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LekaFEV [45]
3 years ago
7

Which statement about the universe is correct​

Physics
1 answer:
likoan [24]3 years ago
7 0

Answer:

where's the statement?

Explanation:

how can i help without a statement.

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Two electrodes connected to a 9.0 v battery are charged to ±45 nc. What is the capacitance of the electrode?
mart [117]

Answer:

5\cdot 10^{-9} F

Explanation:

The capacitance of the electrode is given by:

C=\frac{Q}{V}

where

C is the capacitance

Q is the charge on the electrode

V is the potential difference

In this problem, we have

Q=45 nC=45\cdot 10^{-9} C

V = 9.0 V

Substituting into the equation, we find

C=\frac{45\cdot 10^{-9}C}{9.0 V}=5\cdot 10^{-9} F (5 nF)

6 0
3 years ago
Short essay your teacher have provided you​
Zinaida [17]
I don’t understand....?
6 0
3 years ago
Jack travelled 360 km at an average speed of 80 km/h. Elaine
diamong [38]

Answer:

Average speed of Elain = 60 km/h

Explanation:

Total Distance covered by Jack = 360km

Average Speed of Jack = 80 km/h

Time taken by Jack to complete his journey = Distance / Average speed = 360 km / 80 km/h

Time taken by Jack to complete his journey = 4.5 hours

As it is given the both Jack and Elain travelled the same amount of distance:

Total distance travelled by Elain = 360 km

It is given that Elain took 1.5 hourse more than Jack to cover the distance, so Time taken by Elain to cover the distance is = 4.5 hours + 1.5 hours = 6 hours

Average speed of Elain = Distance/ time = 360 km / 6 hours

Average speed of Elain = 60 km/h

3 0
3 years ago
A force of constant magnitude pushes a box up a vertical surface, as shown in the figure.
Ray Of Light [21]

The work done on the box by the applied force is zero.

The work done by the force of gravity is 75.95 J

The work done on the box by the normal force is 75.95 J.

<h3>The given parameters:</h3>
  • Mass of the box, m = 3.1 kg
  • Distance moved by the box, d = 2.5 m
  • Coefficient of friction, = 0.35
  • Inclination of the force, θ = 30⁰

<h3>What is work - done?</h3>
  • Work is said to be done when the applied force moves an object to a certain distance

The work done on the box by the applied force is calculated as;

W = Fd cos(\theta)\\\\W = (ma)d \times cos(\theta)

where;

a is the acceleration of the box

The acceleration of the box is zero since the box moved at a constant speed.

W = (0) d \times cos(30)\\\\W = 0 \ J

The work done by the force of gravity is calculated as follows;

W = mg \times d\\\\W = 3.1 \times 9.8 \times 2.5 \\\\W = 75.95 \ J

The work done on the box by the normal force is calculated as follows;

W = (F_n) \times d\\\\W = (mg + F sin\theta) \times d\\\\W = (mg + 0) \times d\\\\W = mgd\\\\W = 3.1 \times 9.8 \times 2.5\\\\W = 75.95 \ J

Learn more about work done here: brainly.com/question/8119756

8 0
2 years ago
The forces exerted on an object are shown.
damaskus [11]

The reaper watches. Satan watches. We all watch. As we fade. Into oblivion.

8 0
3 years ago
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