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max2010maxim [7]
2 years ago
14

4. The 50-kg crate shown in Fig. rests on a horizontal surface for which the coefficient of

Physics
1 answer:
laila [671]2 years ago
5 0

Answer:

5.057 m/s^2

Explanation:

Force of kinetic friction = .3  = F /normal force

  .3 = F /(50* 9.81)     F of friction = 147.15

Net force =   400 - 147.5  =  252.85  N

F = m * a

252.85 = 50 * a       a = 5.057 m/s^2

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If it requires 3.0 J of work to stretch a particular spring by 2.1 cm from its equilibrium length, how much more work will be re
-BARSIC- [3]

Answer:

Work done = 13605.44

Explanation:

Data provided in the question:

For elongation of 2.1 cm (0.021 m) work done by the spring is 3.0 J

The relation between Energy (U) and the elongation (s) is given as:

U = \frac{1}{2}kx^2   ................(1)

where,

k is the spring constant

on substituting the valeus in the above equation, we get

3.0 = \frac{1}{2}k\times0.021^2

or

k = 13605.44 N/m

now

for the elongation x = 2.1 + 4.1 = 6.2 cm = 0.062 m

using the equation 1, we have

U = \frac{1}{2}\times13605.44\times (0.062)^2

or

U = 26.149 J

Also,

Work done = change in energy

or

W = 26.149 - 3.0 = 23.149 J

4 0
3 years ago
Please help!!! what is the main point of paragraph 3?
Nikitich [7]
There’s no picture:(
8 0
3 years ago
Read 2 more answers
A 25 kg mass is accelerated by a force at a rate of 5 m/s2. What is the magnitude of the force that accelerates the man?
svlad2 [7]
Force=mass*acceleration
F=ma
F=25*5
F=100 N
8 0
3 years ago
The wavelength of sound wave is 30 m. Frequency is 7 Hz. What is the wave speed
MaRussiya [10]

Answer:23hz

Explanation:

6 0
3 years ago
During lightning strikes from a cloud to the ground, currents as high as 2.50×10^4 Amps can occur and last for about 40.0 micros
dangina [55]

Answer:

1 C

Explanation:

The intensity of electric current is defined as

I=\frac{q}{t}

where

I is the current

q is the amount of charge transferred

t is the time interval during which the charge is transferred

For the lightning in this problem, we have

I=2.50\cdot 10^4 A is the current

t=40.0 \mu s = 40.0\cdot 10^{-6} s is the time interval

Solving the formula for q, we find the amount of charge transferred:

q=I t = (2.50\cdot 10^4 A)(40.0\cdot 10^{-6}s)=1 C

6 0
3 years ago
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