An alfisol is most closely related to _______

A skier with a 65kg mass skies down a 30 degree incline hill. The coefficient of friction is 0.1.

RUDIKE [14]

Part a.
Refer to the diagram shown below.

m = 65 kg, the mass of the skier
θ = 30°, the angle of the incline
μ = 0.1, the coefficient of friction
W = mg , the weight of the skier
N = mg cosθ, the normal reaction
F = mg sinθ, the component of the weight acting down the hill
R = μN, the frictional force resisting motion down the plane
a = the acceleration down the plane.

The weight is
W = mg = (65 kg)*(9.8 m/s) = 637 N

Part b.
The normal force is
N = (637 N)*cos(30) = 551.658 N

The frictional force is
R = 0.1(551.658 N) = 55.1658 N

Part c.
Calculate the component of the weight acting down the hill.
F = mg sin(30) = (637 N)*sin(30) = 318.5 N

The equation of motion is
ma = F - R
(65 kg)*(a m/s²) = 318.5 - 55.1658 N
a = 263.3342/65 = 4.05 m/s²

Answers:
N = 551.7 N
a = 4.05 m/s²

4 0

4 years ago

Please help, it's grade 7 science. :)

xeze [42]

Answer:

The air in the soccer ball in cold weather will decrease slightly in size and it becomes flat. The air in the soccer ball in hot weather will seem flat because the low preasure leads to lower bounce in the ball.

The metal door frame in cold weather contracts and the wood contracts more in the winter. The metal door frame in hot weather thermal blowing can occur on the outer surface of the metal door frame. Hopefully that is what you were looking for have a good day.

4 0

3 years ago

Read 2 more answers

A cylindrical rod 21.5 cm long with a mass of 1.20 kg and a radius of 1.50 cm has a ball of diameter of 6.90 cm and a mass of 2.

Sergeeva-Olga [200]

Answer

given,

length of rod = 21.5 cm = 0.215 m

mass of rod (m) = 1.2 Kg

radius, r = 1.50

mass of ball, M = 2 Kg

radius of ball, r = 6.90/2 = 3.45 cm = 0.0345 m

considering the rod is thin

$I = \dfrac{1}{3}M_{rod}L^2 + [\dfrac{2}{5}M_{ball}R^2+M_{ball}(R+L)^2]$

$I = \dfrac{1}{3}\times 1.2 \times 0.215^2 + [\dfrac{2}{5}\times 2 \times 0.0345^2+2\times (0.0345 +0.215)^2]$

I = 0.144 kg.m²

rotational kinetic energy of the rod is equal to

$KE = M_{rod}g\dfrac{L}{2} + M_{ball}g(L+R)^2$

$KE = 1.2 \times 9.8 \times \dfrac{0.215}{2} + 2\times 9.8\times (0.215+0.0345)^2$

KE = 6.15 J

b) using conservation of energy

$K_f + U_f = K_i + U_i + \Delta E$

$\dfrac{1}{2}I\omega^2+ 0=0 + 6.15+0$

$\dfrac{1}{2}\times 0.144 \times \omega^2= 6.15$

ω = 9.25 rad/s

c) linear speed of the ball

v = r ω

v = (L+R )ω

v = (0.215+0.0345) x 9.25

v =2.31 m/s

d) using equation of motion

v² = u² + 2 g h

v² = 0 + 2 x 9.8 x 0.248

v = √4.86

v =2.20 m/s

speed attained by the swing is more than free fall

% greater = $\dfrac{2.31-2.20}{2.20}\times 100$

= 5 %

speed of swing is 5 % more than free fall

6 0

3 years ago

How is Newtons second law apply to the egg drop ?

kotykmax [81]

Answer:

Newton's Second Law is applied because of the acceleration caused by natural forces as the egg is plummeting to the earth. And the amount of acceleration the egg has will be largely affected by the amount of force Mr. Baker uses to hurl the egg to the ground.

Explanation:

hope that helps

3 0

3 years ago

A mechanic pushes a 2.30 ✕ 103-kg car from rest to a speed of v, doing 4,800 J of work in the process. During this time, the car

Illusion [34]

The horizontal force applied is 160 N while the velocity is 2.03 m/s.

<h3>What is the speed of the car?</h3>

The work done by the car is obtained as the product of the force and the distance;

W = F x

F = ?

x = 30.0 m

W = 4,800 J

F = 4,800 J/30.0 m

F = 160 N

But F = ma

a = F/m

a = 160 N/2.30 ✕ 10^3-kg

a= 0.069 m/s

Now;

v^2 = u^2 + 2as

u = 0/ms because the car started from rest

v = √2as

v = √2 * 0.069 * 30

v = 2.03 m/s

Learn more about force and work:brainly.com/question/758238

#SPJ1

5 0

2 years ago

An alfisol is most closely related to ________.