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3 years ago

6

A body of mass 2kg moves round a circle of radius 5m with a constant speed of 10m/s. Calculate the force toward the centre of th

e circle

1 answer:

antiseptic1488 [7]3 years ago

5 0

Answer:

F = 5

Explanation:

F = m x v^2/r = 2 x 10^2/5 =200/5 = 40 (N)

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Which object will be the easiest for a magnet to pull? a 1-gram piece of paper, a 2-gram eraser, a 3-gram steel paper clip, a 4-

Vlad1618 [11]

Steel paper clip because it can be moved by the magnet and it is lighter than the iron nail

6 0

3 years ago

Can anyone help me? (physics)

Masja [62]

Answer:

The initial velocity of the golf is 15.7 m/s.

The direction of the golf is 57°.

Explanation:

The following data were obtained from the question:

Time of flight (T) = 2.7 secs

Range (R) = 23 m

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =.?

Direction (θ) =.?

T = 2U Sine θ /g

2.7 = 2 × U × Sine θ /9.8

Cross multiply

2.7 × 9.8 = 2 × U × Sine θ

26.46 = 2 × U × Sine θ

Divide both side by 2 × Sine θ

U = 26.46 /2 Sine θ

U = 13.23 / Sine θ ... (1)

R = U² Sine 2θ /g

23 = U² Sine 2θ / 9.8

U = 13.23 / Sine θ

23 = (13.23/ Sine θ)² Sine 2θ / 9.8

23 = (175.0329 / Sine² θ) × Sine 2θ / 9.8

23 = 17.8605/Sine² θ × Sine 2θ

Recall:

Sine 2θ = 2SineθCosθ

23 = 17.8605/ Sine² θ × 2SineθCosθ

23 = 17.8605/ Sine θ × 2Cosθ

23 = 35.721 Cos θ /Sine θ

Cross multiply

23 × Sine θ = 35.721 Cos θ

Divide both side by 23

Sine θ = 35.721 Cos θ /23

Sine θ = 1.5531 × Cos θ

Divide both side by Cos θ

Sine θ /Cos θ = 1.5531

Recall:

Sine θ /Cos θ = Tan θ

Sine θ /Cos θ = 1.5531

Tan θ = 1.5531

Take the inverse of Tan

θ = Tan¯¹ (1.5531)

θ = 57°

Therefore, the direction of the golf is 57°

Thus, the initial velocity can be obtained as follow:

U = 13.23 / Sine θ

θ = 57°

U = 13.23 / Sine 57

U = 13.23/0.8387

U = 15.7 m/s

Therefore, the initial velocity of the golf is 15.7 m/s

8 0

3 years ago

Determine the thrust produced if 1.5 x 10^3 kg of gas exits the combustion chamber each second, with a speed of 4.00 x 10^3 m/s.

ozzi

Answer:

The thrust is $6\times 10^6\ N$

Explanation:

Given that,

Mass of gas, $m=1.5\times 10^3\ kg$

The rate at which the gas is expelling, $\dfrac{dv}{dt}=4\times 10^{3}\ m/s$

We need to find the thrust produced by the gas.

We know that force is equal to the rate of change of momentum. So,

$F=\dfrac{p}{t}$

Also, p = mv

$F=\dfrac{mv}{t}$

So,

$F=1.5\times 10^3\times 4\times 10^3\\\\F=6\times 10^6\ N$

So, the thrust is $6\times 10^6\ N$

3 0

3 years ago

An iron ball is dropped at a height of 10 m from the surface of the moon.

galben [10]

Answer:

3.51s

Explanation:

There are many students who can not get answers step by step and on time

So there are a wats up group where you can get help step by step and well explained by the trusted experts.

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3 years ago

Objects P and Q are placed in a bowl of water. P floats and Q sinks. Which one or more of the following can be deduced from this

Lady bird [3.3K]

The correct option that can be deduced for both Object P and Q is Option b) I and II only

To solve this question correctly, we need to understand the concept of density and it relation to mass and volume.

<h3>What is Density?</h3>

Density is a physical property of an object and can be expressed by using the relation:

$\mathbf{Density = \dfrac{mass}{volume}}$

From the given parameters, we are being told that:

P → floats
Q → sinks

This implies that Q has a greater density that P. Since Q has a greater density than P, Q will be heavier since it will have greater mass.

However, Q will not be denser than water because if that happens, P will be have a greater density which is untrue in this scenario.

Therefore, we can conclude that:

1. Q is heavier than P
II. 1cm³ of Q has a greater mass than 1cm³ of P

Learn more about density here:

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6 0

3 years ago