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jasenka [17]
3 years ago
6

A 63.0-kg man is riding an escalator in a shopping mall. The escalator moves the man at a constant velocity from ground level to

the floor above, a vertical height of 4.20 m. What is the work done on the man by (a) the gravitational force and (b) the escalator
Physics
1 answer:
Oliga [24]3 years ago
8 0

Answer:

Explanation:

Given that,

Mass of man =Mm=63kg

W=mg

W=63×9.81

W=618.03N

g=9.81m/s²

The escalator moves at a constant velocity, then this shows that it is not accelerating, then, a=0m/s²

Height escalator traveled

H=4.2m.

We assume that the weight of the escalator is negligible.

a. Work done by gravity

Work done by gravity is given as

Work done(gravity)=mgh

Since the work is done against gravity, then, g is negative

Then,

Work done(gravity)=-mgh

Work done(gravity)=-63×9.81×4.2

Work done(gravity)=-2595.726J

b. Work done by escalator

Using equation of motion to know the force pulling the escalator upward

ΣF = ma, but a=0

ΣF = 0

Only two force is acting on the in y axis, the Normal force and the weight

N-W=0

N=W

Since W=618.03N

Then, N=618.03

The normal is the force pulling the escalator upward

Then, the work done by escalator is given as

Work done=Force × distance

Work done=618.03×4.2

Work done= 2595.76J

Work done by escalator is 2595.76J

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Paraphin [41]

1.47x10^5 Joules  
The gravitational potential energy will be the mass of the object, multiplied by the height upon which it can drop, multiplied by the local gravitational acceleration. And since it started at the top of a 60.0 meter hill, halfway will be at 30.0 meters. So  
500 kg * 30.0 m * 9.8 m/s^2 = 147000 kg*m^2/s^ = 147000 Joules.  
Using scientific notation and 3 significant figures gives 1.47x10^5 Joules.
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3 years ago
Which equation describes the line containing the points (-2, 3) and (1, 2)​
snow_lady [41]

Answer:

y =  \frac{ - 1}{3}x +  \frac{7}{3}

Explanation:

\frac{y - 3}{2 - 3}  =  \frac{x + 2}{1 + 2}  \\  \ - y + 3 =  \frac{x + 2}{3}  \\  y =   \frac{ - 1}{3}x  +  \frac{7}{3}

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Find the torque required for the shaft to transmit 40 kW when (a) The shaft speed is 2500 rev/min. (b) The shaft speed is 250 re
Luba_88 [7]

Answer:

(a) 152.85 Nm

(b) 1528.5 Nm

Explanation:

According to the formula of power

P = τ ω

ω = 2 π f

(a) f = 2500 rpm = 2500 / 60 = 41.67 rps

So, 40 x 1000 = τ x 2 x 3.14 x 41.67

τ = 152.85 Nm

(b) f = 250 rpm = 250 / 60 = 4.167 rps

So, 40 x 1000 = τ x 2 x 3.14 x 4.167

τ = 1528.5 Nm

3 0
3 years ago
Difference between kilogram and kilometre in points​
skelet666 [1.2K]

Answer:

Kilogram(kg) is the SI unit for mass while kilometre(km) is a unit for length. They are both similar in that they are 10^3 of a unit, thus kilo. As kilogram represents mass, it is a measure of how much matter is present in an object. While kilometre is a measure of distance/how long or short an object is.

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3 years ago
A space shuttle sits on the launch pad for 2.0 minutes, and then goes from rest to 4600 m/s in 8.0 minutes. Treat its motion as
SpyIntel [72]

Answer:

a.) a = 0 ms⁻²

b.) a = 9.58 ms⁻²

c.) a = 7.67 ms⁻²

Explanation:

a.)

    Acceleration (a) is defined as the time rate of change of velocity

                       a = \frac{v_{2} - v_{1} } {t}  

Given data

 Final velocity = v₂ = 0 m/s

 Initial velocity = v ₁ = 0 m/s

  As the space shuttle remain at rest for the first 2 minutes i.e there is no change in velocity so,

                 a = 0 ms⁻²

b.)

     Given data

As the space shuttle start from rest, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 8 min = 480 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

             a = \frac{4600 - 0 } {480}

                     a = 9.58 ms⁻²

c.)

    Given data

As the space shuttle is at rest for first 2 min then start moving, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 10 min = 600 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

             a = \frac{4600 - 0 } {600}

                     a = 7.67 ms⁻²

8 0
3 years ago
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