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DanielleElmas [232]

3 years ago

11

A feather is dropped on the moon from a height of 1.40 meters. The acceleration due to gravity on the moon is 1.67 m/s2. Determi

ne the time for the feather to fall on the surface of the moon.

1 answer:

RoseWind [281]3 years ago

7 0

Distance = (1/2) (acceleration) (time)²

1.4m = (0.835 m/s²) (time)²

(time)² = (1.4/0.835) s²

<em>time = 1.295 s</em>

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Use the values from PRACTICE IT to help you work this exercise. Suppose the same two vehicles are both traveling eastward, the c

Mariulka [41]

Answer:

A. $v_{3}=12.17m/s$

B. $v_{car}=6.3m/s\\v_{truck}=-6.3m/s$

C. ΔK $=-4.13x10^3J$

Explanation:

From the exercise we know that the car and the truck are traveling eastward. I'm going to name the car 1 and the truck 2

$v_{1}=5.79m/s\\m_{1}=102kg\\v_{2}=18.5m/s\\m_{2}=103kg$

A. Since the two vehicles become entangled the final mass is:

$m_{3}=102kg+103kg=205kg$

From linear momentum we got that:

$p_{1}=p_{2}$

$m_{1}v_{1}+m_{2}v_{2}=m_{3}v_{3}$

$v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3} }=\frac{(102kg)(5.79m/s)+(103kg)(18.5m/s)}{(205kg)}$

$v_{3}=12.17m/s$

B. The change in velocity of both vehicles are:

For the car

$v_{car}=v_{f}-v_{o}=12.17m/s-5.79m/s=6.38m/s$

For the truck

$v_{truck}=12.17m/s-18.5m/s=-6.3m/s$

C. The change in kinetic energy is:

ΔK= $K_{2}-K_{1} =\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2})$

ΔK= $\frac{1}{2}(205)(12.17)^{2}-(\frac{1}{2}(102)(5.79)^{2}+\frac{1}{2}(103)(18.5)^{2})=-4.13x10^{3}J$

ΔK $=-4.13x10^{3}J$

6 0

3 years ago

Earth attracts a person with a gravitational force of 7.0 × 102 newtons. What is the magnitude of the force with which the indiv

Anuta_ua [19.1K]

This is a good time to review Newton's 3rd law of motion:
"For every action, there is an equal and opposite reaction."

Gravitational force always acts in pairs.
Whatever force the Earth attracts something with,
the thing attracts the Earth with exactly the same force.

If Earth attracts a person with a gravitational force of <span><span>7.0 × 10² </span>newtons,
the person attracts Earth with a gravitational force of 7.0 × 10² newtons.

Your weight on Earth is the same as the Earth's weight on you !
</span>

5 0

3 years ago

Consider a block on a spring oscillating on a frictionless surface. The amplitude of the oscillation is 11 cm, and the speed of

IRISSAK [1]

Answer:

The angular frequency of the block is ω = 5.64 rad/s

Explanation:

The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.

Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.

The angular frequency of the oscillation ω is

ω = v/r

ω = 62 cm/s ÷ 11 cm

ω = 5.64 rad/s

So, the angular frequency of the block is ω = 5.64 rad/s

6 0

3 years ago

Point P and point charge Q are separated by a distance R. The electric field at point P has magnitude E. How could the magnitude

Angelina_Jolie [31]

<span>We know , E = kQ/r^2 where q = charge and r is separation between point and point charge. Now, At P, E= kQ/r^2 Since, Q can't be changed, we can do that by varying r 2E = 2kq/r^2 2E = kq/ (r/ sqrt2)^2 Hence, if we bring Q closer such that distance between P and Q becomes r/ sqrt 2, E will get doubled.</span>

5 0

3 years ago

A bar magnet was placed underneath a sheet of paper where a pile of iron filings sits. In the presence of the energy stored in t

Naya [18.7K]

Answer: 1. The field energy will increase

2. The energy increases, and the lines of force are denser

3. It points toward the field of earths magnetic poles

4. 1 and 2 only

5. 2, 4, 1, 3

Explanation: just took it

4 0

2 years ago