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Vikentia [17]
2 years ago
7

A harmonic oscillator starts with an amplitude of 20.0 cm. After 10.0 s, the amplitude decreases to 15.0 cm. If the linear dampi

ng coefficient is 2.00 Ns/m, how much mass is oscillating
Physics
1 answer:
tia_tia [17]2 years ago
4 0

Answer: the amount of mass is oscillating is 34.8 kg

Explanation:

Given that;

amplitude A = 20.0 cm

time t = 10 s

amplitude decreases x = 15.0 cm

damping coefficient b = 2.00 N.s/m

amount of mass is oscillating = ?

we know that; amplitude can be expressed as;

x = Ae^-(∝t)

we substitute

15 = 20e^-∝(10)

∝ = 0.02877 s⁻¹

Hence mass m will be;

m = b/2∝

we substitute

m = (2 N.s/m) / ( 2 × 0.02877 s⁻¹)

m = 34.8 kg

Therefore the amount of mass is oscillating is 34.8 kg

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A car drives around a curve with radius 400 m at a speed of 32 m/s. The road is banked at 7.0 degree. The mass of the car is 150
Ronch [10]

Answer:

The magnitude of the centripetal force to make the turn is 3,840 N.

Explanation:

Given;

radius of the cured road, r = 400 m

speed of the car, v = 32 m/s

mass of the car, m = 1500 kg

The magnitude of the centripetal force to make the turn is given as;

F_c = \frac{mv^2}{r}

where;

Fc is the centripetal force

F_c = \frac{mv^2}{r} \\\\F_c = \frac{(1500)(32)^2}{400}\\\\F_c = 3,840 \ N

Therefore, the magnitude of the centripetal force to make the turn is 3,840 N.

3 0
3 years ago
A student coils a copper wire around a bar magnet. What action will cause the device to generate electricity?
kolbaska11 [484]

I hope the wire is not wound too tightly around the bar magnet.
The device will generate electrical energy when the bar magnet
is moving in or out of the coil of wire.

6 0
3 years ago
When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa
MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2

In parallel

\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}

\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0

Solving the above quadratic equation

\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}

\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

6 0
3 years ago
I push a box with mass 10 kg across a 15 m room. If I apply a force of 20N to do this how much work is being done on the box?
devlian [24]
You could use the formula

W=Fd

F(force)=20N

D(distance/displacement) =15m

W=(20N)(15m)

W= 300 J
8 0
3 years ago
You and a friend are playing with a bowling ball to demonstrate some ideas of Rotational Physics. First, though, you want to cal
RideAnS [48]

Answer:

K_{total} = 19.4 J

Explanation:

The total kinetic energy that is formed by the linear part and the rotational part is requested

         K_{total} = K_{traslation}  + K_{rotation}

let's look for each energy

linear

        K_{traslation} = ½ m v²

rotation

        K_{rotation} = ½ I w²

the moment of inertia of a solid sphere is

       I = 2/5 m r²

we substitute

       K_{total} = ½ mv² + ½ I w²

           

angular and linear velocity are related

           v = w r

we substitute

           K_{total} = ½ m w² r² + ½ (2/5 m r²) w²

           K_{total} = m w² r² (½ + 1/5)

           K_{total} = \frac{7}{10} m w² r²

let's calculate

           K_{total} = \frac{7}{10}   6.40 16.0² 0.130²

           K_{total} = 19.4 J

6 0
2 years ago
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