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Bogdan [553]
3 years ago
14

Which situation would require the MOST work? A) 20 kg weight lifted 6 m B) 25 kg weight lifted 3 m C) 25 kg weight lifted 6 m D)

50 kg weight lifted 1 m
Physics
1 answer:
Reil [10]3 years ago
5 0

Work= force (N) x distance (m)

F= mass (kg) x acceleration (gravity; m/s^2)

for this question, your formula would be

Work= mass (kg) x acceleration (gravity) x distance (m)

a. F=20kg x 9.81 m/s^2 x6 m = 1177.2 J

b. F=25kg x 9.81 m/s^2 x3 m = 735.75 J

c. F=25kg x 9.81 m/s^2 x6 m = 1471.5 J

d. F=50kg x 9.81 m/s^2 x1 m = 490.5 J


Tip:

  1. If this was a timed test, you could save some time by just multiplying the mass (kg) in the question by the distance because 9.81 is a constant in the formula, you can ignore it (+you're not asked for the final answer). c would still be the largest number
  2. it would also help if you noticed:
  • the distance in c is half that of b but the mass is the same.  eliminate b because c is obviously bigger
  • a and c have the same distance but 25 is greater than 20; eliminate a

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It has a 10 electrons. Since it's atomic number is 11 it must have 11 protons. Also, given that it has a +1 charge, it has one less electron than protons since they have equal but opposite charges. The number of protons is the mass number minus the atomic number = 23-11= 12 neutrons.
5 0
2 years ago
A car driving at 20 m/s accelerates continuously 2m/s2​ ​ for 3 seconds. What is its final velocity
Lesechka [4]

Answer: V= u+ at

V= final velocity

u=initial velocity

a=acceleration

t=time taken

V= 20 + 2*3

V= 26m/s

Explanation:

5 0
2 years ago
A stone is dropped from a cliff. What will be its speed when it was fallen 100 m?
Mars2501 [29]

Answer:

final velocity will be44.72m/s

Explanation:

HEIGHT=h=100m

vi=0m/s

vf=?

g=10m/s²

by using third equation of motion for bodies under gravity

2gh=(vf)²-(vi)²

evaluating the formula

2(10m/s²)(100m)=vf²-(0m/s)²

2000m²/s²=vf²

√2000m²/s²=√vf²

44.72m/s=vf

6 0
3 years ago
Read 2 more answers
If an object accelerates from rest, what will its velocity be after 1.3 s if it has a constant acceleration of 9.1 m/s^2?
HACTEHA [7]

\text{Given that,}\\\\\text{Initial velocity,} ~v_0 = 0~ \text{m~s}^{-1}\\\\\text{Time,  t = 1.3~sec}\\\\\text{Acceleration, a = 9.1 m s}^{-2}\\\\\\\\\text{Velocity,}\\\\v = v_0  +at\\\\\implies v = 0 + 9.1 \times 1.3 = 11.83~~ \text{m~s}^{-1}

5 0
2 years ago
A 21 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar
Georgia [21]

Answer:

a)15 N

b)12.6 N

Explanation:

Given that

Weight of block (wt)= 21 N

μs = 0.80 and μk = 0.60

We know that

Maximum value of static friction given as

Frs = μs m g = μs .wt

by putting the values

Frs= 0.8 x 21 = 16.8 N

Value of kinetic friction

Frk= μk m g = μk .wt

By putting the values

Frk= 0.6 x 21 = 12.6 N

a)

When T = 15 N

Static friction Frs= 16.8 N

Here the value of static friction is more than tension T .It means that block will not move and the value of friction force will be equal to the tension force.

Friction force = 15 N

b)

When T= 35 N

Here value of tension force is more than maximum value of static friction that is why block will move .We know that when body is in motion then kinetic friction will act on the body.so the value of friction force in this case will be 12.6 N

Friction force = 12.6 N

8 0
3 years ago
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