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3 years ago

6

A(n) 0.47 kg softball is pitched at a speed of 10 m/s. The batter hits it back directly at the pitcher at a speed of 29 m/s. The

bat acts on the ball for 0.019 s. What is the magnitude of the impulse imparted by the bat to the ball? Answer in units of N Â· s.

1 answer:

nadya68 [22]3 years ago

8 0

Answer:

18.33 Ns

Explanation:

As the pitch back speed has the opposite direction as before, the change in velocity would be

$\Delta v = v_2 - v_1 = 29 - (-10) = 39 m/s$

So the change in momentum of the ball would be the product of its velocity change and its mass

$\Delta p = \Delta v m = 39 * 0.47 = 18.33 kgm/s$

This is equals to the impulse acted on the ball by the bat, which is 18.33 Ns

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Coherent light of wavelength 540 nm passes through a pair of thin slits that are 3.4 × 10-5 m apart. At what angle away from the

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Answer: $1.8\°$

Explanation:

The diffraction angles $\theta_{n}$ when we have a slit divided into $n$ parts are obtained by the following equation:

$dsin\theta_{n}=n\lambda$ (1)

Where:

$d=3.4(10)^{-5}m$ is the width of the slit

$\lambda=540 nm=540(10)^{-9}m$ is the wavelength of the light

$n$ is an integer different from zero.

Now, the second-order diffraction angle is given when $n=2$ , hence equation (1) becomes:

$dsin\theta_{2}=2\lambda$ (2)

Now we have to find the value of $\theta_{2}$ :

$sin\theta_{2}=\frac{2\lambda}{d}$ (3)

Then:

$\theta_{2}=arcsin(\frac{2\lambda}{d})$ (4)

$\theta_{2}=arcsin(\frac{2(540(10)^{-9}m)}{3.4(10)^{-5}m})$ (5)

Finally:

$\theta_{2}=1.8\°$ (6)

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3 years ago

2. What do you understand by balanced and unbalanced force

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Answer:

forces that are equal in size and opposite in direction. Balanced forces do not result in any change in motion. unbalanced. forces: forces applied to an object in opposite directions that are not equal in size. Unbalanced forces result in a change in motion.

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2 years ago

How much work is done when an engine generates 400 Watts of power in 25 seconds?

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Answer:

10000 J or 10 KJ

Explanation:

power = workdone/time taken

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workdone = 400 * 25

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3 years ago

Read 2 more answers

The total energy of a block—spring system is 0.18 J. The amplitude is 14.0 cm and the maximum speed is 1.25 m/s. Find: (a) the m

algol13

a) The mass is 0.23 kg

b) The spring constant is 1.25 N/m

c) The frequency is 1.42 Hz

d) The speed of the block is 1.08 m/s

Explanation:

a)

We can find the mass of the block by applying the law of conservation of energy: in fact, the total mechanical energy of the system (which is sum of elastic potential energy, PE, and kinetic energy, KE) is constant:

$E=PE+KE=const.$

The potential energy is given by

$PE=\frac{1}{2}kx^2$

where k is the spring constant and x is the displacement. When the block is crossing the position of equilibrium, x = 0, so all the energy is kinetic energy:

$E=KE_{max}=\frac{1}{2}mv_{max}^2$ (1)

where

m is the mass of the block

$v_{max}=1.25 m/s$ is the maximum speed

We also know that the total energy is

$E=0.18 J$

Re-arranging eq.(1), we can find the mass:

$m=\frac{2E}{v_{max}^2}=\frac{2(0.18)}{(1.25)^2}=0.23 kg$

b)

The maximum speed in a spring-mass system is also given by

$v_{max} =\sqrt{\frac{k}{m}} A$

where

k is the spring constant

m is the mass

A is the amplitude

Here we have:

$v_{max}=1.25 m/s$ is the maximum speed

m = 0.23 kg is the mass

A = 14.0 cm = 0.14 m is the amplitude

Solving for k, we find the spring constant

$k=\frac{v_{max}^2}{A^2}m=\frac{1.25^2}{0.14^2}(0.23)=18.3 N/m$

c)

The frequency in a spring-mass system is given by

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k is the spring constant

m is the mass

In this problem, we have:

k = 18.3 N/m is the spring constant (found in part b)

m = 0.23 kg is the mass (found in part a)

Substituting and solving for f, we find the frequency of the system:

$f=\frac{1}{2\pi}\sqrt{\frac{18.3}{0.23}}=1.42 Hz$

d)

We can solve this part by using the law of conservation of energy; in fact, we have

$E=PE+KE=\frac{1}{2}kx^2 + \frac{1}{2}mv^2$

Where v is the speed of the system when the displacement is equal to x.

We know that the total energy of the system is

E = 0.18 J

Also we know that

k = 18.3 N/m is the spring constant

m = 0.23 kg is the mass

Substituting

x = 7.00 cm = 0.07 m

We can solve the equation to find the corresponding speed v:

$v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.18)-(18.3)(0.07)^2}{0.23}}=1.08 m/s$

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