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3 years ago

1. A rocket is launched straight up into the air. If its entire flight takes 5 seconds....

Physics

1 answer:

pashok25 [27]3 years ago

8 0

Answer:

1)49m/s

2)122.5m

Explanation:

The expression below can be use to calculate the initial velocity from law of motion

V2 - V1= at

Where V2= final velocity

V1= initial velocity( from rest)

a=acceleration

t= time= 5 seconds

But V2=0

V1=gt

Where g= acceleration due to gravity= 9.8m/s^2

V1= 9.8×5

V1=49m/s

Then we can now calculate the maximum height the rocket reaches by using below formula

V^2=2gh

h(max)=v^2/2g

Where h= maximum height

V2= initial velocity

=49^2/(2×9.8)

=2401/19.6

=122.5m

Hence the maximum height it reaches vis 122.5m

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Water is discharged from a pipeline at a velocity v (in ft/sec) given by v = 1306p(1/2), where p is the pressure (in psi). If th

shepuryov [24]

Answer:

$a=38.5 ft/sec^{2}$

Explanation:

Note that acceleration is the rate change of velocity i.e

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Since the velocity is giving as a variable dependent on the pressure, we have to differentiate implicitly both side with respect to time,i.e

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if we substitute value for the pressure as giving in the question and also since the rate change of pressure is 0.354psi/sec, we have

$a=653*0.1667*0.354\\$

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3 years ago

Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges

Nostrana [21]

Answer:

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

$E_1 = E_2$

Let the position where net field is zero will lie at distance "r" from q1

$\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}$

now we will have

$\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}$

now square root both sides

$\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}$

now we have

$\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1$

so we have

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

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3 years ago

A rocket is fired at a 45° angle, what is the direction of the horizontal velocity vector at the peak height?

Bess [88]

Answer:

B: Horizontally to the left

Explanation:

Horizontal velocity is always constant throughout the entire trajectory of the rocket and acts in the horizontal direction in which the rocket was launched. This is because gravity only acts in the downwards vertical direction.

So in order words at peak height, horizontal velocity is in the horizontal direction in which the rocket was launched.

So if it was to the left, then direction is left but if right, then direction is right.

Looking at the options, the most appropriate will be:

Horizontally to the left

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Answer:

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