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LenKa [72]
2 years ago
10

A football player kicks a ball with the force of 50N. Find the impulse on the ball if his foot stays in contact with the footbal

l for 0.01s
Physics
1 answer:
seropon [69]2 years ago
6 0

Answer:

Impulse  = 0.5Ns

Explanation:

F = 50N

t = 0.01s

Impulse = ?

Impulse = Force * t,

Impulse = 50N * 0.01s = 0.5Ns

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kiruha [24]
It’s red shift and blue shift because This only occurs when the frequency of the wave is made longer or shorter due to the movement of the source relative to the observer
4 0
3 years ago
Solve for velocity in A car has traveled 456 meters in 45 seconds while driving east on Broad Street. What is the velocity of th
OLEGan [10]

Answer:

10.13

Explanation:

take note that velocity is distance over time (v=\frac{d}{t}) so you do 456(distance)÷45(time)= 10.13

3 0
4 years ago
How do you find the velocity after a collision
Evgen [1.6K]

Answer:

In a collision, the velocity change is always computed by subtracting the initial velocity value from the final velocity value. If an object is moving in one direction before a collision and rebounds or somehow changes direction, then its velocity after the collision has the opposite direction as before.

Explanation:

7 0
4 years ago
The perihelion of the comet TOTAS is 1.69 AU and the aphelion is 4.40 AU. Given that its speed at perihelion is 28 km/s, what is
dybincka [34]

Answer:

The speed at the aphelion is 10.75 km/s.

Explanation:

The angular momentum is defined as:

L = mrv (1)

Since there is no torque acting on the system, it can be expressed in the following way:

t = \frac{\Delta L}{\Delta t}

t \Delta t = \Delta L

\Delta L = 0

L_{a} - L_{p} = 0

L_{a} = L_{p}   (2)

Replacing equation 1 in equation 2 it is gotten:

mr_{a}v_{a} =mr_{p}v_{p} (3)

Where m is the mass of the comet, r_{a} is the orbital radius at the aphelion, v_{a} is the speed at the aphelion, r_{p} is the orbital radius at the perihelion and v_{p} is the speed at the perihelion.          

From equation 3 v_{a} will be isolated:    

v_{a} = \frac{mr_{p}v_{p}}{mr_{a}}

v_{a} = \frac{r_{p}v_{p}}{r_{a}}   (4)    

Before replacing all the values in equation 4 it is necessary to express the orbital radius for the perihelion and the aphelion from AU (astronomical units) to meters, and then from meters to kilometers:

r_{p} = 1.69 AU x \frac{1.496x10^{11} m}{1 AU} ⇒ 2.528x10^{11} m

r_{p} = 2.528x10^{11} m x \frac{1km}{1000m} ⇒ 252800000 km

r_{a} = 4.40 AU x \frac{1.496x10^{11} m}{1 AU} ⇒ 6.582x10^{11} m

r_{p} = 6.582x10^{11} m x \frac{1km}{1000m} ⇒ 658200000 km  

     

Then, finally equation 4 can be used:

v_{a} = \frac{(252800000 km)(28 km/s)}{(658200000 km)}

v_{a} = 10.75 km/s

Hence, the speed at the aphelion is 10.75 km/s.

       

8 0
3 years ago
Because quantum mechanics is physics that describes the interactions of very small objects (i.e. molecules, atoms, and electrons
stira [4]
Ha! Lot of words but the question itself is easy.
The answer is 2.5 times 10 to the 5th power.
The main part of the numbers has the decimal point placed after the first digit.
Then for what number of power, you just count the number of decimal places moved.
I hope this helps you.
5 0
3 years ago
Read 2 more answers
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