All categories

3 years ago

A sled is slowing down at the bottom of a snowy hill.

Physics

1 answer:

Kitty [74]3 years ago

5 0

Answer:

A sled and its rider are moving at a speed of along a horizontal stretch of snow, as Figure 4.24a illustrates. The snow exerts a kinetic frictional force on the runners of the sled, so the sled slows down and eventually comes to a stop. The coefficient of kinetic friction is 0.050. What is the displacement x of the sled?

You might be interested in

A man is trying to push a 250 N dresser across his carpeted bedroom. He applies a force of 20 N and the carpet provides a fricti

11111nata11111 [884]

Answer:

The correct option is;

B. Subtract vectors A from C; $F_{Net}$ = 12 N

Explanation:

The given parameters are;

The weight of the dresser = 250 N

The force applied by the man = 20 N = C

The frictional force provided by the carpet = 8 N = The component of the weight resisting motion = A

Therefore, the net force tending to put the dresser in motion, $F_{Net}$ , is given as follows;

$F_{Net}$ = The force applied by the man, C - The frictional force provided by the carpet which is the component of the weight resisting motion, A

$F_{Net}$ = C - A = 20 N - 8 N = 12 N

$F_{Net}$ = 12 N

The resulting net force, $F_{Net}$ = 12 N.

3 0

3 years ago

Wagon wheel. While working on your latest novel about settlers crossing the Great Plains in a wagon train, you get into an argum

OverLord2011 [107]

Answer:

I = 16.7kgm²

Explanation:

Since, Torque is given by,

\tau = F*r = I*\alpha

here, I = Moment of inertia = ??

\alpha = angular acceleration of wheel = a/r

F = tangential tension acting on the wheel = T

a = acceleration of bag of sand = 2.95 m/s^2

r = radius of wheel = d/2 = 120/2 = 60 cm = 0.60 m

from force balance on sand bag,

mg - T = m*a

T = m*(g-a)

m = mass of sand bag = 20 kg

So, I = T*r/\alpha = m*(g-a)*r/(a/r)

Using known values:

I = 20*(9.81 - 2.95)*0.60/(2.95/0.60) = 16.74

I = 16.7 kgm² = Moment of inertia of wheel experimentally

also, Moment of inertia of wheel theoretically(I') = M*r²

given, M = mass of wheel = 70 kg

I' = 70*0.60²= 25.2 kgm² = Moment of inertia of wheel theoretically

7 0

3 years ago

You are moving into an apartment and take the elevator to the 6th floor. Suppose your weight is 660 N and that of your belonging

Ivan

Answer:

Explanation:

Total weight

My weight+weight of belongings

660+1100=1760N.

a. Work done by the elevator to travel a total height of 15.2m

Using newton law of motion

ΣF = ma

There are only two forces acting upward, the weight and the reaction by the elevator

Also note it is moving at constant velocity then, a=0

N - W=0

Then, N=W

N=1760N

So, workdone is given as

Wordone, =force × distance

Work done=1760×15.2

W=26,752J

W=26.752KJ

b. Work done on me alone is still need to go through the same process but will remove the weight of the belonging

Therefore,

Weight now = 660N

And using the same equation of motion

ΣF = ma

Comstant velocity, a=0

N - W=0

N=W

N=660N

Then, workdone

W=F×d

W=660×15.2

W=10,032J

W=10.032KJ

6 0

4 years ago

Read 2 more answers

g When a thin-filament light bulb is connected to two 1.6 V batteries in series, the current is 0.075 A. What is the resistance

larisa [96]

Answer:

R = 42.67 ohms

Explanation:

It is given that, a thin-filament light bulb is connected to two 1.6 V batteries in series, the current is 0.075 A.

It means that when two batteries are connected in series, then the total voltage is 3.2 volts

Let R is the resistance of the glowing thin-filament bulb. So, using Ohm's law we get :

$V=IR\\\\R=\dfrac{V}{I}$

So,

$R=\dfrac{3.2}{0.075}\\\\R=42.67\ \Omega$

So, the resistance of the bulb is 42.67 ohms.

5 0

3 years ago

Which classification of property does not affect density?

Aleks [24]

Intensive property is the classification of property that does not affect density

5 0

3 years ago