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2 years ago

12

A fan that is switched on for 1 minute uses 500 W usefully but also wastes 300 W through the emission of sound and heat. What's

the efficiency of the fan?

1 answer:

koban [17]2 years ago

6 0

Explanation:

Efficiency is defined as the ratio between the useful output over the total amount consumed. $E = \frac{useful}{total}$

The fan does 500W of useful work while wasting 300 W. The total power consumption is 800 W (500 + 300).

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1. Two identical bowling balls of mass M and radius R roll side by side at speed v0 along a flat surface. Ball 1 encounters a ra

UNO [17]

Answer:

1/2

Explanation:

We need to make a couple of considerations but basically the problem is solved through the conservation of energy.

I attached a diagram for the two surfaces and begin to make the necessary considerations.

Rough Surface,

We know that force is equal to,

$F_r = mgsin\theta$

$F_r = \mu N$

$F_r = \mu mg cos\theta$

Matching the two equation we have,

$\mu N = \mu mg cos\theta$

$\mu = tan\theta$

Applying energy conservation,

$\frac{1}{2}mv^2_0+\frac{1}{2}I_w^2 = F_r*d+mgh_1$

$\frac{1}{2}mv^2_0+\frac{2}{5}mR^2\frac{V_0^2}{R^2} = F_r*d+mgh_1$

$\frac{1}{2}mv^2_0+\frac{mv_0^2}{5} = mgsin\theta \frac{h_1sin\theta}+mgh_1$

$\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_1+gh_1$

$h_1 = \frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})$

Frictionless surface

$\frac{1}{2}mv_0^2+\frac{1}{2}I\omega^2 = mgh_2$

$\frac{1}{2}m_v^2+\frac{1}{2}\frac{2}{5}mR^2\frac{v_0^2}{R^2} =mgh_2$

$\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_2$

$h_2 = \frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})$

Given the description we apply energy conservation taking into account the inertia of a sphere. Then the relation between $h_1$ and $h_2$ is given by

$\frac{h_1}{h_2} = \frac{\frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})}{\frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})}$

$\frac{h_1}{h_2} = \frac{1}{2}$

8 0

2 years ago

a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres

shusha [124]

Answer:

Part a) When collision is perfectly inelastic

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b) When collision is perfectly elastic

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

$mv_m = (m + M)v$

so we have

$v = \frac{mv_m}{m + M}$

now we know that in order to complete the circle we will have

$v = \sqrt{5Rg}$

$\frac{mv_m}{m + M} = \sqrt{5Rg}$

now we have

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b)

Now we know that collision is perfectly elastic

so we will have

$v = \frac{2mv_m}{m + M}$

now we have

$\sqrt{5Rg} = \frac{2mv_m}{m + M}$

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

6 0

2 years ago

A pendulum is made of a small sphere of mass 0.250 kg attached to a lightweight string 1.20 m in length. As the pendulum swings

forsale [732]

Answer:v=2 m/s

Explanation:

Given

Length of string L=1.2 m

mass of pendulum m=0.25 kg

maximum inclination with vertical \theta =34

vertical Rise of Pendulum from its mean position is given by

$\Delta h=L(1-\cos \theta )$

Conserving Energy at top and bottom point

Potential Energy of sphere is converted into kinetic energy of sphere

$mgL(1-\cos \theta )=\frac{mv^2}{2}$

$v=\sqrt{2gL(1-\cos \theta )}$

$v=\sqrt{2\times 9.8\times 1.2(1-\cos 34)}$

$v=\sqrt{4.021}$

$v=2 m/s$

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2 years ago

1. A 930-kg car traveling 56 km/h comes to a complete stop in 2.0 s. What is the

Juli2301 [7.4K]

The force exerted on the car during this stop is 6975N

<u>Explanation:</u>

Given-

Mass, m = 930kg

Speed, s = 56km/hr = 56 X 5/18 m/s = 15m/s

Time, t = 2s

Force, F = ?

F = m X a

F = m X s/t

F = 930 X 15/2

F = 6975N

Therefore, the force exerted on the car during this stop is 6975N

6 0

2 years ago

Imagine you derive the following expression by analyzing the physics of a particular system: M= (mv2r)(mGr2). Simplify the expre

alex41 [277]

Answer:

The simplified expression is $M = \frac{v^2 r}{G}$

Explanation:

From the question we are told that

$M = \frac{ \frac{m v^2}{r} }{\frac{ mG}{r^2 } }$

So simplifying we have

$M = \frac{m v^2}{r} * \frac{r^2 }{ mG }$

$M = \frac{v^2 r}{G}$

Thus the simplified formula is $M = \frac{v^2 r}{G}$

3 0

2 years ago