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3 years ago

15

Ken Griffey, Jr's warehouse shot in the 1933 home run derby travelled 93 feet per second for 5 seconds. How far did he hit the b

all?

1 answer:

Ghella [55]3 years ago

3 0

Answer:

465 feet because 93*5 = 465, btw that was 1993 not 1933

Explanation:

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What is the easiest way to increase the magnetic force acting on the rotor in an induction motor?

Schach [20]

Answer:

Explanation:

Magnets are of two major forms namely the permanent magnet and the temporary magnets. Temporary magnets magnetizes and demagnetize easily while permanent magnets does not magnetizes and demagnetize easily.

This permanents magnets are applicable in loudspeakers, generators, induction motor etc.

To increase the

The following will tend to increase the magnetic force acting on the rotor in an induction motor.

1. Increasing the strength of the bar magnet. Increase in strength of the magnet will lead to increase in the magnetic force acting on the rotor.

2. Increase in the magnetic line of force also known as the magnetic flux around the magnet will also increase the magnetic force acting on the rotor.

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3 years ago

"Giant Swing", the seat is connected to two cables as shown in the figure (Figure 1) , one of which is horizontal. The seat swin

Bingel [31]

The horizontal force is m*v²/Lh, where m is the total mass. The vertical force is the total weight (233 + 840)N.

<span>Fx = [(233 + 840)/g]*v²/7.5 </span>

<span>v = 32.3*2*π*7.5/60 m/s = 25.37 m/s </span>

<span>The horizontal component of force from the cables is Th + Ti*sin40º and the vertical component of force from the cable is Ta*cos40º </span>

<span>Thh horizontal and vertical forces must balance each other. First the vertical components: </span>

<span>233 + 840 = Ti*cos40º </span>

<span>solve for Ti. (This is the answer to the part b) </span>

<span>Horizontally </span>

<span>[(233 + 840)/g]*v²/7.5 = Th + Ti*sin40º </span>

<span>Solve for Th </span>

<span>Th = [(233 + 840)/g]*v²/7.5 - Ti*sin40º </span>

<span>using v and Ti computed above.</span>

3 0

3 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

3 years ago

Devise an exponential decay function that fits the given data, then answer the accompanying questions. Be sure to identify the

7nadin3 [17]

Answer:

22145.27733 ft

124984.76055 ft

Explanation:

The equation of pressure is

$P=P_0e^{-kh}$

where,

$P_0$ =Atmospheric pressure = 800 mbar

k = Constant

h = Altitude = 35000 ft

$P=\dfrac{1}{3}P_0$

$\dfrac{1}{3}P_0=P_0e^{-k35000}\\\Rightarrow \dfrac{1}{3}=e^{-k35000}\\\Rightarrow 3=e^{k35000}\\\Rightarrow ln3=k35000\\\Rightarrow k=\dfrac{ln3}{35000}\\\Rightarrow k=3.13\times 10^{-5}$

Now

$P=\dfrac{1}{2}P_0$

$ln2=kh\\\Rightarrow h=\dfrac{ln2}{k}\\\Rightarrow h=\dfrac{ln2}{3.13\times 10^{-5}}\\\Rightarrow h=22145.27733\ ft$

The altitude will be 22145.27733 ft

$P=0.02P_0$

$0.02P_0=P_0e^{-kh}\\\Rightarrow 0.02=e^{-3.13\times 10^{-5}h}\\\Rightarrow ln0.02=-3.13\times 10^{-5}h\\\Rightarrow h=\dfrac{ln0.02}{-3.13\times 10^{-5}}\\\Rightarrow h=124984.76055\ ft$

The elevation is 124984.76055 ft

6 0

3 years ago

What is the reason for the widespread use of fins on surfaces? Group of answer choices Decreasing the rate of heat transfer from

kompoz [17]

Answer:

ask your member

Explanation:

kayo lang din makakasagot nyan

4 0

2 years ago