Complete Question
An oil tanker has collided with a smaller vessel, resulting in an oil spill in a large, calm-water bay of the ocean. You are investigating the environmental effects of the accident and need to know the area of the spill. The tanker captain informs you that 18000 liters of oil have escaped and that the oil has an index of refraction of n = 1.1. The index of refraction of the ocean water is 1.33. From the deck of your ship you note that in the sunlight the oil slick appears to be blue. A spectroscope confirms that the dominant wavelength from the surface of the spill is 485 nm. Assuming a uniform thickness, what is the largest total area oil slick
Answer:
The largest total area of the oil slick [tex]A = 8.257 *10^{9} \ m^2[/tex]
Explanation:
From the question we are told that
The volume of oil the escaped is [tex]V = 18000 \ L[/tex]
The refractive index of oil is [tex]n_o = 1.1[/tex]
The refractive index of water is [tex]n_w = 1.33[/tex]
The wavelength of the light is [tex]\lambda = 485 \ nm = 485 * 10^{-9} \ m[/tex]
Generally the thickness of the oil for condition of constructive interference between the oil and the water is mathematically represented as
[tex]d = m *\frac{\lambda}{2n_w}[/tex]
Where is the order of interference of the light and it value ranges from 1, 2, 3,...n
It is usually take as 1 unless stated otherwise by the question
substituting value
[tex]d = 1 * \frac{485 *10^{-9}}{2 * 1.1}[/tex]
[tex]d = 218 nm[/tex]
The are can be mathematically evaluated as
[tex]A = \frac{V}{d}[/tex]
Substituting values
[tex]A = \frac{18000}{218*10^{-8}}[/tex]
[tex]A = 8.257 *10^{9} \ m^2[/tex]
A 0.28-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 500 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = + 0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is x = -0.052 m, find the acceleration of the block.
Answer:
The block will accelerate at 92.86m/s²
Explanation:
The acceleration of a simple harmonic motion of a spring is expressed as
a= - kx/m
Where k = spring constant
x= displacement
m= mass of block
Given data
Spring constant k = 500N/m
Displacement x= - 0.052m
Mass of block m= 0.28kg
Pluging this parameters into the expression for acceleration we have
a= - 500*(-0.052)/0.28
a= 26/0.28
a= 92.86m/s²
Answer:
The block will accelerate at 92.86m/s²
Explanation:
slab of ice floats on water with a large portion submerged beneath the water surface. The slab is in the shape of a rectangular solid. The volume of the slab is 20 m3 and the surface area of the top is 14 m2. The density of ice is 900 kg/m3 and sea water is 1030 kg/m3. No need to show work. a) Calculate the percentage of the volume of the ice that is submerged. b) Calculate the height, in meters, of the portion of the ice that is submerged. c) Calculate the buoyant force acting on the ice. d) Assume a polar bear has a mass of 400 kg. Calculate the maximum number of polar bears that could be supported by the slab without the slab sinking below the surface of the water.
Answer:
a) [tex]\%V = 87.36\,\%[/tex], b) [tex]x = 1.248\,m[/tex], c) [tex]F_{B} = 176488.341\,N[/tex], d) Six polar bears.
Explanation:
a) The slab of ice is modelled by the Archimedes' Principles and the Newton's Laws, whose equation of equilibrium is:
[tex]\Sigma F =\rho_{w}\cdot g \cdot A \cdot x-\rho_{i}\cdot g\cdot V = 0[/tex]
The height of the ice submerged is:
[tex]\rho_{w}\cdot A \cdot x = \rho_{i}\cdot V[/tex]
[tex]x = \frac{\rho_{i}\cdot V}{\rho_{w}\cdot A}[/tex]
[tex]x = \frac{\left(900\,\frac{kg}{m^{3}}\right)\cdot (20\,m^{3})}{\left(1030\,\frac{kg}{m^{3}} \right)\cdot (14\,m^{2})}[/tex]
[tex]x = 1.248\,m[/tex]
The percentage of the volume of the ice that is submerged is:
[tex]\%V = \frac{(1.248\,m)\cdot (14\,m^{2})}{20\,m^{3}} \times 100\,\%[/tex]
[tex]\%V = 87.36\,\%[/tex]
b) The height of the portion of the ice that is submerged is:
[tex]x = 1.248\,m[/tex]
c) The buoyant force acting on the ice is:
[tex]F_{B} = \left(1030\,\frac{kg}{m^{3}} \right)\cdot (1.248\,m)\cdot (14\,m^{2})\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]F_{B} = 176488.341\,N[/tex]
d) The new system is modelled after the Archimedes' Principle and Newton's Laws:
[tex]\Sigma F = -n\cdot m_{bear}\cdot g-\rho_{i}\cdot V \cdot g + \rho_{w}\cdot V\cdot g = 0[/tex]
The number of polar bear is cleared in the equation:
[tex]n\cdot m_{bear} = (\rho_{w} - \rho_{i})\cdot V[/tex]
[tex]n = \frac{(\rho_{w}-\rho_{i})\cdot V}{m_{bear}}[/tex]
[tex]n = \frac{\left(1030\,\frac{kg}{m^{3}} - 900\,\frac{kg}{m^{3}} \right)\cdot (20\,m^{3})}{400\,kg}[/tex]
[tex]n = 6.5[/tex]
The maximum number of polar bears that slab could support is 6.
The slab of ice floating in water is analyzed in terms of the volume submerged, height of submerged portion, buoyant force, and maximum number of polar bears it can support.
Explanation:a) Percentage of the volume submerged:The volume of the submerged portion can be calculated using the formula:
Volume submerged = Volume of ice × (Density of ice / Density of water)
The percentage of the volume submerged can then be calculated by dividing the volume submerged by the total volume of the ice and multiplying by 100.
The height can be calculated by dividing the volume submerged by the surface area of the top.
The buoyant force is equal to the weight of the water displaced by the ice. It can be calculated using the formula:
Buoyant force = Volume of submerged portion × Density of water × Gravitational acceleration
The maximum number of polar bears that can be supported by the slab can be calculated by dividing the buoyant force by the weight of a single polar bear. To prevent the slab from sinking, the buoyant force must be equal to or greater than the weight of the polar bear.
