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dalvyx [7]
3 years ago
9

Which of the following is an example of applying a force?

Physics
1 answer:
maw [93]3 years ago
4 0
The correct response would be C. Carpenter hammering a nail. The carpenter is applying a force as he or she is hitting the surface of the nail with the hammer.
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Consider a system to be two train cars traveling toward each other. What is the total momentum of the system before the train ca
Brut [27]

Let say the two train cars are of masses m_1 and m_2

now if the speed of two cars are v_1 and v_2

then we can say that the momentum of two cars before they collide is given by

P = m_1v_1 - m_2v_2

here two cars are moving in opposite direction so we can say that the net momentum is subtraction of two cars momentum.

Now since in these two car motion there is no external force on them while they collide

So the momentum of two cars are always conserved.

hence we can say that the final momentum of two cars will be same after collision as it is before collision

P = m_1v_1 - m_2v_2

5 0
3 years ago
Read 2 more answers
How many turns are in its secondary coil, if its input voltage is 120 V and the primary coil has 210 turns
Tasya [4]

Complete Question

How many turns are in its secondary coil, if its input voltage is 120 V and the primary coil has 210 turns.

The output from the secondary coil is  12 V

Answer:

The value  is  N_s  =  21 \  turns

Explanation:

From the equation we are told that

   The input voltage is  V_{in}  = 120 \ V

   The number of turns of the primary coil is N_p =  210 \  turn

    The output from the secondary is V_o =  12V

From the transformer equation

   \frac{N_p}{V_{in}}  =\frac{N_s}{V_o}

Here N_s is the number of turns in the secondary coil

=> N_s  =  \frac{N_p}{V_{in}}  *  V_s

=>N_s  =  \frac{210}{120}  *  12

=>N_s  =  21 \  turns

4 0
3 years ago
A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is
BlackZzzverrR [31]

Answer:

the penny loses contact at the piston's highest point.

f = 2.5 Hz

Explanation:

Concepts and Principles  

1- Newton's Second Law: The net force F on a body with mass m is related to the body's acceleration a by  

∑F = ma                                                          (1)  

2- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed w and the amplitude A as follows:  

a_max = -w^2A                                                (2)  

3- The angular frequency w of a wave is related to the frequency f by:  

w = 2π f                                                              (3)  

Given Data  

- The amplitude of the piston is: A = (4.0 cm) ( 1/ 100 cm)=  0.04 m.  

- The frequency of oscillation of the piston is steadily increased.

Required Data

<em>In part (a), we are asked to determine the point at which the penny first loses contact with the piston.  </em>

<em>In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle.  </em>

Solution  

(a)  

The free-body diagram in Figure 1 shows the forces acting on the penny; mi is the gravitational force exerted by the Earth on the penny andrt is the normal contact force exerted by the piston on the penny.  

figure 1 is attached

Apply Newton's second law from Equation (1) in the vertical direction to the penny:  

∑F_y -mg= ma        

Solve for n=m(g+a) The penny loses contact with the surface of the oscillating piston when the normal force n exerted by the piston is zero. So  

0 = m(g + a)

a = —g  

Therefore, the penny loses contact with the piston when the piston starts accelerating downwards. The piston first acceleratesdownward at its highest point and hence the penny loses contact at the piston's highest point.

(b)  

The maximum acceleration of the penny at the highest point of the piston is found from Equation (2):  

a = —w^2A  

where a = —g at the highest point. So  

g = w^2A  

Solve for w:  

w =√g/A

Substitute for w from Equation (3):

2πf =  √g/A

Solve for f :  

f = 1/2π√g/A

Substitute numerical values:  

f = 1/2π√9.8 m/s^2/0.04

f = 2.5 Hz

6 0
3 years ago
According to Boyle’s Law, when pressure is increased...
klasskru [66]

Boyle's law states that when you "shrink"(change its size) a container the pressure increases so the best answer would be D-volume is decreased.

4 0
2 years ago
The cylinder in the picture is rotating at 500 RPMs. The friction coefficients between the cylinder and block B are static=0.5 a
Tju [1.3M]

  • Coefficient of static friction = 0.5
  • Coefficient of Kinetic friction = 0.3
  • Angular velocity = 500 RPMs

<h3>The Radius of the System</h3>

Let R be the radius of cylinder

m_a + m_b = 4 + 3 = 7kg

The angular velocity is 500 RPMs

\omega ^2 = \frac{500 * 2\pi}{60} rad/s\\N = (M_a + M_b)\omega ^2 R

The normal force

f = \mu N = (M_a + M_b) g\\\mu (M_a + M_b) \omega ^2 R = M_a + M_b\\R = \frac{1}{\mu \omega ^2 R}\\\mu_s = 0.5\\R = \frac{1}{0.5 * (\frac{500 * 2\pi}{60})^2 }\\R = 0.0073m\\R = 7.3mm

Since the radius is very little for two block to execute circular motion so system will slide down.

Learn more on coefficient of static friction here;

brainly.com/question/11841776

brainly.com/question/25772665

brainly.com/question/26400616

8 0
2 years ago
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