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3 years ago

11

HELP EASY PHYSICS QUESTION 1.A ball at rest is dropped from the top of an inclined plane at a height of 50 cm. What is the speed

of the ball when it reaches the bottom of the plane? a.3.13 m/s b.4.9 m/s c.9.8 m/s d.31.3 m/s 2. A 25 kg crate is moved through a distance of 4 m by a horizontal force of 60 N. What is the work done by the frictional force in moving the crate if the coefficient of kinetic friction is 0.2? a.240 J b.-240 J c.-196 J d.196 J

1 answer:

dexar [7]3 years ago

7 0

Answer:

45

Explanation:

it,s because of addition

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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 30.0 kg box hangs from a rope. What is the tension (in N) in the rope if the box has an initial velocity of 2.0 m/s and is slo

hjlf

Answer:

360 N

Explanation:

m = 30kg u = 2 m/s a = -2m/s/s

Since the object has an initial velocity of 2 m/s and acceleration of -2 m/s/s

the object will come to rest in 1 second but the force applied in that one second can be calculated by:

F = ma

F = 30 * -2

F = -60 N (the negative sign tells us that the force is acting downwards)

Now, calculating the force applied on the box due to gravity

letting g = -10m/s/s

F = ma

F = 30 * -10

F = -300 N (the negative sign tells us that the force is acting downwards)

Now, calculating the total downward force:

-300 + (-60) = -360 N

<em></em>

<em>Hence, a downward force of 360 N is being applied on the box and since the box did not disconnect from the rope, the rope applied the same amount of force in the opposite direction</em>

Therefore tension on the force = <u>360 N</u>

5 0

4 years ago