What instrument is used to expand burr holes?

katovenus [111]

Speed is a derived term that is the result when distance or displacement is divided by unit time. Speed is a scalar unit which means it only includes magnitude not direction, thus always positive. The answer thus are the three choices given above.

4 0

4 years ago

Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w

gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

$U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}$

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

At x = 1.00 m

$u(1) = \frac{-2}{1} + \frac{4}{1^2}$

= 4J

Kinetic energy = (1/2)mv²

$= \frac{1}{2} * 4(3)^2$

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

$u(5) = \frac{-2}{5} + \frac{4}{5^2}$

$= \frac{4-10}{25} = \frac{-6}{25} J$

= -0.24J

Kinetic energy =

$\frac{1}{2} * 4Vf^2$

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

$Vf = \sqrt{frac{22.4}{2}$

= 3.33 m/s

b) magnitude of force when x = 5.0m

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

$\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}$

$= \frac{2}{x^2} - \frac{8}{x^3}$

At x = 5.0 m

$\frac{2}{5^2} - \frac{8}{5^3}$

$F = \frac{2}{25} - \frac{8}{125}$

= 0.016N

8 0

4 years ago

What is the magnetic force on a proton that is moving at 4.5 107 m/s to the right through a magnetic field that is 1.6 T and poi

pickupchik [31]

The answer is letter C. 1.2 10^-11 N up
Solution:
F= Bqvsin(theta)
theta = sin 90 = 1
F= 1.4 T * 1.6x10^-19 * 5.2x10^7 ms^-1
F= 1.16 x 10^-11 N
Then the direction is upward.

6 0

3 years ago

Read 2 more answers

Which of the following units are used to describe acceleration?

Pavel [41]

Answer:

m/s2 m/s

Explanation:

7 0

3 years ago

A camera lens used for taking close-up photographs has a focal length of 22.0 mm. The farthest it can be placed from the film is

Arada [10]

Answer:

p = 6.64 cm

Explanation:

For this exercise we use the equation of the constructor

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image, respectively

They tell us the focal length f = 2.2 cm and that the image as far as it can go is q = 3.29 cm, let's find the position of the object that creates this image

$\frac{1}{p} = \frac{1}{f} - \frac{1}{q}$

1 / p = 1 / 2.2 - 1/3.29

1 / p = 0.15059

p = 6.64 cm

therefore the farthest distance from the object is 6.64 c

3 0

3 years ago