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3 years ago

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The path the moon's umbra traces across earth's surface is the path of totality.What would you see if you were standing in the p

ath of totality?

1 answer:

ivann1987 [24]3 years ago

4 0

You would see the darkest part of the moon

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Is a scientist who studies climate change. As part of a larger experiment, her work requires her to travel. Her company makes su

Alex777 [14]

We know that whoever she is is traveling to Antarctica or elsewhere
in the south polar region. June is the beginning of Winter there, with
zero to extremely short daylight.

But we still don't know her name.

3 0

3 years ago

Can someone please solve this

Svetllana [295]

Answer: i can see properly

Explanation:

7 0

3 years ago

A train slows down as it rounds a sharp horizontal turn, going from 94.0 km/h to 46.0 km/h in the 17.0 s that it takes to round

Svetllana [295]

Answer:

1.41 m/s^2

Explanation:

First of all, let's convert the two speeds from km/h to m/s:

$u = 94.0 km/h \cdot \frac{1000 m/km}{3600 s/h} = 26.1 m/s$

$v=46.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=12.8 m/s$

Now we find the centripetal acceleration which is given by

$a_c=\frac{v^2}{r}$

where

v = 12.8 m/s is the speed

r = 140 m is the radius of the curve

Substituting values, we find

$a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2$

we also have a tangential acceleration, which is given by

$a_t = \frac{v-u}{t}$

where

t = 17.0 s

Substituting values,

$a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2$

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

$a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2$

6 0

3 years ago

Read 2 more answers

A bird lands on a bare copper wire carrying a current of 51

nirvana33 [79]

The resistance of the piece of wire is
$R= \frac{\rho L}{A}$
where
$\rho = 1.68 \cdot 10^{-8}\Omega m$ is the resistivity of the copper
$L=5.1 cm=0.051 m$ is the length of the piece of wire
$A=0.13 cm^2 = 0.13 \cdot 10^{-4} m^2$ is the cross sectional area of the wire
By substituting these values, we find the value of R:
$R= \frac{\rho L}{A}=6.6 \cdot 10^{-5} \Omega$

Then, by using Ohm's law, we find the potential difference between the two points of the wire:
$V=IR=(51 A)(6.6 \cdot 10^{-5} \Omega )=3.4 \cdot 10^{-3} V$

7 0

3 years ago

) A steel guitar string with a diameter of 1.00 mm is stretched between supports 80.0 cm apart. The temperature is 0.0°C. (a) Fi

ladessa [460]

Answer: a. Mass per unit length =0.0245kg/m

b. Tension =2.45x10^-8N

C. Tension = 2.45 x10^-8N

Fundamental frequency =200Hz

Explanation:

7 0

3 years ago