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Rashid [163]
2 years ago
15

When capacitors are connected in parallel, they have the same:.

Physics
1 answer:
Gre4nikov [31]2 years ago
3 0

Answer:

THEY HAVE THE SAME VOLTAGE

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Statement or best answers each question.
Setler79 [48]

Answer:

See Explanation

Explanation:

1. A. Five

2. Both B. and C. say the same thing so I'm a bit confused but the answer is approximately 6,000 degrees C

3. I think B. It is under tremendous pressure

4. D. Both A. and B.

Hope this helps

8 0
3 years ago
Why must 0 s always be clearly marked on a motion map?
Ainat [17]
The answer would be latter e
5 0
3 years ago
An object is placed 18 cm in front of spherical mirror.if the image is formed at 4cm to the right of the mirror, calculate it's
ivolga24 [154]
1) Focal length

We can find the focal length of the mirror by using the mirror equation:
\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i} (1)
where 
f is the focal length
d_o is the distance of the object from the mirror
d_i is the distance of the image from the mirror

In this case, d_o = 18 cm, while d_i=-4 cm (the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
\frac{1}{f}= \frac{1}{18 cm}- \frac{1}{4 cm}=- \frac{7}{36 cm}
from which we find
f=- \frac{36}{7} cm=-5.1 cm

2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.

3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.

4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
r=2f=2 \cdot 5.1 cm=10.2 cm
3 0
3 years ago
Help me with this question please
vichka [17]

Answer:

Its true i'm pretty sure

Explanation:

7 0
3 years ago
Read 2 more answers
A bike first accelerates from 0.0m/s to 4.5m/s in 4.5 s, the continues at this constant speed for another 6.0 s. What is the tot
Alex_Xolod [135]

Answer:

37.125 m

Explanation:

Using the equation of motion

s=ut+0.5at^{2} where s is distance, u is initial velocity, t is time and a is acceleration

<u>Distance during acceleration</u>

Acceleration, a=\frac {V_{final}-V_{initial}}{t} where V_{final} is final velocity and V_{initial} is initial velocity.

Substituting 0.0 m/s for initial velocity and 4.5 m/s for final velocity, acceleration will be

a=\frac {4.5 m/s-0 m/s}{4.5 s}=1 m/s^{2}

Then substituting u for 0 m/s, t for 4.5 s and a for 1 m/s^{2} into the equation of motion

s=0*4.5+ 0.5*1*4.5^{2}=0+10.125 =10.125 m

<u>Distance at a constant speed</u>

At a constant speed, there's no acceleration and since speed=distance/time then distance is speed*time

Distance=4.5 m/s*6 s=27 m

<u>Total distance</u>

Total=27+10.125=37.125 m

3 0
3 years ago
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