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laila [671]
3 years ago
9

What charges are needed in the objects to attract both objects?

Physics
2 answers:
kvasek [131]3 years ago
8 0

Answer:

both

Explanation:

Verizon [17]3 years ago
4 0

Answer:

a neutral object will atract the both object

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9. Name the two different symbols used to store all digital data on microchips in binary code. Explain how only two options can
elena-s [515]

Answer:

it got me confused for a bit but then i realized the answer was 1 and 0

Explanation:

ever hear of the phrase 1011000001010101001010100100001001000111? yeah ap.ex is talking about that

4 0
3 years ago
In physics how much energy is required to raise the temp of 9.78kg of water from 40.82c to 52.07c? The specific heat of water in
Sati [7]

Answer:

469.6KJ

Explanation:

Heat energy required can be calculated using the formula

H = mc∆t where

m is the mass of the water

c is the specific heat capacity of the water

∆t is the change in temperature of the water

Given m = 9.78kg

c = 4186j/kg-c.

∆t = 52.07°C - 40.82°C

∆t = 11.25°C

H = 9.78 × 4186 × 11.25

H = 460,564.65Joules

= 460.6KJ

7 0
3 years ago
Read 2 more answers
What angle is formed by the sun, the earth, and the moon during an eclipse?.
Andrew [12]

Answer:

The Sun-Earth-Moon system happens to exhibit a striking geometric coincidence, which we examine in the first problem. PROBLEM 1. To an observer on Earth, the Sun and the Moon subtend almost the same angle in the sky. The average angle is 0.52 degrees for the Moon and 0.53 degrees for the Sun.

4 0
3 years ago
A loop circuit has a resistance of R1 and a current of 1.8 A. The current is reduced to 1.6 A when an additional 3.8 Ω resistor
Allisa [31]

Answer:

Value of R_1=30.4ohm

Explanation:

We have given

In first case resistance is R_1 and current is 1.8 A

Let the potential difference is v

So 1.8=\frac{v}{R_1}----eqn 1

In second case resistance is R_1+3.8 and current is 1.6 A and potential difference will be as it is a series connection

So 1.6=\frac{v}{R+3.8}----eqn 2

From eqn 1 and eqn 2

1.8R_1=1.6R_1+6.08

R_1=30.4ohm

6 0
3 years ago
Read 2 more answers
Electronic flash units for camera contain a capacitor for storing the energy used to produce the flash. In one such unit, the fl
nadezda [96]

Answer:

420J

Explanation:

Power is the time rate of change in energy. Power is the ratio of energy to time. The S.I unit of power is in watts.

Given that the flash lasts for 1/675 s, power output is 2.7 * 10⁵ W. Hence:

Power = Energy / time

Substituting:

2.7 * 10⁵ W = Energy / (1/675)

Energy = 2.7 * 10⁵ W * 1/675 = 400J

Therefore the energy emitted as light is 400J.

Since the conversion of electric energy to light is 95% efficient, hence the energy stored as electrical energy is:

Energy(capacitor) = 5% of 400J + 400J = 0.05*400 + 400

Energy(capacitor) = 420J

8 0
3 years ago
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