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4 years ago

If you double the velocity of a moving object, how is it's momentum affected?

1 answer:

8 0

Well momentum is = to Mass*Velocity so let's use an example to figure this out

If I weighed 50kg and I was jogging at 3m/s then I broke into a run at 6m/s how will me momentum be affected?
3m/s*50kg=150
6m/s*50kg=300

So as you can see by doubling the velocity you also double the momentum

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A position vector with magnitude 10 m points to the right and up. its x-component is 6.0 m. part a what is the value of its y-co

lidiya [134]

The position vector can be transcribed as:

A = 6 i + y j

i points in the x-direction and j points in the y-direction.

The magnitude of the vector is its dot product with itself:

|A|2 = A·A

102 = (6 i + y j)•(6 i+ y j) Note that i•j = 0, and i•i = j•j = 1

100 = 36 + y2

64 = y2

get the square root of 64 = 8

The vertical component of the vector is 8 cm.

3 0

3 years ago

A machine, modeled as a simple spring-mass system, oscillates in simple harmonic motion. Its acceleration is measured to have an

ANTONII [103]

$a=5000\dfrac{mm}{s}=5\dfrac{m}{s}$

$f=10{Hz}\Longrightarrow t=\dfrac{1}{10}s$

$a_{max}=\dfrac{50\frac{m}{s}}{\frac{s}{10}}\cdot\dfrac{1}{\sqrt{2}}=\dfrac{50\dfrac{m}{s^2}}{\sqrt{2}}\approx\boxed{35.4\dfrac{m}{s^2}}$

Hope this helps.

r3t40

5 0

3 years ago

An experiment is conducted on a long straight wire of diameter d. A constant current is sent through the wire and the magnetic f

soldi70 [24.7K]

Answer:

Explanation:

To solve the exercise it is necessary to apply the concepts related to the Magnetic Field described by Faraday.

The magnetic field is given by the equation:

$B = \frac{\mu_0 I}{2\pi d}$

Where,

$\mu =$ Permeability constant

d = diameter

I = Current

For the given problem we have a change in the diameter, twice that of the initial experiment, therefore we define that:

$B_1 = \frac{\mu_0 I}{2\pi d}$

$B_2 = \frac{\mu_0 I}{2\pi 2d}$

The ratio of change between the two is given by:

$\frac{B_2}{B_1} = \frac{\frac{\mu_0 I}{2\pi d}}{\frac{\mu_0 I}{2\pi 2d}}$

$\frac{B_2}{B_1} = \frac{d}{2d}$

$\frac{B_2}{B_1} = \frac{1}{2}$

$B_2 = B_1 \frac{1}{2}$

Therefore the correct answer is D.

4 0

3 years ago

For a charged particle, a constant magnetic field can be used to change

Leona [35]

The direction of motion of the charge

5 0

3 years ago

A ray of light crosses a boundary between two transparent materials. The medium the ray enters has a larger index of refraction.

Liula [17]

The wavelength of the light decreases as it enters into the medium with the greater index of refraction. The wavelength of the light remains constant as it transitions between materials.

7 0

3 years ago