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bazaltina [42]
3 years ago
14

A ball traveling at 15 m/s hits a bat with a force of 200N. How much force does the bat (moving at 20m/s)

Physics
1 answer:
just olya [345]3 years ago
8 0

Answer:

200 N

Explanation:

Given that,

A ball traveling at 15 m/s hits a bat with a force of 200 N.

We need to find the force that the bat moving at 20 m/s hit the ball with.

We know that, this probelm is based on Newton's third law of motion. The force that the ball exerting on bat should be equal to the force that the bat exerting in the ball but in opposite direction.

It would mean that the ball hits the ball with a force of 200 N. Hence, the correct option is (a).

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An object has a velocity of 8 m/s and a kinetic energy of 480 J. What is the mass of the object? (Formula: )
andrezito [222]

Answer:

1/2mv^2 a(7.5 b(15 kg c(60 kg d(120 kg. nevermind i found the answer its (15 kg) because to solve for m its m= K2/v squared.

Explanation:

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3 years ago
What is the difference between a distance time graph and a speed time graph?
Feliz [49]

Answer:

speed-time graph is the acceleration of the particle or straight lines acceleration is constant

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3 years ago
Four charges with equal magnitudes of 10.6 × 10-12 C are placed at the corners of a rectangle. The lengths of the sides of the r
cricket20 [7]

Answer:

Figure a. E_net = 99.518 N/C

Figure b. E_net = 177.151 N / C

Explanation:

Given:

- Attachment for figures missing in the question.

- The dimensions for rectangle are = 7.79 x 3.99 cm

- All four charges have equal magnitude Q = 10.6*10^-12 C

Find:

Find the magnitude of the electric field at the center of the rectangle in Figures a and b.

Solution:

- The Electric field generated by an charged particle Q at a distance r is given by:

                                         E = k*Q / r^2

- Where, k is the coulomb's constant = 8.99 * 10^9

Part a)

- First we see that the charges +Q_1 and +Q_3 produce and electric field equal but opposite in nature. So the sum of Electric fields:

                                 E_1 + E_3 = 0

- For Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Hence, the net Electric Field at center of the rectangle can be given as:

                                  E_net = E_2 + E_4

                                  E_2 = E_4

                                  E_net = 2*E = 2*k*Q / r^2

- The distance r from each corner to mid-point of the rectangle is constant. It can be evaluated by Pythagoras Theorem as follows:

                                  r = sqrt ( (7.79/200)^2 + (3.99/200)^2 )

                                  r = sqrt ( 1.9151*10^-3 ) = 0.043762 m

- Plug the values in the E_net expression developed above:

                                  E_net = 2*(8.99*10^9)*(10.6*10^-12) / 1.9151*10^-3

                                 E_net = 99.518 N/C

Part b)

- Similarly for Figure b, for Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Also, Charges -Q_1 and +Q_3, they are equal in nature but act in the same direction towards the negative charge -Q_1. These Electric fields are equal in magnitude to what we calculated in part a).

- To find the vector sum of two Electric Fields E_1,3 and E_2,4 we see the horizontal components of each cancels each other out. While the vertical components E_1,3 and E_2,4 are equal in magnitude and direction.

Hence,

                                  E_net = 2*E_part(a)*cos(Q)

- Where, Q is the angle between resultant, vertical in direction, and each of the electric field. We can calculate Q using trigonometry as follows:

                                  Q = arctan ( 3.99 / 7.79 ) = 27.12 degrees.

- Now, compute the net electric field E_net:

                                  E_net = 2*(99.518)*cos(27.12)

                                  E_net = 177.151 N / C

               

5 0
3 years ago
What’s an example of contact friction
kotegsom [21]

Answer:

a nightstand on a lamp table

Explanation:

5 0
3 years ago
A 1.5 kg spherical ball is has a radius of 50 cm is rotating with angular velocity of 12 revolutions per minute. Determine the r
kykrilka [37]

Answer:

K.E = 0.0075 J

Explanation:

Given data:

Mass of the ball = 1.5 kg

radius, r = 50 cm = 0.5 m

Angular speed, ω = 12 rev/min = (12/60) rev/sec = 0.2 rev/sec

Now,

the kinetic energy is given as:

K.E = K.E=\frac{1}{2}I\omega^2

where,

I is the moment of inertia = mr²

on substituting the values, we get

K.E=\frac{1}{2}\times1.5\times0.5^2\times0.2^2

or

K.E = 0.0075 J

3 0
3 years ago
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