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3 years ago

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Which are the components of a typical refracting telescope? hints?

1 answer:

Ivenika [448]3 years ago

4 0

<span>covex object and concave eyepiece.</span>

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A 80 kg skier grips a moving rope that is powered by an engine and is pulled at constant speed to the top of a 25 degrees hill.

zloy xaker [14]

Answer:

$P=28.085\,hp$

Explanation:

Given that:

mass of 1 skier, $m=80kg$

inclination of hill, $\theta=25^{\circ}$

length of inclined slope, $l=220m$

time taken to reach the top of hill, $t=2.3 min= 138 s$

coefficient of friction, $\mu=0.15$

<em>Now, force normal to the inclined plane:</em>

$F_N=m.g.cos\theta$

$F_N=80\times 9.8\times cos25^{\circ}$

$F_N=710.54\,N$

<em>Frictional force:</em>

$f=\mu.F_N$

$f=0.15\times 710.54$

$f=106.58\,N$

<em>The component of weight along the inclined plane:</em>

$W_l=m.g.sin\theta$

$W_l=80\times 9.8\times sin25^{\circ}$

$W_l=331.33\,N$

<em>Now the total force required along the inclination to move at the top of hill:</em>

$F=f+W_l$

$F=106.58+331.33$

$F=437.91\,N$

<em>Hence the work done:</em>

$W=F.l$

$W=437.91\times 220$

$W=96340.80\,J$

<em>Now power:</em>

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

$P=\frac{96340.80}{138}$

$P=698.12\,W$

<u>So, power required for 30 such bodies:</u>

$P=30\times 698.12$

$P=20943.65\,W$

$P=\frac{20943.65}{745.7}$

$P=28.085\,hp$

8 0

3 years ago

A hoop, a solid disk, and a solid sphere, all with the same mass and the same radius, are set rolling without slipping up an inc

myrzilka [38]

Answer:

The sphere

Explanation:

Because it has a smaller inertia (I) value in the explanation in the attached file

6 0

3 years ago

Is this a scam/virus?

kaheart [24]

Answer:

Yes, I'm pretty sure it is. That's why I don't click on it!

6 0

3 years ago

In an RL series circuit, an inductor of 3.54 H and a resistor of 7.76 Ω are connected to a 26.6 V battery. The switch of the cir

podryga [215]

Answer:

Energy stored in inductor will be 20.797 J

Explanation:

We have given inductance L = 3.54 H

And resistance R = 7.76 ohm

Battery voltage V = 26.6 VOLT

After very long time means at steady state inductor behaves as short circuit

So current $i=\frac{V}{R}=\frac{26.6}{7.76}=3.427Amp$

Now energy stored in inductor $E=\frac{1}{2}Li^2=\frac{1}{2}\times 3.54\times 3.427^2=20.797J$

So energy stored in inductor will be 20.797 J

4 0

4 years ago

A 70.0-kg skier is sliding at 4 m/s when they slide down a 2m high hill. At the bottom of the hill they run into a large 2800 N/

Katarina [22]

Answer:

$\Delta x=245\ mm$

Explanation:

Given:

mass of skier, $m=70\ kg$

initial velocity of skier, $u=4\ m.s^{-1}$

height of the hill, $h=2\ m$

spring constant, $k=2800\ N.m^{-1}$

<u>final velocity of skier before coming in contact of spring:</u>

Using eq. of motion:

$v^2=u^2+gh$

$v^2=4^2+9.8\times 2$

$v=5.9666\ m.s^{-1}$

<u>Now the time taken by the skier to reach down:</u>

$v=u+gt$

$5.9666=4+9.8\ t$

$t=0.2007\ s$

<u>Now we calculate force using Newton's second law:</u>

$F=\frac{dp}{dt}$

$F=\frac{m(v-u)}{t}$

$F=\frac{70\times(5.9666-4)}{0.2007}$

$F\approx686\ N$

<u>∴Compression in spring before the skier momentarily comes to rest:</u>

$\Delta x=\frac{F}{k}$

$\Delta x=\frac{686}{2800}$

$\Delta x=0.245\ m$

$\Delta x=245\ mm$

4 0

3 years ago