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sveta [45]
3 years ago
9

An object is falling under the influence of gravity in presence of air resistance . There are two forces acting on it. (i) What

are these two forces? (ii)Explain why will the object start moving with uniform velocity after some time?
Physics
1 answer:
MatroZZZ [7]3 years ago
8 0

Answer:

1 the forces are:friction, gravitational pull

2.i guess due to uniform change in displacement per unit time therefore uniform acceleration

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To introduce you to the concept of escape velocity for a rocket. the escape velocity is defined to be the minimum speed with whi
Mama L [17]
A projectile fired upward from the Earth's surface will usually slow down, come momentarily to rest, and return to Earth. For a certain initial speed, however it will move upward forever, with its speed gradually decreasing to zero just as its distance from Earth approaches infinity. The initial speed for this case is called escape velocity. You can find the escape velocity v for the Earth or any other planet from which a projectile might be launched using conservation of energy. The projectile of mass m leaves the surface of the body of mass M and radius R with a kinetic energy Ki = mv²/2 and potential energy Ui = -GMm/R. When the projectile reaches infinity, it has zero potential energy and zero kinetic energy since we are seeking the minimum speed for escape. Thus Uf = 0 and Kf = 0. And from conservation of energy,
Ki + Ui = Kf + Uf
mv²/2 -GMm/R = 0
∴ v = √(2GM/R) 

This is the expression for escape velocity. 
3 0
2 years ago
Read 2 more answers
I need help with #40 need answer and show me what you multiplied please
ale4655 [162]
(0.5)×(0squared)×(3)=(1.5j)
3 0
2 years ago
what will be the restoring force if a spring with a spring constant of 45 newtons per meter is pulled 0.30 meters in the downwar
Anna35 [415]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Below are the choices the can be found elsewhere:

A.) 14 newtons upward 
<span>B.) 45 newtons upward </span>
<span>C.) 67 newtons upward </span>
<span>D.) 130 newtons upward </span>
<span>E.) 150 newtons upward
</span>
The answer is A.) 14 newtons upward


7 0
3 years ago
g A lighter ball A with momentum PA=5.0 kg∙m/s in positive x-direction approaches a heavier ball B at rest before the collision.
Svet_ta [14]

Answer:

7kgm/s

Explanation:

Using the law of conservation of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Let P1A and P1B be the initial momentum of the bodies A and B respectively

Let P2A and P2B be the final momentum of the bodies A and B respectively after collision.

Based on the law:

P1A+P2A = P1B + P2B

Given P1A = 5kgm/s

P2A = 0kgm/s(ball B at rest before collision)

P2A = -2.0kgm/s (negative because it moves in the negative x direction)

P2B = ?

Substituting the values in the equation gives;

5+0 = -2+P2B

5+2 = P2B

P2B = 7kgm/s

3 0
3 years ago
Harry and ron set up this experiment with a glider, whose mass they have measured to be 565 g, and seven washers hanging from th
svetlana [45]
Let's call m=565~g=0.565~kg the mass of the glider and m_w=7\cdot12~g =84~g=0.084~kg the total mass of the seven washers hanging from the string. 
The net force on the system is given by the weight of the hanging washers:
F_{net} = m_w g
For Newton's second law, this net force is equal to the product between the total mass of the system (which is m+m_w) and the acceleration a:
F_{net}=(m+m_w)a
So, if we equalize the two equations, we get
m_w g = (m+m_w)a
and from this we can find the acceleration:
a= \frac{m_w g}{(m+m_w)} = \frac{0.084~kg \cdot 9.81~m/s^2}{(0.565~kg+0.084~kg)}=1.27~m/s^2
5 0
3 years ago
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