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elixir [45]
3 years ago
15

A 78−kg skier is sliding down a ski slope at a constant velocity. The slope makes an angle of 21° above the horizontal direction

. (a) Neglecting any air resistance, what is the force of kinetic friction acting on the skier? (b) What is the coefficient of kinetic friction between the skis and the snow?
Physics
1 answer:
Feliz [49]3 years ago
8 0

Answer:

274N 0.41

Explanation:

As he is sliding down in a constant speed then the force that accelerates him (weight) and the force that slows his down (friction) are equal.

then

<em>friction=mass x gravity x sin(21)</em>

Fr=78kg x 9.8m/s2 x sin(21)=274N

<em>friction= coefficient of kinetic friction x normal force of from the slope</em>

Fr= u x 78kg x 9.8m/s2 x cos(21)=274N

Fr= u x 78kg x 9.8m/s2 x cos(21)=274Nu=274/677=0.41

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8 0
3 years ago
. Consider a fully extended arm that is rotating about the shoulder such that with a shoulder-to-hand length of 30 cm. If the ar
Rainbow [258]

Answer:

13.309 m/s²

Explanation:

Length from shoulder to hand, l = 30 cm = 0.3 m

initial velocity, u = 1 m/s

final velocity, v = 2.5 m/s

time, t = 3 s

Let the tangential acceleration is a.

by using first equation of motion

v = u + at

2.5 = 1 + 3 a

a = 0.5 m/s²

Let the centripetal acceleration is a'.

a' = v'²/l

a' = 2 x 2 / 0.3

a' = 13.3 m/s²

The tangential acceleration and the centripetal acceleration are both perpendicular to each other. So, the net acceleration is given by

A=\sqrt{a^{2}+a'^{2}}

A=\sqrt{0.5^{2}+13.3^{2}}

A = 13.309 m/s²

3 0
3 years ago
In a lever, the effort arm is two times as long as the load arm. The resultant force will be
jasenka [17]
Answer is B. 

In a lever, the effort arm is 2 times as a long as the load arm. The resultant force will be twice the applied force.

Hope it helped you.

-
Charlie
5 0
3 years ago
Read 2 more answers
Which term best describes the basic unit that makes up all matter?
neonofarm [45]

Answer:

C. Atoms

Explanation:

The basic unit of matter and the smallest, indivisible unit of a chemical element. It comprises a nucleus (neutrons + protons) that is surrounded by a cloud of electrons.

6 0
2 years ago
We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top
notka56 [123]

Answer:

A,)FD= 114.1N

B)Torque=798.5Nm

Explanation:

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top of the tree exerts a horizontal force, and thus a torque that can topple the tree if there is no opposing torque. Suppose a tree's canopy presents an area of 9.0 m^2 to the wind centered at a height of 7.0 m above the ground. (These are reasonable values for forest trees.)

If the wind blows at 6.5 m/s, what is the magnitude of the drag force of the wind on the canopy? Assume a drag coefficient of 0.50 and the density of air of 1.2 kg/m^3

B)What torque does this force exert on the tree, measured about the point where the trunk meets the ground?

A)The equation of Drag force equation can be expressed below,

FD =[ CD × A × ρ × (v^2/ 2)]

Where CD= Drag coefficient for cone-shape = 0.5

ρ = Density

Area of of the tree canopy = 9.0 m^2

density of air of = 1.2 kg/m^3

V= wind velocity= 6.5 m/s,

If we substitute those values to the equation, we have;

FD =[ CD × A × ρ × (v^2/ 2)]

F= [ 0.5 × 9.0 m^2 × 1.2 kg/m^3 ( 6.5 m/s/ 2)]

FD= 114.1N

B) the torque can be calculated using below formula below

Torque= (Force × distance)

= 114.1 × 7

= 798.5Nm

8 0
3 years ago
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