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Nikitich [7]
3 years ago
6

The drill used by most dentists today is powered by a small air-turbine that can operate at angular speeds of 350000 {\rm rpm} .

These drills, along with ultrasonic dental drills, are the fastest turbines in the world-far exceeding the angular speeds of jet engines. Suppose a drill starts at rest and comes up to operating speed in 2.0s .How many revolutions does the drill bit make as it comes up to speed? (Rev)Express your answer using two significant figures.
Physics
1 answer:
damaskus [11]3 years ago
6 0

Answer:

5833.33

Explanation:

\alpha = Angular acceleration

\theta = Number of revolutions

\omega_i = Initial angular speed = 0

t = Time taken = 2 s

Final angular speed

\omega_f=\dfrac{350000}{60}=5833.33\ rps

From the equation of rotational motion we have

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{5833.33-0}{2}\\\Rightarrow \alpha=2916.665\ rev/s^2

\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \theta=0\times t+\dfrac{1}{2}\times 2916.665\times 2^2\\\Rightarrow \theta=5833.33\ rev

The number of revolutions is 5833.33

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A French submarine and a U.S. submarine move toward each other during maneuvers in motionless water in the North Atlantic. The F
Mandarinka [93]

Answer:

f_{U}= 1019.72hz

Explanation:

From the equation we are told that:

Velocity of French sub V_{U}= 43.00km/h

Velocity of U.S. sub at V_{F}|=64.00 km//h

French Wave Frequency  F_{F}=1000Hz

Velocity of wave  V_{s}=5470 km/h

Generally the equation for Signal's frequency as detected by the U.S. is mathematically given by

Doppler effect

 f_{U} = \frac{ f_F (vs + v_{U})}{(v_s - v_F) }

 f_{U}= \frac{ 1000 x ( 5470+64)}{(5470-43)}

 f_{U}= 1019.72hz

5 0
3 years ago
Which of the following is NOT an example of the 3rd law of motion (for every Action, there is an equal but opposite reaction?
Vlad1618 [11]

Answer:

C a basketball player pushes into another one and they both fall to the left

Explanation:

I believe his is the answer because I don't see any force and not enough reaction

6 0
2 years ago
Read 2 more answers
Consider three capacitors C1, C2, and C3 and a battery. If
VLD [36.1K]

Answer:

Charge on C₁ = charge on all the three capacitors in series with it = 7.5 μC

Explanation:

Since the same voltage in the battery is used for the entire rundown,

From this information "only C₁ is connected to the battery, the charge on C₁ is 30.0 μC",

Q = C₁V = 30 μC

V = (30/C₁)

the series combination of C₂ and C₁ is connected across the battery, the charge on C₁ is 15.0 μC

The charge on both capacitors are the same and equal to 15 μC (because they are in series)

Q = (Ceq) V = 15 μC

(Ceq) = (15/V) μF

The voltage is still the same as in the first connection process

V = (30/C₁)

(Ceq) = (15/V) μF

(Ceq) = 15 ÷ (30/C₁)

(Ceq) = 15 × (C₁/30) = 0.5 C₁

(1/Ceq) = (2/C₁)

For series connection

(1/Ceq) = (1/C₁) + (1/C₂)

(2/C₁) = (1/C₁) + (1/C₂)

(2/C₁) - (1/C₁) = (1/C₂)

(1/C₁) = (1/C₂)

C₁ = C₂

C₂ = C₁

C₃, C₁, and the battery are connected in series, resulting in a charge on C₁ of 10.0 μC.

The charge on both capacitors are the same and equal to 10 μC (because they are in series)

Q = (Ceq) V = 10 μC

(Ceq) = (10/V) μF

The voltage is still the same as in the first connection process

V = (30/C₁)

(Ceq) = (10/V) μF

(Ceq) = 10 ÷ (30/C₁)

(Ceq) = 10 × (C₁/30) = 0.333 C₁

(1/Ceq) = (3/C₁)

For series connection

(1/Ceq) = (1/C₁) + (1/C₃)

(3/C₁) = (1/C₁) + (1/C₃)

(3/C₁) - (1/C₁) = (1/C₃)

(2/C₁) = (1/C₃)

C₁ = 2C₃

C₃ = (C₁/2)

C₁, C₂, and C₃ are connected in series with one another and

with the battery, what is the charge on C₁

The charge on C₁ is the same as the charge on all the capacitors and equal to Q,

Q = (Ceq) V

(1/Ceq) = (1/C₁) + (1/C₂) + (1/C₃)

Substituting for C₂ and C₃

C₂ = C₁ and C₃ = (C₁/2)

(1/C₂) = (1/C₁) and (1/C₃) = (2/C₁)

(1/Ceq) = (1/C₁) + (1/C₁) + (2/C₁)

(1/Ceq) = (4/C₁)

Ceq = (C₁/4)

Q = (Ceq) V = (C₁/4) V

But recall that V = (30/C₁) from the first connection

Q = (C₁/4) (30/C₁)

Q = (30/4) = 7.5 μC

Hope this helps!

6 0
3 years ago
What is the energy of a<br> baby who weighs 20 N<br> sitting in a high chair 1.5<br> m high?
8090 [49]

Answer:

gravitational potential energy

5 0
3 years ago
Calcula la energia potencial de Tamara que tiene una mas de 50kg y se encuentra, 80 metros arriba en el borde de un edificio
Ilya [14]

Responder:

39200 J

Explicación:

Dado:

Masa de Tamara (m) = 50 kg

Altura a la que se encuentra Tamara (h) = 80 m

Aceleración debido a la gravedad (g) = 9.8 m / s²

La energía potencial de un objeto de masa 'm' ubicada a una altura 'h' sobre el suelo se da como:

U=mgh

Ahora, conecte los valores dados y resuelva la energía potencial. Esto da,

U=50\times 9.8\times 80\\\\U=39200\ J

Por lo tanto, la energía potencial de Tamara ubicada a una altura de 80 m es 39200 J.

8 0
4 years ago
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