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Mariulka [41]
3 years ago
8

In a Little League baseball game, the 145 g ball enters the strike zone with a speed of 17.0 m/s .The batter hits the ball, and

it leaves his bat with a speed of 24.0 m/s in exactly the opposite direction.
What is the magnitude of the impulse delivered by the bat to the ball?
If the bat is in contact with the ball for 1.4 ms , what is the magnitude of the average force exerted by the bat on the ball?
Physics
1 answer:
Deffense [45]3 years ago
8 0

Answer:

2.465Ns;1.76N

Explanation:

From Newton's Second Law:

There is a conservation of Momentum hence:

Momentum before the impact equals that after the impact.

Hence.

1. The Impulse is same as the change in Momentum which is;

0.145 ×17 =2.465Ns { note: 145g in kg is 0.145}

2. From Newton's third Law; force is the rate of change of impulse expressed mathematically as;

F = m×v / t

Where m×v is impulse and t is time given.

F = 2.465/ 1.4 = 1.76N

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Ay tawaging bayani sa makabagong panahon?
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Answer:

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7 0
2 years ago
Two bicycle tires are set rolling with the same initial speed of 3.30 m/s along a long, straight road, and the distance each tra
vredina [299]

Answer:

At low pressure- \mu_{k}=0.02315

At high pressure- \mu_{k}=0.00445

Explanation:

Initial speed, V_{i}=3.3 m/s

Final speed, V_{f}=3.3/2= 1.65 m/s

Net horizontal force due to rolling friction F_{net}=\mu_{k} mg where m is mass, g is acceleration due to gravity, \mu_{k} is coefficient of rolling friction

From kinematic relation, V_{f}^{2}- V_{i}^{2}=2ad

For each tire,

V_{f}^{2}- V_{i}^{2}=2\mu_{k}gd

Making \mu_{k} the subject

\mu_{k}=\frac {V_{f}^{2}- V_{i}^{2}}{2gd}

Under low pressure of 40 Psi, d=18 m

\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*18}=-0.02315

Therefore, \mu_{k}=0.02315

At a pressure of 105 Psi, d=93.7

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Therefore, \mu_{k}=0.00445

4 0
3 years ago
A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The
Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

\sum {F = ma}

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg

Here, {\mu _{\rm{s}}} is the coefficient of static friction, mm is mass of the object, and g is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass {m_1}

Substitute  for in the expression of maximum static friction {F_{\rm{f}}} = {\mu _{\rm{s}}}mg

{F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute  for , [/tex]{m_1}[/tex] for in the equation .

{F_{\rm{f}}} = {m_1}a

Substitute {\mu _{\rm{s}}}{m_1}g for {F_{\rm{f}}} in the equation {F_{\rm{f}}} = {m_1}a

{\mu _{\rm{s}}}{m_1}g = {m_1}a

Rearrange for a.

a = {\mu _{\rm{s}}}g

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right)a

Substitute for in the above equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

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Answer:

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The rate of reaction when the enzyme is saturated with substrate is the maximum rate of reaction, is referred to as Vmax.

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Please kindly check attachment for the step by step solution of the given problem.

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3 years ago
A cheetah can go from the state of rest to running at 20m/s in just two seconds. What is the Cheetahs average acceleration
babymother [125]

acceleration = change in velocity/change in time

so...

a = 20 m/s / 2 seconds

a = 10


hope that helps :)

P.S. found this from Brainly User, sometimes all you have to do is search to find the answer.

7 0
3 years ago
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