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3 years ago

20 pts !!!!

Physics

1 answer:

garri49 [273]3 years ago

4 0

Am not really sure but what i see is D

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A box of mass m is held at rest on a frictionless surface with force F up the ramp. The ramp has an angle

frutty [35]

Answer:

Explanation:

8 0

3 years ago

Read 2 more answers

Consider massive gliders that slide friction-free along a horizontal air track. Glider A has a mass of 1 kg, a speed of 1 m/s, a

mamaluj [8]

Answer:

0.167m/s

Explanation:

According to law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision. The bodies move with a common velocity after collision.

Given momentum = Maas × velocity.

Momentum of glider A = 1kg×1m/s

Momentum of glider = 1kgm/s

Momentum of glider B = 5kg × 0m/s

The initial velocity of glider B is zero since it is at rest.

Momentum of glider B = 0kgm/s

Momentum of the bodies after collision = (mA+mB)v where;

mA and mB are the masses of the gliders

v is their common velocity after collision.

Momentum = (1+5)v

Momentum after collision = 6v

According to the law of conservation of momentum;

1kgm/s + 0kgm/s = 6v

1 =6v

V =1/6m/s

Their speed after collision will be 0.167m/s

6 0

3 years ago

If 5 mm of rain falls in a 100 m2 field, what volume of rain, in m3, fell in the field?

Nina [5.8K]

The volume of rain that fells in the field is simply given by the area of the field, which is

$A=100 mm^2$

multiplied by the height of rain that fell, which is

$h=5.0 mm$

Therefore, the volume is

$V=hA=(5 mm)(100 mm^2)=500 mm^3$

7 0

3 years ago

The electric charge on an atom is neutral because the number of protons is the same as the number of _____and each of their indi

tester [92]

The blank should be electrons, I think.

4 0

3 years ago

A car takes off from rest takes of from rest and covers a distance of 80m on a straight road in 10s.Calculate the magnitude of i

hodyreva [135]

Initial velocity (u) = 0 m/s [the car was at rest]
Distance (s) = 80 m
Time (t) = 10 s
Let the magnitude of acceleration be a.
By using the equation of motion, $s = ut + \frac{1}{2} a {t}^{2}$ we get, $80 = 0 \times 10 + \frac{1}{2} \times a \times {10}^{2} \\ = > 80 = \frac{1}{2} \times 100a \\ = > 80 = 50a \\ = > a = \frac{80}{50} \\ = > a = 1.6$

Answer:

The magnitude of its acceleration is 1.6 m/s^2.

Hope you could get an idea from here.

Doubt clarification - use comment section.

4 0

3 years ago