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maria [59]
3 years ago
11

A constant force of 2.5 N to the right acts on a 4.5 kg mass for 0.90 s.

Physics
1 answer:
Alborosie3 years ago
8 0

Answer:

(a) v_f=0.5\frac{m}{s}

(b) v_f=-11\frac{m}{s}

Explanation:

(a) Since a constant external force is applied to the body, it is under an uniformly accelerated motion. Using the following kinematic equation, we calculate the final velocity of the mass  if it is initially at rest(v_0=0):

v_f=v_0+at\\v_f=at(1)

According to Newton's second law:

F=ma\\a=\frac{F}{m}(2)

Replacing (2) in (1):

v_f=\frac{F}{m}t\\v_f=\frac{2.5N}{4.5kg}(0.9s)\\v_f=0.5\frac{m}{s}

(b) In this case we have v_0=-11.5\frac{m}{s}. So, we use the final velocity equation:

v_f=v_0+at\\v_f=v_0+\frac{F}{m}t\\v_f=-11.5\frac{m}{s}+\frac{2.5N}{4.5kg}(0.9s)\\v_f=-11\frac{m}{s}

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Explanation:

{\bold{\sf{\underline{Understanding \: the \: concept}}}}

✠ This question says that there is an object and its mass is 2 kg ; it's raised to a height 10 m and possess potential energy of 200 J. Now this question ask us to find the kinetic energy and the potential energy at a height 4 metre.

\bold{↬{   }}{\bold{\sf{\underline{Given \: that}}}}

✰ Mass = 2 kilograms

✰ Raised height = 10 metres

✰ Posses potential energy = 200 Joules

\bold{↬{   }}{\bold{\sf{\underline{To \: find}}}}

✰ Kinetic energy at a height 4 metre

✰ Potential energy at a height 4 metre

{\bold{\sf{\underline{Solution}}}}

✰ Kinetic energy at a height 4 metre = 120 J

✰ Potential energy at a height 4 metre = 80 J

{\bold{\sf{\underline{Using \: concepts}}}}

✰ Potential energy formula.

{\bold{\sf{\underline{Using \: formula}}}}

✰ Potential energy = mgh

{\bold{\sf{\underline{We \: also \: write \: these \: as}}}}

✰ Potential energy as P.E

✰ Mass as m

✰ Joules as J

✰ Height as h

✰ Raised height as g

{\bold{\sf{\underline{Full \: solution}}}}

<h3>✠ Let us find the Potential energy.</h3>

↦ Potential energy = mgh

↦ Potential energy = 2 × 10 × 4

↦ Potential energy = 20 × 4

↦ Potential energy = 80 J

<h3>✠ Now according to the question let us find the kinetic energy</h3>

↦ Kinetic energy = Posses potential energy - Finded potential energy

↦ Kinetic energy = 200 J - 80 J

↦ Kinetic energy = 120 Joules

4 0
3 years ago
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Two equal-magnitude forces are applied to a door at the doorknob. The first force is applied perpendicular to the door, and the
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Answer:

in first case the torque is maximum.

Explanation:

Torque is defined as the product of force and the perpendicular distance.

τ = F x d x Sinθ

In case A: the angle between force vector and the distance vector is 90 so torque is

τ = F x d

In case B: the angle between force vector and the distance is 30°.

τ = F x d x Sin30

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So the torque is maximum in first case.

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B. insulator

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You and your friends are having a discussion about weight. He/she claims that he/she weighs less on the 100th floor of a buildin
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Answer:

if the weight theoretically decreases at this height, but in a fraction of 10⁻⁵, which is not appreciable in any scale, therefore, the reading of the scale in the two places is the same.

Explanation:

The weight of a person in the force with which the Earth attracts the person, therefore can be calculated using the law of universal attraction

          F = G m M / r²

Where m is the mass of the person, M the masses of the earth

Let's call the person's weight at ground level as Wo and suppose the distance to the center of the Earth is Re

            W₀ = G m M / Re²

In the calculation of the weight of the person on the 100th floor the only thing that changes is the distance

          r = Re + 100 r₀

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         r = Re + 250

We substitute

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The value of Re is 6.37 10⁶ m, so we can take it out as a factor and perform a serial expansion of the remaining fraction

      W = G m M / Re² (1+ 250 / Re)²

      (1 + 250 / Re)⁻² = 1 + (-2) 250 / Re + (-2 (-2-1)) / 2 (250 / Re)² +….

The value of the expression is

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We can see that the quadratic term is very small, which is why we despise it, we substitute in the weight equation

      W = G m M / Re² (1 - 78.5 10⁻⁶)

Remains

     W = Wo (1 - 7.85  10⁻⁵)

We can see that if the weight theoretically decreases at this height, but in a fraction of 10⁻⁵, which is not appreciable in any scale, therefore, the reading of the scale in the two places is the same.

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3 years ago
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