17. A force of 150N is applied at an angle of 60°

Two forces act on an object. The first force has a magnitude of 15.0 N15.0 N and is oriented 58.0°58.0° counterclockwise from th

SIZIF [17.4K]

Answer:

F = 33.12 N

Explanation:

given,

first force = 15 N is oriented at 58.0° counterclockwise

second force = (−19.0 N,18.5 N)

x-component of first force = 15 cos 58.0° = 7.95 N

y-component of second force = 15 sin 58.0° = 12.72 N

second component

now,

F netx = 7.95 - 19 = -11.05 N

F nety = 12.72 + 18.5 = 31.22 N

$F=\sqrt{(-11.05)^2+31.22^2}$

F = 33.12 N

$tan \theta = -\dfrac{31.22}{11.05}$

θ = -70.51°

3 0

4 years ago

5. George walks to a friend's house. He walks 750 meters North, then realizes he walked too far.

dedylja [7]

Answer:

Average speed: approximately $76.9\; {\rm m\cdot s^{-1}}$ .

Average velocity: approximately $38.5\; {\rm m \cdot s^{-1}}$ (to the north.)

Explanation:

Consider an object that travelled along a certain path. Distance travelled would be equal to the length of the entire path.

In contrast, the magnitude of displacement is equal to distance between where the object started and where it stopped.

In this question, the path George took required him to travel $750\; {\rm m} + 250\; {\rm m} = 1000\; {\rm m}$ in total. Hence, the distance George travelled would be $1000\; {\rm m}$ . However, since George stopped at a point $(750\; {\rm m} - 250\; {\rm m}) = 500\; {\rm m}$ to the north of where he started, his displacement would be only $500\; {\rm m}$ to the north.

Divide total distance by total time to find the average speed.

Divide total displacement by total time to find average velocity.

The total time of travel in this question is $13\; {\rm s}$ .. Therefore:

$\begin{aligned}\text{average speed} &= \frac{\text{total distance}}{\text{total time}} \\ &= \frac{1000\; {\rm m}}{13\; {\rm s}} \\ &\approx 76.9\; {\rm m\cdot s^{-1}}\end{aligned}$ .

$\begin{aligned}\text{average velocity} &= \frac{\text{total displacement}}{\text{total time}} \\ &= \frac{500\; {\rm m}}{13\; {\rm s}} && \genfrac{}{}{0px}{}{(\text{to the north})}{}\\ &\approx 38.5\; {\rm m\cdot s^{-1}} && (\text{to the north})\end{aligned}$ .

3 0

2 years ago

Seawater is full of moving molecules that possess kinetic energy. Could we extract some of this energy

tigry1 [53]

I don’t think soooooo

8 0

3 years ago

Precipitation tends to be _______ on the _______ side of a mountain because water vapor _______ as it rises there.

Lera25 [3.4K]

Answer:

Higher, Windward side, Condenses

Explanation:

The Windward side refers to that side of a mountain that faces the direction from which the wind is blowing. In this direction, the moisture containing hot air blowing from a distant place moves upward and strikes the mountain at a greater height, where the air mass is thin and the temperature is relatively cold. As the temperature and pressure decrease with altitude, the hot uprising air cools and gradually condenses. This results in the occurrence of high precipitation over this region i.e. the windward side of the mountain.

Therefore, the precipitation is always higher on the windward side of a mountain as the hot air undergoes condensation at greater height as it rises upward.

6 0

3 years ago

Infrared radiation has frequencies from 3.0×1011 to 3.0×1014 Hz, whereas the frequency region for microwave radiation is 3.0×108

igor_vitrenko [27]

Answer:

1. The speed of infrared radiation is the same as microwave radiation.

2. The wavelength of infrared radiation is lower than microwave radiation.

Explanation:

1. The speed of infrared radiation is the same as microwave radiation. This is because both infrared radiation and microwave radiation are electromagnetic waves and all electromagnetic waves move at the same speed, the speed of light.

2. Frequency and wavelength has an inverse relationship. This means that the higher the frequency, the lower the wavelength.

Since the frequency range of infrared waves is higher than that of microwaves, the wavelength range of infrared waves is lower than that of microwaves.

We can prove this by using the maximum frequency value of each wave to calculate their corresponding wavelengths.

Speed of light, c, is given as:

c = λf

Where λ is wavelength and f is frequency.

Wavelength is therefore:

λ = c/f

For infrared wave, the maximum frequency is 3 * 10^14 Hz, hence the corresponding wavelength is:

λ = (3 * 10^8) / (3 * 10^14)

λ = 10^(-6) m

For microwave, the maximum frequency is 3 * 10^11 Hz, hence the corresponding wavelength is:

λ = (3 * 10^8) / (3 * 10^11)

λ = 10^(-3) m

Hence, the wavelength of infrared waves is lower than the wavelength of microwaves.

8 0

3 years ago