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AveGali [126]
3 years ago
6

Gary needs to move a chair that is 20 pounds at a distance of 5 feet. How much work will he need to produce?

Physics
2 answers:
Bas_tet [7]3 years ago
8 0

answer- 100 ft-lbs

Hope I helped

tangare [24]3 years ago
3 0
<h2>Hello!</h2>

The answer is:

The answer is the first option, 100 ft-lbs.

<h2>Why?</h2>

To calculate the work done by an object over a distance, we need to multiply the applied force by the amount of movement (distance).

So, to calculate the work that Gary needs to do to move the chair, we need to use the following equation:

Work=Force*Distance

We are given,

Force=20lb\\Distance=5ft

Work=F*D\\Work=(20lb)*(5ft)=100ft-lbs

Hence, the answer is the first option, 100 ft-lbs.

Have a nice day!

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A wheel has a radius of 5.9 m. How far (path length) does a point on the circumference travel if the wheel is rotated through an
posledela

Answer:

(a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

Explanation:

Given that,

Radius = 5.9 m

(a). Angle \theta=30°

We need to calculate the angle in radian

\theta=30\times\dfrac{\pi}{180}

\theta=0.523\ rad

We need to calculate the path length

Using formula of path length

Path\ length =angle\times radius

Path\ length=0.523\times5.9

Path\ length =3.09\ m

(b). Angle = 30 rad

We need to calculate the path length

Path\ length=30\times5.9

Path\ length=177\ m

(c). Angle = 30 rev

We need to calculate the angle in rad

\theta=30\times2\pi

\theta=188.4\ rad

We need to calculate the path length

Path\ length=188.4\times5.9

Path\ length =1111.56\ m

Hence, (a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

8 0
3 years ago
What is the kinetic energy of a vehicle that has a mass of 3,500 kg and is moving at 40 m/s
stiv31 [10]

Answer:2800000j

Explanation:

For us to know the kinetic energy of the vehicle,

Where m is the mass

And v is the velocity

Then, K.E=1/2mv^2

While, K.E=1/2×3500×40^2

Therefore, our answer will now be

K.E=2800000j

5 0
3 years ago
A battery-operated car utilizes a 12.0 V system. Initially, the car is at rest at the base of a 195 m high hill. Some time later
ale4655 [162]

Answer:

140265.8 C = 1.403 × 10⁵ C

Explanation:

The battery's electric potential energy is used to account for the kinetic and potential work done in moving the car up this hill.

Potential work required to move the 757 kg car up a vertical height of 195 m = mgh

P.E = 757 × 9.8 × 195 = 1446627 J

Kinetic work done = (1/2)(m)(v²)

K.E = (1/2)(757)(25²) = 236562.5 J

Total work done in moving the car up that height = 1446627 + 236562.5 = 1683189.5 J

And this would be equal to the potential of the battery.

For the battery, potential difference = (electric potential energy)/(charges moved)

ΔV = ΔU/q

q = ΔU/ΔV

ΔU = 1683189.5 J

ΔV = 12.0 V

q = 1683189.5/12 = 140265.8 C

7 0
3 years ago
Read 2 more answers
Please help fast, 30 Points, Brainliest Answer
maksim [4K]

A.) Cubic meters is your answer

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hope this helps

4 0
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Question 8, is it A or B?
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Answer:

c

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