Ask question

Ask question

All categories

3 years ago

9

A pitcher throws a 0.15 kg baseball so that it crosses home plate horizontally with a speed of 10 m/s. It is hit straight back a

t the pitcher with a final speed of 24 m/s. Assume the direction of the initial motion of the baseball to be positive.

1 answer:

Maru [420]3 years ago

4 0

Answer:

-5.1 kg m/s

Explanation:

Impulse is the change in momentum.

Change in momentum= final momentum - initial momentum=m $v_{2}$ +m $v_{1}$

Plugging in the values= -0.15*24 - (0.15*10) (The motion towards the pitcher is negative as the initial motion is considered to be positive)

Impulse=-5.1 kg m/s (-ve means that it is the impulse towards the pitcher)

You might be interested in

Which nucleus completes the following equation?

natka813 [3]

the answer is C

IGNORW irritating but not even on the golden bath bath and

3 0

3 years ago

A potential difference of 107 mV exists between the inner and outer surfaces of a cell membrane. The inner surface is negative r

sergij07 [2.7K]

Answer:

The workdone is $W = 1.712 *10^{-20 } \ J$

Explanation:

From the question we are told that

The potential difference is $V = 107 mV = 107 *10^{-3} \ V$

Generally the charge on $Na^{+}$ is $Q_{Na^{+}} = 1.60 *10^{-19 } \ C$

Generally the workdone is mathematically represented as

$W = Q_{Na^{+}}V$

=> $W = 1.60 *10^{-19 } * 107 *10^{-3}$

=> $W = 1.712 *10^{-20 } \ J$

8 0

3 years ago

A ball is dropped from rest at the top of a 6.10 m

natita [175]

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

$\frac{v_1}{v}$ = e ( coefficient of restitution ) = $\frac{1}{\sqrt{10} }$

and

$\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }$

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

$e = \sqrt{\frac{h_1}{6.1} }$

$e = \sqrt{\frac{h_2}{h_1} }$

So on

$e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }$

$(\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}$ = .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

3 0

3 years ago

While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)

ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration= $2.40m/s^{2}$ , c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = $2.40m/s^{2}$

time = 12.0 s

initial position = 0

c)

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

$x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}$

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

$x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}$

after cancelling the pertinent units, I get that my answer will be given in meters. So I get:

$x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}$

which solves to:

$x=172.8m$

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

$a=\frac{V_{f}-V_{0} }{t}$

Next, since I need to find the final velocity, I can solve the equation just for that, I can start by multiplying both sides by t so I get:

$at=V_{f}-V_{0}$

and finally I can add $V_{0}$ to both sides so I get:

$V_{f}=at+V_{0}$

and now I can proceed and substitute the known values:

$V_{f}=at+V_{0}$

$V_{f}=(2.40\frac{m}{s^{2}}} (12s)+0$

which solves to:

$V_{f}=28.8m/s$

8 0

3 years ago

Read 2 more answers

A futuristic design for a car is to have a large solid disk-shaped flywheel within the car storing kinetic energy. The uniform f

Aleks04 [339]

A futuristic design for a car is to have a large solid disk-shaped flywheel within the car storing kinetic energy. The uniform flywheel has mass 370 kg with a radius of 0.500 m and can rotate up to 320 rev/s. Assuming all of this stored kinetic energy could be transferred to the linear velocity of the 3500-kg car, find the maximum attainable speed of the car.

5 0

3 years ago