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vredina [299]
2 years ago
7

A baseball pitcher throws a ball at 90.0 mi/h in the horizontal direction. How far does the ball fall vertically by the time it

reaches home plate, which is a horizontal distance of 60.5 ft away? (Express your answer in units of feet. Neglect air resistance.)
Physics
1 answer:
Lisa [10]2 years ago
3 0

Answer:

Vertical distance=  3.3803ft

Explanation:

First with the speed of the ball and the distance traveled horizontally we can determine the flight time to reach the plate:

Velocity= (90 mi/h) × (1 mile/5280ft) = 475200ft/h

Distance= Velocity × time⇒ time= 60.5ft / (475200ft/h) = 0.00012731h

time=  0.00012731h × (3600s/h)= 0.458316s

With this time we can determine the distance traveled vertically taking into account that its initial vertical velocity is zero and its acceleration is that of gravity, 9.81m/s²:

Vertical distance= (1/2) × 9.81 (m/s²) × (0.458316s)²=1.0303m

Vertical distance= 1.0303m × (1ft/0.3048m) = 3.3803ft

This is the vertical distance traveled by the ball from the time it is thrown by the pitcher until it reaches the plate, regardless of air resistance.

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1. When the fission of uranium-235 is carried out, about 0.1 percent of the mass of the reactants is lost during the reaction. W
DerKrebs [107]

The correct answer to the question is that the lost mass has been converted into energy.

EXPLANATION:

From Einstein's theory, we know that energy and mass are inter convertible .

When some amount of mass is lost, same amount of energy equivalent to mass is produced.

Let us consider m is the mass lost during any reaction. Hence, the amount of energy produced will be-

                                 Energy E = mc^2                

Here, c is the velocity of light i.e c = 3\times 10^8\ m/s

As per the question, uranium-235 undergoes fission. The amount of mass defect is 0.1 %.

The mass defect is defined as the difference between mass of reactants and products. During the fission, energy is produced.

The energy produced in this reaction is nothing else than the energy equivalent to mass defect.  Approximately 199.5 Mev of energy equivalent to this mass defect is produced in this reaction.

6 0
2 years ago
Read 2 more answers
An archer tests various arrowheads by shooting arrows at a pumpkin that is suspended from a tree branch by a rope, as shown to t
erik [133]

Answer:

Bounce 1 ,  pass 3,   emb2

Explanation:

(By the way I am also doing that question on College board physics page) For the Bounce arrow, since it bumps into the object and goes back, it means now it has a negative momentum, which means a larger momentum is given to the object. P=mv, so the velocity is larger for the object, and larger velocity means a larger kinetic energy which would result in a larger change in the potential energy. Since K=0.5mv^2=U=mgh, a larger potential energy would have a larger change in height which means it has a larger angle θ with the vertical line. Comparing with the "pass arrow" and the "Embedded arrow", the embedded arrow gives the object a larger momentum, Pi=Pf (mv=(M+m)V), it gives all its original momentum to the two objects right now. (Arrow and the pumpkin), it would have a larger velocity. However for the pass arrow, it only gives partial of its original momentum and keeps some of them for the arrow to move, which means the pumpkin has less momentum, means less velocity, and less kinetic energy transferred into the potential energy, and means less change in height, less θangle.  So it is  Bounce1, pass3, emb2.  

6 0
2 years ago
a small bar magnet is suspended horizontally by a string. When placed in a uniform horizontal magnetic field, it will
Reil [10]

Answer:

It will neither translate in the opposite direction nor .rotate so as to be at right angles, it will also neither rotate so as to be vertical direction

6 0
2 years ago
Two stunt drivers drive directly toward each other. At time t=0 the two cars are a distance D apart, car 1 is at rest, and car 2
lesantik [10]

Answer: Hello there!

We know this:

The distance between the cars at t= 0 is D.

car 2 has an initial velocity of v0 and no acceleration.

car 1 has no initial velocity and a acceleration of ax that starts at  t = 0

then we could obtain the acceleration of the car 1 by integrating the acceleration over the time; this is v(t) = ax*t where there is not a constant of integration because the car 1 has no initial velocity.

Because the cars are moving against each other, we want to se at what time t they meet, this is equivalent to see:  

position of car 1 + position of car 2 = D

and in this way we could ignore constants of integration :D

for the position of each car we integrate again:  

P1(t) = (1/2)ax*t^2 and P2(t) = v0t

v0t + (1/2)ax*t^2 = D

v0t + (1/2)ax*t^2  - D = 0

now we can solve it for t using the Bhaskara equation.

t = \frac{-v0 +\sqrt{v0^{2} + 4*(1/2)ax*D } }{2(1/2)ax} =\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax}

that we cant solve witout knowing the values for v0, D and ax. But you could replace them in that equation and obtain the time, where you must remember that you need to choose the positive solution (because this quadratic equation has two solutions).

Now we want to know the velocity of car 1 just before the impact, this can be calculated by valuating the time in the as the time that we just found in the velocity equation for the car 1, this is:

v(\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax}) = ax*\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax} = {-v0 +\sqrt{v0^{2} + 2ax*D }

where again, you need to replace the values of v0, D and ax.

7 0
2 years ago
To determine the y-component of a projectile’s velocity, what operation is performed on the angle of the launch?
koban [17]

<em>To determine the y component of velocity of a projectile </em><u><em>sine </em></u><em>operation is performed on the angle of launch.</em>

<u>Answer:</u> <em>sine</em>

<u>Explanation:</u>

Thus a_x=0,a_y=g

The initial velocity u can be resolved along two directions.

Along the X direction initial velocity = u cos θ

Along y direction initial velocity= u sin θ

From the equation of motion v= u+at

Thus velocity along x direction v_x=u cos θ

Velocity along y direction v_y= u sinθ -gt

Sign of g is negative.

3 0
2 years ago
Read 2 more answers
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