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Hitman42 [59]
2 years ago
15

A 0.60-kilogram softball initially at rest is hit with a bat. The ball is in contact with the bat for 0.20 second and leaves the

bat with a speed of 25 meters per second. What is the magnitude of the average force exerted by the ball on the bat?
a. 15 N
b. 3.0 N
c. 75 N
d. 8.3 N
Physics
1 answer:
Alona [7]2 years ago
8 0

Answer:

c. 75 N

Explanation:

Force: This can be defined as the product of mass and the acceleration of a body. The S.I unit of Force is Newton (N).

From Newton's Fundamental equation of kinematic,

F = ma ............................................. Equation 1

Where F = Force exerted by the ball on the bat, m = mass of the ball, a = acceleration.

From motion,

a = (v-u)/t....................................... Equation 2

Where v = final velocity, u = initial velocity, t = time.

Given: v = 25 m/s, u = 0 m/s ( at rest), t = 0.20 s.

Substitute into equation 2

a = (25-0)/0.20

a = 125 m/s².

Also given: m = 0.60 kg.

Substitute into equation 1

F = 0.6(125)

F = 75 N.

Hence the force exerted by the ball on the bat = 75 N,

The right option is c. 75 N

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 Similarly, the 235g particle, y-component of the total momentum in that direction is given by p_(fy) = p_(1y) + p_(2y) + p_(3y) = 0.
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2 years ago
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A car of mass 1100kg moves at 24 m/s. What is the braking force needed to bring the car to a halt in 2.0 seconds? N
LenaWriter [7]

13200N

Explanation:

Given parameters:

Mass = 1100kg

Velocity = 24m/s

time = 2s

unknown:

Braking force = ?

Solution:

The braking force is the force needed to stop the car from moving.

   Force  =  ma = \frac{mv}{t}

  m is the mass of the car

  v is the velocity

  t is the time taken

  Force = \frac{1100 x 24}{2} = 13200N

Learn more:

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2 years ago
(1 point) Which of the following statements are true?A.The equation Ax=b is referred to as a vector equation.B.If the augmented
Anvisha [2.4K]

Answer:

A. False

B. False

C. True

D. True

E. True

F. True

Explanation:

A. The equation Ax=b is referred to as a matrix equation and not vector equation.

B. If the augmented matrix [ A b ] has a pivot position in every​ row then equation Ax=b may or may not be consistent. It is inconsistent if [A b] has a pivot in the last column b and it is consistent if the matrix A has a pivot in every row.

C. In the product of Ax also called the dot product the first entry is a sum of products. For example the the product of Ax where A has [a11 a12 a13] in the first entry of each column and the corresponding entries in x are [x1 x2 x3] then the first entry in the product is the sum of products i.e.  a11x1 + a12x2 +a13x3

D. If the columns of mxn matrix A span R^m, this states that every possible vector b in R^m is a linear combination of the columns which makes the equation consistent. So the equation Ax=b has at least one solution for each b in R^m.

E. It is stated that a vector equation x1a1 + x2a2 + x3a3 + ... + xnan = b has the same solution set as that of the linear system with augmented matrix [a1 a2 ... an b]. So the solution set of linear system whose augmented matrix is [a1 a2 a3 b] is the same as solution set of Ax=b if A=[a1 a2 a3]  and b can be produced by linear combination of a1 a2 a3 iff the solution of linear system corresponding to [a1 a2 a3 b] takes place.

F.  It is true because lets say b is a vector in R^m which is not in the span of  the columns. b cannot be obtained for some x which belongs to R^m as b = Ax. So Ax=b is inconsistent for some b in R^m and has no solution.

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2 years ago
In a car crash, large accelerations of the head can lead to severe injuries or even death. A driver can probably survive an acce
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Answer:

14.7 m/s

Explanation:

a = acceleration experienced by driver's head = 50 g = 50 x 9.8 m/s² = 490 m/s²

v₀ = initial speed of the driver = 0 m/s

v = final speed of the driver after 30 ms

t = time interval for which the acceleration is experienced = 30 ms = 0.030 s

Using the equation

v = v₀ + a t

Inserting the values

v = 0 + (490) (0.030)

v = 14.7 m/s

6 0
2 years ago
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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

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Total momentum = m×√(2² + 3·7²)

                             = m× 4·21 kgm/s

3 0
2 years ago
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