Alisa, a skateboarder, is riding down a hill. If Alisa and the skateboard are considered a single object, what forces are acting on her? Check all that apply.

Answers

Answer 1

Answer: gravity

normal force

friction

Explanation:

A force is refered to as the interaction which when unopposed, will lead to a change in an objects motion. The velocity of an object will be changed when force is applied.

Since Alisa s riding down a hill and she and the skateboard are considered to be a single object, then the forces that are acting on her include gravity, normal force and friction.


Related Questions

A spring extends by 10cm when a mass of 100g is attached to it. What is the spring constant?​

Answers

Answer:

10N/m

Explanation:

Using F=kx

F=mg

k=mg/x

k=0.1*10/0.1 (kg*m/s^2)/m

10N/m


[tex]what \: is \: light \: {?} [/tex]

Answers

Light is a form of electromagnetic radiation with a wavelength which can be detected by the human eye. It is a small part of the electromagnetic spectrum and radiation given off by stars like the sun. Animals can also see light. The study of light, known as optics, is an important research area in modern physics.

Answer:

Light is a form of energy which produces the sensation of sight .

hope it is helpful to you

The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actual depth of the liquid in the tank?​

Answers

[tex] \orange{\underline{\huge{\bold{\textit{\green{\bf{QUESTION}}}}}}}[/tex]

The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actual depth of the liquid in the tank?

[tex]{\bold{\blue{GIVEN}}}[/tex]

REFRACTIVE INDEX = 1.3

APPARENT DEPTH = 7.7 cm

[tex]{ \bold{\green{To \: Find}}}[/tex]

REAL DEPTH OF THE OBJECT.

[tex]{\red{FORMULA \: \: USED }}[/tex]

[tex]Reflective \: Index = \frac{Real \: Depth}{Apparent \: Depth } [/tex]

[tex] \huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}[/tex]

Refractive Index = 1.3

Apparent Depth = 7.7 cm

Putting the values in the formula:-

[tex]Reflective \: Index = \frac{Real \: Depth}{Apparent \: Depth } \\ \\ 1.3 = \frac{Real \: Depth}{7.7 \: cm} \\ \\ 1.3 \times 7.7 = Real \: Depth \\ \\ 10.01 \: \: cm = Real \: Depth[/tex]

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