0.00432 km = ____

Answer each question yes or no. Assume the motions and currents mentioned are along the x axis and fields are in the y direction

Sergeeva-Olga [200]

(a) Does an electric field exert a force on a stationary charged object? YES ( F = Eq)

(b) Does a magnetic field do so?- NO ( F= qvB)

(c) Does an electric field exert a force on a moving charged object? YES

(d) Does a magnetic field do so? YES ( F = qvB)

(e) Does an electric field exert a force on a straight current-carrying wire? ( NO)

(f) Does a magnetic field do so? Yes

(g) Does an electric field exert a force on a beam of moving electrons? Yes

(h) Does a magnetic field do so? YeS

To know more about magnetic field visit : brainly.com/question/10353944

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6 0

2 years ago

PLEASE HELP ASAPP!!!

Sati [7]

Answer:

hello

Explanation:

for average speed the total distance is important so.

and you must convert min to hour 15 min is 1/4 or 0.25 hour

$s = \frac{x}{t} = \frac{12}{0.25} = 48 \frac{km}{h}$

for average velocity the straight line distance is important so

$v = \frac{d}{t} = \frac{8}{0.25} = 32 \frac{km}{h}$

hope you understand

7 0

2 years ago

When have you experienced an increase in kinetic<br> energy within a system?

Mars2501 [29]

Answer:

If a man starts running on a boat with an acceleration a with respect to the boat, there is no external force that acts on the Boat+Man system

8 0

3 years ago

An object is released from rest and falls a distance h during the first second of time. How far will it fall during the next sec

Viefleur [7K]

Answer:

E. 3h

Explanation:

We know that

u = 0 m/s.

velocity after t = 1s

v = u+gt = 0+9.81 x 1s= 9.81 m/s

distance covered in 1st sec

= =>> ut+0.5 x g x t²

=>>0 + 0.5x 9.81 x 1 = 4.90m

Let 4.90 be h

distance travelled in 2nd second will now be used

So velocity after t = 1s

=>>1 x t+ 0.5 x g x t²

=>9.81x 1 + 0.5 x 9.81 x 1 = 3 x 4.90

So since h= 4.90

Then the ans is 3x h = 3h

3 0

3 years ago

A block is given a very brief push up a 20.0 degree frictionless incline to give it an initial speed of 12.0 m/s.(a) How far alo

Orlov [11]

Explanation:

(a) Net force acting on the block is as follows.

$F_{net} = -mg Sin (\theta)$

or, ma = -mg Sin (\theta)[/tex]

a = $-g Sin (\theta)$

= $-9.8 \times Sin (20^{o})$

= -3.35 $m/s^{2}$

According to the kinematic equation of motion,

$v^{2} - v^{2}_{o} = 2as$

Distance traveled by the block before stopping is as follows.

s = $\frac{v^{2} - v^{2}_{o}}{2a}$

= $\frac{(0)^{2} - (12.0)^{2}_{o}}{2 \times -3.35}$

= 21.5 m

According to the kinematic equation of motion,

v = $v_{o} + at$

0 = $12.0 m/s + \frac{1}{2} \times -3.35 m/s^{2} \times t$

$t_{1}$ = 7.16 sec

Therefore, before coming to rest the surface of the plane will slide the box till 7.16 sec.

(b) When the block is moving down the inline then net force acting on the block is as follows.

$F_{net} = -mg Sin (\theta)$

ma = $mg Sin (\theta)$

a = $g Sin (\theta)$

= $9.8 m/s^{2} \times Sin (20^{o})$

= 3.35 $m/s^{2}$

Kinematics equation of the motion is as follows.

s = $v_{o}t + \frac{1}{2}at^{2}$

21.5 m = $0 + \frac{1}{2} \times 3.35 m/s^{2} \times t^{2}$

$t_{2}$ = $\sqrt{\frac{2 \times 21.5 m}{3.35 m/s^{2}}}$

= 3.58 sec

Hence, total time taken by the block to return to its starting position is as follows.

t = $t_{1} + t_{2}$

= 7.16 sec + 3.58 sec

= 10.7 sec

Thus, we can conclude that 10.7 sec time it take to return to its starting position.

3 0

4 years ago

0.00432 km = _____ mm