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4 years ago

15

A simple pendulum is 1.15 meters long and has a period of 6.29 seconds. The pendulum is on an unknown planet. What is the accele

ration of gravity of the Unknown Planet

1 answer:

Nastasia [14]4 years ago

7 0

Answer:

The value of the acceleration of gravity of the Unknown Planet = 1.14 $\frac{m}{s^{2} }$

Explanation:

length of the pendulum (L)= 1.15 m

Time period (T)= 6.29 seconds

We know that time period of a simple pendulum is given by

⇒ T = 2 $\pi$ × $\sqrt\frac{L}{g}$

put the values in the above formula we get

⇒ T = 2 $\pi$ × $\sqrt \frac{1.15}{g}$

⇒ 6.29 = 2 $\pi$ × $\sqrt \frac{1.15}{g}$

By solving the above equation we get

⇒ g = 1.14 $\frac{m}{s^{2} }$

This is the value of the acceleration of gravity of the Unknown Planet.

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Calculate the location xcm of the center of mass of the Earth-Moon system. Use a coordinate system in which the center of the Ea

Jet001 [13]

Answer:

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

Explanation:

Let suppose that planet and satellite can be treated as particles. The masses of Earth and Moon ( $m_{E}$ , $m_{M}$ ) are $5.972\times 10^{24}\,kg$ and $7.349\times 10^{22}\,kg$ , respectively. The distance between centers is 384,403 kilometers. The location of the center of mass can be found by using weighted averages:

$\bar x = \frac{x_{E}\cdot m_{E}+x_{M}\cdot m_{M}}{m_{E}+m_{M}}$

If $x_{E} = 0\,km$ and $x_{M} = 384,403\,km$ , then:

$\bar x = \frac{(0\,km)\cdot (5.972\times 10^{24}\,kg)+(384,403\,km)\cdot (7.349\times 10^{22}\,kg)}{5.972\times 10^{24}\,kg+7.349\times 10^{22}\,kg}$

$\bar x = 4.673\,km$

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

8 0

3 years ago

particle of mass 59 g and charge 51 µC is released from rest when it is 32 cm from a second particle of charge −14 µC. Determine

AleksAgata [21]

Answer:

The initial acceleration of the 59g particle is $1062.7\frac{m}{s^{2}}$

Explanation:

Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:

$F=ma$ (1)

We already know the mass of the particle so we should find the electric force on it to use on (1), the magnitude of the electric force between two charged objects by Columb's law is:

$F=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}$

with q1 and q2 the charge of the particles, r the distance between them and k the constant $k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}$ . So:

$F=(9.0\times10^{9})\frac{\mid (51\times10^{-6})(-14\times10^{-6})\mid}{0.32^{2}}$

$F=62.7 N$

Using that value on (1) and solving for a

$a=\frac{F}{m}=\frac{62.7}{0.059}=1062.7\frac{m}{s^{2}}$

4 0

4 years ago

Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 60 and 3

Scilla [17]

Explanation:

Given that,

Angle by the normal to the slip α= 60°

Angle by the slip direction with the tensile axis β= 35°

Shear stress = 6.2 MPa

Applied stress = 12 MPa

We need to calculate the shear stress applied at the slip plane

Using formula of shear stress

$\tau=\sigma\cos\alpha\cos\beta$

Put the value into the formula

$\tau=12\cos60\times\cos35$

$\tau=4.91\ MPa$

Since, the shear stress applied at the slip plane is less than the critical resolved shear stress

So, The crystal will not yield.

Now, We need to calculate the applied stress necessary for the crystal to yield

Using formula of stress

$\sigma=\dfrac{\tau_{c}}{\cos\alpha\cos\beta}$

Put the value into the formula

$\sigma=\dfrac{6.2}{\cos60\cos35}$

$\sigma=15.13\ MPa$

Hence, This is the required solution.

3 0

3 years ago

Two horses are side by side on a carousel. Which has a greater tangential speed the one closer to the center or the one farther

photoshop1234 [79]

Answer:

The horse father from the center has a greater tangential speed. Although both horses complete one circle in the same time period, the one farther from the center covers a greater distance during that same period.

Explanation:

8 0

3 years ago

Pascal's Principle states that (a) If we apply pressure to a fluid in a sealed container, the pressure will be felt undiminished

nika2105 [10]

Answer:

a) If we apply pressure to a fluid in a sealed container, the pressure will be felt undiminished at every point in the fluid and on the walls of the container.

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area(A2) of the piston.

P=F/A

P1=P2

F1/ A1= F2/ A2

F2= F1* A2/ A1

The pressure acting on one side is transmitted to all the molecules of the liquid because the liquid is incompressible.

In an incompressible liquid, the volume and amount of mass does not vary when pressure is applied.

5 0

3 years ago