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3 years ago

1) Displacement is to velocity as ____________ is to acceleration. A) direction B) speed C) time D) velocity

Physics

1 answer:

vivado [14]3 years ago

7 0

Answer: C) time

Explanation:

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A pickup truck has a width of 79.0 in. If it is traveling north at 42 m/s through a magnetic field with vertical component of

bekas [8.4K]

Answer:

The magnitude of the induced Emf is $0.003371V$

Explanation:

The width of the truck is given as 79inch but we need to convert to meter for consistency, then

The width= 79inch × 0.0254=2.0066 metres.

Now we can calculate the induced Emf using expresion below;

Then the $induced EMF= B L v$

Where B= magnetic field component

L= width

V= velocity

=(40*10^-6) × (42) × (2.0066)

=0.003371V

Therefore, the magnitude emf that is induced between the driver and passenger sides of the truck is 0.003371V

8 0

4 years ago

Kingda Ka has a maximum speed of 57.2 m/s. Determine the height of the hill on this roller coaster. Explain to get full credit,

kodGreya [7K]

This question involves the concepts of the law of conservation of energy, kinetic energy, and potential energy.

The height of the hill is "166.76 m".

<h3>LAW OF CONSERVATION OF ENERGY:</h3>

According to the law of conservation of energy at the highest point of the roller coaster ride, that is, the hill, the whole (maximum) kinetic energy of the roller coaster is converted into its potential energy:

$Maximum\ Kinetic\ Energy\ Lost = Maximum\ Potential\ Energy\ Gain\\\\
\frac{1}{2}mv_{max}^2=mgh\\\\
h=\frac{v_{max}^2}{2g}$

where,

h = height of the hill = ?

$v_{max}$ = maximum velocity = 57.2 m/s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

$h=\frac{(57.2\ m/s)^2}{2(9.81\ m/s^2)}\\\\
$

Learn more about the law of conservation of energy here:

brainly.com/question/101125

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3 years ago

Using Boyle's law, predict what will happen to a balloon that an ocean diver takes to a pressure of 202 kPa.

Savatey [412]

Answer:

Balloon that an ocean diver takes to a pressure of 202 k Pa will get reduced in size that is the volume of the balloon will get reduced

Explanation:

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3 years ago

How<br> are<br> earthquakes measured quantitatively

AlexFokin [52]

Answer:Intensity: The severity of earthquake shaking is assessed using a descriptive scale – the Modified Mercalli Intensity Scale. Magnitude: Earthquake size is a quantitative measure of the size of the earthquake at its source. The Richter Magnitude Scale measures the amount of seismic energy released by an earthquake.

Explanation:

4 0

3 years ago

Rank the wavelengths of the following quantum particles from the largest to the smallest. If any have equal wavelengths, display

goldfiish [28.3K]

The wavelengths of the following quantum particles from the largest to the smallest is (d) > (a) = (e) > (b) > (c)

De Broglie proposed that because light has both wave and particle properties, matter exhibits both wave and particle properties. This property has been explained as the dual behavior of matter.
From his observations, de Broglie derived the relationship between the wavelength and momentum of matter. This relationship is known as de Broglie's relationship.

De Broglie's relationship is given by $\lambda=\frac{h}{mv}$ .

This can be written as $\lambda=\frac{h}{p}$ .....(1) where λ is known as de Broglie wavelength and p is momentum , h = Plank’s constant .

As we know that mass of proton is greater than electron and photon .

(a)For photon , the momentum is given by $p=\frac{E}{c}$ ...(2) where c is the speed is the speed of light .

Putting E = 3eV in equation (2) , we get

$p=\frac{3\times 1.6\times10^-^1^9}{3\times 10^8} \\p=1.6\times10^-^2^7Js/m$

Putting this value of p in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}}{1.6\times10^{-27}}\\\lambda=4.13\times10^{-7}$

(b) As we know that kinetic energy is given by

$K.E=\frac{1}{2} mv^2\\\\2K.E = mv^2\\2K.E\times m = m^2v^2\\2K.E\times m = (mv)^2\\2K.E\times m = p^2\\\\\sqrt{2K.E\times m }=p$ ...(3)

Where mass of electron is $9.1\times10^-^3^1 kg$ .

Putting K.E = 3eV in equation (3) , we get

$\sqrt{2K.E\times m }=p\\p=\sqrt{2\times3\times1.6\times10^{-19} \times 9.31\times10^{-31} }\\p=\sqrt{89.376\times10^{-50}} \\p=9.45\times10^{-25}Js/m$

Putting this value of p in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}}{9.45\times10^{-25}}\\\lambda=0.7005\times10^{-9}$

$\sqrt{2K.E\times m }=p\\p=\sqrt{2\times3\times1.6\times10^{-19} \times 1.67\times10^{-27} }\\p=\sqrt{16.032\times10^{-46}} \\p=4.003\times10^{-23}Js/m$