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Elza [17]
3 years ago
8

Identical marbles are released from the same height on each of the following four frictionless ramps . Compare the speed of the

marbles at the end of each ramp. Explain your reasoning
Physics
1 answer:
Montano1993 [528]3 years ago
3 0

Since there is no friction on the ramp so there is no loss of energy in terms of frictional loss

So here we can use mechanical energy conservation law in this case

so here we will have

U_i = KE_f

here we know that

U_i = mgh

also we know that

K_f = \frac{1}{2}mv^2

now we will have

mgh = \frac{1}{2}mv^2

now by solving above equation we have

v = \sqrt{2gh}

so here as we know all identical balls are released from same height so all balls must have same speed at the end

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Answer:

Models,Mathematics

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Explanation:

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3 years ago
A tomato of mass 0.18 kg is dropped from a tall bridge. If the tomato has a speed of 11 m/s just before it hits the ground, what
kondor19780726 [428]
The kinetic energy of the tomato is : 

K.E =  1/2 mv^2

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Hope this helps
7 0
3 years ago
If you weigh 690 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun an
lana [24]

Answer:

Explanation:

Given that,

Mass of star M(star) = 1.99×10^30kg

Gravitational constant G

G = 6.67×10^−11 N⋅m²/kg²

Diameter d = 25km

d = 25,000m

R = d/2 = 25,000/2

R = 12,500m

Weight w = 690N

Then, the person mass which is constant can be determined using

W =mg

m = W/g

m = 690/9.81

m = 70.34kg

The acceleration due to gravity on the surface of the neutron star is can be determined using

g(star) = GM(star)/R²

g(star) = 6.67×10^-11 × 1.99×10^30 / 12500²

g (star) = 8.49 × 10¹¹ m/s²

Then, the person weight on neutron star is

W = mg

Mass is constant, m = 70.34kg

W = 70.34 × 8.49 × 10¹¹

W = 5.98 × 10¹³ N

The weight of the person on neutron star is 5.98 × 10¹³ N

5 0
3 years ago
The neutron is located in the what part of the atom?
kolbaska11 [484]

Answer:

The neutron can be found in the nucleus of the atom with the proton.

4 0
3 years ago
Read 2 more answers
What is the accletation of a 1500kg with a net force of 7500 N
matrenka [14]
Acceleration formulae is:
a=Fnet/mass
According to the question
a=7500N/1500kg
a=5m/s sq.
3 0
3 years ago
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