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melamori03 [73]
3 years ago
15

What is the farthest planet from earth?

Physics
1 answer:
IRINA_888 [86]3 years ago
7 0
The answer is pluto. just look it up
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A mass attached to a 50.0 cm long string starts from rest and is rotated 40 times in one minute before reaching a final angular
ch4aika [34]

To solve this problem it is only necessary to apply the kinematic equations of angular motion description, for this purpose we know by definition that,

\theta = \frac{1}{2}\alpha t^2 +\omega_0 t + \theta_0

Where,

\theta = Angular Displacement

\alpha =Angular Acceleration

\omega_0 = Angular velocity

\theta_0 =Initial angular displacement

For this case we have neither angular velocity nor initial angular displacement, then

\theta = \frac{1}{2}\alpha t^2

Re-arrange for \alpha,

\alpha = \frac{2\theta}{t^2}

Replacing our values,

\alpha = \frac{2(40rev*\frac{2\pi rad}{1rev})}{60^2}

\alpha = 0.139rad/s

Therefore the ANgular acceleration of the mass is 0.139rad/s^2

4 0
3 years ago
a steel block has a mass of 40g.it is in the form of a cube. each edge is 1.74cm long. calculate the density
Vinil7 [7]

Answer:

d ≈ 7,6 g/cm³  

Explanation:

d = m/V = 40g/5,27cm³ ≈ 7,6 g/cm³

V = l³ = (1.74cm)³ ≈ 5,27 cm³

3 0
3 years ago
Two wheels are identical but wheel b is spinning with twice the angular speed of wheel
Murljashka [212]
<span>If two wheels are exactly the same but spin at different speeds, wheel b is twice te speed of wheel a, it is possible to find the ratio of the magnitude of radial acceleration at a singular point of the rim on wheel be to the spot is four.</span>
6 0
3 years ago
From this diagram which of the following can you conclude
Volgvan

Answer: A

Explanation:

I say opposites attract

3 0
3 years ago
Read 2 more answers
Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
3 years ago
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