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ohaa [14]
3 years ago
13

a 0.500 kg mass is oscillating on a spring with k=330 N/m. the total energy of its oscillation is 3.24 J .what is the amplitude

of the oscillation? (unit=m)
Physics
2 answers:
juin [17]3 years ago
6 0

Answer:

anyone know this or will i have to get my brother

Alekssandra [29.7K]3 years ago
5 0

Answer:

0.140

Explanation:

Give credit to those comment

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Can someone answer this question quickly??
masya89 [10]

Answer:

Lithium and Sodium

Explanation:

Lithium and Sodium is in the same group 1A

7 0
3 years ago
Four capacitors with capacitance 3.0 pF, 2.0 pF, 5.0 pF and X pF are connected in series to each other. If the equivalent capaci
bagirrra123 [75]

Answer:

<h2>A. 6pF</h2>

Explanation:

If unknown capacitance C1, C2, C3 and C4 are connected in series to one another, their equivalent capacitance of the circuit will be expressed as shown

\frac{1}{C_t} = \frac{1}{C_1} +\frac{1}{C_2} +\frac{1}{C_3} +\frac{1}{C_4} \\

Given the capacitance's 3.0 pF, 2.0 pF, 5.0 pF and X pF connected in series to each other. If the equivalent capacitance of the circuit is 0.83 pF, then to get X, we will apply the formula above;

\frac{1}{0.83} = \frac{1}{3.0} +\frac{1}{2.0} +\frac{1}{5.0} +\frac{1}{C_4} \\\\\\1.205 = 0.333+0.5+0.2+\frac{1}{C_4} \\\\1.205 = 1.033 + \frac{1}{C_4} \\\\\frac{1}{C_4}  = 1.205-1.033\\\\\frac{1}{C_4}  = 0.172\\\\C_4 = \frac{1}{0.172}\\ \\C_4 = 5.8pF\\\\

C₄ ≈ 6pF

Hence the value of the X capacitor is approximately 6pF

8 0
3 years ago
Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface
FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

\tau=I\alpha     (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

\tau=Fd        (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

Fd=I\alpha\\\\I=\frac{Fd}{\alpha}

Finally, you replace the values of all parameters in the previous equation for I:

I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2

The moment of inertia of the door around the hinges is 2 kgm^2

3 0
3 years ago
A package of aluminum foil contains 50. Ft2 of foil, which weighs approximately 7.0 oz . Aluminum has a density of 2.70 g/cm3. W
kumpel [21]

We know density = Mass / Volume

So Volume = Mass/Density

Volume =  Area * Thickness

Area = 50 ft^2 = 4.645152 m^2\\ \\Mass = 7 oz = 0.198447 kg\\ \\Density = 2.70 g/cm^3 = 2700 kg/m^3\\ \\Substituting \\ \\4.645152 *t = \frac{0.198447}{2700} \\ \\t = 1.58*10^{-5}m = 1.58*10^{-2}mm

So the approximate thickness of the foil in millimeters = 1.58*10^{-2}mm

6 0
3 years ago
1The density of an of an object with a volume of 60.0 and mass of 400.0g is______
Kobotan [32]
D is the answer
To get the density you divide mass by volume
So the equation is 400/60=d
4 0
3 years ago
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