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swat32
3 years ago
14

51.Shoveling snow can be extremely taxing because the arms have such a low efficiency in this activity. Suppose a person shoveli

ng a footpath metabolizes food at the rate of 800 W. (a) What is her useful power output? (b) How long will it take her to lift 3000 kg of snow 1.20 m? (This could be the amount of heavy snow on 20 m of footpath.) (c) How much waste heat transfer in kilojoules will she generate in the process?
Physics
1 answer:
Komok [63]3 years ago
3 0

Complete question is;

Shoveling snow can be extremely taxing since the arms have such a low efficiency in this activity. Suppose a person shoveling a sidewalk metabolizes food at the rate of 800 W. (The efficiency of a person shoveling is 3%.)

(a) What is her useful power output? (b) How long will it take her to lift 3000 kg of snow 1.20 m? (This could be the amount of heavy snow on 20 m of footpath.) (c) How much waste heat transfer in kilojoules will she generate in the process?

Answer:

A) P_out = 24 W

B) t = 1470 s

C) Q = 1140.72 KJ

Explanation:

We are given;

Input Power; P_in = 800 W

Efficiency; η = 3% = 0.03

A) Formula for efficiency is;

η = P_out/P_in

Making P_out the subject, we have;

P_out = η•P_in

P_out = 0.03 × 800

P_out = 24 W

B) We know that;

Power = work done/time taken

Thus;

P_out = mgh/t

We are given;

m = 3000 kg

h = 1.20 m

Thus, time is;

t = (3000 × 9.8 × 1.2)/24

t = 1470 s

C) amount of heat wasted is calculated from;

Q = (P_in - P_out)t

Q = (800 - 24) × 1470

Q = 1,140,720 J

Q = 1140.72 KJ

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2. ¿Puede haber trabajo en un sistema si no hay movimiento
Andru [333]

Answer:

d. No, porque la ecuación de trabajo lo define.

Explanation:

En Física, el trabajo realizado se puede definir como la cantidad de energía transferida cuando un objeto o cuerpo se mueve a lo largo de una distancia debido a la acción de una fuerza externa.

Matemáticamente, el trabajo realizado viene dado por la fórmula;

W = F * d

<u>Dónde;</u>

  • W es el trabajo realizado.
  • F representa la fuerza que actúa sobre un cuerpo.
  • d representa la distancia recorrida por el cuerpo.

Por lo tanto, podemos deducir de la definición de trabajo y su fórmula que el trabajo se realiza cuando un objeto (cuerpo) se mueve una distancia o experimenta cualquier forma de desplazamiento mientras transfiere energía.

7 0
3 years ago
A 0.145 kg baseball is thrown with a velocity of 25.0 m/s. How much work was done on the baseball to bring it from rest to 25.0
sergey [27]

Answer:

45.31 J

Explanation:

We are given that

Mass of baseball , m=0.145 kg

Initial velocity, u=0

Final velocity, v=25 m/s

We have to find the work done on the baseball to bring it from rest to 25 m/s

We know that

Work done = Change in kinetic energy

Work done, W=\frac{1}{2}m(v^2-u^2)

Using the formula

Work done, W=\frac{1}{2}(0.145)((25)^2-0)

Work done=\frac{1}{2}(0.145)(625)

Work done, W=45.31 J

Hence, the work done on the baseball to bring it from rest to 25 m/s

=45.31 J

3 0
3 years ago
A cord is wrapped around the rim of a solid uniform wheel 0.300 m in radius and of mass 8.00 kg. A steady horizontal pull of 34.
disa [49]

Answer:

A. α = 94.4 rad/s

B. a = 28.32 m/s

C. N = 34N

D. α = 94.4 rad/s

    a = 28.32 m/s

     N =  44.4  N

Explanation:

part A:

using:

∑T = Iα

where T is the torque, I is the moment of inertia and α is the angular momentum.

firt we will find the moment of inertia I as:

I = \frac{1}{2}MR^2

Where M is the mass and R is the radius of the wheel, then:

I = \frac{1}{2}(8 kg)(0.3 m)^2

I = 0.36 kg*m^2

Replacing on the initial equation and solving for α, we get::

∑T = Iα

Fr = Iα

34 N =  0.36α

α = 94.4 rad/s

part B

we need to use this equation :

a = αr

where a is the aceleration of the cord that has already been pulled off and r is the radius of the wheel, so replacing values, we get:

a = (94.4)(0.3 m)

a = 28.32 m/s

part C

Using the laws of newton, we know that:

N = T

where N is the force that the axle exerts on the wheel part and T is the tension of the cord

so:

N = 34N

part D

The anly answer that change is the answer of the part D, so, aplying laws of newton, it would be:

-Mg + N +T = 0

Then, solving for N, we get:

N = -T+Mg

N = -34 + (8 kg)(9.8)

N =  44.4  N

6 0
3 years ago
A marble rolls off a tabletop 1.15 m high and hits the floor at a point 4 m away from the tables edge in the
GREYUIT [131]

A marble rolls off a tabletop 1.15 m high and hits the floor at a point 4 m away from the edge of the table in the horizontal direction,

  • t= 0.45 seconds.
  • V=2.22m/s
  • VT=4.95 m/s

This is further explained below.

<h3>What is its speed when it hits the floor...?</h3>

Generally, the equation for motion is mathematically given as

S= ut + 0.5at²

Therefore

y = Voy t + 0.5gt^2

1 = 0.5x 98 x 6²

1=4.9t^2

t=\sqrt{0.2041 }

t= 0.45 seconds.

b) Horizontal motions are uniform.

V=Horizontal displacement/time

V=1/0.45

V=2.22m/s

C)

Vx: 2.22 m/s At bottom,

Vy² = Voy² + 2as

Vy² = 2x95x1

Vy² = 19.6

Total velocity

VT=\sqrt{( 2.22 m/)^2+19.6}

VT=4.95 m/s

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4 0
2 years ago
7. A 78-kg skydiver has a speed of 62 m/s at an altitude of 870 m above the ground.
miv72 [106K]

The kinetic energy, potential energy and total mechanical energy possessed by the skydriver are 1.5 × 10⁵ Joule, 6.7 × 10⁵ Joule and 8.2 × 10⁵ Joule respectively.

To find the answer, we need to know about the expression of kinetic energy and potential energy.

<h3>What are the expressions of kinetic energy and potential energy?</h3>
  • Mathematically, kinetic energy= 1/2 × mass × velocity²
  • Gravitational potential energy near the earth surface= mass × g × height on the earth surface

<h3>What's the kinetic energy, potential energy and total mechanical energy of the 78kg skydriver at 870 m on earth surface with 62 m/s velocity?</h3>
  • Kinetic energy= 1/2 × 78 × 62² = 1.5 × 10⁵ Joule
  • Potential energy= 78×9.8×870= 6.7× 10⁵ Joule
<h3>What's the total mechanical energy?</h3>
  • Mechanical energy= kinetic energy+ potential energy
  • 1.5 × 10⁵ Joule + 6.7× 10⁵ Joule = 8.2× 10⁵ Joule

Thus, we can conclude that the kinetic energy, potential energy and total mechanical energy possessed by the skydriver are 1.5 × 10⁵ Joule, 6.7 × 10⁵ Joule and 8.2 × 10⁵ Joule respectively.

Learn more about the kinetic energy, potential energy and mechanical energy here:

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8 0
2 years ago
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