What kind of deformation does a cube of Jell-O exhibit when it jiggles?

You have been abducted by aliens and find yourself on a strange planet. Fortunately, you have a meter stick with you. You observ

defon

Answer:

$36.4276\ m/s^2$

Explanation:

m = Mass of stick

L = Length of stick = 1 m

h = Center of mass of stick = $\dfrac{1}{2}=0.5\ m$

g = Acceleration due to gravity

T = Time period = 0.85 s

Time period is given by

$T=2\pi\sqrt{\dfrac{I}{mgh}}$

Moment of inertia is given by

$I=m\dfrac{L^2}{3}$

$T=2\pi\sqrt{\dfrac{m\dfrac{L^2}{3}}{mgh}}\\\Rightarrow T=2\pi\sqrt{\dfrac{L^2}{3gh}}\\\Rightarrow g=\dfrac{4\pi^2L^2}{3hT^2}\\\Rightarrow g=\dfrac{4\pi^2\times 1^2}{3\times 0.5\times 0.85^2}\\\Rightarrow g=36.4276\ m/s^2$

The acceleration of gravity on this planet is $36.4276\ m/s^2$

0 0

2 years ago

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A 5 kg object near Earth's surface is released from rest such that it falls a distance of 10 m. After the object falls 10 m, it

makkiz [27]

Answer:D

Explanation:

Given

mass of object $m=5 kg$

Distance traveled $h=10 m$

velocity acquired $v=12 m/s$

conserving Energy at the moment when object start falling and when it gains 12 m/s velocity

Initial Energy $=mgh=5\times 9.8\times 10=490 J$

Final Energy $=\frac{1}{2}mv^2+W_{f}$

$=\frac{1}{2}\cdot 5\cdot 12^2+W_{f}$

where $W_{f}$ is friction work if any

$490=360+W_{f}$

$W_{f}=130 J$

Since Friction is Present therefore it is a case of Open system and net external Force is zero

An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .

4 0

2 years ago

In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses m1 and m2. To simplify the anal

Dafna1 [17]

Answer:

a) p = m1 v1 + m2 v2 , b) dp / dt = m1 a1 + m2 a2 , c) It is equivalent to force

dp / dt = 0

Explanation:

In this problem we have two blocks and the system is formed by the two bodies.

Part A. Initially they ask us to find the moment of the whole system

p = m1 v1 + m2 v2

Part B.

Find the derivative

dp / dt = m1 dv1dt + m2 dv2 / dt

dp / dt = m1 a1 + m2 a2

Part C.

Let's analyze the dimensions

m a = [kg] [m / s2] = [N]

It is equivalent to force

Part d

Acceleration is due to a net force applied

Part e

The acceleration of block 1 is due to the force exerted by block 2 during the moment change

Part f

Force of block 1 on block 2

True f12 = m1a1 f21 = m2a2

Part g

By the law of action and reaction are equal magnitude F12 = f21

Part H

dp / dt = 0

Isolated system F12 = F21 and the masses are constant. The total moment is only redistributed

7 0

2 years ago

A 64.5 kg person steps off a 129 kg rowboat with a force of 34.0 N. What is the force that is applied to the person by the rowbo

Arada [10]

34.0 N which is the force that applied to the person by rowboat.

3 0

2 years ago

A typical machine tests the tensile strength of a sheet of material cut into a standard size of 5.00 centimeters wide by 10.0 ce

Minchanka [31]

Answer:

Explanation:

tensile strength is stress that is needed to break the wire made of the material .

Here force required to break the sheet of material = 233 N

cross sectional area of the foil = breadth x thickness

= 5 x 10⁻² x 15 x 10⁻⁶ m²

= 75 x 10⁻⁸ m²

breaking stress = force / cross sectional area

= 233 / 75 x 10⁻⁸

= 3.1 x 10⁸ Pa .

Tensile strength = 3.1 x 10⁸ Pa .

4 0

2 years ago