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3 years ago

8

a rocket initially at rest on the ground lifts off vertically with a constant acceleration of 2.0 x 10^1 meters per second^2. Ho

w long will it take the rocket to reach an altitude of 9.0 x 10^3 meters ?

1 answer:

xenn [34]3 years ago

7 0

Here's the formula for the distance covered by an accelerating body in some amount of time ' T '. This formula is incredibly simple but incredibly useful. It pops up so often in Physics that you really should memorize it:

D = 1/2 a T²

Distance = (1/2)·(acceleration)·(time²)

This question gives us the acceleration and the distance, and we want to find the time.

(9,000 m) = (1/2) (20 m/s²) (time²)

(9,000 m) = (10 m/s²) (time²)

Divide each side by 10 m/s²:

(9,000 m) / (10 m/s²) = (time²)

900 s² = time²

Square root each side:

<em>T = 30 seconds</em>

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The answer to your question is Metal

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Which theory best explains the present arrangement of continents oceans and landforms on earth? A the Pangaea theory B The conti

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Plate Tectonic Theory

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3 years ago

A train is moving in a straight railway where it covered one third of the distance with

Alexxandr [17]

Answer:

<em>The average speed of the train is 45 km/h</em>

Explanation:

<u>Speed</u>

It's defined as the distance (d) per unit of time (t) traveled by an object. The formula is:

$\displaystyle v=\frac{d}{t}$

Let's call x the total distance covered by the train. It covered d1=1/3x with a speed of v1=25 km/h. The time taken is calculated solving for t:

$\displaystyle t_1=\frac{d_1}{v_1}$

$\displaystyle t_1=\frac{1/3x}{25}$

$\displaystyle t_1=\frac{x}{75}$

Now the rest of the distance:

d2 = x - 1/3x = 2/3x

Was covered at v2=75 km/h. Thus the time taken is:

$\displaystyle t_2=\frac{d_2}{v_2}$

$\displaystyle t_2=\frac{2/3x}{75}$

$\displaystyle t_2=\frac{2x}{225}$

The total time is:

$\displaystyle t_t=\frac{x}{75}+\frac{2x}{225}$

$\displaystyle t_t=\frac{3x}{225}+\frac{2x}{225}$

$\displaystyle t_t=\frac{5x}{225}$

Simplifying:

$\displaystyle t_t=\frac{x}{45}$

The average speed is the total distance divided by the total time:

$\displaystyle \bar v=\frac{x}{\frac{x}{45}}$

Simplifying:

$\boxed{\displaystyle \bar v=45\ km/h}$

The average speed of the train is 45 km/h

5 0

3 years ago

During lightning strikes from a cloud to the ground, currents as high as 2.50×10^4 Amps can occur and last for about 40.0 micros

dangina [55]

Answer:

1 C

Explanation:

The intensity of electric current is defined as

$I=\frac{q}{t}$

where

I is the current

q is the amount of charge transferred

t is the time interval during which the charge is transferred

For the lightning in this problem, we have

$I=2.50\cdot 10^4 A$ is the current

$t=40.0 \mu s = 40.0\cdot 10^{-6} s$ is the time interval

Solving the formula for q, we find the amount of charge transferred:

$q=I t = (2.50\cdot 10^4 A)(40.0\cdot 10^{-6}s)=1 C$

6 0

3 years ago

A baseball has a mass of 0.15 kg and radius 3.7 cm. In a baseball game, a pitcher throws the ball with a substantial spin so tha

mylen [45]

Answer:

Rotational kinetic energy = 0.099 J

Translational kinetic energy = 200 J

The moment of inertia of a solid sphere is $I = \frac{2}{5}mr^2$ .

Explanation:

Rotational kinetic energy is given by

$\text{RKE} = \frac{1}{2}I\omega^2$

where <em>I</em> is the moment of inertia and <em>ω</em> is the angular speed.

For a solid sphere,

$I = \frac{2}{5}mr^2$

where <em>m</em> is its mass and <em>r</em> is its radius.

From the question,

<em>ω</em> = 49 rad/s

<em>m</em> = 0.15 kg

<em>r</em> = 3.7 cm = 0.037 m

$\text{RKE} = \frac{1}{2}\times \frac{2}{5} mr^2\omega^2 = \frac{1}{5} mr^2\omega^2$

$\text{RKE} = \frac{1}{5} (0.15\ \text{kg})(0.037\ \text{m})^2(49\ \text{rad/s})^2 = 0.099\text{ J}$

Translational kinetic energy is given by

$\text{TKE} = \frac{1}{2} mv^2$

where <em>v</em> is the linear speed.

$\text{TKE} = \frac{1}{2} (0.15\ \text{kg})(52\ \text{m/s})^2 = 200\text{ J}$

5 0

3 years ago

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