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3 years ago

Which object has more gravitational potential energy? Use PE = m × g × h, where g = 9.8 meters/second2.

2 answers:

6 0

Gravitational potential energy can be calculated using the formula <span>PE = m × g × h, where g is the gravitational acceleration and is constant hence the energy is dependent directly to mass and the height of the object. Hence more PE is registered when the object is heavier and/or at greater initial height. </span>

saul85 [17]3 years ago

5 0

Answer:35kg

Explanation:

im right kids

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In the figure, a weightlifter's barbell consists of two identical small but dense spherical weights, each of mass 50 kg. These w

kondaur [170]

The moment of inertia is $24.8 kg m^2$

Explanation:

The total moment of inertia of the system is the sum of the moment of inertia of the rod + the moment of inertia of the two balls.

The moment of inertia of the rod about its centre is given by

$I_r = \frac{1}{12}ML^2$

where

M = 24 kg is the mass of the rod

L = 0.96 m is the length of the rod

Substituting,

$I_r = \frac{1}{12}(24)(0.96)^2=1.84 kg m^2$

The moment of inertia of one ball is given by

$I_b = mr^2$

where

m = 50 kg is the mass of the ball

$r=\frac{L}{2}=\frac{0.96}{2}=0.48 m$ is the distance of each ball from the axis of rotation

So we have

$I_b = (50)(0.48)^2=11.5 kg m^2$

Therefore, the total moment of inertia of the system is

$I=I_r + 2I_b = 1.84+ 2(11.5)=24.8 kg m^2$

Learn more about inertia:

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6 0

3 years ago

A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right

V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

g: Gravitational Acceleration

θ: Angle with the vertical

N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

m*g*cos(θ) - 0 = m*vf^2 / R

g*cos(θ) = vf^2 / R

vf^2 = R*g*cos(θ)

vf^2 = 1.45*32.2*cos(34)

vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

ΔK.E = ΔP.E

0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))

vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

vi^2 = 22.744

vi = 4.77 ft/s

4 0

3 years ago

Will give brainliest

NNADVOKAT [17]

I love science! if you need any more help let me know i cant guarantee i can help but i will try!

The correct answer is "Some substances must be dissolved in water before they can be used".

5 0

3 years ago

What is evaluating?

SSSSS [86.1K]

Answer:

D. Drawing a conclusion about something

3 0

3 years ago

The North American Plate is moving west at a rate of approximately 20 mm/yr. How long will it take for New York to move 10° long

scZoUnD [109]

Answer:

55000000 years

Explanation:

Rate of North American Plate moving (Velocity) = 20 mm/yr

Degrees New York has to move = 10° west

New York's longitude = 110 km/°

Distance of New York = 110×10 = 1100 km

= 1100×10⁶ mm

$\text {Time taken by the continental plate}=\frac {\text {Distance of New York}}{\text {Rate of North American Plate moving (Velocity)}}\\\Rightarrow \text {Time taken by the continental plate}=\frac{1100\times 10^6}{20}\\\Rightarrow \text {Time taken by the continental plate}=55\times 10^6\ years$

∴ Time taken by New York to move 10° west is 55000000 years

6 0

3 years ago