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Pavel [41]
3 years ago
12

If a car stops suddenly, you feel "thrown forward." We’d like to understand what happens to the passengers as a car stops. Imagi

ne yourself sitting on a very slippery bench inside a car. This bench has no friction, no seat back, and there’s nothing for you to hold onto.(a) identify all of the forces action on you as thecar travels at a perfectly steady speed on level ground.(b) repeat part A with the car slowing down(c) describe what happens to you as the car slowsdown(d) suppose now that the bench is not slippery. as the carslows down, you stay on the bench and dont slide off. what force isresponsible for you deceleration?
Physics
1 answer:
stellarik [79]3 years ago
5 0
<h2>A)</h2>

The forces acting on you are:

  • The gravitational pull of the Earth (an others Celestial objects)
  • The Normal Force that balance the gravitational pull and points upward.

That is. There is not need for any other force, cause the car its going at constant speed, so the acceleration its zero, as the net force its mass multiplied by acceleration, the net force is zero.

<h3>B)</h3>

The forces acting on you are:

  • The gravitational pull of the Earth (an others Celestial objects)
  • The Normal Force that balance the gravitational pull and points upward.

That is. Again. The car its slowing down. But, it cant apply a force in you against the direction of movement to slow you down.

<h3>C)</h3>

As you can't be slowed down, you will go forward at the same speed you had when the car went steady, and will crash against whatever its in front of you.

D)

The friction force, will pull you against the direction of movement, and slow you down will the car slows down.

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6. Two light bulbs are designed for use at 120 V and are rated at 75 W and 150 W. Which light bulb has the greater filament resi
Nonamiya [84]

\\ \bull\tt\longrightarrow P=\dfrac{V^2}{R}

  • P is power
  • R is resistance

\\ \bull\tt\longrightarrow R=\dfrac{V^2}{P}

Hence

\\ \bull\tt\longrightarrow R\propto V

\\ \bull\tt\longrightarrow R\propto \dfrac{1}{P}

  • Therefore if power is low then resistance will be high.

The first bulb has less power hence it has greater filament resistance.

5 0
3 years ago
A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re
valina [46]

The box has 3 forces acting on it:

• its own weight (magnitude <em>w</em>, pointing downward)

• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)

• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>

• net perpendicular force:

∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

<em>n</em> = <em>w</em> cos(35°)

<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)

<em>n</em> ≈ 120 N

Solve for the mag. of friction:

<em>f</em> = <em>µ</em> <em>n</em>

<em>f</em> = 0.25 (120 N)

<em>f</em> ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>

<em>a</em> ≈ (54.3157 N) / (15 kg)

<em>a</em> ≈ 3.6 m/s²

Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:

<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>

<em>v</em>² = 2 (3.6 m/s²) (3 m)

<em>v</em> = √(21.7263 m²/s²)

<em>v</em> ≈ 4.7 m/s

4 0
3 years ago
A positively charged particle Q1 = +35nC is held fixed at the origin. A second charge of mass m = 3.5ug is floating a distance d
cupoosta [38]

Answer:

Explanation:

Q1 = 35 nC = 35 x 10^-9 C

m = 3.5 micro gram = 3.5 x 10^-9 Kg

d  = 35 cm = 0.35 m

(a) The electrostatic force between the two charges is balanced by the weight of another charge.

F = m g

\frac{1}{4\pi \epsilon _{0}}\frac{Q_{1}Q_{2}}{d^{2}}=mg

Q_{2}=\frac{4\pi \epsilon _{0}mgd^{2}}{Q_{1}}

(b) By substituting the values

Q_{2}=\frac{3.5\times 10^{-9}\times 9.8\times 0.35\timees 0.35}{9\times 10^{9}\times 35\times 10^{9}}

Q2 = 13.34 x 10^-12 C

Q2 = 0.0134 nC

4 0
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Jane placed two bar magnets on her desk next to each other with like poles facing. She then sprinkled some iron filings around t
viktelen [127]

Answer:

B.

Explanation:

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Efficiency = 1000/2000 = 0.5 = 50%
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