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joja [24]
2 years ago
14

Two point charges are separated by 6.4 cm . The attractive force between them is 10 N . Suppose that the charges attracting each

other have equal magnitude. Rearrange Coulomb's law and find the magnitude of each charge. Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
LenaWriter [7]2 years ago
7 0

Answer:

Two point charges are separated by 6.4 cm . The attractive force between them is 10 N .

units.

Explanation:

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A 0.700-kg particle has a speed of 1.90 m/s at point circled A and kinetic energy of 7.20 J at point circled B. (a) What is its
Savatey [412]

Answer:

a). E_{kA}=1.2635 J

b). V_{B}=4.535\frac{m}{s}

c). ΔE_{t}=8.4635 J

Explanation:

ΔE=kinetic energy

a).

E_{kA}=\frac{1}{2}*m*v_{A} ^{2} \\ v_{A}=1.9 \frac{m}{s}\\ m=0.70kg\\E_{kA}=\frac{1}{2}*0.70kg*(1.9 \frac{m}{s})^{2} \\E_{kA}=1.2635 J

b).

E_{kB}=\frac{1}{2}*m*v_{B} ^{2}

V_{B}^{2}=\frac{E_{kB}*2}{m} \\V_{B}=\sqrt{\frac{E_{kB}*2}{m}} \\V_{B}=\sqrt{\frac{7.2J*2}{0.70kg}} \\V_{B}=4.53 \frac{m}{s}

c).

net work= EkA+EkB

E_{t}=E_{kA}+ E_{kB}\\E_{t}=1.2635J+7.2J\\E_{t}=8.4635J

3 0
3 years ago
The mass of 25cm of ivory was found to be 0.045kg. caculate the density of ivory in SI units​
solmaris [256]

Answer:

dgdhvjgfjfdcdrtdgxbhjcdtsdhcbgtdytfgiuftrshdryrgbiuwdbcuhdds

Explanation:

,ajsgfgsdfubcygduhdsbcdshyudbckjdshfyendhdndbhfdbuehundweugdjnbfcjdfhewjdnbewqwioehsajcnbsfhhgjhbgjhhbjdhgwbvdwnjjguhbwqhjbsjwqhshwbdjagdasndhsajuwqbdjasnjashdjaskxnjgjjsklx ms,dhwshjksnjdbhjssnj,

7 0
2 years ago
Substance A has a heat capacity that is much greater than that of substance B. If 10.0 g of substance A initially at 30.0 ∘C is
astra-53 [7]

Explanation:

In a heat exchange, the temperature change is inversely proportional to the specific heat capacity. Since substance A has a heat capacity that is much greater than that of substance B, the temperature change of substance A will be less than the temperature change of substance B. Therefore, the final temperature is closer to that of 30^\circ C than 80^\circ C.

8 0
3 years ago
Ch 31 HW Problem 31.63 10 of 15 Constants In an L-R-C series circuit, the source has a voltage amplitude of 116 V , R = 77.0 Ω ,
Degger [83]

Answer:

a. I = 0.76 A

b. Z = 150.74

c. RL₁ = 34.41  ,  RL₂ = 602.58

d. RL₂ = 602.58

Explanation:

V₁ = 116 V , R₁ = 77.0 Ω , Vc = 364 V ,  Rc = 473 Ω

a.

Using law of Ohm

V = I * R

I = Vc / Rc =  364 V / 473 Ω

I = 0.76 A

b.

The impedance of the circuit in this case the resistance, capacitance and inductor

V = I * Z

Z = V / I

Z = 116 v / 0.76 A

Z = 150.74

c.

The reactance of the inductor can be find using

Z² = R² + (RL² - Rc²)

Solve to RL'

RL = Rc (+ / -) √ ( Z² - R²)

RL = 473 (+ / -)  √ 150.74² 77.0²

RL = 473 (+ / -)  (129.58)

RL₁ = 34.41  ,  RL₂ = 602.58

d.

The higher value have the less angular frequency  

RL₂ = 602.58

ω = 1 / √L*C

ω = 1 / √ 602.58 * 473

f = 285.02 Hz

6 0
3 years ago
A mechanic uses a mechanical lift to raise a car. The car weighs 11,000 N. The lift raises the car 2.5 m.
KengaRu [80]

Potential energy = (weight) x (height)

After the car has been raised 2.5 meters, it has

                 (11,000) x (2.5) = 27,500 Joules

MORE potential energy than it had before it was lifted.

That's the energy that has to come from the work you do to lift it.

Since no mechanical process is ever 100% efficient, the work required
to accomplish this task is <em>at least  27,500 joules</em>.


5 0
3 years ago
Read 2 more answers
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