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3 years ago

6

PLEASEEE HELP!!!!! I HAVE BEEN STRUGGLING FOR 2 DAYS

1 answer:

____ [38]3 years ago

4 0

Answer:

You should still workout 90 min.

The proper time is measured by a single clock in a single place.

The proper time on earth is 90 min.

The clock on the rocket is also in a single place in the frame of the rocket so you still need to workout for 90 min.

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If the 15.0 cm string is shorter than the 30.0 cm string.

Margaret [11]

Answer:

option c) 2 is the right answer

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3 years ago

A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 15 m/sm/s. A passenger accidentally drops his camera

jeka94

To solve this problem we will apply the linear motion kinematic equations. With the information provided we will calculate the time it takes for the object to fall. From that time, considering that the ascent rate is constant, we will take the reference distance and calculate the distance traveled while the object hit the ground, that is,

$h = v_0 t -\frac{1}{2} gt^2$

$-18 = 15*t + \frac{1}{2} 9.8*t^2$

$t = 3.98s$

Then the total distance traveled would be

$h = h_0 +v_0t$

$h = 18+15*3.98$

$h = 77.7m$

Therefore the railing will be at a height of 77.7m when it has touched the ground

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3 years ago

What two major uses does this (H-R)diagram have for astronomers?

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3 years ago

A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a

Zielflug [23.3K]

Answer: $F_{v} =\mu_{k} mg$

Magnitude of the force is 2601.9 N

Explanation:

m = 450 kg

coefficient of static friction μs = 0.73

coefficient of kinetic friction is μk = 0.59

The force required to start crate moving is $F_{s} =\mu_{s} mg$ .

but once crate starts moving the force of friction is reduced $F_{v} =\mu_{k} mg$ .

Hence to keep crate moving at constant velocity we have to reduce the force pushing crate ie $F_{v} =\mu_{k} mg$ .

Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as forces are balanced.

Magnitude of the force

$F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9 N$

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4 years ago

Thermal energy is transferred from the warm Gulf Stream ocean water to the cooler air above it by the process of convection?

LuckyWell [14K]

This is true thermal energy is transferred from the warm gulf stream ocean water

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3 years ago

Read 2 more answers