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Ludmilka [50]
3 years ago
12

In which switching technique, a dedicated

Physics
1 answer:
Aloiza [94]3 years ago
3 0
The answer is Circuit Switching
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Pravat exerts a force of 30 N to lift a bag of groceries 0.5 m. How much work did Pravat do on the bag?
Dahasolnce [82]

Answer:

15 J

Explanation:

Work = Force x Distance

15= 30 x 0.5

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7 0
3 years ago
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What is kinematics . Give two examples​
Wittaler [7]

Answer:

kinematics explains the terms like acceleration, velocity, and position of the objects while in motion.

Some important parameters in kinematics are displacement, velocity, and time.

Explanation:

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2 years ago
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A compound microscope has an objective lens of focal length 1.40 cm and an eyepiece with a focal length of 2.20 cm. The objectiv
olchik [2.2K]

Answer:

magnification is - 159

objective distance is 3.85 cm

Explanation:

Given data

focal length f1 = 1.40 cm

focal length f2 = 2.20 cm

separated d = 19.6 cm

to find out

angular magnification and How far from the objective

solution

we know magnification formula that is

magnification = ( - L / f1 ) (D/f2)

here D = 25 cm put all value

magnification = ( - 19.6 / 1.40 ) (25/2.20)

magnification = - 159

and

now we apply lens formula

i/f = 1/q + 1/p

p = f2 = 2.20

so

q = f2 p / p -f2

q = 1.4(2.20) / ( 2.2 - 1.4 )

q = 3.85 cm

so objective distance is 3.85 cm

3 0
3 years ago
A 0.70-kg basketball dropped on a hardwood floor rises back up to 65% of its original height. (a) If the basketball is dropped f
dem82 [27]

Answer:

Part a)

Loss = 3.6 J

Part b)

Loss = 0.99 J

Part C)

This is loss in terms of thermal energy due to collision with the floor

Explanation:

Part a)

Since we know that the ball rises up by 65% of initial height

so after first bounce it will lose 35% of its initial energy

so we will have

U = mgH

Energy Loss = 0.35 mgH[/tex]

Loss = 0.35(0.70)(9.81)(1.5)

Loss = 3.6 J

Part b)

Energy of the ball after first bounce

U_1 = 0.65 mgH

energy of ball after 2nd Bounce

U_2 = 0.65(0.65 mgH)

energy of the ball after 3rd bounce

U_3 = (0.65)(0.65^2)mgH

U_3 = 0.65^3(0.70)(9.81)(1.5)

U_3 = 2.83 J

Now we will have energy loss in fourth bounce given as

Loss = 0.35 U_3

Loss = 0.35(2.83)

Loss = 0.99 J

Part C)

This is loss in terms of thermal energy due to collision with the floor

7 0
3 years ago
2. A car travels 40 kilometers at an average speed of 80 km/h and then travels 40 kilometers at an average speed of 40 km/h. The
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53.3 km/h is the average amount of the car trip if 80km
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