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Nikolay [14]
3 years ago
12

Does the initial velocity of an object have anything to do with its acceleration? For example, compared to dropping an object, i

f you throw it downward would the acceleration be different after you released it?
Physics
1 answer:
adelina 88 [10]3 years ago
4 0

Answer:

Explanation:

This is an excellent question to get an answer for. It teaches you much about the nature of physics.

The answer is no.

The distance will be quite different. The time might be different in getting to the distance.  But the acceleration will be the same in either case.

How do you know? Look at one of the formulas, say

d = vi * t + 1/2*a * t^2

What does vi do? vi will alter both t and d. if vi = 0 then both d and/or t will be found. But what will "a" do? Is there anything else acting in the up or down line of action? You should answer no.

If vi is not zero, t will be less and d will take less time to get where it is going.

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A 5kg object is moving downward at a speed of 12m/s. If it is currently 2.6m above the ground, what is its potential energy? Use
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4 0
3 years ago
A worker pushes horizontally on a 34 kg crate with a force of magnitude 110 n . the coefficient of static friction between the c
Vadim26 [7]
<span>Static frictional force = 126.91 N 
1st worker force = 110 N 
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4 0
3 years ago
What should be the spring constant k of a spring designed to bring a 1200 kg car to rest from a speed of 85 km/h so that the occ
borishaifa [10]

Answer:

k = 5178.8 N/m

Explanation:

As we know that spring mass system will oscillate at angular frequency given as

\omega = \sqrt{\frac{k}{m}}

now we have

\omega = \sqrt{\frac{k}{1200}}

now the maximum acceleration of the spring block system is at its maximum compression state which is given as

a = \omega^2 A

here A= maximum compression of the spring

so here in order to find maximum compression of the spring we will use energy conservation as we know that initial total kinetic energy of the car will convert into spring potential energy

\frac{1}{2}mv^2 = \frac{1}{2}kA^2

here we know that

v = 85 km/h

v = 85 \times\frac{1000}{3600} = 23.61 m/s

now we have

(1200)(23.61^2) = kA^2

A^2 = \frac{6.68 \times 10^5}{k}

now from above equation of acceleration we have

5.0 g = (\frac{k}{m})\sqrt{\frac{6.68 \times 10^5}{k}}

5.0(9.81) = \sqrt{k}(0.68)

k = 5178.8 N/m

6 0
3 years ago
Select the statement that best describes gravity.
ch4aika [34]

Answer:

Gravity is the force of attraction between two objects; it is dependent upon the mass of the objects and the distance between the objects.

Explanation:

Gravity is defined as the force of attraction between two objects. It depends on the masses of objects and the distance between them. The gravitational force between two bodies is given by the universal law of gravitation. According to this law, the force of gravity is :

F=G\dfrac{m_1m_2}{d^2}

Where

G is the universal gravitational constant

m₁ and m₂ are masses of two bodies

d is the distance between two bodies

Hence, the correct statement that describes gravity is (b)

6 0
3 years ago
Read 2 more answers
Traveling waves are generated on a string fixed at both ends. The string has a length L, a linear mass density m, and a tension
bagirrra123 [75]

Answer: d. I or II

Explanation: A traveling wave has speed that depends on characteristics of a medium. Characteristics like linear density (μ), which is defined as mass per length.

Tension or Force (F_{T}) is also related to the speed of a moving wave.

The relationship between tension and linear density and speed is ginve by the formula:

|v|=\sqrt{\frac{F_{T}}{\mu} }

So, for the traveling waves generated on a string fixed at both ends described above, ways to increase wave speed would be:

1) Increase Tension and maintaining mass and length constant;

2) Longer string will decrease linear density, which will increase wave speed, due to their inversely proportional relationship;

Then, ways to increase the wave speed is

I. Using the same string but increasing tension

II. Using a longer string with the same μ and T.

8 0
3 years ago
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