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Amiraneli [1.4K]
2 years ago
7

A major-league pitcher throws a baseball toward home plate. The ball rotates at 1560 rpm, and it travels the 18.5 meters to the

plate at an average translational speed of 40.2 m/s. How many revolutions does the ball make during this trip?
Physics
1 answer:
DedPeter [7]2 years ago
4 0

Answer:

no of revolutions  = 56.62 rev

Explanation:

given data

rotates = 1560 rpm

travels distance = 18.5 meters

speed = 40.2 m/s

solution

first we convert here rev/min to rev/sec

rotates = 1560 rpm = 1560  × 0.0167 = 26.052 rev/sec

we get here no of revolutions that is

no of revolutions = spin rotate ÷ time    ..........1

here time = distance ÷ speed

time = 18.5 ÷  40.2  

time = 0.4601 s

put value in equation 1 we get

no of revolutions = \frac{26.052}{0.4601}  

no of revolutions  = 56.62 rev

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The absolute pressure in water at a depth of 5m is read to be 145 kPa. Determine (a) the local atmospheric pressure, and (b) the
irga5000 [103]

Answer:

a) 95950 pascals

b) 137642.5 pascals

Explanation:

The absolute pressure (Pabs) on a fluid is:

P_{abs}=P_{gauge}+P_{atm} (1)

With Pgauge the pressure due depth on the fluid and Patm the atmospheric pressure. Pgauge is equal to:

P_{gauge}=\rho gh (2)

with ρ the fluid density, g the gravitational acceleration and h the depth on the fluid. Using (2) on (1) and solving for Patm:

P_{atm}=P_{abs}-P_{gauge}=P_{abs}-\rho_{water} gh

P_{atm}=(145000Pa)-(1000\frac{kg}{m^{3}})(9.81\frac{m}{s^{2}})(5m)

P_{atm}=95950Pa

b) Here we're going to use again (1) but now we have another value of density because it's other liquid, to know that value we should use the fact that specific gravity (S.G) for liquids is the ratio between fluid density and water density:

S.G=\frac{\rho_{fluid}}{\rho_{water}}

\rho_{liquid}=S.G*\rho_{water}

\rho_{liquid}=(0.85)*(1000\frac{kg}{m^{3}})=850\frac{kg}{m^{3}}

so:

P_{abs}=\rho_{liquid} gh+P_{atm}=(850\frac{kg}{m^{3}})(9.81\frac{m}{s})(5m)+95950Pa

P_{abs}=137642.5 Pa

3 0
2 years ago
A device that uses electricity and magnetism to create motion is called a (motor magnet generator) . In a reverse process, a dev
Tems11 [23]

the answers are a. and c.

I hope I help you out!

6 0
2 years ago
Read 2 more answers
Ann and Bob are carrying a 18.5 kg table that is 2.25 m long. A 8.33 kg box sits on the table 0.750 m from Ann. How much lift fo
IgorC [24]

Answer:

F = 118 N

Explanation:

Assume Ann and Bob lift at their respective ends of the table

Sum moments about Bob's position to zero.

Let F be Ann's upward force

F[2.25] - 18.5(9.80)[2.25 / 2] - 8.33(9.80)[0.750] = 0

F = 117.86133333...  

7 0
2 years ago
A 1,160 kg satellite orbits Earth with a tangential speed of 7,446 m/s. If the satellite experiences a centripetal force of 8,95
Juli2301 [7.4K]

Answer:

8.02×10⁵ m

Explanation:

Equation for centripetal force:

F = mv²/r

Solving for r:

r = mv²/F

Given:

F = 8955 N

m = 1160 kg

v = 7446 m/s

r = (1160 kg) (7446 m/s)² / 8955 N

r = 7.182×10⁶ m

The height above the surface is:

h = 7.182×10⁶ m − 6.38×10⁶ m

h = 0.802×10⁶ m

h = 8.02×10⁵ m

4 0
2 years ago
A nucleus whose mass is 3.499612×10^(−25) kg undergoes spontaneous alpha decay. The original nucleus disappears and there appear
Elanso [62]

Answer:

The sum of the kinetic energies of the alpha particle and the new nucleus = (6.5898 × 10⁻¹³) J

Explanation:

Old nucleus ---> New nucleus + alpha particle.

We will use the conservation of energy theorem for extremely small particles,

Total energy before split = total energy after split

That is,

Total energy of the original nucleus = (total energy of the new nucleus) + (total energy of the alpha particle)

Total energy of these subatomic particles is given as equal to (rest energy) + (kinetic energy)

Rest energy = mc² (Einstein)

Let Kinetic energy be k

Kinetic energy of original nucleus = k₀ = 0 J

Kinetic energy of new nucleus = kₙ

Kinetic energy of alpha particle = kₐ

Mass of original nucleus = m₀ = (3.499612 × 10⁻²⁵) kg

Mass of new nucleus = mₙ = (3.433132 × 10⁻²⁵) kg

Mass of alpha particle = mₐ = (6.640678 × 10⁻²⁷) kg

Speed of light = c = (3.0 × 10⁸) m/s

Total energy of the original nucleus = m₀c² (kinetic energy = 0, since it was originally at rest)

Total energy of new nucleus = (mₙc²) + kₙ

Total energy of the alpha particle = (mₐc²) + kₐ

(m₀c²) = (mₙc²) + kₙ + (mₐc²) + kₐ

kₙ + kₐ = (m₀c²) - [(mₙc²) + (mₐc²)

(kₙ + kₐ) = c² (m₀ - mₙ - mₐ)

(kₙ + kₐ) = (3.0 × 10⁸)² [(3.499612 × 10⁻²⁵) - (3.433132 × 10⁻²⁵) - (6.640678 × 10⁻²⁷)]

(kₙ + kₐ) = (9.0 × 10¹⁶)(0.00007322 × 10⁻²⁵) = (6.5898 × 10⁻¹³) J

5 0
2 years ago
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