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Bond [772]
2 years ago
13

Which observation would be evidence that heat was transferred by radiation?

Physics
1 answer:
oksian1 [2.3K]2 years ago
8 0
I think the answer is D
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Look at a photograph of a fault. notice how the right side appears lower than the left side. this happens when pieces of crust a
dybincka [34]
I believe it’s divergent boundary but I might be wrong
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Amount of work done by a rotating object
Oduvanchick [21]
The work done by a rotating object can be calculated by the formula Work = Torque * angle.

This is analog to the work done by the linear motion where torque is analog to force and angle is analog to distance. This is Work = Force * distance.

An example will help you. Say that you want to calculate the work made by an engine that rotates a propeller with a torque of 1000 Newton*meter over 50 revolution.

The formula is Work = torque * angle.

Torque = 1000 N*m

Angle = [50 revolutions] *  [2π radians/revolution] = 100π radians

=> Work = [1000 N*m] * [100π radians] = 100000π Joules ≈ 314159 Joules of work.

 
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2 years ago
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A cat is running at 24 m/s. It then accelerates at 7 m/s2. How long will it take the cat to reach a speed of 49 m/s?
kozerog [31]

Answer:

t should be 3.57 second

Explanation:

Formula used is v = u+at

In which v is final velocity, u is initial velocity, a is acceleration and t is time.

Substitute each of the info given into the formula and calculate.

49 = 24 + (7)t

t = 3.57s

8 0
3 years ago
PLEASE HELP THIS IS DUE LIKE HELP ME PLEASE
marysya [2.9K]

Answer:

At the end points of motion (either side) the velocity must be zero because the velocity is changing from - to + (it can't turn around around without passing thru zero,

The velocity will then increase to the midpoint of the motion.

m g h = 1/2 m v^2    where h is the vertical distance thru which the pendulum travels

5 0
2 years ago
Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,
Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:  

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:  

F = \frac{k*q_1*q_2}{d^2} Formula (1)  

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)  

d: distance between the charges in meters (m)

Equivalence  

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

F_{13} = \frac{k*q_1*q_3}{d_1^2}

F_{23} = \frac{k*q_2*q_3}{d_2^2}

F₁₃ = F₂₃

\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2} We eliminate k and q₃ on both sides

\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}

\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}

\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2} We eliminate 10⁻⁶ on both sides

(1-x)^2 = \frac{7}{5} x^2

1-2x+x^2=\frac{7}{5} x^2

5-10x+5x^2=7 x^2

2x^2+10x-5=0

We solve the quadratic equation:

x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m

x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in  opposite way .

x = 0.458m = 45.8cm

8 0
3 years ago
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