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3 years ago

Which of the following have derived units? A. 56 kg B. 2.5 m C. 87 m/s² D. 60 N

Physics

2 answers:

photoshop1234 [79]3 years ago

6 0

Answer:

The answer is C!!!!!!!

Becuz meters and seconds are derived into m/s²

Explanation:

Plss follow me and Mark as brainlest

Thanks :-)

Leni [432]3 years ago

5 0

Answer:

C. 87 m/s^2

Hope it helps.

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How much power is needed to lift the 200-N object to a height of 4 m in 4 s?

Vitek1552 [10]

Answer: 2000 watts

Explanation:

Given that,

power = ?

Weight of object = 200-N

height = 4 m

Time = 4 s

Power is the rate of work done per unit time i.e Power is simply obtained by dividing work by time. Its unit is watts.

i.e Power = work / time

(since work = force x distance, and weight is the force acting on the object due to gravity)

Then, Power = (weight x distance) / time

Power = (200N x 4m) / 4s

Power = 8000Nm / 4s

Power = 2000 watts

Thus, 2000 watts of power is needed to lift the object.

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3 years ago

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slega [8]

Answer:

Explanation:

A,D,E

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3 years ago

Describe what happens to chromosomes before mitosis.

Brums [2.3K]

Explanation:

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3 years ago

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Lapatulllka [165]

False... I hope that helps ;)

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3 years ago

A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is: