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Masteriza [31]
2 years ago
12

Electric Cars A 12 volt car battery has a capacity of 100 ampere-hours, supplying 2 A of current for 50 hours. How much ENERGY d

oes it store when fully charged? How does this compare to the contents of a 9 gallon petrol tank (1350 MJ) (express as percent)?
Physics
1 answer:
Elina [12.6K]2 years ago
6 0

Answer: a) 4.32 MJ ; b)  0.32%

Explanation: In order to solve this question we have to consider that power consume to full charge the battery is I*V

so we have 100 A-hours* 12 V = 1200 W*hours ( J*hour/s)

To convert to Joules

we have to multiply by 3600 ( seconds per hour) the Total energy stored in  the battery is:

E= 1200 J/s* 3600 s= 4.32 MJ

To compare with the energy of 9 gallon petrol tank  ( 1350 MJ)

%= E electric/ Epetrol= (4.32 MJ/(1350 MJ))*100%=0.32 %

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Trong thí nghiệm về giao thoa sóng trên mặt nước gồm hai nguồn kết hợp S1S2 cách nhau 15 cm với dao động với tần số 30Hz. Tốc độ
mojhsa [17]

Answer:

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2 years ago
If a suspended object A is attracted to a charged object B, can we conclude that A is charged?
spayn [35]
Not necessarily, object A could also be neutral, and becoming a dipole due to object B's charge. A charged object can induce a dipole in a neutral object, and that object would then become attracted without being charged.
8 0
3 years ago
A camera with a 50.0-mm focal length lens is being used to photograph a person standing 3.00 m away. (a) How far from the lens m
kirill [66]

a) 50.8 mm

b) The whole image (1:1)

c) It seems reasonable

Explanation:

a)

To project the image on the film, the distance of the film from the lens must be equal to the distance of the image from the lens. This can be found by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length of the lens

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem:

f = 50.0 mm = 0.050 m is the focal length (positive for a convex lens)

p = 3.00 m is the distance of the person from the lens

Therefore, we can find q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{0.050}-\frac{1}{3.00}=19.667m^{-1}\\q=\frac{1}{19.667}=0.051 m=50.8 mm

b)

Here we need to find the height of the image first.

This can be done by using the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where:

y' is the height of the image

y = 1.75 m is the height of the real person

q = 50.8 mm = 0.0508 m is the distance of the image from the lens

p = 3.00 m is the distance of the person from the lens

Solving for y', we find:

y'=-\frac{qy}{p}=-\frac{(0.0508)(1.75)}{3.00}=-0.0296 m=-29.6mm

(the negative sign means the image is inverted)

Therefore, the size of the image (29.6 mm) is smaller than the size of the film (36.0 mm), so the whole image can fit into the film.

c)

This seems reasonable: in fact, with a 50.0 mm focal length, if we try to take the picture of a person at a distance of 3.00 m, we are able to capture the whole image of the person in the photo.

3 0
2 years ago
The force on a bullet that results from summing all of the forces on that bullet is known as the _____ on the bullet.
statuscvo [17]

Answer:

The force on a bullet that results from summing all of the forces on that bullet is known as the net force on the bullet.

Explanation:

Net force represents the overall force or the total force acting on an object.

Here, the object is bullet and the net force acting on it is the addition of all the forces acting on the bullet.

4 0
2 years ago
Read 2 more answers
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