Ask question

Ask question

All categories

3 years ago

9

As a protest against the umpire's calls, a baseball pitcher throws a ball straight up into the air at a speed of 19.0 m/s. In th

e process, he moves his hand through a distance of 1.10 m. If the ball has a mass of 0.150 kg, find the force he exerts on the ball to give it this upward speed.

1 answer:

andrew11 [14]3 years ago

6 0

Answer:

$F_{player}=26.1N$

Explanation:

Given data

Initial speed vi=0 m/s

Final speed vf=19.0 m/s

Δy=1.10 m

Mass of ball m=0.150 kg

To find

Force

Solution

First we need to find acceleration of the ball.So we can apply:

vf²=vi²+2aΔy

vf²=0+2aΔy

$a=\frac{(19.0m/s)^2}{2*1.10m}\\ a=164.1m/s^2$

Because we get the acceleration of the ball.We can now find the force the players exerts on the ball by using Newtons second law

∑F=F.player -mg=ma

$F_{player}-mg=ma\\F_{player}=mg+ma\\F_{player}=m[g+a]\\F_{player}=0.150kg[9.81+164.1]\\F_{player}=26.1N$

You might be interested in

A 74-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.56 m,

Reptile [31]

Answer:

Work done, W = 1.44 kJ

Explanation:

Given that,

Mass of boy, m = 74 kg

Initial speed of boy, u = 1.6 m/s

The boy then drops through a height of 1.56 m

Final speed of boy, v = 8.5 m/s

To find,

Non-conservative work was done on the boy.

Solution,

The work done by the non conservative forces is equal to the sum of total change in kinetic energy and total change in potential energy.

$W=\dfrac{1}{2}m(v^2-u^2)+(0-mgh)$

$W=\dfrac{1}{2}m(v^2-u^2)-mgh$

$W=\dfrac{1}{2}\times 74\times (8.5^2-1.6^2)-74\times 9.8\times 1.56$

W = 1447.21 Joules

or

W = 1.44 kJ

Therefore, the non conservative work done on the boy is 1.44 kJ.

4 0

3 years ago

A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm ons tionless bearings. Two 500 g blocks fall from above, hit the tum ble s

Verdich [7]

There are mistakes in the question.The correct question is here

A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick. What is the turntable’s angular velocity, in rpm, just after this event?

Answer:

w=50 rpm

Explanation:

Given data

The mass turntable M=2kg

Diameter of the turntable d=20 cm=0.2 m

Angular velocity ω=100 rpm= 100×(2π/60) =10.47 rad/s

Two blocks Mass m=500 g=0.5 kg

To find

Turntable angular velocity

Solution

We can find the angular velocity of the turntable as follow

Lets consider turntable to be disk shape and the blocks to be small as compared to turntable

$I_{turntable}w=I_{block1}w^{i}+I_{turntable}w^{i}+I_{block2}w^{i}$

where I is moment of inertia

$w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\ So\\I_{turntable}=M\frac{r^{2} }{2}\\I_{turntable}=2*(\frac{(0.2/2)}{2} )\\ I_{turntable}=0.01 \\And\\I_{block1}=I_{block2}=mr^{2}\\I_{block1}=I_{block2}=(0.5)*(0.2/2)^{2} \\ I_{block1}=I_{block2}==0.005\\so\\w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\w^{i}=\frac{0.01*(10.47)}{0.005+0.005+0.01} \\w^{i}=5.235 rad/s\\w^{i}=5.235*(60/2\pi )\\w^{i}=50 rpm$

7 0

3 years ago

Assume that, when we walk, in addition to a fluctuating vertical force, we exert a periodic lateral force of amplitude 25 NN at

dexar [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told

The amplitude of the lateral force is $F = 25 \ N$

The frequency is $f = 1 \ Hz$

The mass of the bridge per unit length is $\mu = 2000 \ kg /m$

The length of the central span is $d = 144 m$

The oscillation amplitude of the section considered at the time considered is $A = 75 \ mm = 0.075 \ m$

The time taken for the undriven oscillation to decay to $\frac{1}{e}$ of its original value is t = 6T

Generally the mass of the section considered is mathematically represented as

$m = \mu * d$

=> $m = 2000 * 144$

=> $m = 288000 \ kg$

Generally the oscillation amplitude of the section after a time period t is mathematically represented as

$A(t) = A_o e^{-\frac{bt}{2m} }$

Here b is the damping constant and the $A_o$ is the amplitude of the section when it was undriven

So from the question

$\frac{A_o}{e} = A_o e^{-\frac{b6T}{2m} }$

=> $\frac{1}{e} =e^{-\frac{b6T}{2m} }$

=> $e^{-1} =e^{-\frac{b6T}{2m} }$

=> $-\frac{3T b}{m} = -1$

=> $b = \frac{m}{3T}$

Generally the amplitude of the section considered is mathematically represented as

$A = \frac{n * F }{ b * 2 \pi }$

=> $A = \frac{n * F }{ \frac{m}{3T} * 2 \pi }$

=> $n = A * \frac{m}{3} * \frac{2\pi}{25}$

=> $n = 0.075 * \frac{288000}{3} * \frac{2* 3.142 }{25}$

=> $n = 1810 \ people$

3 0

3 years ago

You are asked to design a spring that will give a 1070 kg satellite a speed of 3.75 m/s relative to an orbiting space shuttle. Y

Dvinal [7]

Answer:

$380697.33\ \text{N/m}$

$0.138\ \text{m}$

Explanation:

m = Mass rocket = 1070 kg

v = Velocity of rocket = 3.75 m/s

a = Acceleration of rocket = 5g

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

The energy balance of the system is given by

$\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2\\\Rightarrow kx=\dfrac{mv^2}{x}\\\Rightarrow kx=\dfrac{1070\times 3.75^2}{x}\\\Rightarrow kx=\dfrac{7250}{x}$

The force balance of the system is given by

$ma=kx\\\Rightarrow m5g=\dfrac{7250}{x}\\\Rightarrow x=\dfrac{7250}{1070\times 5\times 9.81}\\\Rightarrow x=0.138\ \text{m}$

The distance the spring must be compressed is $0.138\ \text{m}$

$k=\dfrac{7250}{x^2}\\\Rightarrow k=\dfrac{7250}{0.138^2}\\\Rightarrow k=380697.33\ \text{N/m}$

The force constant of the spring is $380697.33\ \text{N/m}$ .

4 0

3 years ago

Which of the following has the most potential energy?

user100 [1]

Options A and C are correct but if you compare their masses, person has negligible mass compared to car. So a car at the top of the hill has the most potential energy.

6 0

3 years ago