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Juliette [100K]
3 years ago
6

Canada is the country (dash)of the united States

Physics
1 answer:
lana66690 [7]3 years ago
5 0

I believe it is north.

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What does the slope of a distance-time graph indicate?
lawyer [7]

Answer: It indicates the speed of a object. The steeper the line the greater the speed of the object.

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2 years ago
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It is essential to behave appropriately on social media and be _______ when you post anything.
Firdavs [7]

Answer:

I would say be Mindful.

Explanation:

There could be like a MILLION answers for this. I think that you should personally go with your gut. That would be the best option. I think it's mindful because you really do have to be mindful when you post. Like not posting too much, not posting stuff you're uncomfortable with, not posting when on vacation, etc. So, I think you should be mindful.

5 0
2 years ago
As the distance between the sun and earth decreases, the speed of the planet
mixas84 [53]

Answer:

Explanation:

Increases. The force of gravity is distance dependent. Therefore, a smaller 'r' value will result in a larger force. Net force is proportional to the acceleration, so the planet will increase its speed.

5 0
3 years ago
A commuter train passes a passenger platform at a constant speed of 40.4 m/s. The train horn is sounded at its characteristic fr
mihalych1998 [28]

(a) -83.6 Hz

Due to the Doppler effect, the frequency of the sound of the train horn appears shifted to the observer at rest, according to the formula:

f' = (\frac{v}{v\pm v_s})f

where

f' is the apparent frequency

v = 343 m/s is the speed of sound

v_s is the velocity of the source of the sound (positive if the source is moving away from the observer, negative if it is moving towards the observer)

f is the original frequency of the sound

Here we have

f = 350 Hz

When the train is approaching, we have

v_s = -40.4 m/s

So the frequency heard by the observer on the platform is

f' = (\frac{343 m/s}{343 m/s - 40.4 m/s})(350 Hz)=396.7 Hz

When the train has passed the platform, we have

v_s = +40.4 m/s

So the frequency heard by the observer on the platform is

f' = (\frac{343 m/s}{343 m/s + 40.4 m/s})(350 Hz)=313.1 Hz

Therefore the overall shift in frequency is

\Delta f = 313.1 Hz - 396.7 Hz = -83.6 Hz

And the negative sign means the frequency has decreased.

(b) 0.865 m

The wavelength and the frequency of a wave are related by the equation

v=\lambda f

where

v is the speed of the wave

\lambda is the wavelength

f is the frequency

When the train is approaching the platform, we have

v = 343 m/s (speed of sound)

f = f' = 396.7 Hz (apparent frequency)

Therefore the wavelength detected by a person on the platform is

\lambda' = \frac{v}{f'}=\frac{343 m/s}{396.7 Hz}=0.865m

5 0
3 years ago
Question 4 - If Angelica starts out at 30m/s, and in 16 s speeds up to 84 m/s, what is her acceleration?
Triss [41]

Answer:

C. 3.375 m/s^2

Explanation:

The acceleration of an object can be found using the equation:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time it takes for the velocity to change from u to v

In this problem:

u = 30 m/s is the initial velocity of Angelica

v = 84 m/s is the final velocity

t is the time

Substituting into the equation, we find the acceleration:

a=\frac{84-30}{16}=3.375 m/s^2

4 0
3 years ago
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