Answer:
2.91 x 10¹² sec
Explanation:
d = distance of nearest star, Proxima Centauri = 4.3 ly = 4.3 x 9.46 x 10¹⁵ m
v = speed of new horizon probe = 14 km/hr = 14000 m/s
t = time taken for the new horizon probe to reach nearest star, Proxima Centauri = ?
Using the equation
d = v t
Inserting the values given
4.3 x 9.46 x 10¹⁵ = (14000) t
t = 2.91 x 10¹² sec
Answer:
6.0 m/s vertical and 9.0 m/s horizontal
Explanation:
For the vertical component, we use the formula:
- Sin(34°) = <em>y</em> / 10.8
Then we <u>solve for </u><u><em>y</em></u>:
- 0.559 = <em>y</em> / 10.8
And for the horizontal component, we use the formula:
- Cos(34°) = <em>x</em> / 10.8
Then we <u>solve for </u><u><em>x</em></u><u>:</u>
- 0.829 = <em>x</em> / 10.8
So the answer is " 6.0 m/s vertical and 9.0 m/s horizontal".
When somebody hands you a Celsius°, it's easy to find the equivalent Fahrenheit°.
Fahrenheit° = (1.8 · Celsius°) + 32° .
So 100°C works out to 212°F.
It's also easy to find the equivalent Kelvin. Just add 273.15 to the Celsius.
So now you can see that 100°C is equal to A and D,
and it's less than B .
The only one it's greater than is C .
The scale is a rest scale reads the support for is not enough net force.
-- We're going to be talking about the satellite's speed.
"Velocity" would include its direction at any instant, and
in a circular orbit, that's constantly changing.
-- The mass of the satellite makes no difference.
Since the planet's radius is 3.95 x 10⁵m and the satellite is
orbiting 4.2 x 10⁶m above the surface, the radius of the
orbital path itself is
(3.95 x 10⁵m) + (4.2 x 10⁶m)
= (3.95 x 10⁵m) + (42 x 10⁵m)
= 45.95 x 10⁵ m
The circumference of the orbit is (2 π R) = 91.9 π x 10⁵ m.
The bird completes a revolution every 2.0 hours,
so its speed in orbit is
(91.9 π x 10⁵ m) / 2 hr
= 45.95 π x 10⁵ m/hr x (1 hr / 3,600 sec)
= 0.04 x 10⁵ m/sec
= 4 x 10³ m/sec
(4 kilometers per second)
