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tatiyna
2 years ago
13

A car accelerates at 4 m/s/s from rest. What is the car's velocity after it travels 20 m?

Physics
1 answer:
MrRa [10]2 years ago
3 0
The cars velocity after it travels 20 m is 4 because you have to divide it by the miles that the car is traveling and the 4 miles from the rest; and your finally answer being 5
Answer: 5
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A man standing on top of a 30 m tall building throws a brick downwards with a velocity of 12 m/s. Determine the speed of the bri
loris [4]

Answer:

27.1m/s

Explanation:

Given parameters:

Height of the building  = 30m

Initial velocity  = 12m/s

Unknown:

Final velocity  = ?

Solution:

We apply one of the kinematics equation to solve this problem:

         v²  = u²  + 2gh

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

h is the height

          v²   = 12²  + (2 x 9.8 x 30)

          v  = 27.1m/s

3 0
3 years ago
If the specific gravity of copper is 8.91, it weighs:
denis-greek [22]

Answer:

Option D is the correct answer.

Explanation:

The specific gravity of copper is 8.91.

We have

            \frac{\texttt{Density of copper}}{\texttt{Density of water}}=8.91\\\\\texttt{Density of copper}=8.91\times 1000kg/m^3=8910kg/m^3

We also have

           1 kg = 2.205 lb

            1 m = 3.28 ft

Substituting

            \texttt{Density of copper}=8910kg/m^3=\frac{8910\times 2.205}{3.28^3}=556.76lb/ft^3

Option D is the correct answer.

6 0
2 years ago
6. A suitcase weighing 120 N sits on a counter 1.8 m high
Liono4ka [1.6K]

Answer:

<em>216 J</em>

Explanation:

h = 1.8

a = 9.8

m = 12.2

<em>GPE</em> = <em>HAM</em> = 216

6 0
2 years ago
As you rise upwards in the atmosphere air pressure___
Oduvanchick [21]

Answer:

Gases - Water vapor, Nitrogen, Oxygen etc. Describe what happens to air pressure as you rise upwards in the atmosphere. What causes this change in air pressure? As altitude increases, air pressure will decrease As altitude increases the gas molecules that make up the air spread further apart

7 0
2 years ago
A stationary 12.5 kg object is located on a table near the surface of the earth. The coefficient of static friction between the
Sophie [7]

When the object is at rest, there is a zero net force due the cancellation of the object's weight <em>w</em> with the normal force <em>n</em> of the table pushing up on the object, so that by Newton's second law,

∑ <em>F</em> = <em>n</em> - <em>w</em> = 0   →   <em>n</em> = <em>w</em> = <em>mg</em> = 112.5 N ≈ 113 N

where <em>m</em> = 12.5 kg and <em>g</em> = 9.80 m/s².

The minimum force <em>F</em> needed to overcome <u>maximum</u> static friction <em>f</em> and get the object moving is

<em>F</em> > <em>f</em> = 0.50 <em>n</em> = 61.25 N ≈ 61.3 N

which means a push of <em>F</em> = 15 N is not enough the get object moving and so it stays at rest in equilibrium. While the push is being done, the net force on the object is still zero, but now the horizontal push and static friction cancel each other.

So:

(a) Your free body diagram should show the object with 4 forces acting on it as described above. You have to draw it to scale, so whatever length you use for the normal force and weight vectors, the length of the push and static friction vectors should be about 61.3/112.5 ≈ 0.545 ≈ 54.5% as long.

(b) Friction has a magnitude of 15 N because it balances the pushing force.

(c) The object is in equilibrium and not moving, so the acceleration is zero.

5 0
3 years ago
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