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PolarNik [594]
3 years ago
12

Does a large object traveling slowly or a small object going fast have a greater momentum?

Physics
1 answer:
nekit [7.7K]3 years ago
4 0

Answer:

It depends on the product of their mass and velocity

Explanation:

The momentum of a moving object is defined as the product of its mass and its moving velocity. So a same object would have larger momentum it's traveling faster, or gets heavier, and vice versa. In our case we don't have the exact value of speed and mass of both objects so we can't say which one has a greater momentum.

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Does any one know the answer
valina [46]
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6 0
3 years ago
A satellite that goes around the earth once every 24 hours (86,400 s) is called a geosynchronous satellite. If a geosynchronous
Olegator [25]

Answer:

42244138.951 m

Explanation:

G = Gravitational constant = 6.667 × 10⁻¹¹ m³/kgs²

r = Radius of orbit from center of earth

M = Mass of Earth = 5.98 × 10²⁴ kg

m = Mass of Satellite

The satellite revolves around the Earth at a constant speed

Speed = Distance / Time

The distance is the perimeter of the orbit

v=\frac{2\pi \times r}{24\times 3600}

The Centripetal force of the satellite is balanced by the universal gravitational force

m\frac{v^2}{r}=\frac{GMm}{r^2}\\\Rightarrow \frac{\left(\frac{2\pi \times r}{24\times 3600}\right)^2}{r}=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{r^2}\\\Rightarrow \left(\frac{2\pi \times r}{24\times 3600}\right)^2=6.667\times 10^{-11}\times 5.98\times 10^{24}\\\Rightarrow r^3=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\\\Rightarrow r=\left(\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\right)^{\frac{1}{3}}\\\Rightarrow r=42244138.951\ m

The radius as measured from the center of the Earth) of the orbit of a geosynchronous satellite that circles the earth is 42244138.951 m

6 0
3 years ago
Velocity of B rays ?<br>​
stich3 [128]

Answer:

is from 9 x 107 m/sec to 27 x 107 m/s

6 0
3 years ago
Read 2 more answers
If the pressure exerted by the liquid of density 500kg/m3 is 20,000Pa. What is height of liquid column.
gregori [183]

The height of the liquid column is 4.08 metres.

4 0
2 years ago
A 4 kg textbook sits on a desk. It is pushed horizontally with a 50 N applied force against a 15 N frictional force.
GarryVolchara [31]

a) See free-body diagram in attachment

b) The book is stationary in the vertical direction

c) The net horizontal force is 35 N in the forward direction

d) The net force on the book is 35 N in the forward horizontal direction

e) The acceleration is 8.75 m/s^2 in the forward direction

Explanation:

a)

The free-body diagram of a body represents all the forces acting on the body using arrows, where the length of each arrow is proportional to the magnitude of the force and points in the same direction.

From the diagram of this book, we see there are 4 forces acting on the book:

- The applied force, F = 50 N, pushing forward in the horizontal direction

- The frictional force, F_f = 15 N, pulling backward in the horizontal direction (the frictional force always acts in the direction opposite to the motion)

- The weight of the book, W=mg, where m is the mass of the book and g=9.8 m/s^2 is the acceleration of gravity, acting downward. We can calculate its magnitude using the mass of the book, m = 4 kg:

W=(4)(9.8)=39.2 N

- The normal reaction exerted by the desk on the book, N, acting upward, and balancing the weight of the book

b)

The book is in equilibrium in the vertical direction, therefore there is no motion.

In fact, the magnitude of the normal reaction (N) exerted by the desk on the book is exactly equal to the weight of the book (W), so the equation of motion along the vertical direction is

N-W=ma

where a is the acceleration; however, since N = W, this becomes

a=0

And since the book is initially at rest on the desk, this means that there is no motion.

c)

We said there are two forces acting in the horizontal direction:

- The applied force, F = 50 N, forward

- The frictional force, F_f = 15 N, backward

Since they act along the same line, we can calculate their resultant as

\sum F = F - F_f = 50 - 15 = 35 N

and therefore the net force is 35 N in the forward direction.

d)

The net force is obtained as the resultant  of the net forces in the horizontal and vertical direction. However, we have:

- The net force in the horizontal direction is 35 N

- The net force in the vertical direction is zero, because the weight is balanced by the normal reaction

Therefore, this means that the total net force acting on the book is just the net force acting on the horizontal direction, so 35 N forward.

e)

The acceleration of the book can be calculated by using Newton's second law:

\sum F = ma

where

\sum F is the net force

m is the mass

a is the acceleration

Here we have:

\sum F = 35 N (in the forward direction)

m = 4 kg

Therefore, the acceleration is

a=\frac{\sum F}{m}=\frac{35}{4}=8.75 m/s^2 (forward)

Learn more about forces, weight and Newton's second law:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

8 0
3 years ago
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