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An air balloon is moving upward at a constant speed of 3 m/s. Suddenly a passenger realizes that she left her camera on the ground. A friend picks it up and throws it upward at 20 m/s at the instant the passenger is 5 m above the ground. (10 pts) a) Calculate the time for camera to reach passenger. b) Calculate the position and velocity of the camera when passenger catches it.
Answer:t=0.3253 s
Explanation:
Given
speed of balloon is [tex]u=3\ m/s[/tex]
speed of camera [tex]u_1=20\ m/s[/tex]
Initial separation between camera and balloon is [tex]d_o=5\ m[/tex]
Suppose after t sec of throw camera reach balloon then,
distance travel by balloon is
[tex]s=ut[/tex]
[tex]s=3\times t[/tex]
and distance travel by camera to reach balloon is
[tex]s_1=ut+\frac{1}{2}at^2[/tex]
[tex]s_1=20\times t-\frac{1}{2}gt^2[/tex]
Now
[tex]\Rightarrow s_1=5+s[/tex]
[tex]\Rightarrow 20\times t-\frac{1}{2}gt^2 =5+3t[/tex]
[tex]\Rightarrow 5t^2-17t+5=0[/tex]
[tex]\Rightarrow t=\dfrac{17\pm \sqrt{17^2-4(5)(5)}}{2\times 5}[/tex]
[tex]\Rightarrow t=\dfrac{17\pm 13.747}{10}[/tex]
[tex]\Rightarrow t=0.3253\ s\ \text{and}\ t=3.07\ s[/tex]
There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .
(b)When passenger catches the camera time is [tex]t=0.3253\ s[/tex]
velocity is given by
[tex]v=u+at[/tex]
[tex]v=20-10\times 0.3253[/tex]
[tex]v=16.747\ m/s[/tex]
and position of camera is same as of balloon so
Position is [tex]=5+3\times 0.3253[/tex]
[tex]=5.975\approx 6\ m[/tex]
Final answer:
The time for the camera to reach the passenger is 0.29 seconds. The position of the camera when the passenger catches it is 4.93 meters and the velocity of the camera at that moment is 17 m/s.
Explanation:
To calculate the time for the camera to reach the passenger, we first need to find the time it takes for the passenger to reach the height of the camera. The passenger is moving upward at a constant speed of 3 m/s, so it will take her 5 / 3 = 1.67 seconds to reach the height of the camera. Since the camera was thrown upward at 20 m/s, we can subtract the passenger's upward velocity to find the relative velocity of the camera with respect to the passenger: 20 - 3 = 17 m/s.
Using the relative velocity, we can calculate the time it takes for the camera to reach the passenger as follows: t = distance / relative velocity = 5 / 17 = 0.29 seconds.
b) To calculate the position of the camera when the passenger catches it, we can multiply the relative velocity by the time it takes for the camera to reach the passenger: position = relative velocity * time = 17 * 0.29 = 4.93 meters. The velocity of the camera when the passenger catches it will be the same as the relative velocity: 17 m/s.
A spring gun with a 75 N/m spring constant is loaded with a 5 g foam dart and isaimed vertically. When the spring is compressed by 10 cm and then released, the fireddart rises to a max height of 5 m above the end of the spring gun. Assuming the dartexperiences a constant friction force due to the air, how fast is it traveling when ithas fallen 2 m from its maximum height
Answer:
The speed is [tex]v = 4.425 m/s[/tex]
Explanation:
From the question we are told that
The spring constant is [tex]k = 75 \ N /m[/tex]
The mass of the foam dart is [tex]m = 5 g = \frac{5}{100} = 0.05 \ kg[/tex]
The compression distance is [tex]d = 10 cm = 0.1 m[/tex]
The height which the gun raised the dart is [tex]h = 5 m[/tex]
The change in height is [tex]\Delta h = 2 m[/tex]
The new height is [tex]h_2 = 5 -2 = 3 m[/tex]
Generally from the law of conservation of energy
[tex]E_s = KE[/tex]
Where [tex]E_s[/tex] is the energy stored in spring and it is mathematically represented as
[tex]E_s = \frac{1}{2} k d^2[/tex]
KE is the kinetic energy possessed by the dart when it is being shut and this is mathematically represented as
[tex]KE = \frac{1}{2} mv^2_r[/tex]
So
[tex]\frac{1}{2} k d^2 = \frac{1}{2} mv^2_r[/tex]
Substituting values
[tex]0.5 * 75 * 0.1 = 0.5 * 0.0005 * v^2_r[/tex]
=> [tex]v_r = \sqrt{\frac{0.5 * 75 * 0.1}{0.5 * 0.0005 } }[/tex]
[tex]v_r = 12.25 m/s[/tex]
When the dart is at the maximum height the
let it acceleration due air resistance be z
So by equation of motion
[tex]v^2 = u^2 - 2ah[/tex]
Where v is the velocity at maximum height which is equal to zero
and u is it initial velocity before reaching maximum height which we calculated as [tex]v_r = 12.25 m/s[/tex]
and a is the acceleration due to gravity + the acceleration due to air resistance
So
a = z+g
= 9.8 + z
=> [tex]v^2 = u^2 - 2(9.8 +z)h[/tex]
Substituting values
[tex]0 = 12.25^2 - 2(9.8 +z)h[/tex]
Making z the subject
[tex]z = \frac{ 12.25}{2 * 5} - 9.8[/tex]
[tex]z = \frac{ 12.25}{2 * 5} - 9.8[/tex]
[tex]z = 5 m/s[/tex]
When the dart is moving downward we can mathematically represent the motion as
[tex]v^2 = u^2 + 2ah[/tex]
Since the motion is downward and air resistance is upward we have that
a = g - z
and the the initial velocity u becomes the velocity at maximum height
i.e u = 0
And v is the velocity the dart has when it is moving downward
So
[tex]v^2 = 0 + 2 * (g -z )h[/tex]
Substituting values
[tex]v = \sqrt{0+ 2 (10 - 5) * 2}[/tex]
[tex]v = 4.425 m/s[/tex]
Describe what ballistic stretching is and why it’s harmful. Then, provide at least two examples of how one should properly stretch? (Site 1)
Answer:
When doing ballistic stretching, it is using motion to bounce and stretch your body past its natural range of motion. In doing so you can harm yourself if you don't have a professional to help you as you might tear, damage, or pop your tendons, ligaments, or joints.
Explanation:
100% On Edge for 2021
The ballistic stretch is one of the dynamic stretching exercise, which can damage the tissues and ligaments if not performed properly and under the expert supervision.
