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melamori03 [73]

3 years ago

6

A frictionless pendulum clock on the surface of the earth has a period of 1.00 s. On a distant planet, the length of the pendulu

m must be shortened slightly to have a period of 1.00 s. What is true about the acceleration due to gravity on the distant planet?

1 answer:

Marianna [84]3 years ago

3 0

Answer:

The acceleration due to gravity on the other planet is less than the acceleration due to gravity on Earth

Explanation:

The time period of a pendulum is given by

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

where,

L = Length of the pendulum

g = Acceleration due to gravity

It can be seen that the time is proportional to the length and inversely proportional to the acceleration due to gravity. So, if L is reduced to keep the same time period then g must be also less.

Hence, the acceleration due to gravity on the other planet is less than the acceleration due to gravity on Earth.

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What is the force of a 12 kg rock falling at 9.8m/s/s

velikii [3]

F=W=mg
F=12*9.8

F=117.6Newtons

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Which of the following are vector quantities? Check all that apply.

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3 years ago

An electron traverses a vacuum tube with a length of 2 m in 2 X 10- 4

Maslowich

Answer:

Average speed = 10,000 m/s

Explanation:

Given the following data;

Distance = 2m

Time = 0.0002secs

To find the average speed;

Average speed = distance/time

Average speed = 2/0.0002

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7 0

3 years ago

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

7 0

3 years ago

a vector has an x-component of 19.5m and a y-component of 28.4m. Find the magnitude and direction of the vector

Delicious77 [7]

Answer:

magnitude=34.45 m

direction= $55.52\°$

Explanation:

Assuming the initial point P1 of this vector is at the origin:

P1=(X1,Y1)=(0,0)

And knowing the other point is P2=(X2,Y2)=(19.5,28.4)

We can find the magnitude and direction of this vector, taking into account a vector has a initial and a final point, with an x-component and a y-component.

For the magnitude we will use the formula to calculate the distance $d$ between two points:

$d=\sqrt{{(Y2-Y1)}^{2} +{(X2-X1)}^{2}}$ (1)

$d=\sqrt{{(28.4 m - 0 m)}^{2} +{(19.5 m - 0m)}^{2}}$ (2)

$d=\sqrt{1186.81 m^{2}}$ (3)

$d=34.45 m$ (4) This is the magnitude of the vector

For the direction, which is the measure of the angle the vector makes with a horizontal line, we will use the following formula:

$tan \theta=\frac{Y2-Y1}{X2-X1}$ (5)

$tan \theta=\frac{24.8 m - 0m}{19.5 m - 0m}$ (6)

$tan \theta=\frac{24.8}{19.5}$ (7)

Finding $\theta$ :

$\theta= tan^{-1}(\frac{24.8}{19.5})$ (8)

$\theta= 55.52\°$ (9) This is the direction of the vector

3 0

3 years ago