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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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Delicious77 [7]

Answer: It should be the 3rd option down!

Explanation:

7 0
2 years ago
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A stone is dropped from the
ICE Princess25 [194]
  • Height=h=500m
  • Acceleration=g=10m/s^2
  • Initial velocity=u=0
  • Speed of sound=c=340m/s
  • TIME TAKEN BY STONE TO HIT WATER=t
  • Time taken by sound to hear back=T

Now

\\ \sf\longmapsto h=ut+\dfrac{1}{2}gt^2

\\ \sf\longmapsto h=0t+\dfrac{1}{2}10t^2

\\ \sf\longmapsto 500=5t^2

\\ \sf\longmapsto t^2=100

\\ \sf\longmapsto t=10s

Now

\\ \sf\longmapsto h=cT

\\ \sf\longmapsto T=\dfrac{h}{c}

\\ \sf\longmapsto T=\dfrac{500}{340}

\\ \sf\longmapsto T=1.47\approx 1.5s

Total time:-

\\ \sf\longmapsto T_{net}=t+T=10+1.5=11.5s

8 0
2 years ago
Tires of a Bigfoot truck has a diameter of 2.2 m. If it rotates 60 revolutions find distance travel on the road.
Firlakuza [10]

Answer:

s = 414.7 m\\

Explanation:

The relationship between the linear distance covered by an object and its angular displacement is given by the following formula:

s = rθ

where,

s = distance traveled on road = ?

r  radius of tires = diameter/2 = 2.2 m/2 = 1.1 m

θ = angular displacement = (60 rev)(2π rad/1 rev) = 377 rad

Therefore,

s = (1.1 m)(377 rad)\\s = 414.7 m

8 0
3 years ago
A rabbit is hopping along at an approximately constant speed of 3.9 m/s. The rabbit passes a crouched cat ready to chase the rab
Firlakuza [10]

Answer:

t  = 7,8 s

Explanation:

From the instant, the rabbit passes the cat. The cat star running acceleration of 0,5 m/s² .

When the cat arrives at the speed of 3,9 m/s the cat catches the rabbit

Then for the cat arrives at 3,9 m/s nedds

v = vo + a*t     vo  = 0  then   v = a*t

3,9 ( m/s) = 0,5 ( m/s² ) * t

t  = 7,8 s

v  =  3,9 m/s =

4 0
2 years ago
a u tube contains a liquid of an unknown density an oil of density is poured into the right arm of the tube until the oil column
scoray [572]

Answer:

The answer is "1155\ \frac{kg}{m^3}"

Explanation:

Please find the complete question in the attached file.

p = p_0 + ?gh

pi = pressure only at two liquids' devices

PA = pressure atmosphere.

1 = oil density

2 = uncertain fluid density

h_1 = 11 \ cm\\\\h_2= 3 \ cm

The pressures would be proportional to the quantity 11-3 = 8 cm from below the surface at the interface between both the oil and the liquid.

\to p_A + ?2g(h_1 - h_2) = p_A + ? 1gh_1\\\\\to ?2 = \frac{?1h_1}{(h_1 - h_2)} \\\\

       = \frac{840 \frac{kg}{m^3}}{\frac{11}{8}} \\\\= 1155\ \frac{kg}{m^3}

8 0
3 years ago
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