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3 years ago

A 2-kg book falls off a shelf. It hits a student traveling 2 m/s. How much kinetic energy does the book have?

Physics

1 answer:

Gekata [30.6K]3 years ago

3 0

4 J

1/2 mv^2 is ke formula

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The potential in a region of space due to a charge distribution is given by the expression V = ax2z + bxy − cz2 where a = −3.00

Elis [28]

Answer:

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = ( 48.00 \frac{V}{m} , 0 , - 144.00 \frac{V}{m} )$

Explanation:

We know that the relationship between the electric field $\vec{E}(\vec{r})$ and the potential $V(\vec{r})$ is given by

$\vec{E} ( \vec{r}) = - \vec{\nabla} V(\vec{r})$

So, for our potential:

$V(r) = a x^2 z + b x y - c z^2$

the electric field is :

$\vec{E} ( \vec{r}) = - \vec{\nabla} ( a x^2 z + b x y - c z^2 )$

$\vec{E} ( \vec{r}) = - ( \frac{ \partial }{\partial x}, \frac{ \partial }{\partial y} , \frac{ \partial }{\partial z}) ( a x^2 z + b x y - c z^2 )$

$\vec{E} ( \vec{r}) = - ( \frac{ \partial }{\partial x} ( a x^2 z + b x y - c z^2 ) , \frac{ \partial }{\partial y} ( a x^2 z + b x y - c z^2 ) , \frac{ \partial }{\partial z} ( a x^2 z + b x y - c z^2 ))$

$\vec{E} ( \vec{r}) = - ( 2 a x z + b y , b x , a x^2 - 2 c z )$

This is the our electric field. At vector point

$\vec{r} = (0, -8.00 \ m, - 8.00 \ m)$

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = - ( 2 a * 0 * (-8.00 \ m) + b (-8.00 \ m) , b * 0 , a 0^2 - 2 c (-8.00 \ m) )$

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = - ( 0 - b 8.00 \ m , 0 , 0 + 2 c 8.00 \ m )$

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = ( b 8.00 \ m , 0 , - 2 c 8.00 \ m )$

Knowing

$b= 6.00 \frac{V}{m^2}$

and

$c=9.00 \frac{V}{m^2}$

the electric field is

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = ( 6.00 \frac{V}{m^2} * 8.00 \ m , 0 , - 9.00 \frac{V}{m^2} * 2* 8.00 \ m )$

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = ( 48.00 \frac{V}{m} , 0 , - 144.00 \frac{V}{m} )$

7 0

2 years ago

A truck is traveling east at 80 km/h. At an intersection 32 km ahead, a car is traveling north at 50 km/h. How long after this m

nirvana33 [79]

The time elapsed when the vehicles are closest to each other is 20 min.

The given parameters:

Speed of the truck, u = 80 km/h
Distance, d = 32 km
Speed of the car, v = 50 km/h

<h3>Principles of relative speed</h3>

The time elapsed when the cars are close to each other is calculated by applying the principles of relative speed.

$(V_r) t = d\\\\V_r^2 = 50^2 + 80^2\\\\V_r =\sqrt{50^2 + 80^2} \\\\V_r = 94.34 \ km/h$

$94.34 t = 32\\\\t = \frac{32}{94.34} \\\\t = 0.34 \ hr\\\\t \approx 20 \min$

Thus, the time elapsed when the vehicles are closest to each other is 20 min.

Learn more about relative velocity here: brainly.com/question/24430414

3 0

2 years ago

PLEASE HELP!!! URGENTTT

Ludmilka [50]

Answer:

For the first one, its "attract"

4 0

2 years ago

Plz help me answer this its 1 question will give brainliest

kolbaska11 [484]

Circularity system........…….......

6 0

3 years ago

Read 2 more answers

Your chances of getting into a collision when talking on a cell phone _________: A. Double B. Triple C. Quadruple D. Remain the

Anit [1.1K]

Answer:

C. Quadruple

Explanation:

¨Drivers who are talking on the phone, even on a hands-free device, are up to four times more likely to be involved in a crash.¨

I hope this helps! Have a great day!

3 0

2 years ago