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Zielflug [23.3K]

3 years ago

5

At a certain instant an object is moving to the right with speed 1.0 m/s and has a constant acceleration to the left of 1.0 m/s2

. At what later time will the object momen- tarily be at rest?

1 answer:

professor190 [17]3 years ago

5 0

Explanation:

Given that,

Initial speed of the object, u = 1 m/s

Acceleration of the object, $a=-1\ m/s^2$

We need to find the time when the object comes at rest (v=0). Let it is given by :

$t=\dfrac{v-u}{a}$

$t=\dfrac{0-1\ m/s}{-1\ m/s^2}$

t = 1 second

So, the object will comes to rest at 1 second. Hence, this is the required solution.

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Answer:

Explanation:

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3 years ago

At a given point in its orbit, the moon is 2.4 x 10^5 miles from earth. How long does it take light from a source on earth to re

Mice21 [21]

Answer:

Total time= 2.56 s

Explanation:

Given that

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8 0

4 years ago

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Consider two identical microscope slides in air illuminated with monochromatic light. the bottom slide is rotated (counterclockw

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Answer:

Explanation:ruhfu

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4 years ago

A solenoidal coil with 25 turns of wire is wound tightly around another coil with 300 turns. The inner solenoid is 25.0 cm long

joja [24]

Answer:

$5.65487\times 10^{-8}\ Wb$

$1.17\times 10^{-5}\ H$

-0.020475 V

Explanation:

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$N_1$ = Number of turns of coil = 25

$N_2$ = Number of turns of coil 2 = 300

$\frac{di_2}{dt}$ = Rate of current increased = $1.75\times 10^3\ A/s$

d = Diameter = 2 cm

r = Radius = $\frac{d}{2}=\frac{2}{2}=1\ cm$

A = Area = $\pi r^2$

Magnetic field in the solenoid is given by

$B=\mu_0\frac{N_2}{l}I\\\Rightarrow B=4\pi\times 10^{-7}\frac{300}{0.25}\times 0.12\\\Rightarrow B=0.00018\ T$

Magnetic flux is given by

$\phi=BA\\\Rightarrow \phi=0.00018\times \pi\times 0.01^2\\\Rightarrow \phi=5.65487\times 10^{-8}\ Wb$

The average magnetic flux through each turn of the inner solenoid is $5.65487\times 10^{-8}\ Wb$

Mutual inductance is given by

$L=\frac{N_1\phi}{i_1}\\\Rightarrow L=\frac{25\times 5.65487\times 10^{-8}}{0.12}\\\Rightarrow L=1.17\times 10^{-5}\ H$

The mutual inductance of the two solenoids is $1.17\times 10^{-5}\ H$

Induced emf is given by

$V=-L\frac{di_2}{dt}\\\Rightarrow V=-1.17\times 10^{-5}\times 1.75\times 10^3\\\Rightarrow V=-0.020475\ V$

The emf induced in the outer solenoid by the changing current in the inner solenoid is -0.020475 V

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Kazeer [188]

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