Answer:
42.11 years old
Explanation:
Given that:
In 2000, a 20-year-old astronaut left Earth to explore the galaxy; her spaceship travels at 2.5 x 10^8 m/s. She returns in 2040
To find her age we use:
Δtm is time interval for the observer stationary relative to the sequence of
events = 2040 - 2000 = 40 years
Δts is is the time interval for an observer moving with a speed v relative to the sequence of event
v = velocity = 2.5 x 10^8 m/s
c = speed of light = 3 x 10^8 m/s
Here age in 2000 is 20 year, therefore when she appear she would be 20 year + 22.11 year = 42.11 years old
Acceleration = (change in speed) / (time for the change)
Change in speed = (ending speed) - (starting speed)
= (7.0 m/s) - (0.8 m/s) = 6.2 m/s
Time for the change = 15 seconds
Acceleration = (6.2 m/s) / (15 sec)
= (6.2/15) m/s²
= 0.413 m/s²
Seriously? It’s that hard to use a calculator??
Answer:
hellooooo :) ur ans is 33.5 m/s
At time t, the displacement is h/2:
Δy = v₀ t + ½ at²
h/2 = 0 + ½ gt²
h = gt²
At time t+1, the displacement is h.
Δy = v₀ t + ½ at²
h = 0 + ½ g (t + 1)²
h = ½ g (t + 1)²
Set equal and solve for t:
gt² = ½ g (t + 1)²
2t² = (t + 1)²
2t² = t² + 2t + 1
t² − 2t = 1
t² − 2t + 1 = 2
(t − 1)² = 2
t − 1 = ±√2
t = 1 ± √2
Since t > 0, t = 1 + √2. So t+1 = 2 + √2.
At that time, the speed is:
v = at + v₀
v = g (2 + √2) + 0
v = g (2 + √2)
If g = 9.8 m/s², v = 33.5 m/s.
Answer:
The temperature is 2541.799 K
Explanation:
The formula for black body radiation is given by the relation;
Q = eσAT⁴
Where:
Q = Rate of heat transfer 56.6
σ = Stefan-Boltzman constant = 5.67 × 10⁻⁸ W/(m²·k⁴)
A = Surface area of the cube = 6×(3.72 mm)² = 8.3 × 10⁻⁵ m²
e = emissivity = 0.288
T = Temperature
Therefore, we have;
T⁴ = Q/(e×σ×A) = 56.6/(5.67 × 10⁻⁸ × 8.3 × 10⁻⁵ × 0.288) = 4.174 × 10¹⁴ K⁴
T = 2541.799 K
The temperature = 2541.799 K.
