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4 years ago

Enter the expression

Physics

1 answer:

vlabodo [156]4 years ago

6 0

I see the word "when..." kind of fading out at the end of the first line.
Whatever comes after it may be important.

If you're just supposed to copy the expression into the box,
then the problem is that you left the 'e' out of it.

I'm guessing that you're supposed to enter whatever the expression becomes
when either N₀ or ' t ' has some special value that's in the first line.

Just taking a wild guess here . . . . .

If it's "Enter the expression ..... , when t=0 ." ,
then the correct answer in the box is N₀ .

But that's just a wild guess. As I pointed out, you cut off
the picture in the middle of the word 'when', and I've got
a hunch that there's something important after it.

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While hiking in the mountains, tanya observed a narrow and steep valley with a stream at the bottom of it. what most likely caus

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Erosion over thousands of millennia from the stream or river in the valley

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4 years ago

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3 years ago

In the formula for magnetic deflection, FM is equal to qvB. Match each symbol in this formula with what it represents.

slava [35]

Answer:

1. $F_m$ -Deflection force

2.q-Charge

3.v-Velocity of the charge

4.B-Perpendicular magnetic field strength

Explanation:

We are given that magnetic deflection

$F_m=qvB$

We have to match each symbol in formula with what it represents.

In this formula

$F_m=$ Deflection force

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v=Velocity of charge

B=Perpendicular magnetic field strength

Therefore,

1. $F_m$ -Deflection force

2.q-Charge

3.v-Velocity of the charge

4.B-Perpendicular magnetic field strength

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3 years ago

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If two bodies have the same kinetic energy do they have the same momentum or not and why?

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3 years ago

A box of mass 20kg is pulled up an inclined plane by a force of 285N. Given that the value of the incline angle is 30 degrees an

docker41 [41]

Given :

Mass of box , m = 250 kg.

Force applied , F = 285 N.

The value of the incline angle is 30°.

the coefficient of dynamic friction is $\mu=0.72$ .

To Find :

The speed with which the box is moving with, assuming it takes 4 seconds to reach the top of the incline.

Solution :

Net force applied in box is :

$F=285 - mgsin\ \theta - \mu mg cos \ \theta\\ \\F=285-mg( sin \ \theta - \mu cos\ \theta)\\\\F=285 - 20\times 10( \dfrac{1}{2}+0.72\times \dfrac{\sqrt{3}}{2})\\\\F=60.29\ N$

Acceleration , $a=\dfrac{F}{m}=\dfrac{60.29}{20}=3.01\ m/s^2$ .

By equation of motion :

$v=u+at\\\\v=0+3.01\times 4\\\\v=12.04\ m/s$

Therefore, the speed of box is 12.04 m/s.

Hence, this is the required solution.

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3 years ago