required power

[tex]P =\dfrac{1}{f}[/tex]

[tex]P =\dfrac{1}{-0.40}[/tex]

       P = -2.5 D

b) If the lens is used the Person's near point

The lens should form an upright, virtual image at near point from the distant object should be q = - 18 cm = -0.18 m

    [tex]p = \dfrac{qf}{q-f}[/tex]

    [tex]p = \dfrac{(-0.18)(-0.4)}{-0.18-(-0.4)}[/tex]

     p = 32.72 cm

The person's near point is 32.72 cm

Answer:

Please refer to the attachment below since we need to prove that the period of motion is 2π*(sqrt(h/g))

Explanation:

Answer:

The Proof for T=2π (sqrt(h/g) for a floating block exhibiting SHM is shown in the pictures attached below

Explanation:

The net torque on the circular object, considering the three forces acting on it, is calculated to be 1.46 N × m in an anti-clockwise direction.

To find the net torque on the circular object about its axle, we will first consider each of the three forces acting on it separately and determine the torque produced by each force. Torque (Τ) is defined by the equation Τ = r  ×  F  × sin(θ), where r is the radius at which the force is applied, F is the magnitude of the force, and θ is the angle between the force and the direction of the radius.

The 12 N force acts on the outer radius, so r = 0.26 m (26 cm converted to meters). Since the force is perpendicular to the radius, θ = 90 degrees, and the sin(90) = 1. Therefore, the torque from this force is Τ = 0.26 m × 12 N × 1 = 3.12 N × m.

The 16 N force also acts on the outer radius at 90 degrees to the radius, so its torque is Τ = 0.26 m × 16 N × 1 = 4.16 N   × m.

The 26 N force acts at an angle 30 degrees below the horizontal, so it makes an angle of 60 degrees with the radius. Hence, the torque from this force is Τ = 0.26 m × 26 N  × sin(60) = 5.82 N × m (rounded to two decimal places).

To find the net torque, we need to consider the direction of each torque. Both the 12 N and 16 N forces produce torque in the same direction (let's say clockwise), whereas the 30 degree component of the 26 N force produces torque in the opposite direction (counter-clockwise).

Net torque = Torque from 12 N + Torque from 16 N - Torque from 26 N = 3.12 N × m + 4.16 N × m - 5.82 N × m = 1.46 N   × m (anti-clockwise).

To develop this problem we will start using the concept of maximum speed for this type of systems. The maximum velocity can be described as the product between the Amplitude and the Angular velocity. At the same time, said angular velocity can be found through the relationship between linear and "angular wavenumber" velocity. The Angular wavenumber is a wave number defined as the number of radians per unit distance. Finally with the value of the angular velocity found we will proceed to find the maximum speed.

The maximum speed is given by

[tex]v_{max} = A\omega[/tex]

Here,

A = Amplitude

[tex]\omega[/tex]= Angular velocity

The angular velocity can be described as the number of radians per unit distance

[tex]\omega = vk[/tex]

[tex]\omega = v (\frac{2\pi}{\lambda})[/tex]

[tex]\omega = 112(\frac{2\pi}{12.16})[/tex]

[tex]\omega =57.8714rad/s[/tex]

Then,

[tex]v_{max} = 0.12 *57.8714[/tex]

[tex]v_{max} = 6.94m/s[/tex]

Therefore the maximum speed a point on the medium moves as this wave passes is 6.94m/s

Answer:

The calculated vectors are:

[tex]\vec{A}-\vec{B}=(3,-5,-4)[/tex]

[tex]\vec{B}-\vec{C}=(-5,4,0)[/tex]

[tex]-\vec{A}+\vec{B}-\vec{C}=(-6,4,3)[/tex]

[tex]3\vec{A}-2\vec{C}=(-3,-2,-11)[/tex]

Explanation:

To operate with vectors, you sum or rest component to component. To multiply scalars with vectors, you distribute the scalar with each component of the vector. These are the following rules you must apply in these cases:

[tex]\vec{V}+\vec{W}=(V_1,V_2,V_3)+(W_1,W_2,W_3)=(V_1+W_1,V_2+W_2,V_3+W_3)[/tex] (1)

[tex]\vec{V}-\vec{W}=(V_1,V_2,V_3)-(W_1,W_2,W_3)=(V_1-W_1,V_2-W_2,V_3-W_3)[/tex] (2)

[tex]\alpha\cdot\vec{V}=\alpha\cdot(V_1,V_2,V_3)=(\alpha\cdot V_1,\alpha\cdot V_2,\alpha\cdot V_3)[/tex] (3)

The operations in these cases are:

[tex]\vec{A}-\vec{B}=(1,0, -3)-(-2,5,1)=(3,-5,-4)[/tex]

[tex]\vec{B}-\vec{C}=(-2,5,1)-(3,1,1)=(-5,4,0)[/tex]

[tex]-\vec{A}+\vec{B}-\vec{C}=-(1,0, -3)+(-2,5,1)-(3,1,1)=(-6,4,3)[/tex]

[tex]3\vec{A}-2\vec{C}=3(1,0, -3)-2(3,1,1)=(3,0, -9)-(6,2,2)=(-3,-2,-11)[/tex]

Answer:

Part A: (3, -5, -4)

Part B: (-5, 4, 0)

Part C: (-6, 5, 3)

Part D: (-3, -2, -11)

Part E: (17, -12, -6)

Explanation:

This problem involves addition and subtraction of vectors. This can be done by adding and subtracting the respective components of each vector as the case may be.

The full descriptive solution can be found in the attachment below.

Answer:

The charge carried by the droplet is [tex]1.330\times10^{-19}\ C[/tex]

Explanation:

Given that,

Distance =8.4 cm

Time = 0.250 s

Suppose tiny droplets of oil acquire a small negative charge while dropping through a vacuum in an experiment. An electric field of magnitude [tex]5.92\times10^4\ N/C[/tex] points straight down and if the mass of the droplet is [tex]2.93\times10^{-15} kg[/tex]

We need to calculate the acceleration

Using equation of motion

[tex]s=ut+\dfrac{1}{2}at^2[/tex]

Put the value into the formula

[tex]8.4\times10^{-2}=0+\dfrac{1}{2}\times a\times(0.250)^2[/tex]

[tex]a=\dfrac{8.4\times10^{-2}\times2}{(0.250)^2}[/tex]

[tex]a=2.688\ m/s^2[/tex]

We need to calculate the charge carried by the droplet

Using formula of electric filed

[tex]E=\dfrac{F}{q}[/tex]

[tex]q=\dfrac{ma}{E}[/tex]

Put the value into the formula

[tex]q=\dfrac{2.93\times10^{-15}\times2.688}{5.92\times10^4}[/tex]

[tex]q=1.330\times10^{-19}\ C[/tex]

Hence, The charge carried by the droplet is [tex]1.330\times10^{-19}\ C[/tex]

When you slide a coin across the floor, it slows down and eventually stops due to the force of friction. The friction converts the coin's kinetic energy into thermal energy, resulting in an increase in temperature.

When you slide a coin across the floor, it eventually slows down and stops due to the force of friction acting on it. Friction is a force that opposes the motion of objects in contact, and it causes the coin to lose kinetic energy, slowing it down. As the coin slows down, its kinetic energy is converted into thermal energy, increasing the temperature of the coin and the surface it slides on. This is why a sensitive thermometer shows an increase in temperature when the coin slides.