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hammer [34]
3 years ago
6

A car has a unibody-type frame and is supported by four suspension springs, each with a force constant of 29600 n/m. the combine

d mass of the car's frame and everything inside it (the engine, the seats, the passengers, etc.) is 1090 kg. because of worn-out shockabsorbers, the car vibrates up and down every time it is driven over a pothole. what is the frequency of this vibration? answer in units of hz.
Physics
1 answer:
Dafna1 [17]3 years ago
4 0
I’m done with my homework e ee was a good night I got to go to
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A proton is a subatomic particle that carries a ___________ charge
Dovator [93]

The answer would be a positive charge

4 0
3 years ago
The smallest detail visible with ground-based solar telescopes is about 1 arc second. How large a region (in km) does this repre
mestny [16]

Answer:

x = 727.5 km

Explanation:

With the conditions given using trigonometry, we can find the tangent

       tan θ = CO / CA

With CO the opposite leg and CE is the adjacent leg which is the distance from the Tierral to Sun

   

        D =150 10⁶ km (1000m / 1 km)

        D = 150 10⁹ m.

We must take the given angle to radians.

       1º  = 3600 arc s  

       π rad = 180º

       θ = 1 arc s (1º / 3600 s arc) (pi rad / 180º) =

       θ = 4.85 10⁻⁶ rad

That angle is extremely small, so we can approximate the tangent to the angle

     

       θ = x / D

       x = θ D

       x = 4.85 10-6  150 109

       x = 727.5 103 m

       x = 727.5 km

4 0
3 years ago
A 6.00-mH solenoid is connected in series with a 5.0-μF capacitor and an AC source. The solenoid has internal resistance 3.0 Ω w
son4ous [18]

Answer:

5773.50269 Hz

23 A

Explanation:

L = Inductance = 6 mH

C = Capacitance = 5 μF

R = Resistance = 3 Ω

\epsilon = Maximum emf = 69 V

Resonant angular frequency is given by

\omega=\dfrac{1}{\sqrt{LC}}\\\Rightarrow \omega=\dfrac{1}{\sqrt{6\times 10^{-3}\times 5\times 10^{-6}}}\\\Rightarrow \omega=5773.50269\ Hz

The resonant angular frequency is 5773.50269 Hz

Current is given by

I=\dfrac{\epsilon}{R}\\\Rightarrow I=\dfrac{69}{3}\\\Rightarrow I=23\ A

The current amplitude at the resonant angular frequency is 23 A

7 0
3 years ago
Your answer should be precise to 0.1 m/s. Use a gravitational acceleration of 10 m/s/s. At it lowest point, a pendulum is moving
saw5 [17]

Explanation:

It is given that,

Speed, v₁ = 7.7 m/s

We need to find the velocity after it has risen 1 meter above the lowest point. Let it is given by v₂. Using the conservation of energy as :

\dfrac{1}{2}mv_1^2=\dfrac{1}{2}mv_2^2+mgh

v_2^2=v_1^2-2gh

v_2^2=(7.7)^2-2\times 10\times 1

v_2=6.26\ m/s

So, the velocity after it has risen 1 meter above the lowest point is 6.26 m/s. Hence, this is the required solution.

4 0
3 years ago
Read 2 more answers
Can you find the magnetic field based on force? a straight segment of wire 35.0 cm long carrying a current of 1.40 a is in a uni
Lostsunrise [7]
The force exerted by a magnetic field on a wire carrying current is:
F=ILB \sin \theta
where I is the current, L the length of the wire, B the magnetic field intensity, and \theta the angle between the wire and the direction of B.

In our problem, the force is F=0.20 N. The current is I=1.40 A, while the length of the wire is L=35.0 cm=0.35 m. The angle between the wire and the magnetic field is 53 ^{\circ}, so we can re-arrange the formula and substitute the numbers to find B:
B= \frac{F}{IL \sin \theta}= \frac{0.20 N}{(1.40 A)(0.35 m)(\sin 53^{\circ})}=0.51 T
3 0
3 years ago
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