The stretching activity that utilizes the momentum of body to achieve greater range of motion and flexibility, is known as Ballistic stretching. It is one of the intense stretching method that involves the bouncing movements to force the body beyond the normal range of motion.
This can be harmful if an athlete do not have a professional trainer to train for the cause. This may cause tear, damage of tendons, ligaments, or joints.
Following are the ways to perform a perform a proper stretch:
One should balance its body weight by standing on its feet together.The bending of knees should be done in a steady manner, and before this proper warm ups are needed to be done.It is not required to start with higher intensity, one can go with 5-10 repetitions for initial days, after that the repetitions can be increased gradually.Thus, we can conclude that the ballistic stretch is one of the dynamic stretching exercise, which can damage the tissues and ligaments if not performed properly and under the expert supervision.
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atoms with electronegativity differences higher than 1.7 generally form ionic bonds. true or false
Answer: True
Explanation:
A covalent bond is defined as the bond which is formed when there is sharing of electrons and the atoms have electronegative difference between the elements less than 1.7. Example: [tex]H_2[/tex]
Ionic bond is formed when there is complete transfer of electron from a highly electropositive metal to a highly electronegative non metal. The electronegative difference between the elements is more than 1.7. Example: [tex]NaCl[/tex]
QUESTIONS
A 800 kg person sprints up a 700 meter high stairway in 4.00 seconds flat. What is their power output in watts?
Answer:
Power;P = 1372000 W
Explanation:
We are given;
Mass of person;m = 800 kg
Distance sprinted;h = 700m
Time taken;t = 4 seconds
Now, the formula for Power is;
Power = Work done/Time taken
Where work done = Force x Distance
Now, Force is expressed as;
F = mg
Where m is mass and g is acceleration due to gravity = 9.8 m/s²
Thus, plugging in the relevant values into power equation gives;
Power;P = 800 x 9.8 x 700/4
Power;P = 1372000 W
A 12 A fuse is placed in a parallel circuit that has two branches. 8 A flows in branch 1 and 6 A flows in branch 2. This fuse
1. will blow because the total current in this circuit is 14 A which is greater than 12 A.
2. will blow because a 12 A fuse only allows 6 A to flow in each branch.
3. will not blow because each branch is less than 12 A.
4. will not blow because the average current is 7 A which is less than 12 A.
Answer:
1. will blow because the total current in this circuit is 14 A which is greater than 12 A.
Explanation:
According to Kirchoff current law (KCL) which states that the total current flowing in a circuit is equal to the sum of the individual branch current.
If the supply current is greater than the sum of the individual branch current, then the load will collapse or blow off.
In the question given, the total current of the fuse is 12A
Sum of branch currents = current in branch 1 + current in branch 2
= 8A+6A
= 14A
As we can see that the supply current is lower than the sum of the branch current, this will cause the fuse to blow because some of the branch current will be sent back on the fuse and thereby causing the fuse to blow.
An electron in a vacuum is first accelerated by a voltage of 81700 V and then enters a region in which there is a uniform magnetic field of 0.508 T at right angles to the direction of the electron’s motion. The mass of the electron is 9.11 × 10−31 kg and its charge is 1.60218 × 10−19 C. What is the magnitude of the force on the electron due to the magnetic field? Answer in units of N.
Answer:
Magnetic force is equal to [tex]1.37\times 10^{-11}N[/tex]
Explanation:
We have given electron is accelerated with a potential difference of 81700 volt.
Magnetic field B = 0.508 T
Angle between magnetic field and velocity [tex]\Theta =90^{0}[/tex]
Mass of electron [tex]m=9.11\times 10^{-31}kg[/tex]
Charge on electron [tex]e=1.6\times 10^{-19}C[/tex]
By energy conservation.
[tex]\frac{1}{2}mv^2=qV[/tex]
[tex]\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=1.6\times 10^{-19}\times 81700[/tex]
[tex]v=169.4\times 10^6m/sec[/tex]
Magnetic force on electron
[tex]F=qvBsin\Theta[/tex]
[tex]F=1.6\times 10^{-19}\times 169.4\times 10^6\times 0.508\times sin90^{\circ}[/tex]
[tex]=1.37\times 10^{-11}N[/tex]
Answer:
Explanation:
After acceleration under potential difference , velocity v acquired can be calculated by the following expression
V e = 1/2 m v² ;
V is potential under which electron with mass m and charge e is accelerated to velocity v .
81700 x 1.60218 x 10⁻¹⁹ = .5 x 9.11 x 10⁻³¹ x v²
v² = 28737 x 10¹²
v = 169.52 x 10⁶ m /s
Force = Bev , B is magnetic field , e is charge on lectron and v is its velocity
= .508 x 1.60218 x10⁻¹⁹ x 169.52 x 10⁶
= 128 x 10⁻¹³ N.
calcular la longitud de un péndulo que oscila a 10 Hz en santa fe de bogota, sabiendo que en esta ciudad la aceleracion de la gravedad es de 978 cm/s2.
Answer:
[tex]L=2.48*10^{-3} m[/tex]
Explanation:
The period equation for a pendulum is given by:
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex]
and we know that T = 1/f, where f is the frequency, so we will have:
[tex]\frac{1}{f}=2\pi \sqrt{\frac{L}{g}}[/tex]
Now, we just need to solve this equation for L.
[tex]\frac{1}{2\pi f}=\sqrt{\frac{L}{g}}[/tex]
[tex]L=\frac{g}{(2\pi f)^{2}}[/tex]
g is the gravity in Bogota (g=9.78 m/s^{2})f is 10 HzL is the lenght of the pendulum[tex]L=\frac{9.78}{(2\pi*10)^{2}}[/tex]
[tex]L=2.48*10^{-3} m[/tex]
I hope it helps you!
A baseball is thrown a distance of 18m. What is its speed if it takes 0.5 seconds to cover the distance?
Answer:36m/s
Explanation:
Distance=18m
time=0.5 seconds
speed=distance ➗ time
Speed=18 ➗ 0.5
Speed=36m/s
The question is about calculating the speed at which a baseball was thrown given that it traveled a distance of 18 meters in 0.5 seconds. By using the speed formula (speed = distance/time), we find that the baseball was thrown at a speed of 36 meters per second.
Explanation:To calculate the speed of the baseball, we need to use the formula for speed which is speed = distance/time. Here, the distance covered by the baseball is 18m and the time taken is 0.5 seconds.
