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2 years ago

Teams A and B are in a tug-of-war challenge. Team A wins the challenge. What can be said about Team A?

Physics

1 answer:

DanielleElmas [232]2 years ago

4 0

Team A created a larger net force opposite in direction to the lesser net force of Team B.

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12)A black body is heated from 27°C to 127° C. The ratio of their energies of radiations emitted will be

Nat2105 [25]

Answer:

$81:256$ .

Explanation:

Let $T$ denote the absolute temperature of this object.

Calculate the value of $T$ before and after heating:

$T(\text{before}) = 27 + 273 = 300\; \rm K$ .

$T(\text{after}) = 127 + 273 = 400\; \rm K$ .

By the Stefan-Boltzmann Law, the energy that this object emits (over all frequencies) would be proportional to $T^4$ .

Ratio between the absolute temperature of this object before and after heating:

$\displaystyle \frac{T(\text{before})}{T(\text{after})} = \frac{3}{4}$ .

Therefore, by the Stefan-Boltzmann Law, the ratio between the energy that this object emits before and after heating would be:

$\displaystyle \left(\frac{T(\text{before})}{T(\text{after})}\right)^{4} = \left(\frac{3}{4}\right)^{4} = \frac{81}{256}$ .

4 0

2 years ago

Suppose that you have been chosen for a space mission to a distant planet. Due to the length of time you'll be away from Earth y

Contact [7]

Answer:

I should be active for 15 hours to meet the physical activity requirement.

Explanation:

Since time dilates in moving objects, we use the formula t = t₀/√(1 - β²) where t = time in space vehicle, t₀ = time on earth = 9 hours and β = v/c where v = speed of space vehicle = 0.8c.

So, t = t₀/√(1 - β²)

t = 9/√(1 - (v/c)²)

= 9/√(1 - (0.8c/c)²)

= 9/√(1 - (0.8)²)

= 9/√(1 - (0.64)

= 9/√0.36

= 9/0.6

= 15 hr

So, according to a timer on the space vehicle, I should be active for 15 hours to meet the physical activity requirement.

8 0

3 years ago

A spring is stretched from x=0 to x=d, where x=0 is the equilibrium position of the spring. It is then compressed from x=0 to x=

VARVARA [1.3K]

Answer:

The same amount of energy is required to either stretch or compress the spring.

Explanation:

The amount of energy required to stretch or compress a spring is equal to the elastic potential energy stored by the spring:

$U=\frac{1}{2}k (\Delta x)^2$

where

k is the spring constant

$\Delta x$ is the stretch/compression of the spring

In the first case, the spring is stretched from x=0 to x=d, so

$\Delta x = d-0=d$

and the amount of energy required is

$U=\frac{1}{2}k d^2$

In the second case, the spring is compressed from x=0 to x=-d, so

$\Delta x = -d -0 = -d$

and the amount of energy required is

$U=\frac{1}{2}k (-d)^2= \frac{1}{2}kd^2$

so we see that the amount of energy required is the same.

7 0

3 years ago

While two forces act on it, a particle of mass m=3.2kg is it to move continuously with velocity (3m/s)i - (4m/s)j . one of the f

marta [7]

Mmm tricky.
Since the velocity is constant, I'm going to assume there is no acceleration in any direction. This means there is no net force in the I or J forection!

Since there are 2 forces, both must be equal and opposite in direction to perfectly cancel each other out.

So the opposite of F1 is (-2N)I + (6N)J!

8 0

3 years ago

A road with a radius of 75.0 m is banked so that a car can navigate the curve at a speed of 15.5 m/s without any friction. When

kirza4 [7]

Answer: 0.683

Explanation: The relationship between a car moving along a curve and the frictional force is given below as

us×g = v²/r

Where us = coefficient of static friction =?, g = acceleration due to gravity = 9.8 m/s², v = 22.4 m/s and r = 75m

By substituting the parameters, we have that

us×9.8 = 22.4²/ 75

us ×9.8 = 501.76/75

us ×9.8 = 6.690

us = 6.690/9.8 = 0.683

8 0

2 years ago