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Alex777 [14]
3 years ago
10

Which of the following is an example of a force?

Physics
1 answer:
nydimaria [60]3 years ago
5 0

Answer:

D) Both A and B

Explanation:

When an object exerts either a <em>push</em> or a <em>pull</em> on another object, this is known as a force.

If you have any questions, just feel free to let me know!

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At what angle do the cars move after the collision?
liubo4ka [24]

Answer:

his results in the final angle after the collision of 37.2 degrees basically what we did there is turn the vector into a right triangle. We use sohcahtoa to solve for the angle. Being.

Explanation:

6 0
2 years ago
Which of the following is incorrect for nuclear forces
densk [106]

Answer:

c they obey inverse square law

4 0
3 years ago
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Put the waves in order from shortest to longest wavelength
FromTheMoon [43]

Answer:

b, a, c

Explanation:

The middle one has the shortest wavelength, then it's the top one and the last one has the longest wavelength.

3 0
2 years ago
An electron moves a 2.0x10^6 meter per the second perpendicular to a magnetic field having a flux density of 2 teslas. What is t
Darina [25.2K]

Answer:

F_B = 6.4*10^-13 N

Explanation:

The magnetic force on the electron, generated by the motion of the electron and the magnetic field is given by:

F_B=qv\ X\ B

q: electron charge = 1.6*10^{-19}C

v: speed of the electron = 2.0*10^6 m/s

B: magnitude of the magnetic field = 2T

However, the direction of B and v are perpendicular between them. So, the angle between vectors is 90°. The magnitude of the magnetic force is:

|F_B|=qvBsin90\°=qvB

You replace the values of q, v and B in the last equation:

|F_B|=(1.6*10^{-19}C)(2.0*10^6m/s)(2T)\\\\|F_B|=6.4*10^{-13}\ N

hence, the magnetic force on the electron is 6.4*10^-23 N

6 0
3 years ago
A hot-air balloon is 11.0 m above the ground and rising at a speed of 7.00 m/s. A ball is thrown horizontally from the balloon b
Nesterboy [21]

Answer:

18.6 m/s

Explanation:

h = Initial height of the balloon = 11 m

v_{o} = initial speed of the ball

v_{oy} = initial vertical speed of the ball = 7 m/s

v_{ox} = initial horizontal speed of the ball = 9 m/s

initial speed of the ball is given as

v_{o} = \sqrt{v_{ox}^{2} + v_{oy}^{2}} = \sqrt{9^{2} + 7^{2}} = 11.4 m/s

v_{f} = final speed of the ball as it strikes the ground

m = mass of the ball

Using conservation of energy

Final kinetic energy before striking the ground = Initial potential energy + Initial kinetic energy

(0.5) m v_{f}^{2} = (0.5) m v_{o}^{2} + mgh \\(0.5) v_{f}^{2} = (0.5) v_{o}^{2} + gh\\(0.5) v_{f}^{2} = (0.5) (11.4)^{2} + (9.8)(11)\\(0.5) v_{f}^{2} = 172.78\\v_{f} = 18.6 m/s

7 0
3 years ago
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