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Nastasia [14]
3 years ago
7

Block A, with a mass of 6.0 kg, is sliding across a frictionless track at 65 m/s when it collides with Block B ( 9.0 kg) which i

s initially at rest . The two masses stick together and travel around a vertical loop with an unknown radius. They just barely make it around the vertical loop with an unknown radius.
a) Calculate the centrifugal force acting on the masses at the top of the loop
b) Calculate the velocity of the blocks just after the collision
c) Calculate the radius of the loop so that the blocks just barely make it
d) Calculate the normal force acting on the blocks when they reach the bottom after traveling through the loop
Physics
1 answer:
Mars2501 [29]3 years ago
3 0
The correct answer is a I hope that helped enjoy the rest of your weekend
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The drawing shows three objects rotating about a vertical axis. The mass of each object is given in terms of m0, and its perpend
photoshop1234 [79]

Answer:

I₁ > I₃ > I₂

Explanation:

Taking the pic shown, we have

m₁ = 10m₀

m₂ = 2m₀

m₃ = m₀

r₁ = r₀

r₂ = 2r₀

r₃ = 3r₀

We apply the formula

I = mr²

then

I₁ = m₁r₁² = (10m₀)(r₀)² = 10m₀r₀²

I₂ = m₂r₂² = (2m₀)(2r₀)² = 8m₀r₀²

I₃ = m₃r₃² = (m₀)(3r₀)² = 9m₀r₀²

finally we have

I₁ > I₃ > I₂

7 0
3 years ago
What do you think happens to the temperature when water changes to gas?
Olenka [21]

Answer:

it gets hot and more hot until it turns to gas

Explanation:

5 0
3 years ago
A ball of mass 0.500 kg is carefully balanced on a shelf that is 2.70 m above the ground. What is its gravitational potential en
ratelena [41]

Answer:

The gravitational potential energy of the ball is 13.23 J.

Explanation:

Given;

mass of the ball, m = 0.5 kg

height of the shelf, h = 2.7 m

The gravitational potential energy is given by;

P.E = mgh

where;

m is mass of the ball

g is acceleration due to gravity = 9.8 m/s²

h is height of the ball

Substitute the givens and solve for gravitational potential energy;

PE = (0.5 x 9.8 x 2.7)

P.E = 13.23 J

Therefore, the gravitational potential energy of the ball is 13.23 J.

8 0
3 years ago
1 kg air in a piston-cylinder assembly is heated at constant pressure, resulting the expansion of the volume. the initial temper
rewona [7]

Answer:

57,42 KJ

Explanation:

By a isobaric proces, the expresion for the works in the jpg adjunt. Then:

W = Pa(Vb - Va) = Pa*Vb - Pa*Va ---(1)

By the ideal gases law: PV=RTn

Then, in (1): (remember Pa = Pb)

W =  R*Tb*n - R*T*an = R*n*(Tb - Ta) --- (2)

Since we have 1 Kg air: How much is this in moles?

From bibliography: 28.96 g/mol

Then, in 1 Kg (1000 g) there are:

n =  34,53 mol

Finally, in (2):

W =  (8,3144 J/K.mol)*(34,53 mol)*(500K - 300K) = 51 419,9 J ≈ 57,42 KJ

4 0
3 years ago
Could anyone help with number 9?
Alisiya [41]
The answer would be A
3 0
3 years ago
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