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.

Physics

1 answer:

ozzi3 years ago

3 0

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 $6\times 10^6\ N$

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Mrac [35]

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3 0

3 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.

ikadub [295]

Answer:

b) No acceleration in the vertical

c) 35N

d) 35N

e) $8.75\ m/sec^2$

Explanation:

a) The situation can be shown in the free body diagram shown in the figure below where F is the applied force, Fr is the friction force, W is the weight of the book and N is the normal force exerted vertically up from the desk to the book

b) The vertical movement is restrained by the normal force which opposes to the weight. In absence of any other force, they both are in equilibrium and the net force is zero

c) The net horizontal force acting on the book is the vectorial sum of the applied force and the friction force. Since they lie in the same axis and are opposed to each other:

$Fh_{net}=F-F_r=50 N - 15N=35N$

d) The net force acting on the book is the vector sum of all forces in all axes. The normal and the weight cancel each other in the y-axis, so our resulting force is the x-axis net force, computed as above:

$F_n=35N$ in the x-axis