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Nikolay [14]
2 years ago
6

A ball is let go and falls to the ground. It bounces a few times before

Physics
1 answer:
Ksenya-84 [330]2 years ago
4 0

Answer:

I just want the points but you should pay attention in class jk

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Which of the following statements BEST explains why ball-and-socket joints have the greatest range of motion?
galina1969 [7]

Answer:

its d on edge c:

5 0
3 years ago
A batter hits a 0.140-kg baseball that was approaching him at 19.5 m/s and, as a result, the ball leaves the bat at 44.8 m/s in
Arada [10]

Answer:

5295.3 N

Explanation:

According to law of momentum conservation, the change in momentum of the ball shall be from the momentum generated by the batter force

mv + P = mV

P = mV - mv = m(V - v)

Since the velocity of the ball before and after is in opposite direction, one of them is negative

P = 0.14(44.8 - (-19.5)) = 9 kg m/s

Hence the force exerted to generate such momentum within 1.7ms (0.0017s) is

F = P/t = 9/0.0017 = 5295.3 N

4 0
3 years ago
two large boxes sit side by side on a sidewalk. the box on the left has a mass of 80kg and the box on the right has a mass of 50
garri49 [273]

     The force that prevents motion when the surfaces of two objects come into contact is known as friction. Friction decreases a machine's mechanical advantage, or, to put it another way, reduces the output to input ratio.

<h3>How can I figure out the frictional force?</h3>

        The resistive force of friction (Fr) divided by the normal or perpendicular force (N) pushing the objects together yields the coefficient of friction (fr), which is a numerical value.

The formula fr = Fr/N serves as a representation of it.

Therefore, 100N of force is needed to move an item with a mass of 50 kg.

It will accelerate by 10 m/s2.

If a substance's mass does not change over time, friction cannot affect it. Instead, friction can be affected in a variety of ways by an object's mass.

To  Learn more About Friction, Refer:

brainly.com/question/24338873

#SPJ13

8 0
1 year ago
A 4 kg car with frictionless wheels is on a ramp that makes a 25° angle with the horizontal. The car is connected to a 1 kg mass
stellarik [79]
Refer to the diagram shown below.

W₁ = (4 kg)*(9.8 m/s²) = 39.2 N
W₂ = (1 kg)*(9.8 m/s²) = 9.8 N

The normal reaction on the 4-kg mass is
N = (39.2 N)*cos(25°) = 35.5273 N
The force acting down the inclined plane due to the weight is
F = (39.2 N)*sin(25°) = 16.5666 N

The net force that accelerates the 4-kg mass at a m/²s down the plane is
F - W₂ = (4 kg)*(a m/s²)
4a = 16.5666 - 9.8
a = 1.6917 m/s²

Answer: 1.69 m/s²  (nearest hundredth)

7 0
3 years ago
A 50.0-g object connected to a spring with a force constant of 35.0 n/m oscillates with an amplitude of 4.00 cm on a frictionles
Dimas [21]
A) The total energy of the system is sum of kinetic energy and elastic potential energy:
E=K+U= \frac{1}{2}mv^2 +  \frac{1}{2}kx^2
where
m is the mass
v is the speed
k is the spring constant
x is the elongation/compression of the spring

The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
x=A=4.00 cm = 0.04 m
then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to
E=U= \frac{1}{2}kA^2 =  \frac{1}{2}(35.0 N/m)(0.04 m)^2=0.028 J

b) When the position of the object is 
x=1.00 cm = 0.01 m
the potential energy of the system is
U= \frac{1}{2}kx^2 =  \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J
and so the kinetic energy is
K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J
since the mass is m=50.0 g=0.05 kg, and the kinetic energy is given by
K= \frac{1}{2}mv^2
we can re-arrange the formula to find the speed of the object:
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.026 J}{0.05 kg} }=1.02 m/s

c) The potential energy when the object is at 
x=3.00 cm=0.03 m
is
U= \frac{1}{2}kx^2 =  \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J
Therefore the kinetic energy is
K=E-U=0.028 J-0.016 J = 0.012 J

d) We already found the potential energy at point c, and it is given by
U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J
5 0
3 years ago
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