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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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Mademuasel [1]
I might have did mistake with calculations but this is how you should do.

6 0
3 years ago
As a rocket travels upward from earth, air resistance decreases along with the force of gravity. The rocket's mass also decrease
ddd [48]

Answer:

The decrease of these factors increases the acceleration.

Explanation:

Hi, the decrease of these factors increases the acceleration.

Air resistance is a force opposing the acceleration. So if it decreases, the acceleration increases, because the opposite forces decreases.

The same is applied to the force of gravity, since the rocket travels upward; gravity is also an opposite force.

Finally, if the mass decreases, it means that the rocket becomes lighter and the force acting on the smaller mass causes an increase in the acceleration.

6 0
3 years ago
An automobile starter motor has an equivalent resistance of 0.0500Ω and is supplied by a 12.0-V battery with a 0.0100-Ω internal
alexandr1967 [171]

Answer

given,

resistance = 0.05 Ω

internal resistance of battery = 0.01 Ω

electromotive force = 12 V

a) ohm's law

        V = IR

     and volage

   V = \epsilon - Ir

now,

   IR = \epsilon - Ir

   I(R+r) = \epsilon

   I= \dfrac{\epsilon}{R+r}

inserting the values

   I= \dfrac{12}{0.05+0.01}

      I = 200 A

b) Voltage

   V = I R

   V = 200 x 0.05

   V = 10 V

c) Power

    P = I V

    P = 200 x 10 = 2000 W

d) total resistance = 0.05 + 0.09 = 0.14 Ω

 I= \dfrac{\epsilon}{R+r}

   I= \dfrac{12}{0.14+0.01}

     I = 80 A

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     P = 4 x 80 = 320 W

6 0
3 years ago
Read 2 more answers
Coherent light of frequency 6.37×1014 Hz passes through two thin slits and falls on a screen 88.0 cm away. You observe that the
IgorC [24]

Answer:

The distance between the two slits is 40.11 μm.

Explanation:

Given that,

Frequency f= 6.37\times10^{14}\ Hz

Distance of the screen l = 88.0 cm

Position of the third order y =3.10 cm

We need to calculate the wavelength

Using formula of wavelength

\lambda=\dfrac{c}{f}

where, c = speed of light

f = frequency

Put the value into the formula

\lambda=\dfrac{3\times10^{8}}{6.37\times10^{14}}

\lambda=471\ nm

We need to calculate the distance between the two slits

m\times \lambda=d\sin\theta

d =\dfrac{m\times\lambda}{\sin\theta}

Where, m = number of fringe

d = distance between the two slits

Here, \sin\theta =\dfrac{y}{l}

Put the value into the formula

d=\dfrac{3\times471\times10^{-9}\times88.0\times10^{-2}}{3.10\times10^{-2}}

d=40.11\times10^{-6}\ m

d = 40.11\ \mu m

Hence, The distance between the two slits is 40.11 μm.

7 0
3 years ago
Convert 141.2 kg to lbs and show all units and work​
zavuch27 [327]
Your answer is 311.29271 lbs
5 0
3 years ago
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