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Svetradugi [14.3K]
4 years ago
14

How much power is used by a 12.0-v car battery that draws 0.5 a of current?

Physics
1 answer:
tigry1 [53]4 years ago
3 0
The electric power used by the car battery is given by
P=VI

where
V is the voltage of the battety
I is the current
P is the electric power.

For the battery in our problem, the voltage is V=12.0 V while the current intensity is I=0.5 A, so the power used by the battery is
P=VI=(12.0 V)(0.5 A)=6.0 W
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Temperature has a(n) _____ effect on the pressure and volume of a gas.
Fiesta28 [93]

Temperature has a direct effect on the pressure of a gas.

If the pressure is constant and the gas is free to fill more
or less space, then temperature also has a direct effect on
the volume of the gas.

4 0
4 years ago
How many simple machines are in a wine opener? The kind with the bottle opener at the head and two "arms."
Blababa [14]
It uses the corkscrew to anchor it to the cork and a lever to pull the cork out

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7 0
3 years ago
Read 2 more answers
1. Use Coulomb’s Law (equation below) to calculate the approximate force felt by an electron at point A in the schematic below.
Amiraneli [1.4K]

Answer:

Explanation:

From the data it appears that A is the middle point between two charges.

First of all we shall calculate the field at point A .

Field due to charge -Q ( 6e⁻ ) at A

= 9 x 10⁹ x 6 x 1.6 x 10⁻¹⁹ / (2.5)² x 10⁻⁴

= 13.82 x 10⁻⁶ N/C

Its direction will be towards Q⁻

Same field will be produced by Q⁺ charge . The direction will be away

from Q⁺  towards Q⁻ .

We shall add the field  to get the resultant field  .

= 2 x 13.82 x 10⁻⁶

= 27.64 x 10⁻⁶ N/C

Force on electron put at A

= charge x field

= 1.6 x 10⁻¹⁹ x 27.64 x 10⁻⁶

= 44.22 x 10⁻²⁵ N

8 0
4 years ago
A block is projected up a frictionless inclined plane with initial speed v0 = 7.14 m/s. The angle of incline is θ = 36.5°. (a) H
Wewaii [24]

Answer:

(a)x=4.37m\\\\(b)t=1.225s\\\\(c) v_{f}=7.14m/s

Explanation:

Given data

v_{o}=7.14m/s\\\alpha =36.5^o

For Part (a)

Starting with the -ve acceleration of the body (opposite to the gravitational force)

a=-gSin\alpha \\a=-(9.8m/s^2)Sin(36.5)\\a=-5.83m/s^2

Using equation of motion

v_{f}^2=v_{o}^2+2ax\\(0m/s)^2=(7.14m/s)^2+2(-5.83m/s^2)x\\-(7.14m/s)^2=2(-5.83m/s^2)x\\x=\frac{-(7.14m/s)^2}{2(-5.83m/s^2}\\ x=4.37m

For Part (b)

Using the result in Part (a) we can substitute in other equation of motion to get time t:

x=\frac{1}{2}vt\\ 4.37m=\frac{1}{2}(7.14m/s)t\\ (7.14m/s)t=2*(4.37)\\t=8.744/7.14\\t=1.225s

For Part (c)

At state 2 where vo=0m/s and the acceleration is positive (same direction as the gravitational force)

a=gSin\alpha \\a=(9.8m/s^2)Sin(36.5)\\a=5.83m/s^2\\\\\\v_{f}^2=v_{o}^2+2ax\\v_{f}^2=(0m/s)^2+2(5.83m/s^2)(4.37m)\\v_{f}^2=50.95\\v_{f}=\sqrt{50.95}\\ v_{f}=7.14m/s

4 0
3 years ago
A uniform electric field, with a magnitude of 370 N/C, is directed parallel to the positive x-axis. If the electric potential at
dem82 [27]

Answer:

\Delta U = 0.2072 J

Explanation:

Potential difference between two points in constant electric field is given by the formula

\Delta V = E.\Delta x

here we know that

E = 370 N/C

also we know that

\Delta x = 2.1 - 1.9 = 0.2 m

now we have

\Delta V = 370 (0.2) = 74 V

now change in potential energy is given as

\Delta U = Q\Delta V

\Delta U = (2.80 \times 10^{-3})(74)

\Delta U = 0.2072 J

3 0
4 years ago
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