The answer is B.
The planet cannot be too hot or too cold it has to be the right distance from its sun to maintain life.
<span>95 km/h = 26.39 m/s (95000m/3600 secs)
55 km/h = 15.28 m/s (55000m/3600 secs)
75 revolutions = 75 x 2pi = 471.23 radians
radius = 0.80/2 = 0.40m
v/r = omega (rad/s)
26.39/0.40 = 65.97 rad/s
15.28/0.40 = 38.20 rad/s
s/((vi + vf)/2) = t
471.23 /((65.97 + 38.20)/2) = 9.04 secs
(vf - vi)/t = a
(38.20 - 65.97)/9.04 = -3.0719
The angular acceleration of the tires = -3.0719 rad/s^2
Time is required for it to stop
(0 - 38.20)/ -3.0719 = 12.43 secs
How far does it go?
65.97 - 38.20 = 27.77 M</span>
Given that:
k = 500 n/m,
work (W) = 704 J
spring extension (x) = ?
we know that,
Work = (1/2) k x²
704 = (1/2) × 500 × x²
x = 1.67 m
A spring stretched for 1.67 m distance.
Answer:
T=1.384×10⁶seconds
Explanation:
Given data
p (Intensity)=1.30 kw/m²
E (Energy)=1.8×10⁹ J
A (Area)=1.00 m²
T (Time required)=?
Solution
E=PT ................eq(i)
where E is energy
P is radiation power
T is time
Radiating Power is given as
P=pA
Where p is intensity
A is Area
Put P=pA in eq(i) we get
E=pAT
T=E/pA
