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Snezhnost [94]
3 years ago
11

. A person weighing 750 N gets on an elevator.

Physics
1 answer:
Kobotan [32]3 years ago
5 0

 

F = 750 N  (Force)

d = 10 m  (displacement )

t = 25 s   (time)

L = ?   (Mechanical work )  =  (Energy)

P = ?   (Power)

Solve:

L = F × d = 750 × 10 = 7500 Joules

P = L / t = 7500 / 25 = 300 Watts

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An input for of 80 N is used to lift an object weighing 240 N with a system of pulleys. How far must the rope around the pulleys
sammy [17]

Answer:

4.2 m

Explanation:

Note: If energy is conserved, i.e no work is done against friction

Work input = work output.

Work output = Force output × distance,

Work input = force input × distance moved moved.

Therefore,

input force×distance moved = output force × distance moved........................Equation 1

Given: input force = 80 N, output force = 240 N, output distance = 1.4 m

Let input distance = d

Substitute into equation 1

80×d = 240×1.4

80d = 336

d = 336/80

d = 4.2 m.

Thus the rope around the pulley must be pulled 4.2 m

7 0
3 years ago
In at least 150 words, discuss how Chang's use of personification, metaphor, or connotation express one his themes in "Garden of
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<span>Poet Kuangchi Chang did not remain in China long enough to be "re-educated." Following the Communist takeover he fled to the United States. His poem "Garden of My Childhood" describes China before the revolution as a peaceful, idyllic garden with a violent horde rapidly approaching. A vine, the wind, and the sea are each personified, and each beckons for him to run. It is not until "eons later," when he is "worlds away," that his "running is all done," and he finds himself at his destination: another garden, just like the one he had left behind.</span>
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3 years ago
Two long, straight wires, one above the other, are seperated by a distance 2a2a and are parallel to the x−axisx−axis. Let the +y
kompoz [17]

Answer:

note:

<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</u>

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3 0
3 years ago
A 53.0 kg sled is sliding on snow with μk=0.110. how much friction force does it feel?
Anon25 [30]

Answer:

57N

Explanation:

F_F = \mu F_N = \mu mg = 0.11 \times 53 kg \times 9,8 \frac{m}{s^{2} }

6 0
3 years ago
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A 2.00 kg box slides on a rough, horizontal surface, hits a spring with a speed of 1.90 m/s and compresses it a distance of 10.0
oksian1 [2.3K]

Answer:

Explanation:

Given

mass of box m=2\ kg

speed of box v=1.9\ m/s

distance moved by the box x=10\ cm

coefficient of kinetic friction \mu _k=0.66

Friction  force f_r=\mu_kN

f_r=0.66\times mg

f_r=0.66\times 2\times 9.8=12.936 \N

Kinetic Energy of box will be utilize to overcome friction and rest is stored in spring in the form of elastic potential energy

\frac{1}{2}mv^2=f_r\cdot x+\frac{1}{2}kx^2

\frac{1}{2}\times 2\times 1.9^2=12.936\times 0.1+\frac{1}{2}\times k\times (0.1)^2

3.61-1.2936=0.005\times k

k=463.28\ N/m

3 0
3 years ago
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