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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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What is the resistance in a circuit with a 1.5-v battery and 10 amps<br> of current?
Mariana [72]

Answer:

15

Explanation:

15

3 0
3 years ago
How is work affected when an object is lifted straight up instead of using a ramp? Work will increase.
kifflom [539]
Without friction, the amount of work only depends on the final height,
and is not affected by the route used to get there. 

If the ramp has no friction, then it has no effect on the total amount
of work done.  The work to lift the load straight up is the same.

If the ramp has some friction, then it takes more work to use the ramp
than to lift the load straight up.  Then the work to lift the load straight up
would be less than when the ramp is used.


3 0
3 years ago
Car A is parked on a hill and Car B is driving down a level highway. Which statement best describes the two cars? A) Car A has k
12345 [234]
The answer is b because kinetic energy is the energy of motion, while potential energy is the energy stored in an object because of its position.
5 0
4 years ago
Read 2 more answers
A spherical snowball is melting in such a way that its radius is decreasing at rate of 0.1 cm/min. At what rate is the volume of
andrew11 [14]

Answer:

\frac{dV}{dt}=-78,4\pi \frac{cm^{3} }{min}

Explanation:

Knowing that the volume of a sphere is V=(4/3)πr³ and \frac{dr}{dt}=-0.1\frac{cm}{min}

We must find \frac{dV}{dt}=? when r=14cm

V=(4/3)πr³ ⇒

\frac{dV}{dt}=\frac{4}{3}\pi3r^{2}\frac{dr}{dt}\\\frac{dV}{dt}=4\pi r^{2}(-0.1\frac{cm}{min})

and r=14cm then

\frac{dV}{dt}=4\pi(14cm)^{2}(-0.1\frac{cm}{min})\\\frac{dV}{dt}=4\pi196cm^{2}(-0.1\frac{cm}{min})\\\frac{dV}{dt}=-78,4\pi \frac{cm^{3} }{min}

Note: as you can see the relationship of change of r with respect to t is negative because it is a decrease, and also its volume ratio

4 0
4 years ago
A projectile travels above level ground. The initial vertical component of its velocity is 0.90 m/s and the initial horizontal c
sladkih [1.3K]
<span>at maximum height the final velocity will be 0 
using v=u+at and resolving vertically we get 
v=0.6+(-9.81)t 
v=0.6-9.81t 
0=0.6-9.81t 
9.81t=0.6 
t=0.6/9.81 
t=0.061 to 3sf 

Now we need to resolve horizontally to find the horizontal distance 
using s=ut+1/2at^2 
However we now need the total time taken for the projectile travel and return to the ground. We can assume the time taken for the projectile to reach its maximum height and return to the ground is the same therefore 
the total time is 2 x 0.061=0.122seconds. They'll be now horizontal acceleration in this case scenario therefore 

Hence s=ut+1/2at^2 
since a=0 
s=ut 
s=0.6 x 0.122 
s=0.073m 
</span>
5 0
3 years ago
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