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natka813 [3]
3 years ago
13

What is the SI unit of electric charge

Physics
1 answer:
shusha [124]3 years ago
4 0

Answer:

The SI unit for electric chargeis the C (which is the abbreviation of Coulomb}

Explanation:

The SI unit for electric chargeis the Coulomb. The letter used is the C.

1 C = 1 As

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Determine the total moment of inertia of a merry-go round with 5 children sitting on it. Of the five children, four are seated a
BaLLatris [955]

Answer:

Explanation:

Given that,

We have five children.

Each of mass m =30kg

They sit on a merry go round

Mass of Merry go round M= 150kg

Radius of Merry go round is r =2m

Four children sit at the edge of the merry go round but one child sit at the centre.

The four child that sit at the edge are 2m from the centre of the merry go round but the one at the centre is 0m from the centre

Moment of inertia?

Moment of inertia is given as

I=Σmi•ri²

For the question, the moment of inertia is the combination of inertial of child and the merry go round

I= I(merry go round) + I(four child)+ I(last child)

The merry go round is assumed to be a solid cylinder, so it is going to have the moment of inertia of solid cylinder

Then,

I(merry go round ) =½ Mr²

Also, Four of the child has the same moment of inertia, they are 2m form the centre of the merry go round why the last child has no moment of inertia

I= I(merry go round) + I(four child) +I(last child )

I= ½Mr² + 4mr² + mr'²

I = ½ × 150 ×2² + 4×30×2² + 30×0²

I = 300 +480+0

I = 780 kgm²

7 0
3 years ago
A device that changes mechanical energy to electrical energy by rotating a coil of wire through a magnetic field is called a(an)
tigry1 [53]
A device that changes mechanical energy to electrical energy by rotatin a coil of wire through a magnetic field is called B. Generator
4 0
2 years ago
Read 2 more answers
Thandy is looking at two cells under the microscope.One is a human cheek cell and the other is a leaf mesophyll cell from a plan
Vanyuwa [196]
Cell are small and us human can't see them with our own eyes. it is in possible to see cell without a microscope
7 0
3 years ago
Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1.38 m and finds that it makes 441
Tasya [4]
The period of a simple pendulum is given by:
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
g= \frac{4 \pi^2}{T^2}L (1)

We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.

We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
f= \frac{N}{t}= \frac{441}{1090 s}=0.40 Hz
And its period is the reciprocal of its frequency:
T= \frac{1}{f}= \frac{1}{0.40 Hz}=2.47 s

So now we can use eq.(1) to find the gravitational acceleration of the planet:
g= \frac{4 \pi^2}{T^2}L =  \frac{4 \pi^2}{(2.47 s)^2} (1.38 m) =8.92 m/s^2
3 0
3 years ago
a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro
finlep [7]

Answer:

\theta=53.13^o

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

V_x=V_{ox}=V_ocos\theta

V_y=V_{oy}-gt=V_osin\theta-gt

x=V_{ox}t

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}

\displaystyle y_{max}=\frac{V_{oy}^2}{2g}

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

x_{max}=3y_{max}

\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}

Using the formulas for V_{ox}, V_{oy}:

\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}

Simplifying

4cos\theta sin\theta=3sin^2\theta

Dividing by sin\theta

4cos\theta=3sin\theta

Rearranging

tan\theta=\frac{4}{3}

\theta=arctan\frac{4}{3}

\theta=53.13^o

4 0
3 years ago
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