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VladimirAG [237]
2 years ago
14

Pls helppp :)List several examples of applied force, normal force, and friction that you’ve observed in your life.

Physics
1 answer:
Anon25 [30]2 years ago
5 0

Answer:

an example for applied force is

Explanation:

The applied force is the force applied to the object to either displace it or change its shape.

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Rosný bod závisí od ?
WITCHER [35]

Answer:

what did u say and what language are you speaking in

8 0
3 years ago
Hansel pushes a 2 lb. box 5 feet in 20 seconds. How much work has he done?
Licemer1 [7]
Work = force x distance

You can see time doesn’t matter (if we were talking about power, which is the RATE at which work is performed, that would be a different story).

W = 2 x 5 = 10 foot-pounds of work

Foot-pounds are gross units. Better to work in SI units when you can!

8 0
3 years ago
PLS HEP WILL GIVE BRAINLIEST TO FIRST ANSWER SCIENCEE!!
MAVERICK [17]

Answer:

Claim: The heart pumps more blood throughout the body when one exercises because exercise takes a lot of energy from the body.

Evidence: Heart rate went from 80 bpm to 120 bpm

Reasoning: Doing exercise takes a lot of energy to do, causing the circulatory system to have to work harder and pump more blood throughout the body in order to allow someone to be able to do a task that involves so much movement and energy.

Explanation:

3 0
2 years ago
If a t-shirt gun can fire t-shirts with an initial speed of 15 m/s, what is the maximum distance (along horizontal, flat ground)
kotegsom [21]

Answer:

h = 11.47 m

Explanation:

Initial speed pf the t-shirt gun is 15 m/s

We need to find the maximum distance covered by the t-shirt. It is based on the conservation of energy. The maximum distance covered is given by :

h=\dfrac{u^2}{2g}\\\\h=\dfrac{(15)^2}{2\times 9.8}\\\\h=11.47\ m

So, it will cover a distance of 11.47 m.

7 0
3 years ago
A caris initially at rest starts moving with a constant acceleration of 0.5 m/s2 and travels a distance of 5 m. Find
EleoNora [17]

Answer:

(I)

{ \bf{ {v}^{2} =  {u}^{2}  - 2as }} \\  {v}^{2}  =  {0}^{2}  - (2 \times 0.5 \times 5) \\  {v}^{2}  = 5 \\ { \tt{final \: velocity = 2.24 \:  {ms}^{ - 1} }}

(ii)

{ \bf{v = u + at}} \\ 2.24 = 0 + (0.5t) \\ { \tt{time = 4.48 \: seconds}}

8 0
3 years ago
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