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dangina
8 days ago
15

A rocket starts from rest and moves upward from the surface of the earth. for the first 10.0 s of its motion, the vertical accel

eration of the rocket is given by ay=(2.60m/s3)t, where the +y-direction is upward.
Physics
1 answer:
kicyunya [2.9K]8 days ago
5 0

The rocket's acceleration is described here as

a_y = 2.60* t

now recognizing that

\frac{dv}{dt} = 2.60t

we integrate both sides

\int dv = \int 2.60t dt

v = 2.60\frac{t^2}{2}

v = 1.30 t^2

given that the rocket is accelerating for a duration of t = 10 s

thus, we have

v = 1.30 * 10^2

v = 130 m/s

consequently, after t = 10 s, the rocket will achieve a speed of 130 m/s in an upward direction

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Maru [2969]

The moment the body impacts the ground, two types of Forces are produced: the gravitational pull and the Normal Force. This aligns with Newton's third law, indicating that every action has an equal and opposite reaction. If the downward force of gravity is directed toward the earth, the reactionary force from the block acts upwards, equivalent to its weight:

F = mg

Where,

m = mass

g = gravitational acceleration

F = 5*9.8

F = 49N

Consequently, the answer is E.

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27 days ago
Complete the paragraph to describe the relationship between kinetic energy and braking distance. Use . A car moves at a speed of
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ke prop to v^2

ke1/v1^2=ke2/v2^2

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8 days ago
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Two projectiles are launched with the same initial speed from the same location, one at a 30° angle and the other at a 60° angle
Ostrovityanka [2807]

Answer:

Explanation:

Let us denote the launch angle as \theta _1=30^{\circ}.

\theta _2=60^{\circ}, \theta _1, and \theta _2 are complementary angles which means their ranges are identical.

H_{max}=\frac{u^2(\sin \theta )^2}{2g}

H_{30}=\frac{u^2(\sin 30)^2}{2g}

H_{30}=\frac{u^2}{8g}

Time of flight is indicated as =\frac{2u\sin \theta }{g}.

t_{30}=\frac{u}{g}

For \theta =60^{\circ}

H_{60}=\frac{u^2(\sin 60)^2}{2g}

H_{60}=\frac{3u^2}{8g}

t_{60}=\frac{2u\sin 60}{g}=\frac{\sqrt{3}u}{g}

H_{60}=3\times H_{30}

t_{60}=\sqrt{3}\times t_{30}

7 0
28 days ago
A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of 25.60 rad/s2. During a 4.20-s time inte
Yuliya22 [2962]

We will use the equations of rotational kinematics,

\theta =\theta _{0} + \omega_{0} t+ \frac{1}{2}\alpha t^2             (A)

\omega^2= \omega^2_{0} +2\alpha\theta                                     (B)                                          

Here, \theta and \theta _{0} denote the final and initial angular displacements, respectively, whereas \omega and \omega_{0} represent final and initial angular velocities, and \alpha is the angular acceleration.

We are provided with \alpha = - 25.60 \ rad/s^2, \theta = 62.4 \ rad and t = 4.20 \ s.

By substituting these values into equation (A), we have

62.4 \ rad = 0 + \omega_{0} 4.20 \ s + \frac{1}{2} (- 25.60 \ rad/s^2) ( 4.20)^2 \\\\ \omega_{0} = \frac{220.5+ 62.4 }{4.20} =67.4 \ rad/s

Now, using equation (B),

\omega^2=(67.4 \ rad/s)^2 + 2 (- 25.60 \ rad/s^2)62.4 \ rad \\\\\ \omega = 36.7 \ rad/s

This indicates that the wheel's angular speed at the 4.20-second mark is 36.7 rad/s.

4 0
1 month ago
The amount of electric energy consumed by a 60.0-watt lightbulb for 1.00 minute could lift a
Sav [2826]

Answer:

Explanation:

For a 60W light bulb used for 1 minute:

P = 60 W

t = 1 minute = 60 seconds

This energy is capable of lifting an object weighing 10N.

W = 10N

This indicates conversion of electrical energy into potential energy.

Let's calculate the electrical energy:

Power describes the rate of work done.

Power = Work / time

Thus, work = power × time

Work = 60 × 60

Work = 3600 J

Potential energy calculation:

P.E = mgh

Where the weight is given by:

W = mg

Therefore, P.E = W·h

P.E = 10·h

Thus, we equate:

Potential energy = Electrical energy

P.E = Work

10·h = 3600

Dividing both sides by 10 gives:

h = 3600 / 10

h = 360m

The object can be lifted to a height of 360m.

6 0
1 month ago
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