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Blababa [14]
4 years ago
14

If Chris drives 27 km N and then 40 km E , what is the distance and displacement

Physics
1 answer:
Bas_tet [7]4 years ago
3 0
The answer would be 13.
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Consider the following hypotheses: 12:V ) >(x)) Use rules of inference to prove that the following conclusion follows from th
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Explanation:

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A deep space probe travels in a straight line at a constant speed of over 16,000 m/s. Assuming there is no friction in space, if
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I believe that the answer is d.

Explanation:

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4 years ago
A spring has a length of .200 m when a .300 kg hangs from it,and a length of .750 m when a 1.95 kg hangs from it.what is the for
musickatia [10]

1) 29.5 N/m

2) 0.100 m

Explanation:

1)

The force constant of the spring can be found by using the fact that the force on the spring is proportional to the extension of the spring (Hooke's Law). Therefore, we can write:

\Delta F= k \Delta x

where

\Delta F = F_2 - F_1 is the change in the force on the spring, where

F_1 = m_1 g = (0.300)(9.8)=2.94 N is the force applied when the hanging mass is

m_1 = 0.300 kg

F_2 = m_2 g = (1.95)(9.8)=19.1 N is the force applied when the hanging mass is

m_2 = 1.95 kg

\Delta x=x_2 -x_1 is the change in extension of the spring, where

x_1=0.200 m is the extension of the spring when the hanging mass is 0.300 kg

x_2=0.750 m is the extension of the spring when the hanging mass is 1.95 kg

Solving for k,

k=\frac{F_2-F_1}{x_2-x_1}=\frac{19.1-2.94}{0.750-0.200}=29.5 N/m

2)

When the first mass is hanging on the spring, we have

F_1 = k (x_1 - x_0)

where:

F_1 is the force applied on the spring (the weight of the hanging mass)

k is the spring constant

x_1 is the extension of the spring wrt its natural length

x_0 is the natural length of the spring (the unloaded length)

Here we have

F_1=2.94 N

k = 29.5 N/m

x_1=0.200 m

Solving for x_0, we find:

x_0 = x_1 - \frac{F_1}{k}=0.200 - \frac{2.94}{29.5}=0.100 m

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3 years ago
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The resultant or the net force act on the object cannot be 0. it must have a value but the direction is opposite from the object move direction.

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