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love history [14]
3 years ago
15

Which of the following is an example of increasing friction intentionally? a. waxing skis b. adding grease to gears on a bike c.

throwing sand on an icy driveway d. oiling a squeaky door Please select the best answer from the choices provided
Physics
1 answer:
Aleonysh [2.5K]3 years ago
7 0
C i think that it tell me if its wrong
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Urgente!!!!! <br> Necesito ayuda con esto!!!!
ahrayia [7]
media.discordapp.net/attachments/782414373888458783/826224189828366377/video0.mp4
5 0
3 years ago
Acceleration is always act in the direction of​
hjlf

Answer:

Acceleration acts always in the direction. Of the displacement. Of the initial velocity.

5 0
3 years ago
Describe a situation in which an electron will be affected by an external electric field but will not be affected by an external
irina [24]

Answer:

Explanation:

Situations in which an electron will be affected by an external electric field but will not be affected by an external magnetic field

a ) When an electron is stationary in the electric field and magnetic field , he will be affected by electric field but not by magnetic field. Magnetic field can exert force only on mobile charges.

b ) When the electron is moving parallel to electric field and magnetic field . In this case also electric field will  exert force on electron but magnetic field field will not exert force on electrons . Magnetic field can exert force only on the perpendicular component of the velocity of charged particles.

Situations when electron is affected by an external magnetic field but not by an external electric field

There is no such situation in which electric field will not affect an electron . It will always affect an electron .

6 0
3 years ago
A man pushes a stalled car with a force of 100 N. If the acceleration of the car is 0.75 m/s 2 , what is the mass of the car?
shutvik [7]
125 because f=ma so you would use 100=mx0.75
4 0
3 years ago
A parallel beam of light of wavelength 4.5 x 10^-7 m is incident on a pair of slits that are 5.0 x 10^-4 m apart. The interferen
RSB [31]

Answer:

1.8x10⁻³m

Explanation:

From the question above, the following information was used to solve the problem.

wavelength λ = 4.5x10⁻⁷m

Length L = 2.0 meters

distance d = 5 x 10₋⁴m

ΔY = λL/d

= 4.5x10⁻⁷m (2) / 5 x 10₋⁴m

= 0.00000045 / 0.0005

= 0.0000009/0.0005

= 0.0018

= 1.8x10⁻³m

from the solution above The separation between two adjacent bright fringes is most nearly 1.8x10⁻³m

thank you!

3 0
3 years ago
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