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3 years ago

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A sled of mass m is coasting on the icy surface of a frozen river. While it is passing under a bridge, a package of equal mass m

is dropped straight down and lands on the sled (without causing any damage). The sled plus the added load then continue along the original line of motion. How does the kinetic energy of the (sled + load) compare with the original kinetic energy of the sled? A) It is 1/4 the original kinetic energy of the sled. B) It is 1/2 the original kinetic energy of the sled. C) It is 3/4 the original kinetic energy of the sled. D) It is the same as the original kinetic energy of the sled. E) It is twice the original kinetic energy of the sled.

1 answer:

sweet [91]3 years ago

5 0

Answer:B

Explanation:

Given

mass of Sled is m

another package of mass m is thrown on it

Suppose u be the initial velocity of sled

conserving momentum

$mu=2mv$

$v=\frac{u}{2}$

where v is the final velocity

Initial kinetic energy $=\frac{1}{2}mu^2$

Final Kinetic Energy $=\frac{1}{2}(2m)\cdot (\frac{u}{2})^2$

Final Kinetic Energy $=\frac{1}{2}\times \frac{1}{2}mu^2$

Final kinetic Energy is half of initial

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Answer:

Part a)

$f_w = f_g = 4.57 \times 10^{14} Hz$

Part b)

$\lambda_w = 492 nm$

$\lambda_g = 437.3 nm$

Part c)

$v_w = 2.25 \times 10^8 m/s$

$v_g = 2.0 \times 10^8 m/s$

Explanation:

Part a)

frequency of light will not change with change in medium but it will depend on the source only

so here frequency of light will remain same in both water and glass and it will be same as that in air

$f = \frac{v}{\lambda}$

$f = \frac{3 \times 10^8}{656 \times 10^{-9}}$

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Part b)

As we know that the refractive index of water is given as

$\mu_w = 4/3$

so the wavelength in the water medium is given as

$\lambda_w = \frac{\lambda}{\mu_w}$

$\lambda_w = \frac{656 nm}{4/3}$

$\lambda_w = 492 nm$

Similarly the refractive index of glass is given as

$\mu_w = 3/2$

so the wavelength in the glass medium is given as

$\lambda_g = \frac{\lambda}{\mu_g}$

$\lambda_g = \frac{656 nm}{3/2}$

$\lambda_g = 437.3 nm$

Part c)

Speed of the wave in water is given as

$v_w = \frac{c}{\mu_w}$

$v_w = \frac{3 \times 10^8}{4/3}$

$v_w = 2.25 \times 10^8 m/s$

Speed of the wave in glass is given as

$v_g = \frac{c}{\mu_g}$

$v_g = \frac{3 \times 10^8}{3/2}$

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3 years ago

A wet bar of soap slides down a ramp 9.2 m long inclined at 8.0∘ .

xenn [34]

Answer:

The time taken by the bar to reach the bottom t=4.886s

Given:

Displacement of the bar S=9.2m

Angle of inclination $\theta=8.0^{\circ}$

Coefficient of friction factor $\mu k=0.056$

To find:

How long it takes to reach the bottom ‘t’

<u>Step by Step Explanation:</u>

Solution:

We know that the formula for weight of the soap bar is given as

$F_{g}=m g \sin \theta$

The frictional force acting on this soap bar is determined by

$F_{f}=\mu m g \cos \theta$

To determine the constant acceleration of the bar, we derive as

$F=m a$

Here $F=F_{g}-F_{f}$ and thus

$F_{g}-F_{f}=m a$ $m g \sin \theta-\mu m g \cos \theta=m a$

$a=g \sin \theta-\mu g \cos \theta$

Where $F_{g}$ =Force imparted due to weight

$F_{f}$ =Frictional Force

m=Mass of the bar

g=Acceleration due to gravity

a=Acceleration of the bar

$\sin \theta$ and $\cos \theta$ are the angles involved in the system

If the bar starts from the rest

Equations of motion involved in calculating the displacement of the bar is given as

$s=\frac{1}{2} a t^{2}$ , From this

$a t^{2}=2 s$

$t^{2}=\frac{2 s}{a}$

$t=\sqrt{\frac{2 s}{a}}$

Where s= displacement or length moved by the bar

a=Acceleration of the bar

t=Time taken to reach bottom

Substitute all the known values in the above equation we get

$t=\sqrt{\frac{2 \times 9.2}{a}}$ and we know that

$a=g \sin \theta-\mu g \cos \theta$

$=9.8 \times \sin 8.0-0.056 \times 9.8 \times \cos 8.0$

$=1.364-0.543$

$a=0.821$

$t=\sqrt{\frac{2 \times 9.2}{0.821}}$

$t=\sqrt{\frac{19.6}{0.821}}$

$t=\sqrt{23.87332}$

t=4.886s

Result:

Thus the time taken by the bar to reach the bottom is t=4.886s

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3 years ago