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2 years ago

Please someone help me.

Physics

1 answer:

iren2701 [21]2 years ago

7 0

Answer:

1) The block of wood is at rest.

2) Weight and Normal Reaction (see attached diagram).

Explanation:

Assumptions

The block of wood is a particle (a body whose mass acts at a point, so its dimensions don't matter).
There are no other external forces involved.

Question 1

"At rest" or "resting" means that the object isn't moving.

Therefore, the block of wood is at rest.

Question 2

Forces that are acting on the block of wood:

Weight (W): Due to the particle's mass, m, and the acceleration due to gravity, g: W = mg
(Weight always acts downwards).

Normal Reaction (R or N): The reaction from a surface.
(Normal Reaction is always at 90° to the surface).

See attached diagram.

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327,360 feet
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3 years ago

How do you calculate the volume and density of something that weighs 20 grams?

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4 years ago

1. Which has more kinetic energy, a 40 kg cheetah running at 25 m/s or a 4,000 kg elephant moving at 2 m/s? How much more energy

s2008m [1.1K]

Answer:

1. The elephant has more kinetic energy at this speed and mass. It has 4,500 J more KE.

2. The elephant would have to go at a speed of 2.5 m/s to reach the same KE as the cheetah.

Explanation:

You would use the formula KE=1/2mv^2.

This formula would be filled in and completed twice, once for the elephant and once for the cheetah.

Cheetah:

KE = 1/2 (40) (25) ^2

KE = 12,500 J

Elephant:

KE = 1/2 (4,000) (2) ^2

KE = 8,000 J

This shows that the cheetah has more KE.

Then you would subtract the elephants amount of J from the cheetahs to find the difference.

Difference = 12,500 J - 8,000 J

Difference = 4,500 J

I hoped this helped with the first part :)

For the second part:

To find the speed the elephant would have to run you would fill in and complete the equation once more with different distance results.

KE = 1/2 (4,000) (2.5) ^2

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3 years ago

you stand on a straight desert road at night and observe a vehicle approaching. this vehicle is equipped with two small headligh

inysia [295]

Answer:

The distance that you marginally able to discern that there are two headlights rather than a single light source is 6.084 km

Explanation:

Given:

d = distance = 0.679 m

λ = wavelength of the light = 537 nm = 537x10⁻⁹m

dp = pupil diameter = 4.81 mm = 0.00481 m

Question: What distance, in kilometers, are you marginally able to discern that there are two headlights rather than a single light source, dx = ?

For the separation of the peak from the central maximum it is:

$sin\theta =\frac{\lambda }{d_{p} } =\frac{537x10^{-9} }{0.00481} =1.116x10^{-4}$

In this case, the two small sources of the headlights have the same angle as the images that form inside the eye

$d_{x} =\frac{d}{sin\theta } =\frac{0.679}{1.116x10^{-4} } =6.084x10^{3} m=6.084km$

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4 years ago

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astraxan [27]

Answer:

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Explanation:

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3 years ago