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saw5 [17]
3 years ago
12

A crate feels a 277 n normal force as it sits on the ground. what is its mass?

Physics
2 answers:
Galina-37 [17]3 years ago
6 0

Answer:

<u>28.3 kg </u>

Explanation:

Assuming the ground is level, the normal force equals the weight.

N = mg

277 N = m × 9.8 m/s²

m = 28.3 kg

Umnica [9.8K]3 years ago
5 0

Answer:

28.3 kg

Explanation:

Assuming the ground is level, the normal force equals the weight.

N = mg

277 N = m × 9.8 m/s²

m = 28.3 kg

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To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)

PART A) By definition the frequency in a spring is given by the equation

f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}

Where,

m = mass

k = spring constant

Our values are,

k=1700N/m

m=5.3 kg

Replacing,

f = \frac{1}{2\pi} \sqrt{\frac{1700}{5.3}}

f=2.85 Hz

PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} kY^2

Where,

k = Spring constant

m = mass

y = Vertical compression

v = Velocity

This expression is equivalent to,

kA^2 =mV^2 +ky^2

Our values are given as,

k=1700 N/m

V=1.70 m/s

y=0.045m

m=5.3 kg

Replacing we have,

1700*A^2=5.3*1.7^2 +1700*(0.045)^2

Solving for A,

A^2 = \frac{5.3*1.7^2 +1700*(0.045)^2}{1700}

A ^2 = 0.011035

A=0.105 m \approx 10.5 cm

PART C) Finally, the total mechanical energy is given by the equation

E = \frac{1}{2}kA^2

E=\frac{1}{2}1700*(0.105)^2

E= 9.3712 J

3 0
3 years ago
A rock falls from rest off a cliff and hits the ground in 2 seconds. If there is no air resistance, determine the rock's velocit
mihalych1998 [28]

Answer:

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Explanation:

A Rock falling off a cliff can be modeled as an object starting with  zero velocity moves with constant acceleration for certain period of time, for such motion following equation of motion can be used.

v(t) = v_{0} +at

here in our case v_{0} =0 because object starts off from rest and a = g =9.8m/s^2 is acceleration because of gravity ( Motion under gravity).

and of course t = 2 second.

Now by substituting all this information in equation of motion we get.

v(2s) = 0+9.8m/s^2 *2s = 19.6m/s

that would be the velocity of rock as it would hit the ground.

Note! We have assumed that there is no air resistance.

A rock falling off a cliff can be modeled as an object starting with zero velocity moves with constant acceleration for a certain period of time, for such motion following equation of motion can be used.

here in our case  because object starts off from rest and  is acceleration because of gravity ( Motion under gravity).

and of course t = 2 seconds.

Now by substituting all this information in equation of motion we get.

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that would be the velocity of rock as it would hit the ground.

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Problem 8 I estimate that the Gauss gun (a solenoid) is wound with 500 turns over a distance of 15cm with an average radius of 1
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Answer:

Energy stored, U = 66.6 J

Explanation:

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U=\dfrac{1}{2}LI^2

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L=\mu_o (n/d)^2A d

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L = 0.00148 Henry

U=\dfrac{1}{2}\times 0.00148\times 300^2

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Out of given options, the correct option for the energy stored in the solenoid is 70 J. So, the correct option is (a) "70 J".  

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