A woman on a bridge 100 m high sees a raft floating at a constant speed on the river below.

The vibrations along a transverse wave move in a direction _________.

GrogVix [38]

Answer: perpendicular to it oscillations.

Explanation: A transverse wave is a wave whose oscillations is perpendicular to the direction of the wave.

By perpendicular, we mean that the wave is oscillating on the vertical axis (y) of a Cartesian plane and the vibration is along the horizontal axis (x) of the plane.

Examples of transverse waves includes wave in a string, water wave and light.

Let us take a wave in a string for example, you tie one end of a string to a fixed point and the other end is free with you holding it.

If you move the rope vertically ( that's up and down) you will notice a kind of wave traveling away from you ( horizontally) to the fixed point.

Since the oscillations is perpendicular to the direction of wave, it is a transverse wave

5 0

2 years ago

An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in

Gwar [14]

The final temperature is 83 K.

Explanation:

For an adiabatic process,

$T {V}^{\gamma - 1} = \text{constant}$

$\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}$

Given:-

${T}_{1} = 275 \; K$

${T}_{2} = T \left( \text{say} \right)$

${V}_{1} = V$

${V}_{2} = 6V$

$\gamma = \cfrac{5}{3} \;$ (the gas is monoatomic)

$\therefore \cfrac{T}{275} = {\left( \cfrac{V}{6V} \right)}^{\frac{5}{3} - 1}$

$\Rightarrow \cfrac{T}{275} = {\left( \cfrac{1}{6} \right)}^{\frac{2}{3}}$

T = 275 $\times$ 0.30

T = 83 K.

3 0

3 years ago

How do I answer this question

maria [59]

Answer:

what question

Explanation:

6 0

3 years ago

Read 2 more answers

A motorboat traveling with a current can go 160 km in 4 hours. against the current it takes 5 hours to go the same distance. Fin

MatroZZZ [7]

<h2>Speed of motorboat is 36 km/hr and speed of current is 4 km/hr.</h2>

Explanation:

Let speed of motor boat be m and speed of current be c.

A motorboat traveling with a current can go 160 km in 4 hours.

Distance = 160 km

Time = 4 hours

Speed = m + c

We have

Distance = Speed x Time

160 = (m+c) x 4

m + c = 40 --------------------- eqn 1

Against the current it takes 5 hours to go the same distance.

Distance = 160 km

Time = 5 hours

Speed = m - c

We have

Distance = Speed x Time

160 = (m-c) x 5

m - c = 32 --------------------- eqn 2

eqn 1 + eqn 2

2m = 40 + 32

m = 36 km/hr

Substituting in eqn 1

36 + c = 40

c = 4 km/hr

Speed of motorboat is 36 km/hr and speed of current is 4 km/hr.

3 0

3 years ago

An astronaut lands on a new, recently discovered planet in a different star system. The astronaut measures the acceleration due

Nezavi [6.7K]

Answer:

The radius of the new planet is ~2.04 * 10⁶ m, or 2,041,752 m.

Explanation:

We can use Newton's Law of Universal Gravitation:

$\displaystyle F_g=G\frac{Mm}{r^2}$

Let's look at Newton's 2nd Law:

$F=ma$

We can set these equations equal to each other:

$\displaystyle G\frac{Mm}{r^2} =ma$

The mass of the second mass (astronaut) cancels out. We are left with:

$\displaystyle G\frac{M}{r^2} =a$

We are solving for the radius of the new planet, so we can rearrange the equation:

$\displaystyle r=\sqrt{\frac{GM}{a} }$

Substitute in our known values given in the problem (G = 6.67 * 10⁻¹¹ ; M = 7.5 * 10²³ ; a = 12).

$\displaystyle r =\sqrt{\frac{(6.67\times 10^{-11})(7.5 \times 10^{23}}{12} }$
$r=2.04 \times 10^6$

The radius of the new planet is ~2.04 * 10⁶ m.

3 0

2 years ago