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Sav [38]
3 years ago
10

A packing crate is sitting at rest on an inclined loading ramp. How does the magnitude of the force of static friction compare t

o the other forces acting on the crate?
Physics
1 answer:
eduard3 years ago
8 0

As a crate is resting on the inclined surface of the ramp we have following forces on the crate

1. Weight of the crate vertically downwards

2. Normal force due to ramp perpendicular to the surface

3. static friction force along the ramp upwards

so here we know that crate is at rest

so we can say that

normal force is balanced by component of weight of crate

so we have

N = mg cos\theta

also we can say that friction force is balanced by another component of weight of the crate

F_f = mg sin\theta

so friction force is less than the weight of the crate and it will act opposite to it

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Vesna [10]

Answer:

C

Explanation:

Radiant= list onto solar panels, Electric= solar into power, Radiant= Electric into light

8 0
3 years ago
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I'm walking 1.6m/s to 7-11 and it started to rain so I sped up to 2.7m/s in 1.2
olga nikolaevna [1]

Answer:

Explanation:

a = \frac{v_f-v_0}{t} which is the final velocity minus the initial velocity in the numerator, and the change in time in the denominator.  For us:

a=\frac{2.7-1.6}{1.2} so

a = .92 m/s/s (NOT negative because you're speeding up)

5 0
3 years ago
A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication
koban [17]

Answer:

r = 4.24x10⁴ km.  

     

Explanation:

To find the radius of such an orbit we need to use Kepler's third law:

\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}

<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km.                           </em>                              

From equation (1), r₁ is:

r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}                            

r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}      

r_{1} = 4.24 \cdot 10^{4} km      

Therefore, the radius of such an orbit is 4.24x10⁴ km.

I hope it helps you!

3 0
3 years ago
A football is kicked from the ground with a velocity of 38m/s at an angle of 40 degrees and eventually lands at the same height.
Anastasy [175]

Initially, the velocity vector is \langle 38cos(40^{\circ}),38sin(40^{\circ}) \rangle=\langle 29.110, 24.426 \rangle. At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by 4.9(0.2)^2, so the velocity is \langle 29.110, 24.426-0.196 \rangle = \langle 29.110, 24.23 \rangle.

Converting back to direction and magnitude, we get \langle r,\theta \rangle=\langle \sqrt{29.11^2+24.23^2},tan^{-1}(\frac{29.11}{24.23}) \rangle = \langle 37.87,50.2^{\circ}\rangle

4 0
3 years ago
A bullet fired vertically at a velocity of 36m/s .after 45 the bullet hit the top of a bulid how height is a bulid?​
zzz [600]

Answer:

The height of the building is 8,302.5 m

Explanation:

Given;

velocity of the projectile, u = 36 m/s

time of motion, t = 45 s

Let the upward direction of the bullet be negative,

The height of the building is calculated as;

h = ut - \frac{1}{2} gt^2\\\\h = (36\times 45) - (\frac{1}{2} \times 9.8 \times 45^2)\\\\h = 1620 - 9922.5\\\\h = -8,302.5 \ m\\\\The \ height \ of \ the \ building \ is \ 8,302.5  \ m

3 0
3 years ago
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