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maksim [4K]
3 years ago
8

Why would a skier try to lower his center of gravity?

Physics
1 answer:
Andrew [12]3 years ago
8 0

He would lower his center of gravity so that it would be harder for him to fall, and so that he can stay and become balanced on his ski's.

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A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the
Mamont248 [21]

Answer:

The minimum coefficient of friction required is 0.35.  

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

-F_{f} + F = 0      

-F_{f} + ma = 0      

\mu mg = ma

\mu = \frac{a}{g}

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

v_{f}^{2} = v_{0}^{2} + 2ad

a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}

Where:  

d: is the distance traveled = 46.1 m

v_{f}: is the final speed of the truck = 0 (it stops)      

v_{0}: is the initial speed of the truck = 17.9 m/s

a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}        

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.  

\mu = \frac{a}{g}  

\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}

\mu = 0.35

Therefore, the minimum coefficient of friction required is 0.35.  

I hope it helps you!

4 0
3 years ago
How small are the wavelengths of gamma ray radiation?
WARRIOR [948]

Gamma rays are the highest energy EM radiation and typically have energies greater than 100 keV, frequencies greater than 1019 Hz, and wavelengths less than 10 picometers.

3 0
3 years ago
What is the process of change in a crust due to the motion of hot material underneath?(answer)?
frutty [35]
Im not exactly sure but I think the answer is techtonic plates collide
5 0
3 years ago
Consider two wheels with fixed hubs. The hub cannot move, but the wheel can rotate about it. The hubs are fixed to a stationary
kykrilka [37]

Answer:

Magnitude of force on wheel B is 4 N

Explanation:

Given that

I=mr^2

For wheel A

m= 1 kg

d= 1 m,r= 0.5 m

F=1 N

We know that

T= F x r

T=1 x 0.5 N.m

T= 0.5 N.m

T= I α

Where I is the moment of inertia and α is the angular acceleration

I_A=1 \times 0.5^2\ kg.m^2

I_A=0.25\ kg.m^2

T= I α

0.5= 0.25 α

\alpha = 2\ rad/s^2

For Wheel B

m= 1 kg

d= 2 m,r=1 m

I_B=1 \times 1^2\ kg.m^2

I_B=1 \ kg.m^2

Given that angular acceleration is same for both the wheel

\alpha = 2\ rad/s^2

T= I α

T= 1 x 2

T= 2 N.m

Lets force on wheel is F then

T = F x r

2 = F x 1

So F= 2 N

Magnitude of force on wheel B is 2 N

3 0
3 years ago
A mass of 5kg accelerates at 3m/s/s, how much force was put on it?
goldfiish [28.3K]

Answer:

15N

Explanation:

According to Newton's Second Law of Motion

F = m*a

mass = m = 5Kg

acceleration = a = 3m/s^2

=> F = 5kg * 3m/s^2

=> F = 15 N

5 0
3 years ago
Read 2 more answers
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