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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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Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 25.0 m above water wit
Rasek [7]

Complete question is;

Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 25.0 m above water with an initial speed of 20.0 m/s strikes the water with a final speed of 31.1 m/s, independent of the direction thrown

Answer:

It is proved that the final speed is truly 31.1 m/s

Explanation:

From energy - conservation principle;

E_i = Initial potential energy + Initial Kinetic Energy

Or

E_i = U_i + K_i

Similarly, for final energy

E_f = U_f + K_f

So, expressing the formulas for potential and kinetic energies, we now have;

E_i = (m × g × y_i) + (½ × m × v_i²)

Similarly,

E_f = (m × g × y_f) + (½ × m × v_f²)

We are given;

y_i = 25 m

y_f = 0 m

v_i = 20 m/s

v_f = 31.1 m/s

So, plugging in relevant values;

E_i = m((9.8 × 25) + (½ × 20²))

E_i = 485m

Similarly;

E_f = m((9.8 × 0) + (½ × v_f²)

E_f ≈ ½m•v_f²

From energy conservation principle, E_i = E_f.

Thus;

485m = ½m•v_f²

m will cancel out to give;

½v_f² = 485

v_f² = 485 × 2

v_f² = 970

v_f = √970

v_f ≈ 31.1 m/s

7 0
3 years ago
Explain why the term Nano is used to describe nanotechnology​
Volgvan

Answer:

Nanotechnology means working with materials at the scale of one billionth of a metre.

Explanation:

The prefix “nano” refers to one billionth: it’s part of the scientific scale of measurement. Its science, engineering, and technology are conducted at the nanoscale, which is about 1 to 100 nanometers. Nanoscience and nanotechnology are the study and application of extremely small things. Thus, this describes why the term nano is used to describe nanotechnology.

I hope this helps :)

5 0
3 years ago
A cannonball is fired from a cliff that is 50 meters above the ground. The cannonball is fired horizontally with a speed of 120
tankabanditka [31]

The horizontal distance that the cannonball will travel is 383.33m

In order to find the horizontal distance, we will use the formula for calculating the range expressed as:

R = u\sqrt{\frac{2H}{g} } where:

u is the velocity

H is the height

g is the acceleration due to gravity

R = 120\sqrt{\frac{2(50)}{9.8} }\\R = 120\sqrt{\frac{100}{9.8} } \\R=120\times 3.1943\\R=383.33m

Hence the horizontal distance that the cannonball will travel is 383.33m

Learn more here: brainly.com/question/19028766

4 0
2 years ago
Two blocks are connected by a light string that passes over two frictionless pulleys. The block of mass m2 is attached to a spri
irina1246 [14]

(BELOW YOU CAN FIND ATTACHED THE IMAGE OF THE SITUATION)

Answer:

d=\frac{2g(m1-m2)}{k}

Explanation:

For this we're going to use conservation of mechanical energy because there are nor dissipative forces as friction. So, the change on mechanical energy (E) should be zero, that means:

E_{i}=E_{f}

K_{i}+U_{i}=K_{f}+U_{f} (1)

With K_{i} the initial kinetic energy, U_{i} the initial potential energy, K_{f} the final kinetic energy and U_{f} the final potential energy. Note that initialy the masses are at rest so K_{i} = 0, when they are released the block 2 moves downward because m2>m1 and finally when the mass 2 reaches its maximum displacement the blocks will be instantly at rest so K_{f} =0. So, equation (1) becomes:

U_{i}=U_{f} (2)

At initial moment all the potential energy is gravitational because the spring is not stretched so U_{i}=U_{gi} and at final moment we have potential gravitational energy and potential elastic energy so U_{f}=U_{gf}+U_{ef}, using this on (2)

U_{gi}=U_{gf}+U_{ef} (3)

Additional if we define the cero of potential gravitational energy as sketched on the figure below (See image attached), U_{gi}=0 and we have by (3) :

0= U_{gf}+U_{ef} (4)

Now when the block 1 moves a distance d upward the block 2 moves downward a distance d too (to maintain a constant length of the rope) and the spring stretches a distance d, so (4) is:

0=-m1gd+m2gd+\frac{kd^{2}}{2}

dividing both sides by d

0=-m1g+m2g+\frac{kd}{2}

g(m1-m2)= \frac{kd}{2}

d=\frac{2g(m1-m2)}{k}, with k the constant of the spring and g the gravitational acceleration.

7 0
3 years ago
What is the difference between a chemical change and a physical change? *
slavikrds [6]

Answer:

I do belive that it is B hrs cn I an gn

7 0
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