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Yanka [14]
2 years ago
11

What is the magnitude of the displacement?

Physics
1 answer:
aleksklad [387]2 years ago
8 0

A) 20 m

<h3>What is called magnitude of displacement?</h3>

The magnitude of displacement is defined as the shortest distance between the initial and final position of the object.

For a particle in motion, the magnitude is either less than or equal to the distance travelled.

Also,

Velocity is simply defined as the rate of change of displacement with time.

Mathematically, it can be expressed as:

Velocity = displacement / time

According to the question,

The following data were obtained:

Time = 5 s

Velocity = 4 m/s

Displacement =?

Velocity = displacement / time

4 = d / 5

D = 4 × 5

D= 20 m

Hence,

20 m is the magnitude of the displacement of the object after it travelled for five seconds.

Learn more about magnitude of the displacement here:

brainly.com/question/10919017

#SPJ2

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A 0.5 kg ball moves in a circle that is 0.4 m in radius at a speed of 4.0 m/s. Calculate the force exerted on the ball.
blsea [12.9K]

Answer:

Explanation:

Given a ball of mass m= 0.5kg

The ball moves in as circle of radius r= 0.4m

Speed of the ball is v = 4m/s

Centripetal force is exerted on ball and it is given as

Fc = m•ac

ac is centripetal acceleration and it is given as

ac = v²/r

Then,

Fc = mv²/r

Fc = 0.5 × 4²/0.4

Fc = 20N.

The force exerted on the ball is 20N

5 0
4 years ago
What is capacitance?
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the amount of charge stored per volt

3 0
3 years ago
I help much help on 18!!!
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4 0
3 years ago
A 1500-kg car accelerates from 0 to 25 m/s in 7.0s with negligible friction and air resistance. What is the average power delive
NARA [144]

Answer:

90 hp

Explanation:

Power = work / time

P = ½ (1500 kg) (25 m/s)² / 7.0 s

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4 0
4 years ago
What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of −25.0 nc? g?
Daniel [21]

To calculate the force between two negative charges, we use the formula which is given by the Coulomb`s Law as

F=\frac{kq_{1}q_{2}  }{r^{2} }

Here, q_{1} andq_{2} are the charges on the pith balls, r is the separation between the charges and k is constant and its value is  8.99\times 10^9 N m^2/C^2.

Given  q_{1} =q_{2} =-25 nC=-25\times 10^{-9}C and  r=8 cm=8\times10^{-2} m.

Substituting these values in above formula we get,

F= 8.99\times 10^9 N m^2/C^2\frac{(-25\times 10^{-9}C)(-25\times 10^{-9}C)}{(8\times10^{-2} m)^2} \\\\\ F= 877929.7\times 10^{-9}  N\\\\F=8.8\times10^{-4}N

Thus, the repulsive force between two pith balls is 8.8\times10^{-4}N.

7 0
3 years ago
Read 3 more answers
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