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2 years ago

What is the displacement of the particle in the time interval 5 seconds to 8 seconds?

Physics

2 answers:

Marizza181 [45]2 years ago

7 0

Answer:

A. 160 meters

B. 80 meters

C. 40 meters

D. 20 meters

E. 0 meters

The correct answer is letter E. 0 meters.

Explanation:

Dmitriy789 [7]2 years ago

3 0

The displacement is the vector with

magnitude
   distance between position at 5 sec and position at 8 sec
and direction
   direction from position at 5 sec to position at 8 sec .

The route followed during the time interval is irrelevant.

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hram777 [196]

The answer is D, human.

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If the jet is moving at a speed of 1300 km/h at the lowest point of the loop, determine the minimum radius of the circle so that

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Answer:

R = 2216m and The normal force of the seat on the pilot is 5008N

Explanation:

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2 years ago

Joey drives along the 110 South freeway and notices a mile marker that reads 260 miles. He drives until he reaches the 150 mile

Nostrana [21]

Answer:

85 miles .

Explanation:

Displacement along the 110 South freeway = 260 - 150 = 110 miles

Displacement along the 110 North freeway = 150 - 175 = - 25 miles

Net displacement = 110 - 25 = 85 miles

So Joey's displacement from the 260 mile marker is 85 miles .

8 0

2 years ago

When you push a 1.90-kg book resting on a tabletop, you have to exert a force of 2.10 N to start the book sliding. Once it is sl

zimovet [89]

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3 0

2 years ago

A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9

Andreyy89

Answer:

$d = 0.0306 m$

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

$U + K = constant$

$\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2$

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

$\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0$

$7.05 + 30.5 = \frac{0.266}{r}$

$r = 7.08 \times 10^{-3} m$

so distance moved by the particle is given as

$d = r_1 - r_2$

$d = 0.0377 - 0.00708$

$d = 0.0306 m$

6 0

2 years ago