Substituting these values into the formula, we get speed = 18m / 0.5s = 36 m/s. So, the speed of the baseball is 36 meters per second.
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A block with mass M attached to a horizontal spring with force constant k is moving with simple harmonic motion having amplitude A1. At the instant when the block passes through its equilibrium position, a lump of putty with mass m is dropped vertically onto the block from a very small height and sticks to it. Part APart complete What should be the value of the putty mass m so that the amplitude after the collision is one-half the original amplitude? Express your answer in terms of the variables M, A1, and k. m = 3M Previous Answers Correct Part B For this value of m, what fraction of the original mechanical energy is converted into heat? Express your answer in terms of the variables M, A1, and k.
Answer:
Explanation:
Given
Mass of block is [tex]M[/tex]
spring constant [tex]=k[/tex]
Amplitude is [tex]A_1[/tex]
when putty is placed then amplitude decreases to [tex]\frac{A_1}{2}[/tex]
Initially [tex]\frac{1}{2}kA^2=\frac{1}{2}Mv^2\quad \ldots(i)[/tex]
Conserving momentum
[tex]Mv_o=(m+M)v[/tex]
where [tex]v_o[/tex]=initial velocity
[tex]v=\frac{M}{M+m}v_o[/tex]
Now
[tex]\frac{1}{2}k(\frac{A_1}{2})^2=\frac{1}{2}(M+m)v^2[/tex]
[tex]\frac{1}{2}k(\frac{A_1}{2})^2=\frac{1}{2}(M+m)(\frac{M}{M+m}v_o)^2\quad \ldots(ii)[/tex]
divide (i) and (ii) we get
[tex]\frac{4}{1}=\frac{M}{M+m}\times (\frac{m+M}{m})^2[/tex]
[tex]4=\frac{m+M}{M}[/tex]
[tex]m=3M[/tex]
Fraction of energy converted into heat[tex]=\frac{1}{2}kA_1^2-\frac{1}{2}k(\frac{A_1}{2})^2[/tex]
[tex]=\frac{1}{2}kA_1^2[1-\frac{1}{4}][/tex]
[tex]=\frac{1}{2}kA_1^2[0.75][/tex]
So, [tex]\frac{3}{4}[/tex] fraction is converted into heat energy
A wire loop with 60 turns is formed into a square with sides of length s . The loop is in the presence of a 1.20 T uniform magnetic field B⃗ that points in the negative y direction. The plane of the loop is tilted off the x-axis by θ=15∘ . If i=2.50 A of current flows through the loop and the loop experiences a torque of magnitude 0.0186 N⋅m , what are the lengths of the sides s of the square loop, in centimeters?
Answer:
the length of the sides s is [tex]s = 1.998 \ cm[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 60 \ turn[/tex]
The magnetic field is [tex]B = 1.20 \ T[/tex]
The angle the loop makes with the x-axis [tex]\theta = 15 ^o[/tex]
The current flowing through the loop is [tex]I = 2.50 A[/tex]
The magnitude of the torque is [tex]\tau = 0.0186 \ N[/tex]
the length of the sides of the square is [tex]s[/tex]
Generally, we can represent the torque magnitude as
[tex]\tau = N I A B sin \theta[/tex]
Where A is the area of the square which is mathematically represented as
[tex]A = s^2[/tex]
Substituting this into the formula for torque
[tex]\tau = N I s^2 B sin \theta[/tex]
making s the subject
[tex]s = \sqrt{\frac{\tau }{NIB sin \theta } }[/tex]
Substituting values
[tex]s = \sqrt{\frac{0.0186}{(60) * (2.50) * (1.20) * (sin (15))} }[/tex]
[tex]s = 0.01998 m[/tex]
Converting to centimeters
[tex]s = 0.01998 * 100[/tex]
[tex]s = 1.998 \ cm[/tex]
Answer:
2 cm
Explanation:
To fins the lengths of the sides of the loop you use the following formula for the calculation of the torque experienced by the loop in a magnetic field:
[tex]\tau=NiABsin\theta[/tex]
N: turns of the loop = 60
i: current in the loop = 2.50A
A: area of the loop = s*s
B: magnitude of the magnetic field = 1.20T
θ: angle between the plane of the lop and the direction of B = 15°
BY replacing the values of the torque and the other parameters in (1) you can obtain the area of the loop:
[tex]A=\frac{\tau}{NiBsin\tetha}=\frac{0.0186Nm}{(60)(2.5A)(1.2T)sin15\°}=3.99*10^{-4}m^2[/tex]
but the area is s*s:
[tex]A=\sqrt{s}=\sqrt{3.99*10^{-4}m^2}=0.019m\approx2cm[/tex]
hence, the sides of the square loop have a length of 2cm
A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 2.3 cm from the axis of rotation. (a) Calculate the acceleration of the seed, assuming that it does not slip. (b) What is the minimum value of the coefficient of static friction between the seed and the turntable if the seed is not to slip? (c) Suppose that the turntable achieves its angular speed by starting from rest and undergoing a constant angular acceleration for 0.36 s. Calculate the minimum coefficient of static friction required for the seed not to slip during the acceleration period.
