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Norma-Jean [14]
3 years ago
14

What is the motor effect?

Physics
1 answer:
Aliun [14]3 years ago
8 0
The motor effect is the term used when a current-carrying wire in the presence of a magnetic field experiences a force.
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Convert 8 light years to Astronomical Units
marusya05 [52]

Answer:

505929 AU

Explanation:

As you may know, one light-year is equivalent to approximately 63241.1 Astronomical Units. To get your answer, simply multiply 63241.1 * 8 to get ≈505929 AU

5 0
3 years ago
Consider a small frictionless puck perched at the top of a xed sphere of radius R. If the puck is given a tiny nudge so that it
MakcuM [25]

Answer:

Explanation:

Let the vertical height by which it descends be h . Let it acquire velocity of v .

1/2 mv² = mgh

v² = 2gh

As it leaves the surface of sphere , reaction force of surface  R = 0 , so

centripetal force = mg cosθ where θ is the angular displacement from the vertex .  

mv² / r = mg cosθ

(m/r )x 2gh = mg cosθ

2h / r = cosθ

cosθ = (r-h) / r

2h / r =  r-h / r

2h = r-h

3h = r

h = r / 3

5 0
3 years ago
What is the difference between heat exhaustion and heat stroke?
balu736 [363]
I believe the answer is D, Heat exhaustion involves a lack of sweating, while heat stroke involves extreme sweating. Also just to add the on if heat exhaustion is left untreated then it could turn into a heat stroke.
6 0
3 years ago
Read 2 more answers
Dejamos caer un objeto desde lo alto de una torre y medimos el tiempo que tarda en llegar al suelo que resulta ser de 0,02 minut
Harman [31]

Answer: a) 11.76 m/s  b) 7.056 m

Explanation:

The described situation is as follows:

An object is dropped from the top of a tower and when measuring the time it takes to reach the ground that turns out to be 0.02 minutes.

This situation is related to free fall, this also means we have constant acceleration, hence the equations we will use are:

V_{f}=V_{o}+at (1)  

{V_{f}}^{2}={V_{o}}^{2}+2ad (2)  

Where:  

V_{f} Is the final velocity of the object

V_{o}=0 Is the initial velocity of the object (it was dropped)

a=9.8 m/s^{2} is the acceleration due gravity

d is the height of the tower

t=0.02min=1.2 s is the time it takes to the object to reach the ground

b) Begining with (1):

V_{f}=0+at (3)  

V_{f}=at=(9.8 m/s^{2})(1.2 s) (4)  

V_{f}=11.76 m/s (5)  This is the final velocity of the object

a) Substituting (5) in (2):

(11.76 m/s)^{2}=0+2(9.8 m/s^{2})d (6)  

Clearing d:

d=\frac{(11.76 m/s)^{2}}{2(9.8 m/s^{2})} (7)  

d=7.056 m (8)  This is the height of the tower

4 0
4 years ago
A 120.0 kg crate is placed on a 15.00°
Citrus2011 [14]

F = 2820.1 N

Explanation:

Let the (+)x-axis be up along the slope. The component of the weight of the crate along the slope is -mgsin15° (pointing down the slope). The force that keeps the crate from sliding is F. Therefore, we can write Newton's 2nd law along the x-axis as

Fnet = ma = 0 (a = 0 no sliding)

= F - mgsin15°

= 0

or

F = mgsin15°

= (120 kg)(9.8 m/s^2)sin15°

= 2820.1 N

7 0
3 years ago
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