Ask question

Ask question

All categories

2 years ago

7

Two skiers start at the same place at the top of the hill, and finish at the same place at the bottom. Skier A takes a straight,

smooth route to the finish whereas Skier B takes a curvy, bumpy route to the finish. If you assume that friction is negligible, which of the following statements is true? A) Skier A has the same speed as Skier B at the finish. B) Skier B has the greater speed at the finish C) Skier A has the greater speed at the finish.

1 answer:

Rus_ich [418]2 years ago

6 0

Answer:A

Explanation:

Given

Skier A takes straight, smooth route while Skier B takes curvy, bumpy route to finish.

We know gravity is conservative in nature i.e. work done against it is independent of path followed

So work done by gravity for A and B is same

According to Work energy Theorem Change in the kinetic Energy is equal to work done by all the forces.

$W_g=\frac{1}{2}mv^2$

$W_g=mgh$

where h=vertical height of hill

thus $v=\sqrt{2gh}$

which is same for both A and B

You might be interested in

Two narrow slits 0.02 mm apart are illuminated by light from a CuAr laser (λ = 633 nm) onto a screen. a) If the first fringe is

Masja [62]

Answer:

a) 0.063m

b) 2.72°

c) 3151 fringes

d) 1.87*10^-6m

Explanation:

a) To find the screen distance you use the following formula:

$y=\frac{m\lambda D}{d}\\\\D=\frac{dy}{m\lambda}$

D: screen distance

d: distance between slits

m: order of the fringes

λ: wavelength

By replacing the values of the parameters you obtain:

$D=\frac{(0.02*10^{-3}m)(0.2*10^{-2}m)}{(1)(633*10^{-9}m)}=0.063m$

b) The condition for dark fringes is given by:

$\lambda(m+\frac{1}{2})=dsin\theta$

for the first dark fringe the angle is:

$\theta=sin^{-1}(\frac{\lambda(m+\frac{1}{2})}{d})\\\\\theta=sin^{-1}(\frac{(633*10^{-9}m)(1+\frac{1}{2})}{0.02*10^{-3}m})=2.72\°$

c) the visible number of fringes is given by:

$N=1+2\frac{D}{d}=1+\frac{0.063m}{0.02*10^{-3}}=3151 \ fringes$

d) the wavelength of a laser in which its first order fringe coincides with the third one of the CuAr laser is:

$y=\frac{(3)(633*10^{-9}m)(0.063m)}{0.02*10^{-3}m}=5.98*10^{-3}m\approx0.59cm\\\\\lambda'=\frac{dy}{mD}=\frac{(0.02*10^{-3}m)(0.59*10^{-2}m)}{(1)(0.063m)}=1.87*10^{-6}m$

4 0

2 years ago

Through which Earth layer are S waves not transmitted?

Rus_ich [418]

S-waves<span> passing from the </span>mantle<span> to the core are absorbed because shear </span>waves<span>cannot be </span>transmitted through<span> liquids.

Hope this helps!</span>

4 0

3 years ago

Read 2 more answers

In the photo, a locomotive has broken through the wall of a train station. During the collision, what can be said about the forc

Pavlova-9 [17]

Answer:

Whether the force exerted by the locomotive on the wall was larger

Than the force the locomotive could exert on the wall.

Explanation:

The Newton's third law of motion States that every force have it's equal and opposite reaction force, whose magnitude is the same as the applied force. Therefore the magnitude of these opposite forces will be equal.

So we have;

F12=-F21

F12 is the force in a direction

-F21 is the force in the opposite direction.

Therefore we see that the magnitude of the force the locomotive exerts on the wall is equal to the one the wall exerts on the locomotive. Both magnitudes are equal but in opposite directions.

6 0

3 years ago

A blow-dryer and a vacuum cleaner each operate with a voltage of 120 V. The current rating of the blow-dryer is 10 A, and that o

Katyanochek1 [597]

Answer:

Part a)

P = 1200 Watt

Part b)

P = 480 W

Part c)

$\frac{E_{bd}}{E_{vc}} = 5 : 2$

Explanation:

Part a)

Power consumed by blow dryer is given as

$P = iV$

here supply voltage is given as

$V = 120 volts$

current rating of blow dryer is given as

$i = 10 A$

$P = (120 V)(10 A)$

$P = 1200 W$

Part b)

Power consumed by vacuum cleaner is given as

$P = iV$

here supply voltage is given as

$V = 120 volts$

current rating of vacuum cleaner is given as

$i = 4 A$

$P = (120 V)(4 A)$

$P = 480 W$

Part c)

Energy consumed in a given interval of time is given as

$Energy = power \times time$

now we have to find the ratio of energy consumed in time interval of 11 minutes

so we have

$\frac{E_{bd}}{E_{vc}} = \frac{P_{bd} t}{P_{vc} t}$

now we have

$\frac{E_{bd}}{E_{vc}} = \frac{1200 W\times 11 min}{480 W \times 11 min}$

$\frac{E_{bd}}{E_{vc}} = 5 : 2$

5 0

3 years ago

When is the particle in figure (a) speeding up? (enter your answer using interval notation.)

worty [1.4K]

Answer:

When your mom walks in the room, lol. I don't see a "figure A"

:D

4 0

2 years ago