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butalik [34]
4 years ago
6

You make a straight line when an object is

Physics
2 answers:
swat324 years ago
8 0

Answer:

can you explain more or..,.

Explanation:

Nana76 [90]4 years ago
3 0
When you have two points, if you connect every point between those two points, you have a straight line.
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Which of the following best describes what went wrong with the scientists’ study? an improper experimental procedure the lack of
Rasek [7]
B- lack of control.

the scientists did not have a proper control group.
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3 years ago
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If light has a speed of 122,000 mps in a transparent medium, what is the index of refraction of the medium? A. n = 1.52
castortr0y [4]

Answer:

A.\hspace{3}n=1.52

Explanation:

The refractive index of a medium is a measure to know how much the speed of light within the medium is reduced. It can be calculated with the next equation:

n=\frac{c}{v}   (1)

Where:

c=Speed\hspace{3}of\hspace{3}light\hspace{3}in\hspace{3}vacuum\\n=Refractive\hspace{3}index\\

v=Velocity\hspace{3}of\hspace{3}light\hspace{3}in\hspace{3}the\hspace{3}medium

The speed of light in the vacuum is approximately 300,000 km/s. In order to work with the same units let's do the proper conversion with the velocity of the medium:

122,000\frac{mi}{s} *\frac{1.60934km}{1mi}=196339.48\frac{km}{s}

Finally, replacing the data in (1):

n=\frac{300,000}{196339.48} =1.527965746\approx1.52

3 0
3 years ago
Calculate the amount of momentum of an object with a mass of 10 kg travelling al a
irina [24]

Answer:

50kg.m/s

Explanation:

In order to find momentum you must use the formula P=mv

p= momentum

m=mass

v= velocity

so in other words, momentum= mass times velocity

or in this case, momentum= 10 times 5  :)

7 0
3 years ago
A result of chemical change is
Tcecarenko [31]
We're u can never put it back together
4 0
3 years ago
The potential energy of a 1.7 x 10-3 kg particle is described by U (x )space equals space minus (17 space J )space cos open squa
hichkok12 [17]

Answer:

Explanation:

Given a particle of mass

M = 1.7 × 10^-3 kg

Given a potential as a function of x

U(x) = -17 J Cos[x/0.35 m]

U(x) = -17 Cos(x/0.35)

Angular frequency at x = 0

Let find the force at x = 0

F = dU/dx

F = -17 × -Sin(x/0.35) / 0.35

F = 48.57 Sin(x/0.35)

At x = 0

Sin(0) =0

Then,

F = 0 N

So, from hooke's law

F = -kx

Then,

0 = -kx

This shows that k = 0

Then, angular frequency can be calculated using

ω = √(k/m)

So, since k = 0 at x = 0

Then,

ω = √0/m

ω = √0

ω = 0 rad/s

So, the angular frequency is 0 rad/s

4 0
4 years ago
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