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Hunter-Best [27]
2 years ago
10

Use Newton's laws of motion to explain why it is important that baseballs and softballs each have a small acceptable range of ma

sses.​
Physics
1 answer:
tiny-mole [99]2 years ago
5 0

Explanation:

new non law neutron means neutral then it's important that baseball and softball features small respectable range of masses soft it means that when a ball hits anything hard it comes back by the Newton Law if the baseball is big and the small boy small and then if the contract with each other they ignore triple so when a ball hits the wall if the comeback because of the Mutants and when a big ball if we throw it to the wall it doesn't come that it comes back but in a very low way because it contains less neutrons in it if it is helpful please share with me

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The position of an object is given by x = at3 - bt2 + ct,where a = 4.1 m/s3, b = 2.2 m/s2, c = 1.7 m/s, and x and t are in SI un
Elza [17]

Answer:

The answer to your question is: 15 m/s2

Explanation:

Equation    x = at3 - bt2 + ct

a = 4.1 m/s3

b = 2.2 m/s2

c = 1.7 m/s

First we find  x at t = 4.1 s

x = 4.1(4.1)3 - 2.2(4.1)2 + 1.7(4.1)

x = 4.1(68.921) - 2.2(16.81) + 6.97

x = 282.58 - 36.98 + 6.98

x = 252.58 m

Now we find speed

v = x/t = 252.58/ 4.1 = 61.6 m/s

Finally

acceleration = v/t = 61.6/4.1 = 15 m/s2

6 0
1 year ago
A group of students collected the data shown below while attempting to measure the coefficient of static friction (of course, it
anzhelika [568]

Answer:

0.130

Explanation:

From the given data, the coefficient of static friction for each trial are:

1. 0.053

2. 0.081

3. 0.118

4. 0.149

5. 0.180

6. 0.198

The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198

                                              = 0.779

So that;

the average coefficient of static friction = \frac{sum of coefficient of static friction}{number of trials}

                                              = \frac{0.779}{6}

                                              = 0.12983

The average coefficient of static friction is 0.130

8 0
1 year ago
Calculate the acceleration of the body of mass 3kg on which a force of 42N has been applied for 5s
____ [38]
F=ma

42/3 = a
a = 14m/s^2
5 0
2 years ago
A mass of 0.4 kg hangs motionless from a vertical spring whose length is 0.76 m and whose unstretched length is 0.41 m. Next the
Blizzard [7]
<h3><u>Answer;</u></h3>

= 1.256 m

<h3><u>Explanation;</u></h3>

We can start by finding the spring constant  

F = k*y  

Therefore;  k = F/y = m*g/y

                               = 0.40kg*9.8m/s^2/(0.76 - 0.41)

                               = 11.2 N/m  

Energy is conserved  

Let A be the maximum displacement  

Therefore;  1/2*k*A^2 = 1/2*k*(1.20 - 0.41)^2 + 1/2*m*v^2  

Thus;  A = sqrt((1.20 - 0.55)^2 + m/k*v^2)

               = sqrt((1.20 -0.55)^2 + 0.40/9.8*1.6^2)

                = 0.846 m  

Thus; the length will be 0.41 + 0.846  = 1.256 m

6 0
1 year ago
Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov
frez [133]

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v_{1} be the velocity of ball A along positive x axis

v_{2} be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v_{1}

∴  v_{1} = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v_{2}

2 = u +  v_{2} → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 =  v_{2} - u → equation 2

By adding both equations 1 and 2 we get

v_{2} = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

                         = 8·85×m J

Total momentum = m×√(2² + 3·7²)

                             = m× 4·21 kgm/s

3 0
2 years ago
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