All categories

3 years ago

A private aviation helicopter's main rotor blades rotate at approximately

Physics

1 answer:

Arisa [49]3 years ago

4 0

Answer: 7.5 rev/s

Explanation:

We are given the angular velocity $\omega$ a helicopter's main rotor blades:

$\omega=450 rpm=450 \frac{rev}{min}$

However, we are asked to express this $\omega$ in the International Systrm (SI) units. In this sense, the SI unit for time is second ( $s$ ):

$\omega=450 \frac{rev}{min} \frac{1 min}{60 s}$

$\omega=7.5 \frac{rev}{s}$

You might be interested in

Wyatt is moving a box with a mass of 37 kg a distance of 37 meters. Wyatt did 360 J of work in 2 minutes when moving the box. Wh

pentagon [3]

His power output was 3 Watt (360 Joule/120 seconds). The power output can be calculated by dividing the quantity of work by the amount of second needed for the activity and also by multiplying the force amount with the velocity of the activity. The power output usually used for measuring the ability of machine for doing its job.

7 0

3 years ago

Read 2 more answers

Of all the radio waves, these have the shortest wavelengths and the the highest frequency?

Julli [10]

If by chance you mean any wave, the wave with the shortest wavelength/highest frequency is gamma rays.

4 0

3 years ago

The force required to stretch a Hooke’s-law spring varies from 0 N to 63.5 N as we stretch the spring by moving one end 5.31 cm

Alika [10]

Answer:

Force constant will be 1195.85 N/m

Work done will be 1.6859 J

Explanation:

We have given the force, F = 63.5 N

Spring is stretched by 5.31 cm

So x = 0.0531 m

Force is given , F = 63.5 N

We know that force is given by $F=kx$

So $63.5=k\times 0.0531$

k = 1195.85 N/m

Now we have to find the work done

We know that work done is given by

$W=\frac{1}{2}kx^2=\frac{1}{2}\times 1195.85\times 0.0531^2=1.6859J$

8 0

3 years ago

A charged object traveling 7 m in a uniform electric field of 5 N/C experiences a 4 J increase in Kinetic Energy.

Travka [436]

To solve this problem it is necessary to apply the principles of conservation of Energy in order to obtain the final work done.

The electric field in terms of the Force can be expressed as

$E = \frac{F}{q} \rightarrow F=Eq$

Where,

F = Force

E= Electric Field

q = Charge

Puesto que el trabajo realizado es equivalente al cambio en la energía cinetica entonces tenemos que

KE = W

KE = F*d

In the First Case,

$4 = (qE)d\\q = \frac{4}{Ed}\\q = \frac{4}{5*7}\\q = 0.1142C$

In Second Case,

$KE = q E'd$

$KE = (0.1142)(40)(7)$

$KE = 31.976J$

The total energy change would be subject to,

$\Delta KE = 31.976-4$

$\Delta KE = 27.976J$

Therefore the Kinetic Energy change of the charged object is 27.976J

3 0

2 years ago

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

5 0

3 years ago