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3 years ago

7

A man is walking while riding a train. He says he is moving at 2 mph. A woman standing on a platform at a train station says the

man on the train is moving at 72 mph. Which person is correct?

1 answer:

Semenov [28]3 years ago

6 0

The woman on the platform is correct because it is the pace of the man moving on the train not walking.

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5. Granny Goodwitch boards a bus at the local shopping center. The bus travels an

irina [24]

As BBC ztrrtv B 4 C w.a)z

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3 years ago

A skier moving at 4.75 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220

disa [49]

First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force, $\mu m g$ . For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
$ma=-\mu m g$
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
$a=-(0.220)(9.81 m/s^2)=-2.16 m/s^2$

Now we can use the following relationship to find the distance covered by the skier before stopping, S:
$2aS=v_f^2-v_i^2$
where $v_f=0$ is the final speed of the skier and $v_i=4.75 m/s$ is the initial speed. Substituting numbers, we find:
$S=- \frac{v_i^2}{2a}=- \frac{(4.75 m/s)^2}{2(-2.16 m/s^2)}=5.23 m$

5 0

3 years ago

Why is the curve between 1950 and 1980 relatively flat and centered around zero degrees difference from the baseline? (Hint: how

zimovet [89]

Look at the title of the graph, in small print under it.

Each point is "compared to 1950-1980 baseline". So the set of data for those years is being compared to itself. No wonder it matches up pretty close !

3 0

4 years ago

Delilah does 170 Joules of work in 30 seconds. Adam does 260 Joules of work in 20 seconds. Who was more powerful?

sp2606 [1]

Power = $\frac{Work}{Time}$

Delilah: 170J/30s = 5.66 W

Adam: 260J/20s = 13 W

6 0

3 years ago

The traffic on the freeway is moving at a constant speed of 24 m/sm/s. What distance does the traffic travel while the car is mo

ExtremeBDS [4]

Incomplete question as there is so much information is missing.The complete question is here

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

Answer:

Distance traveled=240 m

Explanation:

Given data

Initial velocity of car v₀=0 m/s

Final velocity of car vf=24 m/s

Distance traveled by car S=120 m

To find

Distance does the traffic travel

Solution

To find the distance first we need to find time, for time first we need acceleration

So

$(V_{f})^{2}=(V_{o})^{2}+2aS\\ So\\a=\frac{(V_{f})^{2}-(V_{o})^{2} }{2S}\\ a=\frac{(24m/s)^{2}-(0m/s)^{2} }{2(120)}\\a=2.4 m/s^{2}$

As we find acceleration.Now we need to find time

So

$V_{f}=V_{i}+at\\t=\frac{V_{f}-V_{i}}{a}\\t=\frac{(24m/s)-(0m/s)}{(2.4m/s^{2} )}\\t=10s$

Now for distance

So

$Distance=velocity*time\\Distance=(24m/s)*(10s)\\Distance=240m$

7 0

4 years ago