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2 years ago

8

2. Turn off the Parallel line and turn on the Line through focal point. Move the light bulb around. What do you notice about the

reflected ray in this situation? __________________________________________________

1 answer:

MArishka [77]2 years ago

8 0

Answer:

The group of light rays is reflected back towards the focal point thereby producing a magnifying effect.

Explanation:

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What is the average force (average with respect to height of the barbell from the ground) exerted by the weightlifter in the pro

Novay_Z [31]

This question is incomplete, the complete question is;

A weightlifter holds a 1,300 N barbell 1 meter above the ground. One end of a 2-meter-long chain hangs from the center of the barbell. The chain has a total weight of 400 N. How much work (in J) is required to lift the barbell to a height of 2 m?

What is the average force (average with respect to height of the barbell from the ground) exerted by the weightlifter in the process?

Answer: Average force exerted by the weightlifter in the process = 1600N

Explanation:

To find Work done to lift a barbell and half of the hanging chain we say;

W₁ = ( 1300N + (1/2 × 400N)) × 1m

W₁ = (1300 + 200) Nm

W₁ = 1500J

now work done to lift the upper half of the chain we say:

W₂ = (1/2 × 400N) × (1/2 × 1m)

W₂ = 200N × 0.5m

W₂ = 100J

So total work done will be

W = W₁ + W₂

W = 1500J + 100J

W = 1600J

To find the average force exerted by the weight lifter, we say;

F = W/D

F = (1600 / 1m) N

F = 1600N

∴Average force = 1600N

6 0

2 years ago

Airplane flight recorders must be able to survive catastrophic crashes. Therefore, they are typically encased in crash-resistant

chubhunter [2.5K]

Answer:

3.2 MN

Explanation:

Given that

Mass of the box, m = 52 kg

Initial velocity of the box, u = 400 m/s

Final velocity of the box, v = 0 m/s

Time taken for the collision, t = 0.0065 s

Using the equation of motion

V = u + at, we turn around and make acceleration, a the subject of the formula. Now we have,

a = (v - u) / t

a = (0 - 400) / 0.0065

a = 61538.5 m/s²

The acceleration(which is negative acceleration or retar.dation actually) is 61538.5 m/s². We then proceed to is this acceleration in the basic Force equation, to get the magnitude of force needed.

Remember,

Force = mass * acceleration

F = ma, we already have our mass and acceleration, all we do is multiply

F = 52 * 61538.5

F = 3200002 N or 3.2 MN

Therefore, the magnitude of the force that acts on the box during collision is 3.2 MN

8 0

2 years ago

What is globalization? A) an organization of people competing against each other in physical activities B) learning about belief

SOVA2 [1]

Globalization would be c

3 0

2 years ago

Read 2 more answers

A sphere has surface area 1.25 m2, emissivity 1.0, and temperature 100.0°C. What is the rate at which it radiates heat into empt

Yuri [45]

The total power emitted by an object via radiation is:
$P=A\epsilon \sigma T^4$
where:
A is the surface of the object (in our problem, $A=1.25 m^2$
$\epsilon$ is the emissivity of the object (in our problem, $\epsilon=1$ )
$\sigma = 5.67 \cdot 10^{-8} W/(m^2 K^4)$ is the Stefan-Boltzmann constant
T is the absolute temperature of the object, which in our case is $T=100^{\circ} C=373 K$

Substituting these values, we find the power emitted by radiation:
$P=(1.25 m^2)(1.0)(5.67 \cdot 10^{-8}W/(m^2K^4)})(373 K)^4=1371 W = 1.4 kW$
So, the correct answer is D.

6 0

2 years ago

At a given location from a source charge, the electric field______________.a. strength is dependent on the source charge and the

noname [10]

Answer:

b. The electric field points away from the source charge, if the source charge is positive

d. The electric field points toward the source charge, if the source charge is negative.

Explanation:

A positive source charge would create an electric field that would exert a repulsive effect upon a positive test charge. Thus, the electric field vector would always be directed away from positively charged objects. On the other hand, a positive test charge would be attracted to a negative source charge. Therefore, electric field vectors are always directed towards negatively charged object.

Also electric field strength depends only on test charge

The correct options include b and d

The electric field points away from the source charge, if the source charge is positive.

Also, the electric field points toward the source charge, if the source charge is negative.

3 0

2 years ago