All categories

2 years ago

Cfare eshte energjia

Physics

2 answers:

Whitepunk [10]2 years ago

5 0

Answer:

what

Explanation:

faust18 [17]2 years ago

4 0

Answer:

jnj

Explanation:

You might be interested in

On the weather map above, what does the "H" represent?

kodGreya [7K]

I honestly don't see anything above. But 'H' on a weather map usually shows the center of a high-pressure system.

7 0

3 years ago

A car traveling on a flat (unbanked) circular track accelerates uniformly from rest with a tangential acceleration of 1.7 m/s^2.

Komok [63]

Answer:

The coefficient of static friction between the car and the track

u=0.572

Explanation:

We don't know the mass of the car or any other information so the acceleration is the reason to solve the friction coefficient

∑ $F=F_{f}=m*a_{t}$

As we know

$F_{f}=u*F_{N}=u*m*g$

Also the center ward direction forces

$F_{fc}=m*a_{c}$

$a_{c}=\frac{v_{t}^2}{r}$

$F_{fc}=m*\frac{v_{t}^2}{r}$

But now vt relation with the tangential acceleration

$v_{t}=2*a_{t}*\frac{\pi }{r}$

replacing

$F_{fc}=m*a_{t}*\frac{2\pi*r}{2r}$

$F_{fc}=m*a_{t}*\pi$

So magnitude of the force can get by

$F_{f}=\sqrt{(m*a_{t}*\pi)^{2}+(m*a_{t})^{2}}$

Get the factor to simplify

$F_{f}=a_{t}*m*\sqrt{(1+\pi^2)}$

$u*m*g=m*a_{t}*(\sqrt{1+\pi^2})$

Solve to u'

$u=\frac{a_{t}}{g}*(\sqrt{1+\pi^2})$

$u=\frac{17\frac{m}{s^2} }{9.8\frac{m}{s^2}}*(\sqrt{1+\pi^2})=0.572$

5 0

3 years ago

Neglecting air, what speed does a rock thrown straight up have to be to reach the edge of our atmosphere: say 100 km? Still negl

Ugo [173]

To solve this problem it is necessary to apply the kinematic equations of movement description, specifically those that allow us to find speed and acceleration as a function of distance and not time.

Mathematically we have to

$v_f^2-v_i^2 = 2ax$

Where,

$v_{f,i} =$ Final velocity and Initial velocity

a = Acceleration

x = Displacement

From the description given there is no final speed (since it reaches the maximum point) but there is a required initial speed that is contingent on traveling a certain distance under the effects of gravity

$0 - v_i^2 = 2(9.8)(100*10^3)$

$v_i = 14*10^2m/s$

Therefore the speed which must a rock thrown straight up is 14*10^2m/s to reach the edge of our atmosphere.

The displacement and gravity traveled are the same, therefore the final speed will be the same but in the opposite vector direction (towards the earth), that is $14 * 10 ^ 2m / s$