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kolbaska11 [484]

3 years ago

12

How much work did the movers do (horizontally) pushing a 43.0-kg crate 10.4 m across a rough floor without acceleration, if the

effective coefficient of friction was 0.60

1 answer:

Ilia_Sergeevich [38]3 years ago

8 0

The crate is in equilibrium. Newton's second law gives

∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0

∑ <em>F</em> (horizontal) = <em>p</em> - <em>f</em> = 0

where

• <em>n</em> = magnitude of the normal force

• <em>mg</em> = weight of the crate

• <em>p</em> = mag. of push exerted by movers

• <em>f</em> = mag. of kinetic friciton, with <em>f</em> = 0.60<em>n</em>

<em />

It follows that

<em>p</em> = <em>f</em> = 0.60<em>mg</em> = 0.60 (43.0 kg) <em>g</em> = 252.84 N

so that the movers perform

<em>W</em> = <em>p</em> (10.4 m) ≈ 2600 J

of work on the crate. (The <em>total</em> work done on the crate, on the other hand, is zero because the net force on the crate is zero.)

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Certain suv ads claim that their vehicles can climb a slope of 45?. what is the minimum coefficient of static friction between t

kotykmax [81]

First, you make a force body diagram to illustrate the problem. Then, apply Newton's laws of motion.

Summation of forces along the y-direction:

F = 0 (bodies at equilibrium) = Normal force - weight*cos45 = 0
Normal force = wcos45

Summation of forcesalong the x-direction:

F = 0 (bodies at equilibrium) = Frictional force - weight*sin45 = 0
x*Normal force - weight*sin45 = 0 (let x be the min coeff of static friction)
x*weight*cos45 = weight*sin 45
x = sin45/cos45 = 1

Therefore, the minimum coefficient of static friction must be 1.

8 0

3 years ago

An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temper

bixtya [17]

Answer:

a. $P=18.61\ W$

b. $\eta_c=0.6976=69.76\%$

c. $\eta=0.4718=47.18\%$

Explanation:

Given:

temperature of the hotter reservoir, $T_H=1250\ K$

temperature of the colder reservoir, $T_C=378\ K$

heat absorbed by the engine, $Q_H=1.42\times 10^{5}\ J$

heat rejected to the cold reservoir, $Q_L=7.5\times 10^4\ J$

time duration of the energy transfer, $t=1\ hr=3600\ s$

<u>Now the work done by the engine:</u>

Using energy conservation,

$W=Q_H-Q_L$

$W=14.2\times 10^4-7.5\times 10^4$

$W=6.7\times 10^4\ J$

a.

<u>Hence the power output:</u>

$P=\frac{W}{t}$

$P=\frac{6.7\times 10^4}{3600}$

$P=18.61\ W$

b.

$\eta_c=1-\frac{T_L}{T_H}$

$\eta_c=1-\frac{378}{1250}$

$\eta_c=0.6976=69.76\%$

c.

now actual efficiency:

$\eta=\frac{W}{Q_H}$

$\eta=\frac{6.7\times 10^4}{1.42\times 10^{5}}$

$\eta=0.4718=47.18\%$

6 0

4 years ago

The launching catapult of the aircraft carrier gives the jet fighter a constant acceleration of 59 m/s2 from rest relative to th

irina [24]

Answer:

Explanation:

We shall find the final velocity of aircraft with respect to aircraft carrier using the following relation.

v² = u² + 2as

v² = 0² + 2 x 59 x 97

v² = 11446

v = 107 m /s

velocity of aircraft carrier = 1.852 x 26 = 48.152 km/h

= 48.152 x 1000 / (60 x 60) m/s

= 13.37 m /s

This velocity of aircraft carrier will be added to the velocity of aircraft .

So absolute velocity of aircraft = 107 m /s + 13.37 m/s

= 120.37 m/s

7 0

3 years ago

I need help PLEASE HEEEEEEEEELPP

lana66690 [7]

Answer:

a = (v2 - v1) / t

From A to B (8 - 4) m/s / 1 s = 4 m / s^2

From A to D ( 7 - 4) m/s / 5 s = .6 m / s^2

Note these equations hold for "uniform" values

They say nothing about the acceleration at intermediate points - the equation just says that his average speed increased from 4 m/s to 7 m/s during a 5 sec period

6 0

3 years ago

A ballast is dropped from a stationary hot-air balloon that is at an altitude of 576 ft. Find (a) an expression for the altitude

Nadya [2.5K]

Answer:

<h2>a) $S = \frac{1}{2}gt^2\\$ </h2><h2>b) 6secs</h2><h2>c) 192ft</h2>

Explanation:

If a ball dropped from a stationary hot-air balloon that is at an altitude of 576 ft, an expression for the altitude of the ballast after t seconds can be expressed using the equation of motion;

$S = ut + \frac{1}{2}at^{2}$

S is the altitude of the ballest

u is the initial velocity

a is the acceleration of the body

t is the time taken to strike the ground

Since the body is dropped from a stationary air balloon, the initial velocity u will be zero i.e u = 0m/s

Also, since the ballast is dropped from a stationary hot-air balloon, the body is under the influence of gravity, the acceleration will become acceleration due to gravity i.e a = +g

Substituting this values into the equation of the motion;

$S = 0 + \frac{1}{2}gt^2\\ S = \frac{1}{2}gt^2\\$

a) An expression for the altitude of the ballast after t seconds is therefore

$S = \frac{1}{2}gt^2\\$

b) Given S = 576ft and g = 32ft/s², substituting this into the formula in (a);

$576 = \frac{1}{2}(32)t^2\\\\\\576*2 = 32t^2\\1152 = 32t^2\\t^2 = \frac{1152}{32} \\t^2 = 36\\t = \sqrt{36}\\ t = 6.0secs$

This means that the ballast strikes the ground after 6secs

c) To get the velocity when it strikes the ground, we will use the equation of motion v = u + gt.

v = 0 + 32(6)

v = 192ft

7 0

4 years ago