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Anika [276]
3 years ago
6

The black board appears black because:

Physics
1 answer:
marishachu [46]3 years ago
8 0

Answer:

I think its C All visible light wavelengths (ROYGBV) are absorbed by the board

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Before Collision Consider a system to be one train car moving toward another train car at rest When the train cars collide, the
erma4kov [3.2K]

Answer:

2,400kg * m/s

Explanation:

You are missing some information in the question but the rest could be found some where else.

The question gives the masses and starting velocity of each car.

Car 1: m = 600kg and sv = 4m/s

Car 2: m 400kg and sv = 0m/s

Find the momentum of both cars.

Car 1: 600 * 4 = 2400

Car 2: 400 * 0 = 0

Add both.

2400 + 0 = 2400

Best of Luck!

7 0
3 years ago
Read 2 more answers
Consider three scenarios in which a particular box moves downward under the pull of gravity :
luda_lava [24]

Answer:

1. True  WA > WB > WC

Explanation:

In this exercise they give work for several different configurations and ask that we show the relationship between them, the best way to do this is to calculate each work separately.

A) Work is the product of force by distance and the cosine of the angle between them

    WA = W h cos 0

   WA = mg h

B) On a ramp without rubbing

     Sin30 = h / L

     L = h / sin 30

     WB = F d cos θ  

     WB = F L cos 30

     WB = mf (h / sin30) cos 30

     WB = mg h ctan 30

C) Ramp with rubbing

    W sin 30 - fr = ma

   N- Wcos30 = 0

   W sin 30 - μ W cos 30 = ma

    F = W (sin30 - μ cos30)

   WC = mg (sin30 - μ cos30) h / sin30

   Wc = mg (1 - μ ctan30) h

When we review the affirmation it is the work where there is rubbing is the smallest and the work where it comes in free fall at the maximum

Let's review the claims

1. True The work of gravity is the greatest and the work where there is friction is the least

2 False. The job where there is friction is the least

3 False work with rubbing is the least

4 False work with rubbing is the least

5 0
3 years ago
NEED ASAP PLEASE HELP
sveta [45]
1. GPE - 40 * 2 * 10 = 800j
3 0
3 years ago
Convert the speed of light 3.0x10^8 m/s to km/day
Aleksandr [31]

Answer: 2.592 \times 10^{8}km/day

1 m = 0.001 km\\ 1 s= 1.157\times10^{-5} days\\ 1 m/s = \frac {0.001}{1.157\times10^{-5}} km/day = 86.4 km/day \\ 3.0\times 10^{8} m/s = 3.0\times 10^{8} m/s \times \frac {86.4 km/day}{1m/s} =2.592 \times 10^{8}km/day



7 0
3 years ago
Read 2 more answers
In the demolition of an old building, a 1,300 kg wrecking ball hits the building at 1.07 m/s2. Calculate the amount of force at
Y_Kistochka [10]

Answer: F = 1391 N

Explanation:

The information given to you are:

Mass M = 1300 kg

Acceleration a = 1.07 m/s^2

The magnitude of the force striking the building will be

F = ma

Where

F = force

Substitute mass M and acceleration a into the formula

F = 1300 × 1.07

F = 1391 N

Therefore, the wrecking ball strikes the building with a force of 1391 N

3 0
3 years ago
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