Answer:
a) [tex]a_{r} = 0.275\,\frac{m}{s^{2}}[/tex], b) [tex]\mu_{s} = 0.028[/tex], c) [tex]\mu_{s} = 0.036[/tex]
Explanation:
a) The linear acceleration of the watermelon seed is:
[tex]a_{r} = \omega^{2}\cdot r[/tex]
[tex]a_{r} = \left[\left(33\,\frac{rev}{min} \right)\cdot \left(2\pi\,\frac{rad}{rev} \right)\cdot \left(\frac{1}{60}\,\frac{min}{s} \right)\right]^{2}\cdot (0.023\,m)[/tex]
[tex]a_{r} = 0.275\,\frac{m}{s^{2}}[/tex]
b) The watermelon seed is experimenting a centrifugal acceleration. The coefficient of static friction between the seed and the turntable is calculated by the Newton's Laws:
[tex]\Sigma F = \mu_{s}\cdot m\cdot g = m\cdot a[/tex]
[tex]a = \mu_{s}\cdot g[/tex]
[tex]\mu_{s} = \frac{a}{g}[/tex]
[tex]\mu_{s} = \frac{0.275\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]\mu_{s} = 0.028[/tex]
c) Angular acceleration experimented by the turntable is:
[tex]\alpha = \frac{\omega-\omega_{o}}{\Delta t}[/tex]
[tex]\alpha = \frac{3.456\,\frac{rad}{s}-0\,\frac{rad}{s} }{0.36\,s}[/tex]
[tex]\alpha = 9.6\,\frac{rad}{s^{2}}[/tex]
The tangential acceleration experimented by the watermelon seed is:
[tex]a_{t} = \left(9.6\,\frac{rad}{s^{2}} \right)\cdot (0.023\,m)[/tex]
[tex]a_{t} = 0.221\,\frac{m}{s^{2}}[/tex]
The linear acceleration experimented by the watermelon seed is:
[tex]a = \sqrt{a_{t}^{2}+a_{r}^{2}}[/tex]
[tex]a = \sqrt{\left(0.221\,\frac{m}{s^{2}} \right)^{2}+\left(0.275\,\frac{m}{s^{2}} \right)^{2}}[/tex]
[tex]a = 0.353\,\frac{m}{s^{2}}[/tex]
The minimum coefficient of static friction is:
[tex]\mu_{s} = \frac{0.353\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]\mu_{s} = 0.036[/tex]
A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the dry steel frying pan is only 0.100 N. Knowing the coefficient of kinetic friction between the two materials (0.3), he quickly calculates the normal force in newtons. What is it
Answer:
Normal force = 3N
Explanation:
We are given;
Coefficient of kinetic friction friction; μ_k = 0.3
Frictional force;F_f = 0.1N
Now,the formula for frictional force is;
F_f = μ_k*N
Where μ_k is coefficient of friction and N is the normal force.
So, making N the subject, we have;
N = μ_k/F_f
Plugging in the relevant values to obtain;
N = 0.3/0.1
N = 3N
A 1.50 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 15.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 C) the ammeter reads 18.8 A, while at 92.0 C it reads 17.4 A. You can ignore any thermal expansion of the rod.
1.Find the resistivity and for the material of the rod at 20 C. (rho= ? Ω*m)2.Find the temperature coefficient of resistivity at 20 C for the material of the rod.α= ? (C)^-1)
The answer to the questions are:
1. The resistivity for the material of the rod at 20 °C (ρ) is
[tex]1.26378\times 10^{-5} \Omega m[/tex] .
2. The temperature coefficient of resistivity at 20 °C for the material of the
rod(α) is [tex]1.1169\times10^{-3}\ ^oC^{-1}\end{aligned}[/tex].
Given to us:
Voltage across the rod, V = 15.0 V
Length of rod, L = 1.5 m = 150 cm,
Diameter, d = 0.55 cm,
[tex]Radius, r= \frac{d}{2}=\frac{0.550}{2}=0.275\ cm[/tex]
[tex]\begin{aligned}Area, A&= \pi r^2\\&=\pi\times (0.275)^2\\&=0.075625\pi\ cm^2\\\end{aligned}[/tex]
Initial temperature, [tex]T_o=20.0\ ^oC[/tex]
current at [tex]T_o[/tex], [tex]I_o= 18.8\ A[/tex]
Final temperature, [tex]T_1=92.0\ ^oC[/tex]
current at [tex]T_1[/tex], [tex]I_1=17.4\ A[/tex]
1.) To find out the resistivity of the rod(ρ),
Resistant of the rod(R),
[tex]\begin{aligned}\\R_o&=\frac{Voltage}{Current(I_o)}\\&=\frac{15.0}{18.8} \\&=0.7979\ \Omega \\\end{aligned}[/tex]
Resistivity of the rod at [tex]20^o\ C[/tex](ρ),
[tex]\begin{aligned}\\\rho&=\frac{RA}{L}\\&=\frac{0.7979\times 0.075625\pi}{150}\\&=0.00126378\ \Omega cm\\&=1.26378\times 10^{-3} \Omega cm\\&=1.26378\times 10^{-5} \Omega m\\\end{aligned}[/tex]
Hence, the resistivity for the material of the rod at 20 °C (ρ) is [tex]1.26378\times 10^{-5} \Omega m[/tex] .
2.) To find out the temperature coefficient of resistivity at 20°C for the material of the rod (α) can be gotten from the equation,
[tex]{R_1} ={R_o}[ 1+\alpha(T_1-T_o)][/tex]
we need resistant of the rod([tex]R_1[/tex]),
[tex]\begin{aligned}\\R_1&=\frac{Voltage}{Current(I_o)}\\&=\frac{15.0}{17.4} \\&=0.862\ \Omega \\\end{aligned}[/tex]
Now, solving to get the value of α
[tex]\begin{aligned}{R_1} &={R_o}[ 1+\alpha(T_1-T_o)]\\0.862&=0.7979[1+\alpha(92-20)]\\\frac{0.862}{0.7979}&= [1+\alpha(72)]\\1.0804&=[1+\alpha(72)]\\0.0804&=\alpha(72)\\\alpha&=0.0011169\ ^oC^{-1}\\\alpha&=1.1169\times10^{-3}\ ^oC^{-1}\end{aligned}[/tex]
Hence, the temperature coefficient of resistivity at 20 °C for the material of the rod(α) is [tex]1.1169\times10^{-3}\ ^oC^{-1}\end{aligned}[/tex].
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To find the resistivity and temperature coefficient of resistivity of the rod material, we need to use Ohm's Law and the formula for resistivity. The resistivity can be calculated using the resistance, area, and length of the rod, while the temperature coefficient of resistivity can be found by comparing resistivities at different temperatures.
Explanation:To find the resistivity of the material of the rod at 20°C, we can use the formula:
Resistivity (ρ) = (Resistance × Area) / (Length)
First, we need to find the resistance of the rod using Ohm's Law:
Resistance (R) = Voltage (V) / Current (I)
Next, we can substitute the values into the formula to find the resistivity. The temperature coefficient of resistivity can be calculated using the equation:
α = (ρ₂ - ρ₁) / (ρ₁ × (T₂ - T₁))
Where ρ₁ and ρ₂ are the resistivities at temperatures T₁ and T₂ respectively.
