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Olenka [21]
3 years ago
9

You are on a new planet, and are trying to use a simple pendulum to find the gravity experienced on this new planet. You find th

at a pendulum having a length of 35 cm has a period of 2 seconds. What is the acceleration due to gravity on this planet?
A. 13.82 m/s^2
B. 345.4 m/s^2
C. 3.45 m/s^2
D. 1.1 m/s^2
Physics
1 answer:
levacccp [35]3 years ago
4 0

To solve this problem it is necessary to apply the concepts related to the Period based on the length of its rope and gravity, mathematically it can be expressed as

T= 2\pi \sqrt{\frac{L}{g}}

g = Gravity

L = Length

T = Period

Re-arrange to find the gravity we have

g = \frac{4\pi^2 L}{T^2}

Our values are given as

L = 0.35m\\T = 2s\\

Replacing we have

g = \frac{4\pi^2 L}{T^2}

g = \frac{4\pi^2 0.35}{2^2}

g = 3.45 m/s^2

Therefore the correct answer is C.

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two cars are moving at constant speeds in a straight line along a major highway. The first is travelling at 20ms^-1 and the seco
Reika [66]

Answer:

t=750s

Explanation:

The two cars are under an uniform linear motion. So, the distance traveled by them is given by:

\Delta x=vt\\x_f-x_0=vt\\x_f=x_0+vt

x_f is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have:

x_0_1=6km\\x_0_2=0

We have x_f_1=x_f_2. Thus, solving for t:

x_0_1+v_1t=x_0_2+v_2t\\x_0_1=t(v_2-v_1)\\t=\frac{x_0_1}{v_2-v_1}\\t=\frac{6*10^3m}{28\frac{m}{s}-20\frac{m}{s}}\\t=750s

8 0
2 years ago
When does a star reach equilibrium?
LenaWriter [7]
A protostar reaches equilibrium when gravity is balanced by the outward force of nuclear fusion. 
6 0
3 years ago
A positively charged particle is in the center of a parallel-plate capacitor that has charge ±Q on its plates. SUppose the dista
slamgirl [31]

Answer:

Stay the same

Explanation:

First of all, let's find how the capacitance of the capacitor changes.

Initially, it is given by

C=\frac{\epsilon_0 A}{d}

where

\epsilon_0 is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

From the formula, we see that the capacitance is inversely proportional to the separation between the plates. In this problem, the distance between the plates is doubled, so the capacitance will be halved:

C' = \frac{1}{2}C

The potential difference across the capacitor is given by

V= \frac{Q}{C}

where

Q is the charge on the plates

C is the capacitance

We see that the voltage is inversely proportional to the capacitance. We said that the capacitance has halved: therefore, the potential difference across the two plates will double:

V' = 2 V

Now we can analyze the electric field between the plates of the capacitor, which is given by

E=\frac{V}{d}

we said that:

- The voltage has doubled: V' = 2 V

- The distance between the plates has doubled: d' = 2 d

therefore, the new electric field will be

E'=\frac{2V}{2d}=\frac{V}{d}=E

So, the electric field is unchanged. And since the force on the particle at the center is directly proportional to the electric field:

F = qE

Then the force on the particle will stay the same.

4 0
3 years ago
Carbon is allowed to diffuse through a steel plate 9.7-mm thick. The concentrations of carbon at the two faces are 0.664 and 0.3
cupoosta [38]

Answer:

844°C

Explanation:

The problem can be easily solve by using Fick's law and the Diffusivity or diffusion coefficient.

We know that Fick's law is given by,

J = - D \frac{\Delta c}{\Delta x}

Where \frac{\Delta c}{\Delta x} is the concentration of gradient

D is the diffusivity coefficient

and J is the flux of atoms.

In the other hand we have, that

D= D_0 e^{\frac{E_d}{RT}}

Where D_0 is the proportionality constant,

R is the gas constant, T the temperature and E_d is the activation energy.

Replacing the value of diffusivity coefficient in Fick's law we have,

J = -D_0 ^{\frac{E_d}{RT}}\frac{\Delta c}{\Delta x}

Rearrange the equation to get the value of temperature,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

We have all the values in our equation.

\Delta c = 0.664-0.339 = 0.325 C. cm^{-1}

\Delta x = 9.7*10^{-3}m

E_d = 82000J

D_0 = 6.5*10^{-7}m^2/s

J = 3.2*10^{-9}m^2/s

R= 8.31Jmol^{-1}K

Substituting,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

T=-\frac{-82000}{(8.31)ln(\frac{3.2*10^{-9}(9.7*10^{-3})}{6.5*10^{-7} (0.325)})}

T=1118.07K=844\°C

4 0
3 years ago
A car is traveling at a constant speed on the highway. Its tires have a diameter of 65.0 cm and are rolling without sliding or s
lesya692 [45]

Answer:

Explanation:

The car is rolling without slipping so Vcm= R×ω = 0.325×49 = 16

4 0
3 years ago
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