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3 years ago

6. You must turn on your headlights

Physics

1 answer:

Umnica [9.8K]3 years ago

8 0

The most logical answer would be C :)

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When a wave enters a new medium from an angle, both the speed and the ________ change

slamgirl [31]

Answer:

B: Amplitude

Explanation:

When a wave travels from one medium to the other from an angle, the things that change are amplitude, wavelength, intensity and velocity.

The frequency doesn't change because the frequency depends upon the source of the wave and not the medium by which the wave is propagated.

8 0

3 years ago

Read 2 more answers

Which statement correctly describes mass-energy equivalence? All energy in the universe will be converted to an equivalent amoun

olchik [2.2K]

The statement 'all energy in the universe is a result of mass being converted into energy' correctly describes mass-energy equivalence.

<h3>What is mass-energy equivalence?</h3>

The expression mass-energy equivalence refers to the proportion of matter that can be converted into energy in the universe.

This mass-energy equivalence is an outcome of process of converting mass into energy.

In conclusion, the statement 'all energy in the universe is a result of mass being converted into energy' correctly describes mass-energy equivalence.

Learn more about mass-energy equivalence here:

brainly.com/question/3171044

#SPJ1

4 0

2 years ago

A satellite is in a circular orbit around Mars, which has a mass M = 6.40 × 1023 kg and radius R = 3.40 ×106 m.

Pepsi [2]

Answer:

a) The orbital speed of a satellite with a orbital radius R (in meters) will have an orbital speed of approximately $\displaystyle \sqrt\frac{4.27 \times 10^{13}}{R}\; \rm m \cdot s^{-1}$ .

b) Again, if the orbital radius R is in meters, the orbital period of the satellite would be approximately $\displaystyle 9.62 \times 10^{-7}\, R^{3/2}\; \rm s$ .

c) The orbital radius required would be approximately $\rm 2.04 \times 10^7\; m$ .

d) The escape velocity from the surface of that planet would be approximately $\rm 5.01\times 10^3\; m \cdot s^{-1}$ .

Explanation:

Since the orbit of this satellite is circular, it is undergoing a centripetal motion. The planet's gravitational attraction on the satellite would supply this centripetal force.

The magnitude of gravity between two point or spherical mass is equal to:

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}}$ ,

where

$G$ is the constant of universal gravitation.
$M$ is the mass of the first mass. (In this case, let $M$ be the mass of the planet.)
$m$ is the mass of the second mass. (In this case, let $m$ be the mass of the satellite.)
$r$ is the distance between the center of mass of these two objects.

On the other hand, the net force on an object in a centripetal motion should be:

$\displaystyle \frac{m \cdot v^{2}}{r}$ ,

where

$m$ is the mass of the object (in this case, that's the mass of the satellite.)
$v$ is the orbital speed of the satellite.
$r$ is the radius of the circular orbit.

Assume that gravitational force is the only force on the satellite. The net force should be equal to the planet's gravitational attraction on the satellite. Equate the two expressions and solve for $v$ :

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}} = \frac{m \cdot v^{2}}{r}$ .

$\displaystyle v^2 = \frac{G \cdot M}{r}$ .

$\displaystyle v = \sqrt{\frac{G \cdot M}{r}}$ .

Take $G \approx 6.67 \times \rm 10^{-11} \; m^3 \cdot kg^{-1} \cdot s^{-2}$ , Simplify the expression $v$ :

$\begin{aligned} v &= \sqrt{\frac{G \cdot M}{r}} \cr &= \sqrt{\frac{6.67 \times \rm 10^{-11} \times 6.40 \times 10^{23}}{r}} \cr &\approx \sqrt{\frac{4.27 \times 10^{13}}{r}} \; \rm m \cdot s^{-1} \end{aligned}$ .

Since the orbit is a circle of radius $R$ , the distance traveled in one period would be equal to the circumference of that circle, $2 \pi R$ .

Divide distance with speed to find the time required.

$\begin{aligned} t &= \frac{s}{v} \cr &= 2 \pi R}\left/\sqrt{\frac{G \cdot M}{R}} \; \rm m \cdot s^{-1}\right. \cr &= \frac{2\pi R^{3/2}}{\sqrt{G \cdot M}} \cr &\approx 9.62 \times 10^{-7}\, R^{3/2}\; \rm s\end{aligned}$ .

Convert $24.6\; \rm \text{hours}$ to seconds:

$24.6 \times 3600 = 88560\; \rm s$

Solve the equation for $R$ :

$9.62 \times 10^{-7}\, R^{3/2}= 88560$ .

$R \approx 2.04 \times 10^7\; \rm m$ .

If an object is at its escape speed, its kinetic energy (KE) plus its gravitational potential energy (GPE) should be equal to zero.

$\displaystyle \text{GPE} = -\frac{G \cdot M \cdot m}{r}$ (Note the minus sign in front of the fraction. GPE should always be negative or zero.)

$\displaystyle \text{KE} = \frac{1}{2} \, m \cdot v^{2}$ .

Solve for $v$ . The value of $m$ shouldn't matter, for it would be eliminated from both sides of the equation.

$\displaystyle -\frac{G \cdot M \cdot m}{r} + \frac{1}{2} \, m \cdot v^{2}= 0$ .

$\displaystyle v = \sqrt{\frac{2\, G \cdot M}{R}} \approx 5.01\times 10^{3}\; \rm m\cdot s^{-1}$ .

5 0

4 years ago

A ball is thrown with an initial velocity of u=(10i +15j) m/s. Whan it reaches the top of it trajectory neglecting air resistanc

liraira [26]

Answer:

v = (10 i ^ + 0j ^) m / s, a = (0i ^ - 9.8 j ^) m / s²

Explanation:

This is a missile throwing exercise.

On the x axis there is no acceleration so the velocity on the x axis is constant

v₀ₓ = 10 m / s

On the y-axis velocity is affected by the acceleration of gravity, let's use the equation

v_y = $v_{oy}$ - g t

$v_{y}^2 = v_{oy}^2 - 2 g (y - y_o)$

at the highest point of the trajectory the vertical speed must be zero

v_y = 0

therefore the velocity of the body is

v = (10 i ^ + 0j ^) m / s

the acceleration is

a = (0 i ^ - g j⁾

a = (0i ^ - 9.8 j ^) m / s²

5 0

3 years ago

A butcher has a beam balance and masses 0.5kg and 2kg.How would he measure 1.5kg of meat on the balance at once

LekaFEV [45]

Answer and Explanation:

The butcher will balance the meat and the 0.5 kg mass on one side of the beam balance and 2 kg mass on the other side of the beam balance.
By so doing, when the beam balances then the amount of meat measured will be exactly 1.5 kg as required. Since the 1.5 kg meat and 0.5 kg mass is equivalent to the 2 kg on the other side of the beam balance.

6 0

3 years ago