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The electron transport chain (ETC), or respiratory chain, is linked to proton movement and ATP synthesis. Select the STATEMENTS that ACCURATELY describe the electron transport chain. (True/False)
a)Electron transfer in the ETC is coupled to proton transfer form the matrix to the intermembrane space.b)Electrons generated by he citric acid cycle in the mitochondrial matrix enter the ETC.c)The outer membrane of mitochondria is readily permeable to small molecules and hydrogen ions.d)Electron carriers in the mitochondrial matrix include ubiquinone (coenzyme Q), FMN, and cytochrome c.e)Prosthetic groups, such as iron-sulfer centers, are directly involved with electron transfer.f)Electron carriers are organized into four complexes of proteins and prosthetic groups.g)The reactions of the ETC take place in the outer membrane of mitochondria.
Final answer:
The electron transport chain (ETC) occurs in the inner mitochondrial membrane and involves protein complexes and mobile carriers to produce ATP. Electrons from the citric acid cycle enter the ETC, leading to proton movement and ATP synthesis. The outer membrane is permeable to small molecules, but the ETC components, including carriers such as ubiquinone and cytochrome c, are located in the inner membrane.
Explanation:
The electron transport chain (ETC) is a critical step in cellular respiration, taking place in the inner mitochondrial membrane of eukaryotic cells. The process involves several protein complexes and mobile carriers that facilitate the transfer and stepwise release of energy from reduced substrates like NADH and FADH₂ to produce ATP via oxidative phosphorylation.
a) True: Electron transfer in the ETC is indeed coupled to proton transfer from the matrix to the intermembrane space.b) True: Electrons generated by the citric acid cycle in the mitochondrial matrix do enter the ETC.c) True: The outer membrane of mitochondria is permeable to small molecules and ions, although hydrogen ions' passage is more specifically regulated by the ETC.d) False: Electron carriers such as ubiquinone and cytochrome c are embedded in the inner mitochondrial membrane, not in the matrix.e) True: Prosthetic groups such as iron-sulfur centers are indeed directly involved with electron transfer within the ETC.f) False: The ETC consists of four, not three, complexes of proteins and prosthetic groups.g) False: The reactions of the ETC take place in the inner mitochondrial membrane, not the outer.The correct statements about the electron transport chain (ETC) are a,b,e,f which include its coupling with proton transfer, the role of electrons from the citric acid cycle, the significance of prosthetic groups, and the organization into four complexes.
To accurately describe the Electron Transport Chain (ETC) and assess the statements given, we will evaluate each statement based on what we know about the ETC.
True. As electrons move through the ETC, energy is used to pump hydrogen ions (H⁺) from the mitochondrial matrix into the intermembrane space, creating a proton gradient.True. High-energy electrons carried by NADH and FADH₂, produced in the citric acid cycle, enter the ETC.False. While the outer membrane is permeable to small molecules due to porins, it is not freely permeable to protons (H+). The inner mitochondrial membrane tightly regulates the transfer of protons.False. Ubiquinone (coenzyme Q) and cytochrome c are located within the inner mitochondrial membrane, not the matrix, though they are involved in the ETC.True. These groups are essential components of the protein complexes and play a crucial role in electron transfers.True. The ETC consists of four main protein complexes (I, II, III, IV) that facilitate the transfer of electrons.False. The reactions occur in the inner mitochondrial membrane, where the protein complexes are embedded.Therefore the correct statements out of the given ones are a, b, e, f.
The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the decibel scale is logarithmic, it changes by an additive constant when the intensity as measured in W/m^2 changes by a multiplicative factor. The number of decibels increases by 10 for a factor of 10 increase in intensity.
A. What is the sound intensity level β, in decibels, of a sound wave whose intensity is 10 times the reference intensity?
B. What is the sound intensity level β, in decibels, of a sound wave whose intensity is 100 times the reference intensity?
Note: Complete Question:
The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the decibel scale is logarithmic, it changes by an additive constant when the intensity as measured in W/m2 changes by a multiplicative factor. The number of decibels increases by 10 for a factor of 10 increase in intensity. The general formula for the sound intensity level, in decibels, corresponding to intensity I is
β=10log(II0)dB,
where I0 is a reference intensity. For sound waves, I0 is taken to be 10−12W/m2. Note that log refers to the logarithm to the base 10.
Part A
What is the sound intensity level β, in decibels, of a sound wave whose intensity is 10 times the reference intensity (i.e., I=10I0)?
Part B
What is the sound intensity level β, in decibels, of a sound wave whose intensity is 100 times the reference intensity (i.e. I=100I0)?
Express the sound intensity numerically to the nearest integer.
Concepts and reason
The concept required to solve this problem is decibel scale of sound intensity.
Use the formula of sound intensity level in decibels and substitute the value of intensity to calculate decibels for all the parts.
Answer:
Find the given 2 attachments for complete solution. Thanks
Multiple-Concept Example 7 and Interactive LearningWare 26.1 provide some helpful background for this problem. The drawing shows a crystalline slab (refractive index 1.665) with a rectangular cross section. A ray of light strikes the slab at an incident angle of 1 = 37.0°, enters the slab, and travels to point P. This slab is surrounded by a fluid with a refractive index n. What is the maximum value of n such that total internal reflection occurs at point P?
Answer:
n = 1.4266
Explanation:
Given that:
refractive index of crystalline slab n = 1.665
let refractive index of fluid is n.
