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3 years ago

Acceleration equation physics

Physics

2 answers:

ella [17]3 years ago

7 0

Answer:

F=ma

dv/dt= a

a is referred to as acceleration

Rudiy273 years ago

7 0

Answer:

Acceleration=change in velocity ➗ time

Explanation:

Acceleration=change in velocity ➗ time

Whereby:

Change in velocity=final velocity - initial velocity

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What is the lift (in newtons) due to Bernoulli's principle on a wing of area 76 m2 if the air passes over the top and bottom sur

AveGali [126]

Answer:

So lift will be 30.19632 N

Explanation:

We have given area of the wing $a=76m^2$

We know that density of air $d=1.29kg/m^3$

Speed at top surface $v_2=290m/sec$ and speed at bottom surface $v_1=150m/sec$

According to Bernoulli's principle force is given by

$F=A\times d\times \frac{v_2^2-v_1^2}{2}=76\times 1.29\times \frac{290^2-150^2}{2}=3019632N$

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3 years ago

The Earth and the Moon are attracted to each other by universal gravitation. The Earth is much more massive than is the Moon. Do

OverLord2011 [107]

Answer:

Earth attract the Moon with a force that is greater.

Explanation:

According to the law of gravitation, the gravitational force between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Mathematically, F1 = Gm1m2/r²... 1

Let m1 be the mass of the earth and m2 be that of the moon

If the Earth is much more massive than is the Moon, the new force of attraction between them will become;

F2= G(2m1)m2/r²

F2 = 2Gm1m2/r² ... (2)

Dividing eqn 1 by 2 we have;

F1/F2 = (Gm1m2/r²)÷(2Gm1m2/r²)

F1/F2 = Gm1m2/r²×r²/2Gm1m2

F1/F2 = 1/2

F2=2F1

This shows that that the earth will attract the moon by a force 2times the initial force of the masses(i.e a much greater force)

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3 years ago

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natulia [17]

Answer:

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3 years ago

Read 2 more answers

A 460 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exert

ollegr [7]

Answer:

The rocket should be launched at a horizontal distance of <u>6.45 m</u> left of the loop.

Explanation:

Given:

Mass of the rocket model (m) = 460 g = 0.460 kg [1 g = 0.001 kg]

Speed of the cart (v) = 3.0 m/s

Thrust force by the rocket engine (F) = 8.5 N

Vertical height of the loop (y) = 20 m

Let the horizontal distance left to the loop for launch be 'x'. Also, let 't' be the time taken by the rocket to reach the loop.

Now, there are two types of motion associated with the rocket- one is horizontal and the other vertical.

So, we will apply kinematics of motion in the two directions separately.

Vertical motion:

Given:

Force acting in the vertical direction is given as:

$F_y=F-mg=8.5-0.46\times 9.8=3.992\ N$

So, acceleration in the vertical direction is given as:

Acceleration = Force ÷ mass

$a_y=\frac{F_y}{m}=\frac{3.992\ N}{0.46\ kg}=8.678\ m/s^2$

Vertical displacement of rocket is same as the height of loop. So, $y=20\ m$

There is no initial velocity in the vertical direction. So, $u_y=0\ m/s$

Now, applying equation of motion in vertical direction. we have:

$y=u_yt+\frac{1}{2}a_yt^2\\\\20=0+\frac{1}{2}\times 8.678t^2\\\\20=4.339t^2\\\\t^2=\frac{20}{4.339}\\\\t=\sqrt{\frac{20}{4.339}}=2.15\ s$

Now, time taken to reach the loop is 2.15 s.

Horizontal motion:

There is no acceleration in the horizontal motion. So, displacement in the horizontal direction is equal to the product of horizontal speed and time.

Also, displacement of the rocket in the horizontal direction is nothing but the horizontal distance of its launch left of the loop. So,

$x=vt\\\\x=3.0\ m/s\times 2.15\ s\\\\x=6.45\ m$

Therefore, the rocket should be launched at a horizontal distance of 6.45 m left of the loop.

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3 years ago