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FrozenT [24]
3 years ago
7

Along the frictionless path you have chosen, the main force(s) acting on the puck after receiving the kick is (are):

Physics
1 answer:
Vaselesa [24]3 years ago
5 0
Gravity, and Normal. Check the comments for why Applied isn't one.
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Two identical silver spheres of mass m and radius r are placed a distance R (sphere 1) and 2R (sphere 2) from the Sun, respectiv
lys-0071 [83]

Answer:

The ratio of T2 to T1 is 1.0

Explanation:

The gravitational force exerted on each sphere by the sun is inversely proporational to the square of the distance between the sun and each of the spheres.

Provided that the two spheres have the same radius r, the pressure of solar radiation too, is inversely proportional to the square of the distance of each sphere from the sun.

Let F₁ and F₂ = gravitational force of the sun on the first and second sphere respectively

P₁ and P₂ = Pressure of solar radiation on the first and second sphere respectively

M = mass of the Sun

m = mass of the spheres, equal masses.

For the first sphere that is distance R from the sun.

F₁ = (GmM/R²)

P₁ = (k/R²)

T₁ = (F₁/P₁) = (GmM/k)

For the second sphere that is at a distance 2R from the sun

F₂ = [GmM/(2R)²] = (GmM/4R²)

P₂ = [k/(2R)²] = (k/4R²)

T₂ = (F₂/P₂) = (GmM/k)

(T₁/T₂) = (GmM/k) ÷ (GmM/k) = 1.0

Hope this Helps!!!

3 0
3 years ago
(15 Points)
oksian1 [2.3K]

The vertical weight carried by the builder at the rear end is F = 308.1 N

<h3>Calculations and Parameters</h3>

Given that:

The weight is carried up along the plane in rotational equilibrium condition

The torque equilibrium condition can be used to solve

We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person

This would lead to:

F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)

F(1cos20)= 197/2(3.10sin20 + 2 cos 20)

Fcos20= 289.55

F= 308.1N

Read more about vertical weight here:

brainly.com/question/15244771

#SPJ1

5 0
2 years ago
A pendulum is transported from sea-level, where the acceleration due to gravity g is 9.80 m/s2, to the bottom of Death Valley. A
tatyana61 [14]

Answer: the value of g in Death Valley is 10.417 m/s²

Explanation:

Given that;

acceleration due to gravity at the point is g = 9.8 m/s²

Lets say the acceleration due to gravity at the bottom of Death valley is g'

as the period of the pendulum is decreased by 3.00%

T' = 0.97 T  

T is the period of the pendulum at sea level  and T' is the period of the pendulum at bottom of Death valley

therefore from the relation

T = 2π√(l/g)

g'/g = T²/T'²

g' = (T²/ (0.97T)²)g

g' = 1.063g

g' = 10.417 m/s²

therefore the value of g in Death Valley is 10.417 m/s²

7 0
3 years ago
The owner of a company that manufactures drinking cups decides it would be impressive to build an inground swimming pool that is
garri49 [273]

Answer:

The depth is 5.15 m.

Explanation:

Lets take the depth of the pool = h m

The atmospheric pressure ,P = 101235 N/m²

The area of the top = A m²

The area of the bottom =  a m²

Given that A= 1.5 a

The force on the top of the pool = P A

The total pressure on the bottom = P + ρ g h

ρ =Density of the  water = 1000 kg/m³

The total pressure at the bottom of the pool =  (P + ρ g h) a

The bottom and the top force is same

(P + ρ g h) a = P A

P a +ρ g h a =  P A

ρ g h a =  P A - P a

h=\dfrac{P ( A-a)}{\rho g a}

h=\dfrac{P ( 1.5 a-a)}{\rho g a}

h=\dfrac{P ( 1.5- 1)}{\rho g}

h=\dfrac{101235 ( 1.5- 1)}{1000\times 9.81}\ m

h=5.15 m

The depth is 5.15 m.

7 0
3 years ago
An​ airplane, flying horizontally at an altitude of 3 miles​, passes directly over an observer. If the constant speed of the air
natima [27]
I’m not sure if this helped you so here you go

6 0
3 years ago
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