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vredina [299]
3 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]3 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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2 years ago
A 2kg box is pushed along a flat frictionless surface with an applied force of 53.91newton j+19.62 j were j is horizontal and j
tamaranim1 [39]

Since the box doesn't move vertically at all, no work is done by the vertical component of the force.

If 53.91 Newtons is the horizontal component of the force (very unclear in the question), then the work done is

<u>Work = (force) x (distance)</u>

Work = (53.91 N) x (10 m)

<em>Work = 539.1 Joules</em>

<u>Power = (work done) / (time to do the work)</u>

Power = (539.1 joules) / (90 sec)

Power = (539.1/90) (joule/sec)

<em>Power = 5.99 watts</em>

=====================================

Note:  If the mass of the box is only 2 kg, and you push it along the surface with a constant force of 53⁺ Newtons, and the surface really is frictionless, then that box is gonna cover a WHOOOOOLE LOT more than 10 meters in 90 seconds.  I get 109,168 meters !

5 0
3 years ago
A small sphere of radius R is arranged to pulsate so that its radius varies in simple harmonic motion between a minimum of R−x a
Colt1911 [192]

Answer:

The intensity of sound wave at the surface of the sphere I =   \frac{ 2\pi^{2}R^{2} f^{2}\sqrt{\rho B}(\triangle R)^{2}}{ d^{2} }

Explanation:

B = Bulk modulus

Intensity, I = \frac{P_{max} ^{2} }{2\sqrt{\rho B} }

The amplitude of oscillation of the sphere is given by:

P_{max} = BkA\\k = \frac{2\pi }{\lambda} \\

A = \triangle R\\

Substitute v and A into Pmax

P_{max} = (2\pi f)\sqrt{\rho B} \triangle R\\ P_{max} ^{2} = 4\pi^{2} *f^{2} \rho B (\triangle R)^{2}

I = \frac{ 4\pi^{2} f^{2} \rho B (\triangle R)^{2}}{2\sqrt{\rho B} }

P_{total} = 4\pi R^{2} I

P_{total} =4\pi R^{2}  \frac{ 2\pi^{2} f^{2} \rho B (\triangle R)^{2}}{\sqrt{\rho B} }

The intensity of the sound wave at a distance  is given by:

I = \frac{P_{total} }{4\pi d^{2} }

I = 4\pi R^{2}  \frac{ 2\pi^{2} f^{2} \rho B (\triangle R)^{2}}{\sqrt{\rho B} } * \frac{1}{4\pi d^{2} } \\I =   \frac{ 2\pi^{2}R^{2} f^{2}\sqrt{\rho B}(\triangle R)^{2}}{ d^{2} }

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3 years ago
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Answer: 2.4×10^-3 v/m

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Strength of electric field (E) =?

The formulae that relates potential difference, eoectiic field strength and distance between plates is given as

v = Ed

By substituting the parameters, we have that

12 = E × 5.0×10^-3

E = 12/ 5.0 × 10^-3

E = 2.4×10^-3 v/m

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