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3 years ago

10

Please help me choose which one the correct answer is.

1 answer:

earnstyle [38]3 years ago

6 0

Answer:more than g I’m 99.9 % sure of it bc if you just drop the ball versus throwing the ball downward which is faster , throwing the ball downward so more than g hope this helps

Explanation:

You might be interested in

What caused the blackout in Canada?

Maru [420]

Answer:

Hello, the tripping of a 230-kilovolt transmission line.

Explanation:

the tripping of a 230-kilovolt transmission line near Ontario, Canada, at 5:16 p.m., which caused several other heavily loaded lines also to fail. Hopefully this helps you find what your looking for!.

7 0

3 years ago

If themass is 50kg, what weight of water is to be displaced to float on water? why

ludmilkaskok [199]

Answer:if youre looking for the weight of the thermas in genral it should be 500n

Explanation:using the formula w=mg

w=500x10

giving us 500 newtons which is the weight.

5 0

1 year ago

During a plane showcase, a pilot makes circular "looping" with a speed equal to the sound speed (340 m/s). However, the pilot ca

Dmitriy789 [7]

Answer:

1472.98 m

Explanation:

Data provided:

Speed of circular looping, v = 340 m/s

Acceleration, a = 8g

here,

g is the acceleration due to the gravity = 9.81 m/s²

Now,

the centripetal acceleration is given as,

$a=\frac{v^2}{r}$

r is the radius of the loop

on substituting the respective values, we get

$8\times9.81=\frac{340^2}{r}$

or

r = 1472.98 m

5 0

4 years ago

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a

S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

$p_{e}$ = momentum of an electron

$v_{e}$ = velocity of an electron

$m_{e} = 9.1\times 10_{- 31} kg$ = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

$\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T$

$}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}$

$}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}$

$v_{e} = 2.86\times 10^{5} m/s$ (2)

Using eqn (2) in (1):

$\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm$

Now, to calculate the de-Broglie wavelength of proton, $\lambda_{e}$ :

$\lambda_{p} = \frac{h}{p_{p}}$

$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

where

$m_{p} = 1.6726\times 10_{- 27} kg$ = mass of proton

$v_{p}$ = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

$\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T$

$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

$}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}$

$v_{p} = 6.674\times 10^{3} m/s$ (4)

Using eqn (4) in (3):

$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

7 0

3 years ago

Sue and Jenny kick a soccer ball at exactly the same time. Sue’s foot exerts a force of 75.9 N to the north. Jenny’s foot exerts

Lady_Fox [76]

Answer:

Fr^2 = 75.9N+105.8N=181.7

<u><em>Fr = </em></u><u><em>181.7N.</em></u>

6 0

3 years ago