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3 years ago

Why is the answer B and not E?

Physics

2 answers:

makkiz [27]3 years ago

8 0

Answer: B

<u>Explanation:</u>

E shows the direction of the ball but the question asks about the direction of the STICK, which is the opposite of the ball.

Margarita [4]3 years ago

5 0

Explanation:

The ball's horizontal velocity changes from vi to 0, so the acceleration is negative, meaning the force component must point to the left.

The ball's vertical velocity changes from 0 to vf, so the acceleration is positive, meaning the force component must point up.

So you're looking for a force that points to the left and up, where the horizontal component is greater than the vertical component.

You might be interested in

The acceleration due to gravity on the Moon's surface is

Molodets [167]

Answer:

50 lb

Explanation:

Given,

The weight of astronaut's life support backpack on Earth (w) = 300 lb

Acceleration due to gravity on Earth (g) = 9.8 m/s²

Acceleration due to gravity on Moon = g'

$g'=\frac{g}{6}$

We know that weight of an object on Earth is,

$w = m\times g$

$m = \frac{w}{g}$

Similarly, weight on Moon will be

$w' = m\times g'$

$w' = \frac{w}{g}\times\frac{g}{6}$

$w' = \frac{300}{6}$

$w' = 50$

Thus the astronaut's life support backpack will weigh 50 lb on Moon.

7 0

3 years ago

He throws a second ball (B2) upward with the same initial velocity at the instant that the first ball is at the ceiling. c. How

Gwar [14]

Answer:

hello your question has some missing parts

A juggler performs in a room whose ceiling is 3 m above the level of his hands. He throws a ball vertically upward so that it just reaches the ceiling.

answer : c) 0.39 sec

d) 2.25 m

e) 1.92 m/sec

Explanation:

The initial velocity of the first ball = 7.67 m/sec ( calculated )

Time required for first ball to reach ceiling = 0.78 secs ( calculated )

Determine how long after the second ball is thrown do the two balls pass each other

Distance travelled by first ball downwards when it meets second ball can be expressed as : d = 1/2 gt^2 = 9.8t^2 / 2

hence d = 4.9t^2 ----- ( 1 )

Initial speed of second ball = first ball initial speed = 7.67 m/sec

3 - d = 7.67t - 4.9t ---- ( 2 )

equating equation 1 and 2

3 = 7.67t therefore t = 0.39 sec

Determine how far the balls are above the Juggler's hands ( when the balls pass each other )

form equation 1 ;

d = 4.9 t^2 = 4.9 *(0.39)^2 = 0.75 m

therefore the height the balls are above the Juggler's hands is

3 - d = 3 - 0.75 = 2.25 m

determine their velocities when the pass each other

velocity = displacement / time

velocity = d / t = 0.75 / 0.39 sec = 1.92 m/sec

7 0

3 years ago

Select the correct answer.

r-ruslan [8.4K]

Answer:

That would be B. Hope this helps!

Explanation:

Orion's Belt or the Belt of Orion, also known as the Three Kings or Three Sisters, is an asterism in the constellation Orion. It consists of the three bright stars Alnitak, Alnilam and Mintaka. Looking for Orion's Belt in the night sky is the easiest way to locate Orion in the sky.

3 0

3 years ago

Read 2 more answers

Q 2 Two mirrors meet at right angles. A ray of light is incident on one at an angle of 30°

serg [7]

a ray of light is incident towards a plane mirror at an angles of 30degrees with the mirror surface. what will be the angles of reflection is 60degree.

3 0

3 years ago

X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to

lys-0071 [83]

Answer:

a) $\Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

b) $\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

c) $E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

Explanation

Part a

For this case we can use the Compton shift equation given by:

$\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

Part b

For this cas we can calculate the wavelength of the phton with this formula:

$\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

$E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

3 0

3 years ago

Read 2 more answers