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2 years ago

6

A rocket moves through outer space a 12,000 m/s. At this time, how much time would be required to travel the distance from Earth

to the Moon, which is 380,000 km?

2 answers:

nexus9112 [7]2 years ago

6 0

12000 m/s = 12 km/s. Now to go 380000 km, it will take some time. How much time is given in the formula 12km/s. You go 12 kilometers every second. So you take $\frac{380000km}{12km/s}$ and that gives you 31,666.666 seconds.

aev [14]2 years ago

4 0

12,000 m = 12 km

   (380,000 km) / (12 km/sec)

   = (380,000 / 12) sec

   = 31,666-2/3 seconds

   =    8 hours 47 minutes 46.7 seconds

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Answer:

See the answer below

Explanation:

<u>Independent variable</u>: Type of drug (Mem-Reen or placebo)

<u>Dependent variable</u>: memories

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The dependent variable is the actual variable that is measured during an experiment. It is the main purpose of setting-up of an experiment.

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Two long straight wires enter a room through a window. One carries a current of 2.9A into the room, while the other carries a cu

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Answer and Explanation:

curents i = 2.9 A

i ' = 4.4 A

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Since from Ampere's law

where μ o = permeability of free space = 4π * 10 ^-7 H / m

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A 1150 kg car is on a 8.70° hill. using x-y axis tilted down the plane, what is the x-component of the weight?

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I assume the x-y axis are tilted such that the x-axis is parallel to the surface of the hill while the y-axis is perpendicular to it.

In this case, the x-component of the weight is given by:
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where
m is the mass of the car
g is the acceleration of gravity
$\theta$ is the angle of the hill

Substituting numbers into the formula, we find
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2 years ago

A 2 kg stone is tied to a 0.5 m string and swung around a circle at a constant angular velocity of 12 rad/s. the angular momentu

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Starting from the angular velocity, we can calculate the tangential velocity of the stone:
$v=\omega r= (12 rad/s)(0.5 m)= 6 m/s$

Then we can calculate the angular momentum of the stone about the center of the circle, given by
$L=mvr$
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.

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2 years ago

A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin

KengaRu [80]

a) 219.8 rad/s

b) $20.0 rad/s^2$

c) $2.9 m/s^2$

d) $7005 m/s^2$

e) Towards the axis of rotation

f) $0 m/s^2$

g) 31.9 m/s

Explanation:

a)

The angular velocity of an object in rotation is the rate of change of its angular position, so

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time elapsed

In this problem, we are told that the maximum angular velocity is

$\omega_{max}=35 rev/s$

The angle covered during 1 revolution is

$\theta=2\pi$ rad

Therefore, the maximum angular velocity is:

$\omega_{max}=35 \cdot 2\pi = 219.8 rad/s$

b)

The angular acceleration of an object in rotation is the rate of change of the angular velocity:

$\alpha = \frac{\Delta \omega}{t}$

where

$\Delta \omega$ is the change in angular velocity

t is the time elapsed

Here we have:

$\omega_0 = 0$ is the initial angular velocity

$\omega_{max}=219.8 rad/s$ is the final angular velocity

t = 11 s is the time elapsed

Therefore, the angular acceleration is:

$\alpha = \frac{219.8-0}{11}=20.0 rad/s^2$

c)

For an object in rotation, the acceleration has two components:

- A radial acceleration, called centripetal acceleration, towards the centre of the circle

- A tangential acceleration, tangential to the circle

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

Here we have

$\alpha =20.0 rad/s^2$

d = 29 cm is the diameter, so the radius is

r = d/2 = 14.5 cm = 0.145 m

So the tangential acceleration is

$a_t=(20.0)(0.145)=2.9 m/s^2$

d)

The magnitude of the radial (centripetal) acceleration is given by

$a_c = \omega^2 r$

where

$\omega$ is the angular velocity

r is the radius of the circle

Here we have:

$\omega_{max}=219.8 rad/s$ is the angular velocity when the fan is at full speed

r = 0.145 m is the distance of the gum from the centre of the circle

Therefore, the radial acceleration is

$a_c=(219.8)^2(0.145)=7005 m/s^2$

e)

The direction of the centripetal acceleration in a rotational motion is always towards the centre of the axis of rotation.

Therefore also in this case, the direction of the centripetal acceleration is towards the axis of rotation of the fan.

f)

The magnitude of the tangential acceleration of the fan at any moment is given by

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

When the fan is rotating at full speed, we have:

$\alpha=0$ , since the fan is no longer accelerating, because the angular velocity is no longer changing

r = 0.145 m

Therefore, the tangential acceleration when the fan is at full speed is

$a_t=(0)(0.145)=0 m/s^2$

g)

The linear speed of an object in rotational motion is related to the angular velocity by the formula:

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular velocity

r is the radius

When the fan is rotating at maximum angular velocity, we have:

$\omega=219.8 rad/s$

r = 0.145 m

Therefore, the linear speed of the gum as it is un-stucked from the fan will be:

$v=(219.8)(0.145)=31.9 m/s$

7 0

2 years ago