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3 years ago

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

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.

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A volleyball player’s hand applies a 39 N force while in contact with a volleyball for 2 seconds. What is the impulse on the bal

nikklg [1K]

Given:
F = 39 N, the force applied
t = 2 s, the time interval in which the force is applied.

By definition, the impulse is
$I = \int_{0}^{2} F(t) dt = \int _{0}^{2} 39dt = 39*2=78 \, N-s$

Answer: 78 N-s

3 0

3 years ago

Read 2 more answers

What can you observe to know that when clear methane gas is burned it is a chemical change?

aalyn [17]

4) Formation of a gas

When methane burns, it produces the gases water vapor and carbon dioxide.

6 0

3 years ago

Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m

mamaluj [8]

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

3 0

3 years ago

At this radius, what is the magnitude of the net force that maintains circular motion exerted on the pilot by the seat belts, th

Ainat [17]

Answer:

Fc=5253 N

Explanation:

Answer:

Fc=5253 N

Explanation:

sequel to the question given, this question would have taken precedence:

"The 86.0 kg pilot does not want the centripetal acceleration to exceed 6.23 times free-fall acceleration. a) Find the minimum radius of the plane’s path. Answer in units of m."

so we derive centripetal acceleration first

ac (centripetal acceleration) = v^2/r

make r the subject of the equation

r= v^2/ac

ac is 6.23*g which is 9.81

v is 101m/s

substituing the parameters into the equation, to get the radius

(101^2)/(6.23*9.81) = 167m

Now for part

( b) there are two forces namely, the centripetal and the weight of the pilot, but the seat is exerting the same force back due to newtons third law.

he net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth is the centripetal force.

Fc (Centripetal Force) = m*v^2/r

So (86kg* 101^2)/(167) =

Fc=5253 N

4 0

3 years ago

If a 1.00 kg body has an acceleration of 2.44 m/s2 at 53Â° to the positive direction of the x axis, then what are (a) the x comp

Ilia_Sergeevich [38]

(a) Fx = 1.464 N

(b) Fy = 1.952 N

Given

Mass = 1kg

Acceleration = 2.44 m/s2

Angle with positive X axis = 53°

As we know

F = ma

By substituting value

F= 1×2.44 N

F= 2.44 N

(a) Component of force in X direction

Fx = F Cosθ

Fx = 2.44 Cos(53°)

Fx = 2.44 × 0.60 = 1.464 N

(b) Component of force in Y direction

Fy = F Sinθ

Fy = 2.44 Sin(53°) = 2.44 × 0.80 = 1.952 N

F(x, y) = 1.464 i + 1.952 j

Thus we got net force.

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6 0

2 years ago