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Fed [463]
3 years ago
8

At what point in its trajectory does a batted baseball have its minimum speed? If air resistance can be neglected, how does this

speed compare with the horizontal component of its velocity at other points?
Physics
1 answer:
lys-0071 [83]3 years ago
8 0

To understand the problem and give a logical explanation we will use the concepts related to the movement of a Projectile.

On this concept the components concerning the speed of an object that is launched in parabolic motion are characterized. There will be two types of speed components, the vertical and the horizontal.

The vertical component of velocity gradually decreases as it is affected by the force of gravity until it becomes zero at the highest point of the trajectory. After this point, it again begins to increase gradually in effect of its movement towards the gravitational attraction of the Earth.

In the case of horizontal force, if the effect of air resistance is neglected, the speed will be the same throughout the trajectory.

Now comparing this with the movement of the baseball, at the highest point, the vertical component will be zero, and the horizontal velocity will remain. Both at the beginning of the trajectory and at the end of it, there will be components of the vertical velocity, which would increase this value. Therefore the minimum speed value will be at the top, when the vertical component is zero, and the horizontal component is maintained

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A 2.3 kg block of copper is heated at atmospheric pressure such that its temperature increases from 6 oC to 90 oC. How much heat
Shtirlitz [24]

Answer:

the  heat absorbed by the block of copper is 74368.476J

Explanation:

Hello!

To solve this problem use the first law of thermodynamics that states that the heat applied to a system is the difference between the initial and final energy considering that the mass and the specific heat do not change so we can infer the following equation

Q=mCp(T2-T1)

Where

Q=heat

m=mass=2.3kg

Cp=0.092 kcal/(kg C)=384.93J/kgK

T2=Final temperatura= 90C

T1= initial temperature=6 C

solving

Q=(2.3kg)(384.93\frac{J}{kgC} )(90C-6C)=74368.476J

the  heat absorbed by the block of copper is 74368.476J

7 0
3 years ago
Describe a situation when you might travel at a high velocity bit with low acceleration
iren [92.7K]

I am sitting in my seat.
I am listening to my mp3 and reading my book.
My eyes are getting heavy.  They start to close.
I try to stay awake, but it's no use.
I am so warm and comfortable and sleepy,
and I have just finished my dinner.
Finally I can't help it.  Resistance is futile.
I give up, and fall deep asleep.
My head rests back against my soft, comfy seat.

My seat is in row 26 on the airplane I'm flying in
to visit my grandmother on the coast.
We are cruising at 560 miles an hour, bearing 280°,
at flight level 320 .
The temperature outside my window is  -60°F .

5 0
3 years ago
A merry-go-round is spinning at a rate of 4.04.0 revolutions per minute. Cora is sitting 0.50.5 m from the center of the merry-g
dsp73

Answer:

angular speed of both the children will be same

Explanation:

Rate of revolution of the merry go round is given as

f = 4.04 rev/min

so here we have

f = \frac{4.04}{60} =0.067 rev/s

here we know that angular frequency is given as

\omega = 2\pi f

\omega = 2\pi(0.067)

\omega = 0.42 rad/s

now this is the angular speed of the disc and this speed will remain same for all points lying on the disc

Angular speed do not depends on the distance from the center but it will be same for all positions of the disc

7 0
3 years ago
When a current flows in an aluminum wire of diameter 2.91 mm 2.91 mm , the drift speed of the conduction electrons is 0.000191 m
charle [14.2K]

Answer:

Number of electrons are flowing per second is 2.42 x 10¹⁹

Explanation:

The electric current flows through a wire is given by the relation :

I=envA   ....(1)

Here I is current, e is electronic charge, v is drift velocity of electrons and A is the Area of the wire.

But electric current is also define as rate of electrons passing through junction times their charge, i.e. ,

I=Ne      ....(2)

Here N is the rate of electrons passing through junction.

From equation (1) and (2).

eN = envA

N=nvA

But area of wire, A=\pi \frac{d^{2} }{4}

Here d is diameter of wire.

So, N = nv\pi \frac{d^{2} }{4}

Substitute 2.91 x 10⁻³ m for d, 0.000191 m/s for v and 6 x 10²⁸ m⁻³ for n in the above equation.

N = 6\times10^{28}\times 0.000191\times\pi \frac{(2.91\times10^{-3} )^{2} }{4}

N = 2.42 x 10¹⁹ s⁻¹  

8 0
3 years ago
A 20-lb force acts to the west while an 80-lb force acts 45° east of north. The magnitu 80 lb 70 lb 067 lb 100 lb
natulia [17]

Answer:

option C

Explanation:

given,

force act on west  = 20 lb

force act at 45° east of north = 80 lb

magnitude of force = ?

∑ F y  =  80 cos 45⁰

    F y =  56.57 lb

magnitude of forces in x- direction

∑ F x = -20 + 80 sin 45⁰

        = 36.57 lb

net force

F = \sqrt{F_x^2+F_y^2}

F = \sqrt{56.57^2+36.57^2}

F = 67.36 lb≅ 67 lb

hence, the correct answer is option C

4 0
3 years ago
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