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Mazyrski [523]
3 years ago
7

How to tell how much work gravity does on something?

Physics
1 answer:
Grace [21]3 years ago
4 0
By dropping a ball and seeing how long it takes to hit the ground or throw a ball up and time it as well
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Rohith has been jogging to the bus stop for 2.0 minutes at 3.5 m/s when he looks at
nikdorinn [45]
Multiply these numbers and there’s your answer
6 0
3 years ago
Is tin magnetic, ferromagnetic or non-magnetic?
drek231 [11]
Tin is very weakly attracted to magnets, but is not ferro magnetic. I would say it is magnetic even though it is very weakly attracted to it.
8 0
3 years ago
Suppose that, while lying on a beach near the equator of a far-off planet watching the sun set over a calm ocean, you start a st
zubka84 [21]

Answer:

R=3818Km

Explanation:

Take a look at the picture. Point A is when you start the stopwatch. Then you stand, the planet rotates an angle α and you are standing at point B.

Since you travel 2π radians in 24H, the angle can be calculated as:

\alpha =\frac{2*\pi *t}{24H}  t being expressed in hours.

\alpha =\frac{2*\pi *11.9s*1H/3600s}{24H}=0.000865rad

From the triangle formed by A,B and the center of the planet, we know that:

cos(\alpha )=\frac{r}{r+H}  Solving for r, we get:

r=\frac{H*cos(\alpha) }{1-cos(\alpha) } =3818Km

6 0
3 years ago
100 kw of power is delivered to the other side of a city by a pair of power lines with the voltage difference of 13014.1 v.
FinnZ [79.3K]
A) The power delivered to the lines is
P_{in}= 100 kW=1 \cdot 10^5 W
And the voltage at which the lines work is
V=13014.1 V
Since the power delivered is the product between the voltage and the current:
P=VI
We can find the current flowing in the lines:
I= \frac{P}{V}= \frac{1 \cdot 10^5 W}{13014.1 V}=7.68 A

b) The voltage change along each line can be found by using Ohm's law:
\Delta V = IR = (7.68 A)(10 \Omega)=76.8 V

c) The power wasted as heat along each line is given by:
P_d = I^2 R = (7.68 A)^2 (10 \Omega) = 590 W
And since we have 2 lines, the total power wasted as heat in both lines is
P_d = 2 \cdot 590 W=1180 W
6 0
3 years ago
an n-type semiconductor is known to have an electron concentration of 5.63 x 1019 m-3. if the electron drift velocity is 113 m/s
Grace [21]

Answer:

1.99581248\ /\Omega m

Explanation:

e = Charge of electron = 1.6\times 10^{-19}\ C

n = Electron concentration = 5.63\times 10^{19}\ /m^{3}

v_d = Drift veloctiy = 113 m/s

E = Electric field = 510 V/m

Electron mobility is given by

\mu=\dfrac{v_d}{E}\\\Rightarrow \mu=\dfrac{113}{510}\\\Rightarrow \mu=0.22156\ m^2/Vs

Conductivity is given by

\sigma=ne\mu\\\Rightarrow \sigma=5.63\times 10^{19}\times 1.6\times 10^{-19}\times 0.22156\\\Rightarrow \sigma=1.99581248\ /\Omega m

The conductivity of this material is 1.99581248\ /\Omega m

4 0
3 years ago
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