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2 years ago

A motorcycle is capable of accelerating at 4.7 m/s^2, starting from rest, how far can it travel in 3 seconds?

Physics

2 answers:

tankabanditka [31]2 years ago

7 0

21.15m should be the answer ^_^

maw [93]2 years ago

5 0

Answer: 21.15m

Explanation:

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True or False? Materials that are good conductors of heat are usually poor conductors of electricity.

alexandr402 [8]

False

HOPE THIS WILL HELP YOU

5 0

2 years ago

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One gram of Uranium averages release 1.01 KJ (10^7) of energy. How much mass could be converted to energy to release this much e

frutty [35]

Answer:

The amount of mass that needs to be converted to release that amount of energy is $1.122 X 10^{-7} kg$

Explanation:

From Albert Einstein's Energy equation, we can understand that mass can get converted to energy, using the formula

$E= \Delta mc^{2}$

where $\Delta m$ = change in mass

c = speed of light = $3 \times 10 ^{8}m/s$

Making m the subject of the formula, we can find the change in mass to be

$\Delta m = \frac{E}{c^{2}}= \frac{1.01 \times 10^{3} \times 10^{7}}{(3 \times 10^{8})^{2}}= 1.122 \times 10 ^{-7}kg$

There fore, the amount of mass that needs to be converted to release that amount of energy is 1.122 X 10 ^-7 kg

5 0

2 years ago

A ray of light incident in water strikes the surface separating water from air making an angle of 10 ° with the normal to the su

labwork [276]

Answer:

$\theta _2 = 13^o$

$\theta _1 =32.94^o$

$\theta_c = 53.05^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta_1 = 10^o$

The refractive index of water is $n_1 = 1.3$

Generally Snell's law is mathematically represented as

$n_1 sin(\theta_1) = n_2 sin(\theta_ 2)$

Here $n_2$ is the refractive index of air with value $n_2 = 1$

$\theta_2$ is the angle of refraction

$\theta _2 = sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ]$

=> $\theta _2 = sin^{-1}[\frac{1.3 * sin(10)}{1} ]$

=> $\theta _2 = 13^o$

Given that the angle should not be greater than $\theta _2 =45^o$ then the angle of incidence will be

$\theta _1 = sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ]$

=> $\theta _1 = sin^{-1}[\frac{1 * sin(45)}{1.3} ]$

=> $\theta _1 =32.94^o$

Generally for critical angle is mathematically represented as

$\theta_c = sin^{-1}[\frac{n_2}{n_1} ]$

=> $\theta_c = sin^{-1}[\frac{1}{1.3} ]$

=> $\theta_c = 53.05^o$

4 0

2 years ago

When setting your schedule, it is sometimes necessary to tell people "No." Please select the best answer from the choices provid

mart [117]

Answer:

True Edg2020

Explanation:

6 0

2 years ago

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Situation: two electrons, each with mass m and charge q, are released from positions very far from each other. With respect to a

Ahat [919]

Answer:

$r_{min}=\frac{kq^2}{5m_ev^2}$

Explanation:

The total kinetic energy of both electrons will be electrostatic potential energy, when the electrons reach the minima distance due to electrostatic repulsion. Then, you have:

$E_{k}=U_E\\\\\frac{1}{2}m_ev_1^2+\frac{1}{2}m_ev_2^2=k\frac{q^2}{r_{min}}$

me: mass of the electron

q: charge of the electron

k: Coulomb's constant

you take into account that v2=3v1=3v and do rmin the subject of the formula:

$\frac{1}{2}m_e[v^2+9v^2]=5m_ev^2=k\frac{q^2}{r_{min}}\\\\r_{min}=\frac{kq^2}{5m_ev^2}$

4 0

3 years ago