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stealth61 [152]
4 years ago
15

A 65.0 kg football player is running at 10.0 m/s downfield when he catches a 0.420 kg football thrown in the same direction but

at 31.0 m/s. How fast will the player and ball move after the catch

Physics
1 answer:
lukranit [14]4 years ago
8 0

Explanation:

Below is an attachment containing the solution.

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Amy uses 20N of force to push a lawn mower 10 meters. How much work does she do? *
babunello [35]

She does 200J .

We know she uses 20N of force and 10m is the distance. We multiply both numbers and we are given our answer of 200J. Hope this was helpful. :)

4 0
4 years ago
A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotat
Natali [406]

centripetal acceleration is given by formula

a_c = \omega^2*R

given that

a_c = 34.1 m/s^2

R  =  5.91 m

now we have

\omega^2 R = 34.1

\omega^2 * 5.91 = 34.1

\omega^2 = 5.77

\omega = 2.4 rad/s

so the ratationa frequency is given by

\omega = 2 \pi f

2.4 = 2 \pi f

f = \frac{2.4}{2\pi}

f = 0.38 Hz

7 0
3 years ago
A 6.7 kg object moves with a velocity of 8 m/s. what is its kinetic energy
blagie [28]

The Correct answer to this question for Penn Foster Students is: 214.4 J

3 0
3 years ago
Read 2 more answers
The mean time between collisions for electrons in room temperature copper is 2.5 x 10-14 s. What is the electron current in a 2
Darya [45]

Answer:

1.87 A

Explanation:

τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s

d = diameter of copper wire = 2 mm = 2 x 10⁻³ m

Area of cross-section of copper wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

E = magnitude of electric field = 0.01 V/m

e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

m = mass of electron = 9.1 x 10⁻³¹ kg

n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³

i = magnitude of current

magnitude of current is given as

i = \frac{Ane^{2}\tau E}{m}

i = \frac{(3.14\times 10^{-6})(8.47\times 10^{28})(1.6\times 10^{-19})^{2}(2.5\times 10^{-14}) (0.01)}{(9.1\times 10^{-31})}

i  = 1.87 A

4 0
3 years ago
If a car has a velocity of 85 km/hr, how long will it take to accelerate to 45 km/hr if the acceleration is -3 km/hr/sec?
weqwewe [10]

Answer:

Time, t = 13.34 seconds.

Explanation:

Given the following data;

Initial velocity, u = 85km/hr to meters per seconds = 85*1000/3600 = 23.61 m/s

Final velocity, v = 45km/hr to meters per seconds = 45*1000/3600 = 12.5 m/s

Acceleration, a = -3 km/hr/sec to meters per seconds square = -3*1000/3600 = -0.833m/s²

To find the time;

Acceleration = (v - u)/t

-0.833 = (12.5 - 23.61)/t

-0.833t = -11.11

t = 11.11/0.833

Time, t = 13.34 seconds.

6 0
3 years ago
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