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shusha [124]
2 years ago
6

Hitting a home run in baseball is difficult, because it requires a lot of force. Most home runs are hit off of fastballs. Which

principle about impact force says that this makes sense, and why? (please help!!)
Physics
1 answer:
OLga [1]2 years ago
7 0

The impact force makes sense because the impulse experienced by the body cause a great change in momentum of the body.

<h3 /><h3>What is impulse?</h3>

This is the force that acts on a body over a given period of time. The impulse experienced by a body is determined as the product of force and time of action.

J = Ft

The change in momentum of a body is equal to the impulse experienced by the body.

ΔP = Ft

Thus, the impact force makes sense because the impulse experienced by the body cause a great change in momentum of the body.

Learn more about impulse here: brainly.com/question/25700778

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The volume of a gas is 200.0 mL at 275 K and 92.1 kPa. Find its volume at STP.
ycow [4]
To solve this question we will use ideal gas equation:
p*V=n*R*T
Where:
p = pressure
V = volume
n = number of moles
R = gas constant
T = temperature

We can rearrange formula to get:
\frac{p*V}{T} =n*R
We are working woth same gas so we can write following formula. Index 1 stands for conditions before change and index 2 stands for conditions after change.
\frac{ p_{1}*V_{1} }{T_{1}} = \frac{ p_{2}*V_{2} }{T_{2}}

We are given:
p1=92.1kPa = 92100Pa
V1=200mL = 0.2L
T1=275K
p2= 101325Pa
T2=273K
V2=?

We start by rearranging formula for V2. After that we can insert numbers:
{ p_{1}*V_{1} *T_{2}} = { p_{2}*V_{2}*T_{1} } \\  \\ V_{2}= \frac{p_{1}*V_{1} *T_{2}}{ p_{2}*T_{1}}  \\  \\ V_{2}=  \frac{92100*0.2*273}{101325*275}  \\  \\ V_{2}= 0.18L=180mL
7 0
4 years ago
Two people are talking at a distance of 3.0 m from where you are and you measure the sound intensity as 1.1 × 10-7 W/m2. Another
Tanya [424]

Answer:I_2=0.618\times 10^{-7} W/m^2

Explanation:

Given

Distance between source and  receiver d_1=3 m

Sound Intensity I_1=1.1\times 10^{-7} W/m^2

Distance of of second observer d_2=4 m

Intensity varies as

I\propto \frac{1}{d^2}

using this

I=\frac{k}{d^2}

\frac{I_1}{I_2}=\frac{d_2^2}{d_1^2}

\frac{1.1\times 10^{-7}}{I_2}=\frac{4^2}{3^2}

I_2=0.75^2\times 1.1\times 10^{-7}

I_2=0.618\times 10^{-7} W/m^2

     

6 0
3 years ago
The field around a solenoid is similar to the field around a bar magnet.<br><br> true or false?
OlgaM077 [116]
The answer would be true :)
8 0
3 years ago
How much work is done when a force of 15.0 newtons is used to lift a box 3.0 meters into the air?
denis-greek [22]

Added potential energy = (mass) x (gravity) x (height)

or

Added potential energy = (weight) x (added height)

If you need to lift a 15N box 3m straight up, you have to increase its potential energy by (15 N) x (3 m) = 45 Joules .

Where is that added potential energy supposed to come from ?  You could use an electric winch, a steam engine, a gasoline-powered motor, thousands of hamsters running on little treadmills that are are connected to the main pulley somehow, or your own arm muscles.  But howEVER you do it, you have to provide <em>45 Joules</em> of WORK in order to increase the potential energy of the box by just that much.  

8 0
3 years ago
Read 2 more answers
Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.​
Tems11 [23]

Answer:

<em>The range is 35.35 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and g=9.8m/s^2 the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

\displaystyle d={\frac  {v_o^{2}\sin(2\theta )}{g}}

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

\displaystyle d={\frac  {20^{2}\sin(2\cdot 30^\circ )}{9.8}}

\displaystyle d={\frac  {400\sin(60^\circ )}{9.8}}

d=35.35\ m

The range is 35.35 m

7 0
3 years ago
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