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Vladimir [108]
2 years ago
6

When the mass of the cylinder increased by a factor of 3, from 1.0 kg to 3.0 kg, what happened to the cylinder’s gravitational p

otential energy? It decreased by a factor of 3. It decreased by a factor of 2. It increased by a factor of 2. It increased by a factor of 3.
Physics
1 answer:
Gnesinka [82]2 years ago
6 0

Answer:

It increased by a factor of 3.

Explanation:

The gravitational potential energy of an object is given by

U=mgh

where

m is the mass

g is the gravitational acceleration

h is the heigth of the object relative to some reference point (for instance, the ground)

As we see from the formula, the gravitational potential energy is directly proportional to the mass, m: therefore, if the mass of the cylinder is increased by a factor 3, then the gravitational potential energy will also increase by a factor 3.

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A uniform cube with mass 0.700 kg and volume 0.0270 m3 is sitting on the floor. A uniform sphere with radius 0.400 m and mass 0.
Sav [38]

Answer:

  44 1/3 cm

Explanation:

The cube has an edge length of ∛0.027 m = 0.3 m, so a center of mass (CoM) 15 cm above the floor.

The sphere's center of mass is 40 cm above the top of the cube, so is 70 cm above the floor. The weighted average of the CoM locations is ...

  ((15 cm)(0.700 kg) +(70 cm)(0.800 kg))/(0.700 kg +0.800 kg)

  = (10.5 kg·cm +56 kg·cm)/(1.500 kg) = 44.333... cm

The center of mass of the two-object system is 44 1/3 cm above the floor.

_____

<em>Comment on the units</em>

We're not familiar with "hcm" as a unit. We presume that you can convert the given answer to the units you desire.

6 0
2 years ago
A force of 500 N is exerted on a baseball by the bat for 0.001 s. What is the change in momentum of the baseball?
GarryVolchara [31]
Answer: Δp = F*Δt = 500N*0.001s = 0.5Ns
3 0
2 years ago
An air-track glider undergoes a perfectly inelastic collision with an identical glider that is initially at rest. what fraction
DiKsa [7]
Refer to the diagram shown below.

The initial KE (kinetic energy) of the system is
KE₁ = (1/2)mu²

After an inelastic collision, the two masses stick together.
Conservation of momentum requires that
m*u = 2m*v
Therefore
v = u/2

The final KE is
KE₂ = (1/2)(2m)v²
       = m(u/2)²
       = (1/4)mu²
      = (1/2) KE₁

The loss in KE is
KE₁ - KE₂ = (1/2) KE₁.

Conservation of energy requires that the loss in KE be accounted for as thermal energy.

Answer:  1/2 

5 0
2 years ago
Read 2 more answers
A guitar string has a linear density of 8.30 ✕ 10−4 kg/m and a length of 0.660 m. the tension in the string is 56.7 n. when the
Sedbober [7]
Ans: Beat Frequency = 1.97Hz

Explanation:
The fundamental frequency on a vibrating string is 

f =   \sqrt{ \frac{T}{4mL} }<span>  -- (A)</span>

<span>here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg </span>


Plug in the values in Equation (A)

<span>so </span>f = \sqrt{ \frac{56.7}{4*5.48*10^{-4}*0.66} }<span> = 197.97Hz </span>

<span>the beat frequency is the difference between these two frequencies, therefore:
Beat frequency = 197.97 - 196.0 = 1.97Hz
-i</span>
3 0
2 years ago
Read 2 more answers
The greatest height reported for a jump into an airbag is 99.4 m by stuntman Dan Koko. In 1948 he jumped from rest from the top
leva [86]

Answer:

44.1613858478 m/s

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

s = Displacement = 99.4

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 99.4+0^2}\\\Rightarrow v=44.1613858478\ m/s

If air resistance was absent Dan Koko would strike the airbag at 44.1613858478 m/s

6 0
1 year ago
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