Physics College

A skier is traveling down a slope that is inclined 30 Â° above the horizontal. The coefficient of kinetic friction between the skier and the slope is 0.10. Which of the following best describes the acceleration of the skier?
answer choices
0
4.0
9.0
10.0

Answers

Answer 1

The skier's acceleration down the slope is 4.0 (Second option)

This kinetic frictional force acts on an object that is in motion.

Here is the formula for kinetic friction:

F = μk x N

Information:

F: magnitude of kinetic friction (N)

μk : coefficient of kinetic friction (N)

N: normal force (N)

In this problem, it is known that the coefficient of kinetic friction is 0.1 and the downhill skier forms an angle of 30°, then:

N = mg cosθ

F = m x a

So:

Resultant force = mg sinθ - 0.1 x mg cosθ

m x a = mg sinθ - 0.1 x mg cosθ

a = m (g sinθ - 0.1 x g cosθ) ÷ m

a = g sinθ - 0.1 x g cosθ

a = (9.8 x 0.5) - (0.1 x 9.8 x 0.866)

a = 4.9 - 0,848

a = 4.052 ≈ 4.0 m/s²

So, the acceleration is 4.0 m/s²

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A Skier Is Traveling Down A Slope That Is Inclined 30 Above The Horizontal. The Coefficient Of Kinetic

Related Questions

A wheel of radios 0.5m initialy at rest with a constant angular acceleration and attains a linear speed of 30 m/s in 5 sec. calculate
a) the angular acceleration?
b) the angular velocit's after 4 sec?
c) the number of revolution at a time of 5sec?

Answers

Answer:

a.) 12[tex]\frac{rad}{s^2}[/tex]

b.) 48[tex]\frac{rad}{s}[/tex]

c.) 23.87 rev

Explanation:

First, convert linear velocity to angular velocity. The angular velocity can be calculated using the relationship v=rω, where v is the linear velocity, r is the radius, and ω is the angular velocity. Rearrange the equation to solve for ω:

[tex]\omega=\frac{v}{r}[/tex]

For this problem, let v=30[tex]\frac{m}{s}[/tex] and r=0.5m. So,

[tex]\omega=\frac{30\frac{m}{s}}{0.5m}\\\omega=60\frac{rad}{s}[/tex]

Next, to solve for angular acceleration use rotational motion equation #1, [tex]\omega_f=\omega_i+\alpha t[/tex]. For this problem, let

[tex]\omega_i=0.0\frac{rad}{s}\\\omega_f=60\frac{rad}{s}\\t=5s[/tex]

So,

[tex]60\frac{rad}{s}=0.0\frac{rad}{s}+\alpha(5s)\\(\frac{1}{5s})60\frac{rad}{s}=\alpha(5s)(\frac{1}{5s})\\12\frac{rad}{s^2}=\alpha[/tex]

Next, use the angular acceleration and the time given in part b in rotational motion equation #1 to solve for the angular velocity at t=4s:

[tex]\omega_f=\omega_i+\alpha t\\\omega_f=0.0\frac{rad}{s}+(12\frac{rad}{s^2})(4s)\\\omega_f=48\frac{rad}{s}[/tex]

To find the number of revolutions, use rotational motion equation #3, [tex]\theta=v_it+\frac{1}{2}\alpha t^2[/tex], to find the angular displacement at t=5s. So,

[tex]\theta=0.0\frac{m}{s}(5s)+\frac{1}{2}(12.0\frac{rad}{s^2})(5s)^2\\\theta=0.0m+150\ rad\\\theta=150\ rad[/tex]

To convert the angular displacement in radians to revolutions, recall that one revolution equals 2π radians. So,

[tex]150\ rad(\frac{1\ rev}{2\pi\ rad})=\frac{75}{\pi}\ rev\approx23.87\ rev[/tex]

which of the following best describes the strength of coriolis deflection where isobars are very close together?

Answers

b) The strength of the apparent deflection varies with speed of Earth's rotation best describes the strength of Coriolis deflection where isobars are very close together.

