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Scrat [10]
3 years ago
8

A skier (down a straight course) is able to enter the starting gate with a speed of 1 m/sec and the average acceleration down th

e hill was 3 m/s2. If the skier covers the course in 10 sec, how fast was he moving at the bottom of the course
Physics
1 answer:
ziro4ka [17]3 years ago
7 0
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Which of the following correctly describes the relationship between current and voltage as the voltage of a battery increases. R
Y_Kistochka [10]

Answer:_COC1\/2+_H\/2O>_HC1+CO\/2

Explanation:

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3 0
3 years ago
_____ is the frictional force needed to slow down an object in motion
tekilochka [14]

Answer

gravity

Explanation:

4 0
4 years ago
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A cart is moving at 55 m/s at an angle of 25° to the ground. Determine the horizontal component.
Elodia [21]

Answer:

<em>The horizontal component of the velocity is 49.85 m/s.</em>

Explanation:

<u>Rectangular Components of a Vector</u>

A 2D vector can be expressed in several forms. The rectangular form gives its two components, one for each axis (x,y). The polar form gives the components as the pair (r,θ) being r the magnitude and θ the angle.

When the magnitude and angle of the vector are given, the rectangular components are calculated as follows:

v_x=v\cos\theta

v_y=v\sin\theta

Where v is the magnitude of the vector and θ is the angle with respect to the x positive direction.

The cart is moving at v=55 m/s at θ=25°, thus:

v_x=55\cos 25^\circ

v_x=49.85\ m/s

The horizontal component of the velocity is 49.85 m/s.

7 0
3 years ago
A stone is thrown horizontally from a 50m high cliff with an initial speed of 15 meters per second. How far will the stone have
IceJOKER [234]

This is a classic case of 'velocity components.'

Imagine a vector for velocity. Now, consider that this vector could be the hypotenuse of a right triangle, with two other sides going along the x and y-axis. These sides of the triangle will have values, and adding them up using the pythagorean theorem will prove that the sum of their squares equals the square of the original vector.

Well, okay, that's nice and all, you may be saying - but how do we solve the actual question?

Let's apply this thought to the question. This vector can have both an x component and y component (essentially, parts of the vector that travel along the x and y-axis).

Now, what could these components be? We know that the stone is thrown perfectly horizontally, meaning that the x-component is quite literally the velocity.

How about the y-component? Since it's thrown at a perfect horizontal, there isn't really any vertical velocity whatsoever. There's only horizontal velocity.

"Great, fantastic! So, what's the importance of figuring out the horizontal and vertical velocities?"

When a stone is in free fall, it experiences a gravitational acceleration. This acceleration from gravity, though, only affects the vertical velocity. Since gravity is vertical as well, it's essentially impossible for the horizontal velocity to be changed at all.

This means that to solve the horizontal distance, we simply need to find the time it takes for the rock to hit the ground and multiply said time by the horizontal velocity.

Since the vertical velocity is the only thing changed by gravity, we can write out an equation that can solve for the time:

∆Y = v_{i}t + \frac{1}{2}gt^{2}

<u>We know that initial vertical velocity is zero, so:</u>

∆Y = v_{i}t + \frac{1}{2}gt^{2}

∆Y = 0t + \frac{1}{2}gt^{2}

∆Y = \frac{1}{2}gt^{2}

<u>We need to solve for t, so let's isolate the variable. Multiply both sides by 2 to get rid of the fraction:</u>

∆Y * 2 =  \frac{1}{2}gt^{2} * 2

2∆Y = gt^{2}

<u>Divide both sides by g:</u>

(2∆Y)/g = \frac{gt^{2} }{g}

<u>Square root both sides:</u>

\sqrt{(2Y)/g} = \sqrt{t^{2} }

t = \sqrt{(2Y)/g}

<u>Input our values for Y and g (Y is the height of the cliff, and g is gravitational acceleration):</u>

t = \sqrt{(2*50)/9.80}

<u>Solve:</u>

t =  \sqrt{(2*50)/9.80}

t = 3.194 (s)

Whew! That was a lot of steps to find the time! Now that we have the time, we can find the horizontal distance the rock travels:

∆x = v_{i}t + \frac{1}{2}at^{2}

<u>The horizontal velocity has no acceleration (gravity is vertical!), so:</u>

∆x = v_{i}t + \frac{1}{2}*0*t^{2}

∆x = v_{i}t

<u>The horizontal velocity is 15 m/s, and the time is 3.194:</u>

∆x = v_{i}t

∆x = 15 * 3.194

∆x = 47.91 (m)

Since we rounded the time, it makes sense that our final answer's a little bit off to the options. The closest one is option B, which is only 0.6m off, a tiny difference that may have come from the test maker's use of '10 m/s^{2}' as the gravitational acceleration (while we stayed as accurate as possible with 9.80) as well as our rounding of the final time.

Option B, the stone will have travelled 47.85 meters.

If you have any questions on how I got to the answer or if you're still confused on any topic I attempted to explain, just ask in the comments and I'll try to answer it to the best of my ability! Good luck!

- breezyツ

7 0
3 years ago
A car traveling at 15 m/s starts to decelerate steadily. It comes to a complete stop in 10 seconds. What is it’s acceleration
tresset_1 [31]

So the initial velocity is 15 m/s, the final velocity is 0 since it's at a complete stop and time is 10 seconds.  Therefore:

Acceleration=\frac{v_{final}-v_{initial}}{t_{final}-t_{initial}} =\frac{0m/s-15m/s}{10s-0s} =-1.5\frac{m}{s^2}

Therefore, the acceleration is -1.5 m/s^2.  The reason it's negative is due to the fact that the vector is going against it's original movement since it's decelerating.

7 0
3 years ago
Read 2 more answers
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