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Artemon [7]
2 years ago
7

Barker is unloading 20kg bottles of water from this delivery truck when one of the bottles tips over and slides down the truck r

amp that is inclined at an angle of 30 degrees to the ground. what amount of force moves the bottle down the ramp?
Physics
1 answer:
Degger [83]2 years ago
8 0
<span>The overall force that is acting on the bottle is gravity. With the incline being 30 degrees the full force of gravity isn't acting on the bottle becuase the ramp isn't allowing the bottle to go straight down. By taking the sin of 30 degrees you find the proportion of gravity that is acting on the bottle to be 4.9 meters per second and the bottle weights 20 kg so the force acting on the bottle is 98 Newtons.</span>
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Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for
Burka [1]

Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

3 0
2 years ago
Suppose your friend claims to have discovered a mysterious force in nature that acts on all particles in some region of space. H
kirill [66]

Answer:

             U = 1 / r²

Explanation:

In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related

             

         F = - dU / dr

this derivative is a gradient, that is, a directional derivative, so we must have

          dU = - F. dr

the esxresion for strength is

         F = B / r³

let's replace

          ∫ dU = - ∫ B / r³  dr

in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product

let's evaluate the integrals

            U - Uo = -B (- / 2r² + 1 / 2r₀²)

To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)

             U = B / 2r²

we substitute the value of B = 2

             U = 1 / r²

5 0
2 years ago
The 20-lb cabinet is subjected to the force F = (3 + 2t) lb, where t is in seconds. If the cabinet is initially moving down the
Andre45 [30]

Answer:

t₁ = 0.95 s

Explanation:

In this chaos we must use the definition of Newton's second law

      F = m a = m dv / dt

      dv = F dt / m

Let's replace and integrate, let's take the upward direction of the plane as positive, the force is positive

       dv = ∫ (3 + 2t) dt / m

       v = (3 t + 2 t²/ 2) /m

Let's evaluate between the lower limit t = 0 v = -6 ft / s (going down) to the upper limit   t = t and v = 0

       0 - (-6) = (3 (t- 0) + (t² -0)) / m

       t² + 3t -6m = 0

Let's look for the mass

      W = mg

      m = W / g

      m = 20/32

      m = 0.625 slug

Let's solve the second degree equation

     t² + 3t -3.75 = 0

     t = (-3 ± √ (32 + 4 1 3.75)) / 2

     t = (-3 ± 4,899) / 2

     t₁ = 0.95 s

     t₂ = -3.95 s

We take the positive time

6 0
2 years ago
Two male moose charge at each other with the same speed and meet on a icy patch of tundra. As they collide, their antlers lock t
Ksenya-84 [330]
Change in velocity of larger moose: (1/3)v - v = -(2/3)v 
<span>change in velocity of small moose: (1/3)v - (-v) = (4/3)v </span>
<span>- (change in velocity of larger moose)/(change in velocity of smaller moose) = 2

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.



</span>
4 0
2 years ago
Read 2 more answers
6) A map in a ship’s log gives directions to the location of a buried treasure. The starting location is an old oak tree. Accord
kiruha [24]

Answer:

Sorry cant find the answer but i hope you got it right and if you didn't you'll still do great. :)

Explanation:

4 0
2 years ago
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