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3 years ago

7

Can some one help me please and thanks

1 answer:

nataly862011 [7]3 years ago

4 0

Answer:

The way in which objects exert forces on each other is described by Newton’s 3rd law of motion

Explanation:

Objects with mass exert forces on each other via the force of gravity. This force is proportional to the mass of the two interacting objects and is inversely proportional to the square of the distance between them. The factors G, M, and rare the same for all masses at the surface of the Earth.

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Write two different unit in which mass is measured.

inessss [21]

Answer:

kilograms and grams

Explanation:

kilograms is the stadard unit for mass according to the SI system.

Grams is another unit for mass.

5 0

3 years ago

A student has to work the following problem: A block is being pulled along at constant speed on a horizontal surface a distance

brilliants [131]

Answer:

D

The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.

Explanation:

Since block moves with constant speed

So, frictional force

f = FCosq

Work done by friction

W = - fd

W = - fd Cos q

The answer may be greater or less than - fdSinq. It depends on the value of q which is less than, or equal to 45°.

6 0

3 years ago

A bicycle has a momentum of 36 kg • m/s and a velocity of 4 m/s. What is the mass of the bicycle? 9 kg 32 kg 40 kg 144 kg

igomit [66]

Answer:

9 kg

Explanation:

Momentum is mass times velocity.

p = mv

36 kgm/s = m (4 m/s)

m = 9 kg

3 0

3 years ago

Read 2 more answers

Maverick and goose are flying a training mission in their F-14. They are flying at an altitude of 1500 m and are traveling at 68

lions [1.4K]

Answer:

The bomb will remain in air for <u>17.5 s</u> before hitting the ground.

Explanation:

Given:

Initial vertical height is, $y_0=1500\ m$

Initial horizontal velocity is, $u_x=688\ m/s$

Initial vertical velocity is, $u_y=0(\textrm{Horizontal velocity only initially)}$

Let the time taken by the bomb to reach the ground be 't'.

So, consider the equation of motion of the bomb in the vertical direction.

The displacement of the bomb vertically is $S=y-y_0=0-1500=-1500\ m$

Acceleration in the vertical direction is due to gravity, $g=-9.8\ m/s^2$

Therefore, the displacement of the bomb is given as:

$S=u_yt+\frac{1}{2}gt^2\\-1500=0-\frac{1}{2}(9.8)(t^2)\\1500=4.9t^2\\t^2=\frac{1500}{4.9}\\t=\sqrt{\frac{1500}{4.9}}=17.5\ s$

So, the bomb will remain in air for 17.5 s before hitting the ground.

5 0

3 years ago

What is the magnitude of the net force acting on a 2.0 x10^3 kilogram car?

Tatiana [17]

Answer:
Acceleration= final velocity - initial velocity / time
= (15 - 0)/ 5
= 3 m/s^2
Force = mass X acceleration
= 2 X 10^3 X 3
= 6 X 10^3 N

7 0

3 years ago