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Stells [14]
3 years ago
5

A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotat

es horizontally about one end while the pilot is seated at the other end. In order to achieve a radial acceleration of 34.1 m/s2 with a beam of length 5.91 m, what rotation frequency is required?
Physics
1 answer:
Natali [406]3 years ago
7 0

centripetal acceleration is given by formula

a_c = \omega^2*R

given that

a_c = 34.1 m/s^2

R  =  5.91 m

now we have

\omega^2 R = 34.1

\omega^2 * 5.91 = 34.1

\omega^2 = 5.77

\omega = 2.4 rad/s

so the ratationa frequency is given by

\omega = 2 \pi f

2.4 = 2 \pi f

f = \frac{2.4}{2\pi}

f = 0.38 Hz

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A car runs into a flying bug. Why does the bug accelerate more in the collision? *
aleksandr82 [10.1K]

Answer:

It has less mass than the car

Explanation:

Both car and bug have the same amount of force acting on them. As the bug mass is much smaller, it sees a greater acceleration

F = ma

5 0
3 years ago
4.72 A full-wave bridge-rectifier circuit with a 1-k load operates from a 120-V (rms) 60-Hz household supply through a 12-to-1 s
melisa1 [442]

Answer:

a) 12.74 V

b) Two pairs of diode will work only half of the cycle

c) 8.11 V

d) 8.11 mA

Explanation:

The voltage after the transformer is relationated with the transformer relationshinp:

V_o=Vrms*\frac{1}{12}\\V_o=10Vrms

the peak voltage before the bridge rectifier is given by:

V_{op}=Vo*\sqrt{2}\\V_{op}=14.14V

The diodes drop 0.7v, when we use a bridge rectifier only two diodes are working when the signal is positive and the other two when it's negative, so the peak voltage of the load is:

V_l=V_{op}-2(0.7)\\V_l=12.74V

As we said before only two diodes will work at a time, because the signal is half positive and half negative,so two of them will work only half of the cycle.

The averague voltage on a full wave rectifier is given by:

V_{avg}=2*\frac{V_l}{\pi}\\V_{avg}=8.11V

Using Ohm's law:

I_{avg}=\frac{V_{avg}}{R}\\\\I_{avg}=8.11mA

7 0
3 years ago
A circular area with a radius of 6.50 cm lies in the xy-plane. What is the magnitude of the magnetic flux through this circle du
White raven [17]

Answer:

a. \phi _B=3.0528\times10^-^3T\ m^2\\ \\b. \phi_B=1.83299\times10^-^3T\ m^2\\\\c.\phi_B=0

Explanation:

#Consider a circular area of radius R=2.98cm in the xy-plane at z=0. This means all the are vector points toward the +ve z-axis.

a. first, find the magnetic flux if the magnetic field has a magnitude of B=0.23T and points toward the +ve z-axis. The angle between the magnetic field and the area is \theta=0. Hence the magnetic flux:-

\phi _B=\int {\bar B} . d\bar A \\=\int BdAcos(\theta)=BAcos(0)=BA\\\\=\pi R^2B=\pi(6.50\times10^-^3m)^2(0.230T)\\=3.0528\times10^-^3 T\ m^2

Hence flux magnitude in +z direction is 3.0528\times10^-^3T \ m^2

b. We now find the magnetic flux when the field has a magnitude of <em>B=0.230T</em> and points at an angle of \theta=53.1\textdegree from the +z direction.

Magnetic flux is calculated as:

\phi _B=\int\bar B \bar dA\\=\int BdAcos (\theta)=BAcos(0)=BA\\=\pi R^2B=\pi(6.50\times 10^-^2m)^2(0.230T)\\=1.83299\times 10^-^3 T \ m^2

Hence the flux at an angle of 53.1\textdegree is 1.83299\times 10^-^3T \ m^2

c. We now need to find the magnetic flux if the field has a magnitue of B=0.230T and points in the direction of +y-direction. As with the previous parts, the magnetic flux will be calculated as:

\phi_B= \int\bar B \times d\bar A\\=\int BdAcos(\theta)\\=BAcos(90\textdegree)\\=0

Hence the magnetic flux in the +y-direction is zero.

3 0
3 years ago
a block suspended from a spring. The spring is stretch 0.5 m. If the spring constant is 500 N/m, what is the weight of the block
zhenek [66]

Answer:250N

Explanation:

Spring constant(K)=500N/m

extension(e)=0.5m

Weight=k x e

Weight=500 x 0.5

Weight=250N

5 0
4 years ago
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bija089 [108]

Answer:

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