angle of incidence θ₁ = 37.0°
Critical angle [tex]\theta _c = sin^{-1} (\frac{n}{n_{slab}} )[/tex]
[tex]sin \theta _ c =\frac{n}{n_{slab}}[/tex]
According to Snell's law of refraction:
[tex]n sin \theta _1 = n_{slab} \ sin \ (90- \theta_c)[/tex]
At point P ; [tex]90 - \theta _2 \leq \theta _c[/tex]
[tex]\theta _2 = 90 - \theta _c[/tex]
Therefore:
[tex]n \ sin \theta_1 = n_{slab} \sqrt{(1-sin^2 \theta _c)} \\ \\ n \ sin \theta_1 = n_{slab} \sqrt{(1- \frac{n}{n_{slab}} )}[/tex]
Then maximum value of refractive index n of the fluid is:
[tex]n = \frac{n_{slab}}{\sqrt{1+ sin^2 \theta _1 } }[/tex]
[tex]n = \frac{1.665}{\sqrt{1+ sin^2 \ 37} }[/tex]
n = 1.4266
A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 59 rad/s. The wheel is run at that angular velocity for 26 s and then power is shut off. The wheel decelerates uniformly at 1.6 rad/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to:
Answer:
Time interval;Δt ≈ 37 seconds
Explanation:
We are given;
Angular deceleration;α = -1.6 rad/s²
Initial angular velocity;ω_i = 59 rad/s
Final angular velocity;ω_f = 0 rad/s
Now, the formula to calculate the acceleration would be gotten from;
α = Change in angular velocity/time interval
Thus; α = Δω/Δt = (ω_f - ω_i)/Δt
So, α = (ω_f - ω_i)/Δt
Making Δt the subject, we have;
Δt = (ω_f - ω_i)/α
Plugging in the relevant values to obtain;
Δt = (0 - 59)/(-1.6)
Δt = -59/-1.6
Δt = 36.875 seconds ≈ 37 seconds
A certain lightbulb has a tungsten filament with a resistance of 26 Ω when cold and 170 Ω when hot. If the equation R = R0 [1 + α ∆T] can be used over the large temperature range involved here, find the temperature of the filament when it is hot. Assume that α , the temperature coefficient of resistivity of tungsten, is 0.0045 (◦C)−1 and that the temperature of the cold filament is 40◦C. Answer in units of ◦C
Answer:
Explanation: The equation that relates resistance of tungsten at different temperatures is as follows
R = R₀ [1 + α ∆T] , R₀ is resistance at lower temperature , R is resistance at higher temperature . α is temperature coefficient of resistivity and ∆T is rise in temperature .
Putting the values
170 = 26 [1 + .0045 ∆T]
∆T = 1230.75
lower temperature = 40◦C
higher temperature = 1230 + 40
= 1270◦C
A 225-kg object and a 525-kg object are separated by 3.80 m. (a) Find the magnitude of the net gravitational force exerted by these objects on a 61.0-kg object placed midway between them. 1 .0000003383 Correct: Your answer is correct. N (b) At what position (other than an infinitely remote one) can the 61.0-kg object be placed so as to experience a net force of zero from the other two objects? 2 1.824 Incorrect: Your answer is incorrect.
Answer:[tex]F_{net}=3.383\times 10^{-7}\ N[/tex]
Explanation:
Given
Mass of first object [tex]m_1=225\ kg[/tex]
Mass of second object [tex]m_2=525\ kg[/tex]
Distance between them [tex]d=3.8\ m[/tex]
[tex]m_3=61\ kg[/tex] object is placed between them
So force exerted by [tex]m_1[/tex] on [tex]m_3[/tex]
[tex]F_{13}=\frac{Gm_1m_3}{1.9^2}[/tex]
[tex]F_{13}=\frac{6.674\times 10^{-11}(225\times 61)}{1.9^2}[/tex]
[tex]F_{13}=2.5374141274×10^{−7}\ N[/tex]
Force exerted by [tex]m_2\ on\ m_3[/tex]
[tex]F_{23}=\frac{Gm_2m_3}{1.9^2}[/tex]
[tex]F_{23}=\frac{6.674\times 10^{-11}(525\times 61)}{1.9^2}[/tex]
[tex]F_{23}=5.920632964\times 10^{-7}\ N[/tex]
So net force on [tex]m_3[/tex] is
[tex]F_{net}=F_{23}-F_{13}[/tex]
[tex]F_{net}=5.920632964\times 10^{-7}-2.5374141274\times 10^{-7}[/tex]
[tex]F_{net}=3.383\times 10^{-7}\ N[/tex]
i.e. net force is towards [tex]m_2[/tex]
(b)For net force to be zero on [tex]m_3[/tex], suppose
So force exerted by [tex]m_1[/tex] and [tex]m_2[/tex] must be equal
[tex]F_{13}=F_{23}[/tex]
[tex]\Rightarrow \frac{Gm_1m_3}{x^2}=\frac{Gm_2m_3}{(3.8-x)^2}[/tex]
[tex]\Rightarrow \frac{m_1}{x^2}=\frac{m_2}{(3.8-x)^2}[/tex]
[tex]\Rightarrow (\frac{3.8-x}{x})^2=\frac{m_2}{m_1}[/tex]
[tex]\Rightarrow \frac{3.8-x}{x}=\sqrt{\frac{525}{225}}[/tex]
[tex]\Rightarrow 3.8-x=1.52752x[/tex]
[tex]\Rightarrow 3.8=2.52x[/tex]
[tex]\Rightarrow x=1.507\ m[/tex]
distant galaxy emits light that has a wavelength of 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm. (a) Decide whether this galaxy is approaching or receding from the earth. Give your reasoning. (b) Find the speed of the galaxy relative to the earth.
Answer:
The speed of the galaxy relative to the Earth is [tex]3.09\times 10^6\ m/s[/tex].
Explanation:
We have,
(a) Wavelength emitted by light at distant galaxy is 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm. It can be seen that the wavelength of light reduces as it reaches Earth. It is called Red shift. As per Doppler's effect, we can say that the galaxy is receding from the Earth.
(b) Let v is the speed of the galaxy relative to the Earth. It can be given by :
[tex]v=c(\dfrac{\lambda'}{\lambda}-1)\\\\v=3\times 10^8\times (\dfrac{438.6 }{434.1 }-1)\\\\v=3\times 10^8\times (\dfrac{438.6}{434.1}-1)\\\\v=0.0103\cdot3\cdot10^{8}\\\\v=3.09\times 10^6\ m/s[/tex]
So, the speed of the galaxy relative to the Earth is [tex]3.09\times 10^6\ m/s[/tex].
The galaxy is receding from the Earth due to redshift. Using the redshift value, we can calculate the speed of the galaxy.
Explanation:The discrepancy in the measured wavelengths of light from a distant galaxy compared to its original wavelength on Earth is indicative of the galaxy moving away from Earth. This phenomenon, known as redshift, occurs when an object is moving away from the observer, causing the wavelength of light to stretch.
To calculate the speed of the galaxy relative to Earth, we can use the equation v = zc, where v is the speed, z is the redshift, and c is the speed of light. By plugging in the given values of 438.6 nm and 434.1 nm for the measured and original wavelengths respectively, we can solve for z. Once we know z, we can calculate the speed of the galaxy.
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Kim holds a pinwheel in the air and says it can be used to model a source of energy. Identify the type of energy Kim's pinwheel models. Explain the type of energy source used.
