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Mnenie [13.5K]
3 years ago
5

Study the motion map shown. what do the circled vectors represent?

Physics
1 answer:
BabaBlast [244]3 years ago
3 0

Answer:

It's option d - Negative acceleration

Explanation:

  • Let's start by demonstrate why <em>it's not option b - Speed : </em>Speed is a scalar quantity so it can not be represented by a vector
  • Let's check that <em>the green  vectors represent velocity</em> (velocity is a vector quantity, velocity is a direction aware, while speed is just a scalar)

  • Now let's  show that the circled vectors are acceleration vectors:

Mathematically position X , velocity V and acceleration A are:

\frac{dX}{dt} = V and \frac{dV}{dt} = A

Where X, V, A are vectors and \frac{d(a)}{dt} = b  indicates the derivate a of a time is equal to b.

So, this show that acceleration is a rate respect of time of velocity ⇒ When acceleration is positive, velocity increments, when acceleration is negative, velocity decrements.

<em>The above explanation correspond to the motion map shown, getting demonstrated that the answer is D - Negative acceleration </em>

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A baseball is hit almost straight up into the air with a speed of 26 m/s . Estimate how high it goes.
Natalka [10]

Answer:

The maximum height of the ball is 34.5 m.

The ball is 5.31 s in the air.

Explanation:

Hi there!

The equations for the height and velocity of the baseball that is hit straight up are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the baseball at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g =  acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at time t.

If we place the origin of the frame of reference at the place where the baseball is hit, then, y0 = 0.

To calculate how high it goes, we have to obtain the time at which the ball is at maximum height. At that point, the velocity is 0. Then using the equation of velocity:

v = v0 + g · t

0 = 26 m/s - 9.8 m/s² · t

-26 m/s / -9.8 m/s² = t

t = 2.65 s

The height at that time will be the maximum height:

y = y0 + v0 · t + 1/2 · g · t²        (y0 = 0)

y = 26 m/s · 2.65 s - 1/2 · 9.8 m/s² · (2.65 s)²

y = 34.5 m

The maximum height of the ball is 34.5 m

If it takes the ball 2.65 s to reach the maximum height it will take another 2.65 s to return to the initial position. Then, the time it will be in the air is (2.65 s + 2.65 s) 5.30 s. However, let´s calculate the time it takes the ball to reach the initial position using the equation for height.

At the initial position y = 0. Then:

y = y0 + v0 · t + 1/2 · g · t²        (y0 = 0)

0 = 26 m/s · t - 1/2 · 9.8 m/s² · t²

0 = t (26 m/s - 1/2 · 9.8 m/s² · t)      (t = 0 when the ball is hit)

0 = 26 m/s - 1/2 · 9.8 m/s² · t

-26 / -4.9 m/s² = t

t = 5.31 s     ( the difference with the 5.30 s obtained above is due to rounding the time to 2.65 s).

The ball is 5.31 s in the air.

Have a nice day!

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3 years ago
The luxury liner Queen Elizabeth 2 has a diesel-electric powerplant with a maximum power of 90 MW at a cruising speed of 31.5 kn
AlexFokin [52]

Answer:

5558643.69 N

Explanation:

F = Force

v = Velocity = 31.5 knots

Converting to m/s

1\ knot=0.514\ m/s

31.5\ knot=31.5\times 0.514\ m/s=16.191\ m/s

Power is given by

P=Fv\\\Rightarrow F=\frac{P}{v}\\\Rightarrow F=\frac{90\times 10^6}{16.191}\\\Rightarrow F=5558643.69\ N

The forward force is exerted on the ship at this highest attainable speed is 5558643.69 N

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