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Akimi4 [234]
2 years ago
8

The power in an electrical circuit is given by the equation P= RR, where /is

Physics
1 answer:
Kobotan [32]2 years ago
4 0

Answer:

The power in an electrical circuit is given by the equation P

The power in an electrical circuit is given by the equation P = I2R, where I is the current flowing through the circuit and R is the resistance of the circui

Explanation:

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The formula for speed is v = s/t. <br> a. True<br> b. False
crimeas [40]
If you mean S is the distance then it is true 
Velocity = Distance / time 
3 0
3 years ago
A car of 1400 kg is subject to multiple forces which produce an acceleration of 3.5 m/s2 directed north. Find the net force.​
Greeley [361]

Answer:

will

Explanation:

3 0
2 years ago
A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o
iVinArrow [24]
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: v=15 m/s
- radius of the hill: r=100 m

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car W=mg (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, m \frac{v^2}{r}, so we can write:
mg-N=m \frac{v^2}{r} (1)
By rearranging the equation and substituting the numbers, we find N:
N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
mg=m \frac{v^2}{r}
from which we find
v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s
7 0
3 years ago
Explain how surface and subsurface events are integral parts of the rock cycle.
vagabundo [1.1K]
Example of surface events are erosion and weathering. Erosion is the carrying of a particle from one place to the other and weathering is the breaking down of particles. These processes help in rock formation because this allows physical changes (grouping together or breaking down) on a certain substance. Subsurface events are those which happened underground such as the flow of underground water which subsequently allow the deposition of minerals, etc. 
3 0
3 years ago
Can someone help me with this real quick?
ch4aika [34]

Answer:

7,546 J

Explanation:

recall that Potential energy is given by

P.E = mgΔh

where m = 70kg (given)

g = 9.8 m/s² (acceleration due to gravity)

Δh = change in height

= distance from top of building to top of car

= height of building - height of car

= (5+8) - 2

= 11m

substituting all these into the equation:

P.E = mgΔh

= 70 x 9.8 x 11

= 7,546 J

4 0
3 years ago
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