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
The planets have (their) moons. So the answer would be planets.
Answer:
Explanation:
extension in the spring = 40.4 - 31.8 = 8.6 cm = 8.6 x 10⁻² m .
kx = mg
k is spring constant , x is extension , m is mass
k x 8.6 x 10⁻² = 7.52 x 9.8
k = 856.93 N/m
= 857 x 10⁻³ KN /m
b ) Both side is pulled by force of 188 N .
Tension in spring = 188N
kx = T
856.93 x = 188
x = .219.38 m
= 21.938 cm
= 21.9 cm .
length of spring = 31.8 + 21.9
= 53.7 cm .
C. Rotations per second
Or normally we'd use Radians Per second
_Brainliest if helped!!
Answer:
0.3 m
Explanation:
Initially, the package has both gravitational potential energy and kinetic energy. The spring has elastic energy. After the package is brought to rest, all the energy is stored in the spring.
Initial energy = final energy
mgh + ½ mv² + ½ kx₁² = ½ kx₂²
Given:
m = 50 kg
g = 9.8 m/s²
h = 8 sin 20º m
v = 2 m/s
k = 30000 N/m
x₁ = 0.05 m
(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²
x₂ ≈ 0.314 m
So the spring is compressed 0.314 m from it's natural length. However, we're asked to find the additional deformation from the original 50mm.
x₂ − x₁
0.314 m − 0.05 m
0.264 m
Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.