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3 years ago

Name the physical quantity measured by the slope of the velocity time graph for uniform motion

Physics

1 answer:

mixas84 [53]3 years ago

8 0

Acceleration. If the line is linear on a velocity time graph, the acceleration does not change, meaning its 0

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Sedimentary rocks are the only rocks that can potentially containa.fossils.c.fractures.b.minerals.d.faults. Please select the be

Elena-2011 [213]

Hello <span>Mr1guy24 </span>

Question: <span>Sedimentary rocks are the only rocks that can potentially contain?
</span>
Answer: Fossils (A)

Reason: This is what makes sedimentary rocks unique

Hope This Helps!
-Chris

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3 years ago

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Which change would create light?

hichkok12 [17]

Answer:

Im not 100% sure but i think the answer is A. An electron in an atom jumping from a lower energy state to a higher one.

Explanation:

lmk if its wrong

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3 years ago

If an experiment involves a large volume of liquid a _______ would most likely be used to hold it.

Grace [21]

The answer is d a beacker

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3 years ago

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The length of the minute hand of the clock is 14cm. Calculate the speed with which the tip of the minute hand moves

Molodets [167]

Speed of the tip of the minute hand=V=0.0244 cm/s

Explanation:

The angular velocity of the minute hand is given by

$\omega= \frac{2\pi}{T}$

T= time period of the minute hand=60 min=3600 s

so ω= 2 π/3600 rad/s

Now linear velocity v= r ω

r= radius of minute hand=14 cm

so v= 14 (2 π/3600)

V=0.0244 cm/s

8 0

3 years ago

A roller coaster car starts from rest at the top of a hill 15 m high and rolls down to ground level. From there it starts into a

Softa [21]

Answer:

955.5N

Explanation:

The normal force is given by the difference between the centripetal force and gravity at the top of the loop:

$F_N = F_C - F_G = m\frac{v^{2} }{r} - mg$

mass m = 65kg

radius of the loop r = 4m

velocity v = ?

g = 9.8 m/s²

To find the centripetal force, you need to find the velocity of the car at the top of the loop.

Use energy conservation:

$E_{tot}=mgh + \frac{1}{2} mv^{2}$

At the top of the hill:

$E_{tot}= mgh_{hill}$

At the top of the loop:

$E_{tot}=mgh_{loo}_p +\frac{1}{2} m v^{2}$

Setting both energies equal and canceling the mass m gives:

$gh_{hill} = gh_{loo}_p + \frac{1}{2} v^{2}$

Solving for v:

$v^{2} = 2g(h_{hill}-h_{loo}_p)$

Using v in the first equation:

$F_N = \frac{2mg(h_{hill}-h_{loo}_p)}{r} - mg$

$F_N = 955.5N$

7 0

3 years ago