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solong [7]
3 years ago
11

A pickup truck starts from rest and maintains a constant acceleration a0. After a time t0, the truck is moving with speed 25 m/s

at a distance of 120 m from its starting point. When the truck has travelled a distance of 60 m from its starting point, its speed is v1 m/s.
Which of the following statements concerning v1 is true?
a. v1< 12.5m/s
b. v1= 12.5m/s
c. v1 >12.5m/s
Physics
1 answer:
MA_775_DIABLO [31]3 years ago
7 0

Answer:

B v1= 12.5m/s

Explanation:

As the acceleration is constant and time taken is also same and distance covered is halfed so the speed will also be halfed.

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Which of the following statements about iron filings placed upon glass resting on top of a bar magnet is false?
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Statements  A,  B,  and  C  are true.

Statement <em> D  is false</em>.

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Terry can ride 30 miles in 2 hours. If his riding speed is
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Answer:

Explanation:

Terry can ride at a speed of

V = 30 miles in 2hours

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Then,

Speed = distane / time

Distance = speed × time

Distance =15 × 1.7

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A body weighs 10newton in air and 9.5 in water. How will it weigh in alchohol if density 0.8gcm-³​
Katen [24]

Answer:

The weight of object in alcohol is, W₀ = 9.6 N

Explanation:

Given,

The weight of body in air, Wₐ = 10 N

The weight of the body in water, Wₓ = 9.5 N

The density of the alcohol, ρ = 0.8 g/cm³

                                                = 800 kg/m³

The weight of body in water = weight of object in air - weight of water displaced

                        9.5 N = 10 N - weight of water displaced

                       Weight of water displaced = 0.5 N

The mass of the displaced water, m = 0.5 / 9.8

                                                            = 0.051 kg

Therefore the volume of the water displaced,

                                              V = m / ρ

                                                  = 0.051 / 1000

                                                   = 5.1 x 10⁻⁵ m³

<em>The volume of water displaced is equal to the volume of alcohol displaced</em>

Therefore, the mass of the alcohol displaced, = V x ρ

                                                                             = 5.1 x 10⁻⁵ x 800

                                                                             = 0.0405 kg

The weight of the alcohol displaced, w = 0.0405 x 9.8

                                                                    = 0.4 N

Therefore,

The weight of body in alcohol = weight of object in air - weight of alcohol displaced

                             W₀ = W - w

                                    = 10 N - 0.4 N

                                     = 9.6 N

Hence, the weight of object in alcohol is, W₀ = 9.6 N

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