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3 years ago

A large rock is motionless on a flat sidewalk. Which statement best explains why the rock remains motionless?

Physics

1 answer:

seraphim [82]3 years ago

6 0

Answer:

There are no forces acting on the rock :)

Explanation:

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Why do physicists use models?

svetoff [14.1K]

To see if the project is actually going to work. Also called prototypes

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3 years ago

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A 55-kg skier starts from rest at the top of a ski jump, point A in Fig. 6–48, and travels down the ramp. If fric- tion and air

andre [41]

Explanation:

The first attachment contains the complete question while the second attachment is the solution.

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4 years ago

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In a parachute, falling steadily at 7.5m/s, how long does it take to reach the ground from 3,000 m high?

photoshop1234 [79]

Answer:

25 minutes

Explanation:

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3 years ago

A package of aluminum foil contains 50. ft2 of foil, which weighs approximately 7.5 oz . Aluminum has a density of 2.70 g/cm3. W

Tems11 [23]

Answer:

The thickness of the foil is 0.017 mm.

Explanation:

Given that,

Weight = 7.5 oz = 212.6175 gm

Density = 2.70 g/cm³

Area of aluminium = 50 ft² = 46451.52 cm²

We need to calculate the thickness of the foil

Using formula of density

$\rho=\dfrac{m}{V}$

$\rho=\dfrac{m}{A\times t}$

$t=\dfrac{m}{A\times \rho}$

Where, A = area

t = thickness

m = mass

Put the value into the formula

$t=\dfrac{212.6175}{46451.52\times2.70}$

$t=0.00170\ cm$

$t=0.017\ mm$

Hence, The thickness of the foil is 0.017 mm.

3 0

4 years ago

We are sending a 30 Mbit file from source host A to destination host B. All links in the path between source and destination hav

Dmitrij [34]

Answer:

$t=1.5\times 10^{-4}\ s$

Explanation:

Given:

file size to be transmitted, $D=30\ Mb$

transmission rate of data, $\dot D=10\ Mb.s^{-1}$

propagation speed, $v=2\times 10^8\ m.s^{-1}$

distance of data transfer, $s=10000\ km=10^4\ m$

<u>Now the delay in data transfer from source to destination for each 10 Mb:</u>

$t'=\frac{s}{v}$

$t'=\frac{10^4}{2\times 10^8}$

$t'=5\times 10^{-5}\ s$

<u>Now this time is taken for each 10 Mb of data transfer and we have 30 Mb to transfer:</u>

So,

$t=3\times t'$

$t=3\times 5\times 10^{-5}$

$t=1.5\times 10^{-4}\ s$

3 0

4 years ago