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just olya [345]
4 years ago
12

calculating power reach each scenario and then answer the questions scenario a 120 j of work is done in 6 seconds scenario b 160

j of work is done at 8 seconds scenario c200 J's of work is done in 10 seconds which scenario uses the most power
Physics
1 answer:
Molodets [167]4 years ago
4 0

Answer: scenario b and scenario c uses most power

Explanation:

Scenario a:

Work=120J

Time=8 seconds

Power=work ➗ time

Power=120 ➗ 8

Power=15

Power=15 watts

Scenario b:

Work=160J

Time=8 seconds

Power=work ➗ time

Power=160 ➗ 8

Power=20

Power =20 watts

Scenario c:

Work=200J

Time=10 seconds

Power= work ➗ time

Power=200 ➗ 10

Power=20

Power=20 watts

Scenario b and scenario c uses most power

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21. A 60 kg student on a scooter is pushed with a constant force of 40 N across a horizontal concrete driveway at a constant spe
DedPeter [7]

The magnitude of the force of friction is 40 N

Explanation:

To solve the problem, we just have to analyze the forces acting on the student and the scooter along the horizontal direction. We have:

- The constant pushing force forward, of magnitude F = 40 N

- The frictional force, acting backward, F_f

Since the two forces are in opposite direction, the equation of motion is

F-F_f = ma

where

m is the mass of the student+scooter

a is the acceleration

However, here the scooter is moving at constant speed: this means that its acceleration is zero, so

a = 0

And therefore,

F - F_f = 0\\F_f = F

which means that the magnitude of the force of friction is also equal to 40 N.

Learn more about force of friction here:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

3 0
3 years ago
A truck accelerating at 0.0083 meters/second2 covers a distance of 5.8 × 104 meters. If the truck's mass is 7,000 kilograms, wha
zhannawk [14.2K]
Work done = force * distance moved (in direction of the force)

force= mass* acceleration 

force=58.1N

58.1*(5.8*10^4)
=3,369,800 J
7 0
3 years ago
Read 2 more answers
As mass increases what happens to the kinetic energy
AnnyKZ [126]
As mass increases kinetic energy also increases; kinetic energy is directly proportional to mass so whatever is done to either affects the other one the same. i hope this helps :)
5 0
3 years ago
a person throws a ball upward into the air with an initial velocity of 20 m/s. calculate (a) how high it goes, and (b) how long
Phantasy [73]

Answer:

a) about 20.4 meters high

b) about 4.08 seconds

Explanation:

Part a)

To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.

a=\frac{Vf-Vi}{t}

-9.8 \frac{m}{s} =\frac{Vf-Vi}{t}

In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:

-9.8  =\frac{0-20}{t}\\t=\frac{20}{9.8} s = 2.04 s

Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:

Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi=20*(2.04)-\frac{1}{2} 9.8*2.04^{2}\\Xf-Xi=20.408 m

Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)

Xf-Xi=0=20*(t)-\frac{1}{2} 9.8*t^{2

To solve for "t" in this quadratic equation, we can factor it out as shown:

0= t(20-\frac{9.8}{2} t)

Therefore there are two possible solutions when each of the two factors equals zero:

1) t= 0 (which is not representative of our case) , and

2) the expression in parenthesis is zero:

0= 20-\frac{9.8}{2} t\\t=\frac{20*2}{9.8} = 4.08 s

7 0
3 years ago
An office building has a 24-volt branch circuit installed for landscape lighting around the front of the building. The circuit w
Arturiano [62]

The circuit was installed in UF cable which requires a minimum burial depth of 6 inches for this circuit.

<h3>UF cable</h3>

UF cable is used as an underground feeder cable to distribute power from an existing building to outdoor equipment. UF cable can also be used as direct burial cable.

For the 24-volt branch circuit installed, the minimum burial depth will be 6 inches.

Thus, the circuit was installed in UF cable which requires a minimum burial depth of 6 inches for this circuit.

Learn more about UF cable here: brainly.com/question/8591560

5 0
2 years ago
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