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Inga [223]
3 years ago
6

A car engine applies a force of 65,000 N, how much work is done by the engine as it pushed a car a distance of 75 m?

Physics
1 answer:
kupik [55]3 years ago
3 0

Answer:

workdone = force \times distance \\  = 65000 \times 75 \\  = 4,875,000 \: J

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QUESTIONS
Sophie [7]

Answer:

Figure xaina ta

Explanation:

Figure samet pathau

5 0
3 years ago
April stands on a flat surface . If she has a mass of 72 what is the nirmal force acting on her
gayaneshka [121]

The normal force acting on April as she stands on a flat surface is 705.6 N.

Normal force of the girl

Fₙ = mg

where;

  • Fₙ is the normal force of the girl
  • m is mass of the girl
  • g is acceleration due to gravity

Fₙ = 72 x 9.8

Fₙ = 705.6 N

Thus, the normal force acting on April as she stands on a flat surface is 705.6 N.

Learn more about normal force here: brainly.com/question/14486416

#SPJ1

7 0
2 years ago
One object has a charge of +5.0 · 10-6 c, and a second object has a charge of +2.0 · 10-6
ruslelena [56]
The electrostatic force between two charged objects is given by
F=k \frac{q_1 q_2}{r^2}
where
k is the Coulomb's constant
q1 is the charge of the first object
q2 is the charge of the second object
r is the separation between the two objects

In our problem:
q_1=+5.0 \cdot 10^{-6} C
q_2 = +2.0 \cdot 10^{-6} C
r=0.5 m
So if we plug these numbers into the equation, we can find the electrostatic force between the two objects:
F=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(5.0 \cdot 10^{-6} C)(2.0 \cdot 10^{-6}C)}{(0.5 m)^2}=0.36 N
4 0
3 years ago
A cable passes over a pulley. Because the cable grips the pulley and the pulley has nonzero mass, the tension in the cable is no
topjm [15]

Answer:

\alpha=214.8 rad/s^2

Explanation:

We are given that

F_1=137 N

F_2=43 N

Net force=F=F_1-F_2=137-43=94 N

Mass,m=1.21 kg

Radius,r=0.723 m

We have to find the magnitude of its angular acceleration.

Moment of inertia ,I=\frac{1}{2}mr^2

Substitute the values

Torque ,\tau=I\alpha

F_{net}\times r=\frac{1}{2}mr^2\alpha

\alpha=\frac{2F_{net}}{mr}

\alpha=\frac{2\times 94}{1.21\times 0.723}

\alpha=214.8 rad/s^2

4 0
3 years ago
Suppose Joe, who can type 40 words per minute, takes 2.5 hours to type his report. To speed things up, his friend Mark offers to
Vika [28.1K]

Explanation:

When Joe works alone, the total number  of words he typed can be given by:

Total words = (40 words per minute) x (60 minutes per hour) x (2.5 hours)

Total words = 6000 words

Now, when Joe and Mark work together, let 'y' be the number of hours for which they both work simultaneously:

Total words = Words Typed by Joe + Words Typed by Mark

6000 = {(40 words per minute) x (60 minutes per hours) x (y hours)} + {(20 words per minute) x (60 minutes per hours) x (y hours)}

6000 = 2400y + 1200y = 3600y

y = 1.67 hours = 1 hour and 40 minutes

Thus, working together simultaneously, Joe and Mark will take 1 hour and 40 minutes to complete the report.

5 0
3 years ago
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