Explanation:
The heart rate of the astronaut is 78.5 beats per minute, which means that the time between heart beats is 0.0127 min. This will be the time t measured by the moving observer. The time t' measured by the stationary Earth-based observer is given by

a) If the astronaut is moving at 0.480c, the time t' is


This means that time between his heart beats as measured by Earth-based observer is 0.0145 min, which is equivalent to 69.1 beats per minute.
b) At v = 0.940c, the time t' is


So at this speed, the astronaut's heart rate is 1/(0.0372 min) or 26.9 beats per minute.
Answer:
10⁴ Pa
Explanation:
When the tank is half filled ,
- Volume = 5m * 4m * 1m = 20m³
Also we know that ,
- Density of water = 10³ kg/ m³
Density = Mass/Volume
10³ kg/m³ = m/20m³
m = 2 * 10⁴ kg
So that ,
Weight = mg
F = 2*10⁴ × 10 N
F = 2 * 10⁵ N
And ,
Pressure = Force/Area
P = 2 *10⁵/ 5 * 4 Pa
P = 10⁴ Pa
What figure??? please show a photo or something so i can help
The acceleration of gravity on or near the Earth's surface is 9.8 m/s² downward.
Is that right ? I don't hear any objection, so I'll assume that it is.
That means that during every second that gravity is the only force on an object,
the object either gains 9.8m/s of downward speed, or it loses 9.8m/s of upward
speed. (The same thing.)
If the rock starts out going up at 14.2 m/s, and loses 9.8 m/s of upward speed
every second, it runs out of upward gas in (14.2/9.8) = <em>1.449 seconds</em> (rounded)
At that point, since it has no more upward speed, it can't go any higher. Right ?
(crickets . . .)
Answer:

Explanation:
= 14 km
= 49 km
Intensity of a wave is inversely proportional to distance

So,

The ratio of the intensities is 