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Anon25 [30]
3 years ago
12

A crane raises a crate with a mass of 75 kg to a height of 10 m. Given that the

Physics
1 answer:
Crazy boy [7]3 years ago
7 0

Answer: C. 7,350 J

Explanation:

Just did it in Apex

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Why does damp soil help excess electrons move
Marina CMI [18]

Explanation:

  • Because where there is water, energy is conducted, and <u>moves </u>the electrons.
  • They also move excess electrons because pure water is a poor conductor

3 0
3 years ago
An 80-kgkg quarterback jumps straight up in the air right before throwing a 0.43-kgkg football horizontally at 15 m/sm/s . How f
vladimir1956 [14]

Answer:

V = 0.0806 m/s

Explanation:

given data

mass quarterback = 80 kg

mass football = 0.43 kg

velocity = 15 m/s

solution

we consider here momentum conservation is in horizontal direction.

so that here no initial momentum of the quarterback

so that final momentum of the system will be 0

so we can say

M(quarterback) ×  V = m(football) × v (football)   ........................1

put here value we get

80 ×  V  = 0.43  × 15

V = 0.0806 m/s

5 0
3 years ago
A force of 20. Newtons to the left exerted on a cart for 10. Seconds. For what period of time must a 50.-newton force to the rig
FinnZ [79.3K]
Impulse = (force) x (time)

The first impulse was (20 N) x (10 sec) = 200 meters/sec

The second one is (50 N) x (time) and we want it equal to the first one, so

(50 N) x (time) = 200 meters/sec

Divide each side by 50N :    Time = 200/50 = <em>4 seconds</em>

By the way, the quantity we're playing with here is the cart's <em>momentum</em>.
6 0
3 years ago
An open pipe, 0.29 m long, vibrates in the second overtone with a frequency of 1,227 Hz. In this situation, the fundamental freq
Andru [333]

Answer:

f = 409 Hz

Explanation:

We have,

Length of the open organ pipe, l = 0.29 m

Frequency of vibration of second overtone, f_2 = 1227 Hz

It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :

f_2=\dfrac{3v}{2l}

v is speed of sound

Let f is the fundamental frequency. It is given by :

f=\dfrac{v}{2l}

The relation between f and f₂ can be written as :

f_2=3f\\\\f=\dfrac{f_2}{3}\\\\f=\dfrac{1227}{3}\\\\f=409\ Hz

So, the fundamental frequency of the pipe is 409 Hz.              

6 0
3 years ago
A 10-cm-long wire is pulled along a u-shaped conducting rail in a perpendicular magnetic field. the total resistance of the wire
Debora [2.8K]

In the above case we can say that power given by external agent to pull the rod must be equal to the power dissipated in the form of heat due to magnetic induction.

Part a)

when we pull the rod with constant speed then power required will be product of force and velocity

here we will have

P = F.v

P = 4 W

v = 4 m/s

now we will have

4 = F*4

F = 1N

So external force required will be 1 N

PART B)

now in order to find magnetic field strength we can say

P = \frac{v^2B^2L^2}{R}

here we know that induced EMF in the wire is E = vBL

so power due to induced magnetic field is given by

P = \frac{E^2}{R}

4 = \frac{4^2*B^2*0.10^2}{0.20}

by solving above equation we will have

B = 2.24 T

5 0
3 years ago
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