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3 years ago

This question was a concept question on my physics test.

Physics

1 answer:

kkurt [141]3 years ago

6 0

Answer:

t√2.

Explanation:

Given that both are rest at initial position.

The time taken by stone A is t

h= 1/2 g t² ----------1

Lets time taken by stone B is t'

2 h= 1/2 g t'² -----------2

From equation 1 and 2

2 h= 1/2 g t'²

2 x 1/2 g t² = 1/2 g t'²

t'²=2 t²

t' = t√2

So the answer is t√2.

e. sqrt(2) .t

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On the way to school, Jed traveled 100m north, 300m east, 100 north, 100m east, and 100m north. A.) Find the total distance trav

Oduvanchick [21]

Answer:

Total distance = 700 m

Displacement = 500 m

Explanation:

Notice that Jed travelled a total of 3 x 100 m = 300 m in the North direction, and 300 m + 100 m = 400 m in the East direction. Therefore the total distance he travelled is: 300 + 400 = 700 m.

But the actual displacement is given by the Pythagorean theorem as the hypotenuse of a right angle triangle of legs 300 m and 400 m:

displacement = $\sqrt{300^2+400^2} =\sqrt{250000} =500\,m$

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3 years ago

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Answer:

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4 years ago

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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.

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3 years ago

What is the name

Molodets [167]

Answer:

<h3>It's called Resistance! </h3>

Explanation:

If the ratio is constant over a wide range of voltages, the material is said to "ohmic" material.

Hope it helps!

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3 years ago

A mover does 422 J of work pushing a crate 8.39 m. How much force did he exert?

kicyunya [14]

Answer:

50.3N

Explanation:

Work done = force x distance

422J. = force x 8.39m

÷8.39 both side to get force

Force is 50.3N to 1 d.p.

Check:

50.3 x 8.39=422.017J

Same as 422J to 1 d.p

4 0

2 years ago

Read 2 more answers