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4 years ago

A girl sitting on a merry-go-round moves counterclockwise through an arc length of 3.15 m.

1 answer:

4 0

Well, you get the movement as the arc in radians times the radius, which we need to find. Doing $\frac{3.15}{1.82}=1.73$ , which is about $\sqrt{3}$ meters.

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Why is the overall charge on an atom neutral?

Oksi-84 [34.3K]

The overall charge on a neutral atom is zero.

A neutral atoms contains equal number of electrons and protons.The charge of a proton and electron is equal in magnitude but opposite in sign. A proton has a charge of 1.6 x 10⁻¹⁹C and the charge of an electron is -1.6 x 10⁻¹⁹C. Thus in a neutral atom, the charge of all the protons gets cancelled with the charge of all the electrons.

Thus a neutral atom has a net zero charge.

5 0

3 years ago

How many hours does it take for a car moving at 65 mi/hr to travel 115 miles?

faltersainse [42]

Formula for time: velocity/distance **Going to use GRESA Given: v=65mi/hr d=115mi --- Required: time=? --- Equation: v/d --- Solution: 65mi/hr / 115mi = 0.57 hr --- Answer: 0.57 hr

8 0

3 years ago

One long wire lies along an x axis and carries a current of 43 A in the positive x direction. A second long wire is perpendicula

qwelly [4]

Answer:

Bnet=1.006*10^-6T

Explanation:

One long wire lies along an x axis and carries a current of 43 A in the positive x direction. A second long wire is perpendicular to the xy plane, passes through the point (0, 5.9 m, 0), and carries a current of 41 A in the positive z direction. What is the magnitude of the resulting magnetic field at the point (0, 1.7 m, 0)?

the magnetic field Bnet= $\sqrt{b1^2+b2^2}$

the magnetic field due this long wire is given by

B1=∨I1/ $(2\pi *R1)$ ..............................1

B2=∨I2/ $(2\pi *R2)$ ............................2

Bnet= $\sqrt{(vI1/2*pi*R1)^2+(vI2/2*pi*R2)^2}$ .......................3

Bnet=v/2*pi $\sqrt{(I1/R1)^2+(i2/R2)^2}$

Bnet=4*pi*10^-7/(2 $\pi$ ) $\sqrt{(43/1.7)^2+(41/29.5)^2}$

Bnet=0.0000002*(641.72)^.5

Bnet=1.006*10^-6T

8 0

4 years ago

At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power ge

Effectus [21]

Answer:

$X=3976.078202kW \approx 3976kW$

Explanation:

From the question we are told that

Wind speed $V_w=10m/s$

Turbine blade diameter $D=90m$

Air density $\tau=1.25kg/m^3$

Mechanical energy $K.E =0.05kJ/kg$

Generally the power generation potential of the wind turbine X is mathematically given as

$X=m'*K.E$

Where

$m'=\tau *\pi*D^2/4*V$

$m'=1.25 *\pi*90^2/4*10$

$m'=79521.56404kg/s \approx 79521.5kg/s$

Therefore

$X=m'*K.E$

$X=79521.56404*0.05$

$X=3976.078202kW \approx 3976kW$

$X=4.0MW$

4 0

3 years ago

Energy transformations in a watefall?

Alexxandr [17]

One of the most common energy transformations is the transformation between potential energy and kinetic energy. In waterfalls such as Niagara Falls, potential energy is transformed to kinetic energy. The water at the top of the falls has gravitational potential energy. As the water plunges, its velocity increases. Thanks to google!!!! LOL :)

6 0

3 years ago