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ANEK [815]
2 years ago
12

You push downward on a box at an angle 25° below the horizontal with a force of 750 N. If the box is on a flat horizontal surfac

e for which the coefficient of static friction with the box is 0.8, what is the mass of the heaviest box you will be able to move?
Physics
1 answer:
RoseWind [281]2 years ago
8 0

Answer:

103.44g

Explanation:

R=f/u

=750/0.8

=937.5N

m = 937.5/10cos25

m = 103.44g

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Answer:

1) a = -1 m/s²

2) v = 12 m/s

Explanation:

Given,

The initial velocity of the object, u = 15 m/s

The final velocity of the object, v = 10 m/s

The time taken by the object to travel is, t = 5 s

Using the first equation of motion

                               <em>v = u + at</em>

                               a = (v - u) / t

Substituting the values

                                a = (10 - 15) / 5

                                    = -1 m/s²

The negative sign indicates the body is decelerating

The acceleration of the object is, a = -1 m/s²

The speed of the object after 2 seconds

From the above equations of motion

                                  v = 15 + (-1) 2

                                     = 12 m/s

Hence, the speed of the object after 2 seconds is, v = 12 m/s

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4 years ago
Describe the conditions necessary for sublimation to occur​ please help
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Answer:

Sublimation occurs more readily when certain weather conditions are present, such as low relative humidity and dry winds. Sublimation also occurs more at higher altitudes, where the air pressure is less than at lower altitudes. Energy, such as strong sunlight, is also needed.

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3 years ago
Sheila weighs 60 kg and is riding a bike. Her momentum on the bike is 340 kg • m/s. The bike hits a rock, which stops it complet
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Answer:

v₂ = 5.7 m/s

Explanation:

We will apply the law of conservation of momentum here:

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Total Initial Momentum = 340 kg.m/s

m₁ = mass of bike

v₁ = final speed of bike = 0 m/s

m₂ = mass of Sheila = 60 kg

v₂ = final speed of Sheila = ?

Therefore,

340\ kg.m/s = m_{1}(0\ m/s) + (60\ kg)v_{2}\\v_{2} = \frac{340\ kg.m/s}{60\ kg}\\\\

<u>v₂ = 5.7 m/s </u>

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