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masya89 [10]
3 years ago
14

The term time dilation refers to which effect?

Physics
2 answers:
RoseWind [281]3 years ago
6 0
Is the apparent loss of time of a moving clock as observed by a stationary observer. Thats what i think.
Strike441 [17]3 years ago
6 0
<span>In the theory of relativity, time dilation is a difference of elapsed time between two events as measured by observers either moving relative to each other or differently situated from a gravitational mass or masses.</span>
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Why do cells undergo mitosis?
avanturin [10]
 <span>Cells undergo mitosis for three main reasons: 
-To repair and/or replace old or damaged cells. 
-During periods of cell and tissue growth 
-When the body needs exact replicas/copies of cells e.g. hair cells and blood cells.</span>
5 0
3 years ago
Read 2 more answers
A chevrolet corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. what
pychu [463]
By definition we have that the final speed is:
 Vf² = Vo² + 2 * a * d
 Where,
 Vo: Final speed
 a: acceleration
 d: distance.
 We cleared this expression the acceleration:
 a = (Vf²-Vo²) / (2 * d)
 Substituting the values:
 a = ((0) ^ 2- (60) ^ 2) / ((2) * (123) * (1/5280))
 a = -77268 mi / h ^ 2
 its stopping distance on a roadway sloping downward at an angle of 17.0 ° is:
 First you must make a free body diagram and see the acceleration of the car:
 g = 32.2 feet / sec ^ 2
 a = -77268 (mi / h ^ 2) * (5280/1) (feet / mi) * (1/3600) ^ 2 (h / s) ^ 2
 a = -31.48 feet / sec ^ 2
 A = a + g * sin (θ) = -31.48 + 32.2 * sin17.0
 A = -22.07 feet / sec ^ 2
 Clearing the braking distance:
 Vf² = Vo² + 2 * a * d
 d = (Vf²-Vo²) / (2 * a)
 Substituting the values:
 d = ((0) ^ 2- (60 * (5280/3600)) ^ 2) / (2 * (- 22.07))
 d = 175.44 feet
 answer:
 its stopping distance on a roadway sloping downward at an angle of 17.0 ° is 175.44 feet
5 0
3 years ago
Read 2 more answers
Calculate the pressure on the ground from an 80 kg woman leaning on the back of one of her shoes with a 1cm diameter heel, and c
rodikova [14]

Answer:

Pressure of woman will be 99.87\times 10^5N/m^2

Pressure of the elephant will be 1716560.50N/m^2

Explanation:

We have given that mass of the woman m = 80 kg

Acceleration due to gravity g=9.8m/sec^2

Diameter of shoes = 1 cm =0.01 m

So radius r=\frac{d}{2}=\frac{0.01}{2}=0.005m

So area A=\pi r^2=3.14\times 0.005^2=7.85\times 10^{-5}m^2

We know that force is given  F = mg

So F=80\times 9.8=784N

Now we know that pressure is given by P=\frac{F}{A}=\frac{784}{7.85\times 10^{-5}}=99.87\times 10^5N/m^2

Now mass of elephant m = 5500 kg

So force of elephant = 5500×9.8 = 53900 N

Diameter = 20 cm

So radius r = 10 cm

So area will be A=3.14\times 0.1^2=0.0314m^2

So pressure will be P=\frac{53900}{0.0314}=1716560.50N/m^2

3 0
3 years ago
How far would a spacecraft moving in a circular orbit 500 km above Pluto's surface travel during that time?
Ilia_Sergeevich [38]

Answer:

The speed of the spacecraft should be 719.35m/s

Explanation:

if the spacecraft is orbiting the planet with a circular orbit, the gravitational force must act as a centripetal force. This means:

F_G=F_c\\\frac{GmM}{d^2}=m\frac{v^2}{d}

In this case, the pluto's mass M is 1.3099·10^22 kg. The radius of the planet R is 1188.3Km and G is the gravitational constant. Therefore:

\displaystyle\frac{G M}{d^2}=\frac{v^2}{d}\\v=\sqrt{\frac{GM}{d} } =\sqrt{\frac{GM}{(R + 500km)} } =719.35m/s

7 0
3 years ago
How much work does it take to slide a box 37 meters along the ground by pulling it with a 217 N force at an angle of 19° from th
-Dominant- [34]

Answer:

W = 7591.56 J

Explanation:

given,

distance of the box, d = 37 m

Force for pulling the box, F = 217 N

angle of inclination with horizontal,θ = 19°

We know,

Work done is equal to product of force and the displacement.

W = F.d cos θ

W = 217 x 37 x cos 19°

W = 7591.56 J

Hence, the work done to pull the box is equal to W = 7591.56 J

8 0
3 years ago
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