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2 years ago

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A 20.0 kg rock is sliding on a rough , horizontal surface at8.00 m/s and eventually stops due to friction .The coefficient ofkin

etic fricction between the rock and the surface is 0.20
what average thermal power is produced as the rock stops?

2 answers:

ra1l [238]2 years ago

3 0

Answer: The power is 156 watt

Explanation:

is in the attachment

Shtirlitz [24]2 years ago

3 0

Explanation:

Below is an attachment containing the solution.

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A truck with a heavy load has a total mass of 7100 kg. It is climbing a 15∘ incline at a steady 15 m/s when, unfortunately, the

Andrej [43]

Answer:

The load has a mass of 2636.8 kg

Explanation:

Step 1 : Data given

Mass of the truck = 7100 kg

Angle = 15°

velocity = 15m/s

Acceleration = 1.5 m/s²

Mass of truck = m1 kg

Mass of load = m2 kg

Thrust from engine = T

Step 2:

⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:

T = (m1+m2)*g*sinθ

⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes m1*gsinθ .

Resultant force on truck is F = T – m1*gsinθ

F causes the acceleration of the truck: F= m*a

This gives the equation:

T – m1*gsinθ = m1*a

T = m1(a + gsinθ)

Combining both equations gives:

(m1+m2)*g*sinθ = m1*(a + gsinθ)

m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ

m2*g*sinθ = m1*a

Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:

m2*g*sinθ = (7100 – m2)*a

m2*g*sinθ = 7100a – m2a

m2*gsinθ + m2*a = 7100a

m2* (gsinθ + a) = 7100a

m2 = 7100a/(gsinθ + a)

m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)

m2 = 2636.8 kg

The load has a mass of 2636.8 kg

6 0

2 years ago

An electric buzzer is activated, then sealed inside a glass chamber. When all of the air is pumped out of the chamber, how is th

Aleksandr-060686 [28]

The sound is increased because sound waves are in fact mech. waves which means the that they can't travel through empty space and thus need a medium to travel through

4 0

2 years ago

Air at 27 ºC and 5 atm is throttled by a valve to 1 atm. If the valve is adiabatic and the change in kinetic energy is negligibl

Citrus2011 [14]

Answer:

27 °C

Explanation:

BY the statement of the question it is clear that it is about an ideal gas - and hence if change in KE is about zero - then there will be no change of temperature.

So, answer is 27 °C

4 0

2 years ago

A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w

cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

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

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

$W=F.x$

The second Newton's law gives us the net force as

$F=m.a$

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

$v_f=v_o+at$

$v_f=at$

Solving for a

$\displaystyle a=\frac{v_f}{t}={5.4}{1}$

$a=5.4\ m/s^2$

The distance traveled in the interval is given by

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

Since vo=0

$\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}$

$x=2.7\ m$

The force is given by

$F=m.a$

We don't know the value of m, so the force is

$F=2.7m$

Computing the work done by the sprinter

$W=F.x=2.7m(5.4)$

$W=14.58m$

The power is finally computed

$\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}$

$P=14.58m$

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

$\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}$

$a=8.8\ m/s^2$

The distance is

$\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}$

$x=3.8\ m$

The net force is

$F=m(8.8)=8.8m$

The work done by the sprinter is now computed as

$W=8.8m(3.8)=33.44m$

At last, the output power is

$\displaystyle P=\frac{33.44m}{0.5}=66.88m$

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0

2 years ago

.. A 15.0-kg fish swimming at 1.10 m>s suddenly gobbles up a 4.50-kg fish that is initially stationary. Ignore any drag effec

stira [4]

Answer:

(a) 0.846 m/s

(b) 2.097J

Explanation:

Parameters given:

Mass of big fish, M = 15 kg

Mass of small fish, m = 4.5 kg

Initial speed of big fish, U = 1.1 m/s

Initial speed of small fish, u = 0 m/s (it is stationary)

(a) We apply the principle of conservation of momentum:

Total initial momentum = Total final momentum

Since both fish have the same final speed, V, (the small fish is in the mouth of the big fish), we have:

MU + mu = (M + m)*V

(15 * 1.1) + (4.5 * 0) = ( 15 + 4.5) * V

16.5 = 19.5V

=> V = 16.5/19.5

V = 0.846 m/s

The speed of the large fish after the meal is 0.846 m/s.

(b) We need to find the change in Kinetic energy of the entire system to find the total mechanical energy dissipated.

Initial Kinetic energy:

KEini = (½ * M * U²) + (½ * m * u²)

KEini = (½ * 15 * 1.1²) + (½ * 4.5 * 0²)

KEini = 9.075 J

Final Kinetic Energy:

KEfin = (½ * M * V²) + (½ * m * V²)

KEfin = (½ * 15 * 0.846²) + (½ * 4.5 * 0.846²)

KEfin = 5.368 + 1.610 = 6.978 J

Change in kinetic energy will be:

KEfin - KEini = 9.075 - 6.978

ΔKE = 2.097 J

The energy dissipated in eating the meal is 2.097 J

5 0

2 years ago