A cook preparing a meal for a group of people is an example of

Sam, whose mass is 78 kg , stands at the top of a 11-m-high, 110-m-long snow-covered slope. His skis have a coefficient of kinet

Valentin [98]

Answer:

v = 8.09 m/s

Explanation:

For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.

Let's calculate the energy

starting point. Higher

Em₀ = U = m gh

final point. To go down the slope

Em_f = K = ½ m v²

The work of the friction force is

W = fr L cos 180

to find the friction force let's use Newton's second law

Axis y

N - W_y = 0

N = W_y

X axis

Wₓ - fr = ma

let's use trigonometry

sin θ = y / L

sin θ = 11/110 = 0.1

θ = sin⁻¹ 0.1

θ = 5.74º

sin 5.74 = Wₓ / W

cos 5.74 = W_y / W

Wₓ = W sin 5.74

W_y = W cos 5.74

the formula for the friction force is

fr = μ N

fr = μ W cos θ

Work is friction force is

W_fr = - μ W L cos θ

Let's use the relationship of work with energy

W + ΔU = ΔK

-μ mg L cos 5.74 + (mgh - 0) = 0 - ½ m v²

v² = - 2 μ g L cos 5.74 +2 (gh)

v² = 2gh - 2 μ gL cos 5.74

let's calculate

v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74

v² = 215.6 -150.16

v = √65.44

v = 8.09 m/s

6 0

3 years ago

Whenever possible, these types of power plants should be built because they generate virtually no pollution. (2 points)

geniusboy [140]

Wind farms, solar collectors, and nuclear power plants generate no pollution.

But wind farms make noise, wind and solar are both hazardous to birds, and there's no known means of disposing of nuclear plant waste that's completely safe.

8 0

3 years ago

Why can you see your own reflected image in a mirror but not on a dry, painted wall?

kaheart [24]

A wall uses diffuse reflection while a mirror uses specular reflection. For example, when parallel light rays enter a mirror, they remain parallel when exiting the mirror, allowing you to see a reflection of the light rays. On the contrary, when incident light rays enter a wall which is painted, the rays scatter, not allowing you to see anything but a painted wall.

7 0

3 years ago

Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t

Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

$v^2=v_0^2+2ax\\$

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

= 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

$a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}$

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

$F_f=ma=(1490kg)(5.55m/s^2)=8271.15N$

Furthermore, you use the relation between the friction force and the friction coefficient:

$F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56$

hence, the friction coefficient is 0.56

6 0

3 years ago

A)

Savatey [412]

Answer:

Distance covered is equal to all the distance traveled.

So for example, if you go from A to B, and then from B to C, the total distance covered is AB + BC.

Displacement is equal to the difference between the final position and the initial position.

So if we go from A to B, the displacement is simply the line AB.

While if we go from A to B, and then from B to C, the displacement will be a segment that directly connects A and C, such that:

displacement = √( (AB)^2 + (BC)^2)

Now, if we want to find the points such that the magnitude of the distance covered is equal to the magnitude of the displacement, we need to look at the pairs that are directly connected by a straight line.

Those are:

A to B ( or B to A)

B to C (or C to B)

C to D (or D to C)

7 0

3 years ago