The Coriolis deflection is a phenomenon of the Earth's rotation that affects the movement of air masses and ocean currents. The strength of the apparent deflection of these air and water masses varies with the speed of the Earth's rotation. Hence, option b is correct.

This is particularly noticeable when isobars, or lines of equal pressure, are very close together. The Coriolis deflection causes the air or water to be deflected to the right in the Northern Hemisphere and to the left in the Southern Hemisphere.

This deflection can result in the formation of cyclonic weather systems and gyres in the ocean.

Although a part of your question is missing, you might be referring to this question:

Which of the following best describes the strength of coriolis deflection where isobars are very close together?

a) Slowly moving objects are deflected more than rapidly moving objects.

b) The strength of the apparent deflection varies with speed of Earth's rotation.

c)Coriolis deflection occurs only along parallels, not meridians.

d) The Coriolis force is zero at the poles, increasing to maximum along the equator.

e)The amount of Coriolis deflection is uniform from equator to poles.

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HELPOOOAOWOWKAKSJWHIndbrhajsjehaha

Answers

In what you need help????

Which property of concave mirror is used in telescope?

Answers

Concave mirrors are widely used in different reflecting telescopes in observational astronomy. The concave mirror in a reflecting mirror collects the light from distant celestial objects. Since the light sources are very far away, the incoming light rays are effectively parallel.

Answer: reflection

Explanation: the concave mirror will gather light from distant objects outside of the Earth's atmosphere. since these celestial objects are so far, the rays of light are parallel, and the mirror will reflect it at the focus point, and the image displayed depends on how far away the object is.

Phillipe is being asked to give testimony. What will this responsibility MOST likely include?
O A. He will donate money to charity.
OB.
O C.
O D.
He will help make peace in his community.
He will take an oath to tell the truth.
He will ignore evidence and say how he feels.
Reset
Next

Answers

He will reportedly swear an oath to speak the truth for this obligation, which MOST likely includes.

What types of proof are there?

Fingerprints, blood test, DNA, a knife, a pistol, and other tangible things are examples of genuine evidence. Real evidence is often allowed into court because it has the power to support or refute a claim of fact during a trial.

Why is testimony crucial in court?

The purpose of evidence is to persuade the jury or judge of the issues involved. Additionally, evidence might be a tangible object or any declarations of fact that were already made at trial to prove the veracity of the subject of the inquiry.

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proton (q 1.6 X10-19 C,m = 1.67 X 10-27 kg) moving with constant velocity enters region containing constant magnetic field that is directed along the z-axis at (x,y) (0,0) as shown The magnetic field extends for distance 0.48 m in the x-direction: The proton leaves the field having velocity vector (5.8 X 105 m/s, 3.2 X 105 m/s).
1) What is the magnitude of the velocity of the proton as it entered the region containing the magnetic field? ______m/s Submit
2) What is R, the radius of curvature of the motion of the proton while it is in the region containing the magnetic field? ______m/s Submit
3) What is h, the co-ordinate of the proton as it leaves the region conating the magnetic field? ______m/s Submit
4)What is Bz, the z-component of the magnetic field? Note that Bz is signed number . ______m/s Submit
5) If the incident velocity were increased_ how would and change if at all? h and would both increase and would both decrease would increase and would decrease would decrease and would increase Neither h nor would change. ______m/s Submit

Answers

Magnetic Field:

A magnetic field is the vector field that depicts how the magnetic force is distributed within or around a certain magnetic substance. The magnetic field, which consists of closed loops, travels from the North Pole to the South Pole.

Given Data:

The charge on the proton is:

q = 1.6 × 10⁻¹⁹ C

The mass of the proton is:

m = 1.67 × 10⁻²⁷ Kg

The distance by which the magnetic field extends is:

D = 0.44 m

The velocity of proton in x direction is:

vₓ = 5.1 × 10⁵ m/s

The velocity of proton in y direction is:

v = 2.7 × 10⁵ m/s

Answer and Explanation:

(1):The expression for the resultant velocity when the proton enter the magnetic field is,

[tex]v = \sqrt{v_{x^2} +v_{y^2}}[/tex]