Answer:
it is kenetic
Explanation: its in motion:D
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.90 s for the boat to travel from its highest point to its lowest, a total distance of 0.700 m . The fisherman sees that the wave crests are spaced a horizontal distance of 5.50 m apart.
How fast are the waves traveling?
What is the amplitude A of each wave?
Answer:
Velocity=1.1m/s
Amplitude=0.35m
Explanation:
Given:
time 't' = 2.9s
wavelength 'λ'= 5.5m
distance 'd'=0.7m
The time period 't' is the time b/w two successive waves. Therefore, the time it takes from the boat to travel from its highest point to its lowest is a half period.
So, T = 2 x 2.9 => 5.8 s
As we know that frequency is the reciprocal of time period, we have
f= 1/T = 1/5.8 =>0.2 Hz
In order to find how fast are the waves traveling, the velocity is given by
Velocity = f λ
V= 0.2 x 5.5 =>1.1m/s
The distance between the boat's highest point to its lowest point is double the amplitude.
Therefore , we can write
Amplitude 'A'= d/2 =>0.7/2 =>0.35m
Which statements represent the rules of significant figures? Check all that apply.
The integers 1, 2, 3, 4, 5, 6, 7, 8, and 9 are always significant.
All numbers to the right of the decimal point are always significant
Zeros to the left of a decimal point and in a number greater than or equal to 10 are always significant.
All zeros are significant numbers.
Zeros in between numbers are always significant.
Answer:
its 1, 3, 5 :) Have a great day/night!!
Explanation:
Significant figures rules are: non-zero digits are always significant; zeros in between non-zero digits are significant; zeros to the right of a decimal point are significant if they follow non-zero digits; zeros to the left of a decimal point are significant only if between non-zero digits or at the end of decimals; trailing zeros without a decimal point are not significant.
Explanation:The following statements represent the rules of significant figures:
The integers 1, 2, 3, 4, 5, 6, 7, 8, and 9 are always significant. Zeros to the right of a decimal point in a number are considered significant only if they follow a non-zero number. Zeros to the left of a decimal point and in a number greater than or equal to 10 are not always significant. They are significant only if they are between non-zero digits or at the end of a decimal number. All zeros are not necessarily significant numbers. The zero(s) located at the end of a number without a decimal point are not significant. Zeros in between numbers (non-zero digits) are always significant.Learn more about Significant Figures here:https://brainly.com/question/33741100
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Two students are holding opposite ends of a spring in a classroom. One student stands on the left end of the classroom and the other stands at the right end. They shake the spring so that a longitudinal wave travels along the spring. In which directions will the longitudinal wave oscillate?
Answer:
Up And Down
Explanation:
In this case, the particles of the medium move parallel to the direction that the pulse moves. This type of wave is a longitudinal wave. Longitudinal waves are always characterized by particle motion being parallel to wave motion.
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Final answer:
In a longitudinal wave on a horizontal spring, the medium oscillates back and forth horizontally, parallel to the direction of the wave's travel.
Explanation:
The question pertains to the behavior of longitudinal waves in a spring. When one student shakes the spring to create a longitudinal wave, the oscillation of the spring occurs in the same direction as the wave's propagation. This means that in a longitudinal wave, the medium—the spring in this case—oscillates parallel to the wave's direction of motion. Therefore, if the spring is held horizontally and the wave travels from left to right, the individual coils of the spring will move left and right along the same horizontal line, compressing and expanding as the wave passes through.
In contrast, with transverse waves, the medium moves perpendicularly to the direction of the wave's travel, such as in the motion of a rope being moved up and down while the wave travels horizontally.
What can help a scientist identify any object in a group of objects?
the object's size
the object's colors
the object's characteristics
the object's shape
Answer:The object characteristics
Explanation:the objects characteristics
Answer:
C- The objects characteristics
Explanation:
I just did it
Two point charges, initially 2.0 cm apart, experience a 1.0 N force. If they are moved to a new separation of 0.25 cm, what is the electric force between them (in N)?
Explanation:
Th electric force between charges is inversely proportional to the square of distance between them. It means,
[tex]F\propto \dfrac{1}{r^2}[/tex]
Initial distance, r₁ = 2 cm
Final distance, r₂ = 0.25 cm
Initial force, F₁ = 1 N
We need to find the electric force between charges if the new separation of 0.25 cm. So,
[tex]\dfrac{F_1}{F_2}=(\dfrac{r_2}{r_1})^2\\\\F_2=\dfrac{F_1r_1^2}{r_2^2}\\\\F_2=\dfrac{1\times 2^2}{(0.25)^2}\\\\F_2=64\ N[/tex]
So, the new force is 64 N if the separation between charges is 64 N.
While Dr. Chesnutt is making breakfast, she turns on her 1060 W toaster, 500 W coffee pot, and 1230 W microwave at the same time. Part Description Answer Save Status A. If all three appliances are connected in parallel across a 120 V power source, what total current would they draw from the source? (include units with answer) Format Check Click here to check your answer 4 pts.100% 13% try penalty 8 try limit # tries: 0
Answer:23.25 A
Explanation:
Given
Rating of toaster [tex]P_1=1060\ W[/tex]
Coffee pot [tex]P_2=500\ W[/tex]
microwave [tex]P_3=1230\ W[/tex]
Voltage applied [tex]V=120\ V[/tex]
if they are connected in parallel then all three operates at same voltage
so their resistance are
[tex]P=\frac{V^2}{R}[/tex]
thus [tex]R_1=\frac{V^2}{P_1}=\frac{120^2}{1060}[/tex]
[tex]R_1=13.58\ \Omega[/tex]
[tex]R_2=\frac{V^2}{P_2}=\frac{120^2}{500}[/tex]
[tex]R_2=28.8\ \Omega[/tex]
[tex]R_3=\frac{V^2}{P_3}=\frac{120^2}{1230}[/tex]
[tex]R_3=11.707\ \Omega[/tex]
and [tex]V=IR[/tex]
where I=current
thus [tex]I_1=\frac{V}{R_1}=\frac{120}{13.58}[/tex]
[tex]I_1=8.83\ A[/tex]
[tex]I_2=\frac{V}{R_2}=\frac{120}{28.8}[/tex]
[tex]I_2=4.16\ A[/tex]
[tex]I_3=\frac{V}{R_3}=\frac{120}{11.707}[/tex]
[tex]I_3=10.25\ A[/tex]
Total current [tex]I=I_1+I_2+I_3=23.24\ A[/tex]