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padilas [110]
3 years ago
8

A model rocket is launched straight upward with an initial speed of 56.5 m/s. It accelerates with a constant upward acceleration

of 1.96 m/s 2 until its engines stop at an altitude of 198.8 m. What is the maximum height reached by the rocket? Answer in units of m.
Physics
1 answer:
marta [7]3 years ago
3 0

Answer:

Maximum height reached by the rocket, h = 202.62 meters            

Explanation:

It is given that,

Initial speed of the model rocket, u = 56.5 m/s

Constant upward acceleration, a=1.96\ m/s^2

Distance traveled by the engine until it stops, d = 198.8 m

Let v is the speed of the rocket when the engine stops. It can be calculated using the third equation of motion as :

v=\sqrt{u^2+2ad}

v=\sqrt{(56.5)^2+2\times 1.96\times 198.8}    

v = 63.02 m/s

At the maximum height, v = 0 and the engine now decelerate under the action of gravity, a = -g. Let h is the maximum height reached by the rocket.

Again using third equation of motion as :

v^2-u^2=-2gh

h=\dfrac{v^2-u^2}{-2g}

h=\dfrac{u^2}{2g}

h=\dfrac{(63.02)^2}{2\times 9.8}

h = 202.62 meters

So, the maximum height reached by the rocket is 202.62 meters. Hence, this is the required solution.

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Answer:

126.56 m

Explanation:

Applying,

-F = ma............. Equation 1

Where F = frictional force, m = mass of the car, a = acceleration.

Note: Frictional force is negative because it act in opposite direction to motion

But,

F = mgμ.......... Equation 2

Where g = acceleration due to gravity, μ = coefficient of friction

Substitute equation 2 in equation 1

-mgμ = ma

a = -gμ.............. Equation 3

From the question,

Given: μ = 0.735

Constant: 9.8 m/s²

Substitute these values in equation 3

a = -9.8×0.735

a = -7.203 m/s²

Finally,

Applying

v² = u²+2as.............. Equation 4

Where v = final velocity, u = initial velocity, s = distance

From the question,

Given: u = 42.7 m/s, v = 0 m/s (to a stop), a = -7.203 m/s²

Substitute these values into equation 4

0² = 42.7²+2(-7.203)s

-1823.29 = -14.406s

s = -1823.29/-14.406

s = 126.56 m

4 0
3 years ago
The focal length of the lens of a simple digital camera is 40 mm, and it is originally focused on a person 25 m away. In what di
ss7ja [257]

Answer:

Explanation:

Here image distance is fixed .

In the first case if v be image distance

1 / v - 1 / -25 = 1 / .05

1 / v = 1 / .05 - 1 / 25

= 20 - .04 = 19.96

v = .0501 m = 5.01 cm

In the second case

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1 / v - 1 / - 4 = 1 / .05

1 / v = 20 - 1 / 4 = 19.75

v = .0506 = 5.06 cm

So lens must be moved forward by 5.06 - 5.01 =  .05 cm ( away from film )

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4 years ago
If the net force on a block is zero
amm1812

If the net force on a block is zero, the block will move at constant velocity

Explanation:

We can answer this question by applying Newton's second law of motion, which states that the net force on an object is equal to the product between its mass and its acceleration:

\sum F = ma (1)

where

\sum F is the net force on the object

m is its mass

a is its acceleration

In this problem, we have a block, and the net force on it is zero:

\sum F = 0

According to eq.(1), this also implies that

a=0

So, the acceleration of the block is zero.

However, acceleration is the rate of change of velocity of a body:

a=\frac{\Delta v}{\Delta t}

where \Delta v is the change in velocity in a time of \Delta t. Since the acceleration is zero, this means that \Delta v=0, and therefore the velocity of the object is constant.

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8 0
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Which phrase best describes a machine?
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Answer:

All of the above.

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All Machines make work easier.

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5 0
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A knight moves on a chessboard two squares up, down, left, or right followedby one square in one of the two directions perpendic
Contact [7]

Answer:

Explanation:

Check attachment for solution.

Generally the movement of the knight is L, i.e (2,1), (1,2),(-1,2),(1,-2) etc.

So using Pythagoras theorem

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x^2+y^2=5

Then, the knight will make one movement when the displacement is √5.

So let take a look at other positions

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Then, for us to have this kind of movement, the knight has to make 3 movements.

When the displacement is 1, then it will make 3 movement.

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When the displacement is 2.

The knight make 2 movement

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