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Vedmedyk [2.9K]

3 years ago

10

A skier starts from rest at the top of a hill that is inclined 10.5° with respect to the horizontal. The hillside is 200 m long,

and the coefficient of friction between snow and skis is 0.075. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier glide along the horizontal portion of the snow before coming to rest?

1 answer:

Svetlanka [38]3 years ago

4 0

Answer:

d) 289.31 m

Explanation:

Energy provided by potential energy = mgh = m x 9.8x 200 sin10.5 = 357.18m

Energy used by friction = μmgcos 10.5 x 200 = .075 x m x 9.8 x cos 10.5 x200 = 144.54 m .

Energy used by friction on plain surface = μmg x d.( dis distance covered on plain ) =.075x m x 9.8 xd = .735 m d

To equate

357.18 m -144.54 m = .735 m d

d = 289.31 m .

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A diode, which allows current to flow in one direction only, consists of two types of semiconductors joined together.

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Answer:

True.

Explanation:

A diode, which allows current to flow in one direction only, consists of two types of semiconductors joined together.

A semiconductor can be defined as a crystalline solid substance that has its conductivity lying between that of a metal and an insulator, due to the effects of temperature or an addition of an impurity. Semiconductors are classified into two main categories;

1. Extrinsic semiconductor.

2. Intrinsic semiconductor.

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3 years ago

A student of weight 652 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude

miskamm [114]

Answer:

Explanation:

Given

Weight of person $W=652\ N$

At highest point Magnitude of the normal force $F=585\ N$

net force at highest point

$F_{net}=W+F_c$

where $F_c=$ centripetal force

$F_c=\dfrac{mv^2}{r}$

$F_{net}=F_{n}=$ Normal Force

$585=652+F_c$

$F_c=-67\ N$

Negative sign shows force is in upward direction

At bottom point centripetal force is towards the bottom

$F_n=F_c+W$

$F_n=652+67$

$F_n=719\ N$

8 0

3 years ago

A 125 g pendulum bob hung on a string of length 35 cm has the same period as when the bob is hung from a spring and caused to os

Bingel [31]

Answer:

k = 3.5 N/m

Explanation:

It is given that the time period the bob in pendulum is the same as its time period in spring mass system:

$Time\ Period\ of\ Pendulum = Time\ Period\ of\ Spring-Mass\ System\\2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{m}{k}$

$\frac{l}{g} = \frac{m}{k}\\\\ k = g\frac{m}{l}$

where,

k = spring constant = ?

g = acceleration due to gravity = 9.81 m/s²

m = mass of bob = 125 g = 0.125 kg

l = length of pendulum = 35 cm = 0.35 m

Therefore,

$k = (9.81\ m/s^2)(\frac{0.125\ kg}{0.35\ m})\\\\$

<u>k = 3.5 N/m</u>

4 0

3 years ago

FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine

Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

$\tau=Fd=mg\cdot R$

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

$\tau=I\alpha$

therefore, equating this to the above equation gives

$mg\cdot R=I\alpha$

solving for alpha gives

$\alpha=\frac{mgR}{I}$

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

$\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}$ $\boxed{\alpha=1.47s^{-2}}$

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

$\theta=\frac{1}{2}\alpha t^2$

Therefore, ar t = 4.2 s, the above gives

$\theta=\frac{1}{2}(1.47)(4.2)^2$

$\boxed{\theta=12.97}$

So how many revolutions is this?

To find out we just divide by 2 pi:

$\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}$ $\boxed{\#\text{rev}=2.06}$

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

$\omega^2=w^2_0+2\alpha(\Delta\theta)_{}$

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

$\omega^2=0+2(1.47)(12.97)$ $w^2=38.12$ $\boxed{\omega=6.17.}$

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

3 0

1 year ago

Susan is making an electromagnet in her science class today. First, she takes a nail and winds coils of copper wire around it th

matrenka [14]

Answer:

Electrical

Explanation:

She uses a battery, which is electrical.

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5 0

3 years ago

Read 2 more answers