I think it’s D................
Answer:
Stretch in the spring = 0.1643 (Approx)
Explanation:
Given:
Mass of the sled (m) = 9 kg
Acceleration of the sled (a) = 2.10 m/s
²
Spring constant (k) = 115 N/m
Computation:
Tension force in the spring (T) = ma
Tension force in the spring (T) = 9 × 2.10
Tension force in the spring (T) = 18.9 N
Tension force in the spring = Spring constant (k) × Stretch in the spring
18.9 N = 115 N × Stretch in the spring
Stretch in the spring = 18.9 / 115
Stretch in the spring = 0.1643 (Approx)
Answer:
Θ=0.01525 rad
or
Θ=0.87°
Explanation:
Given data
wavelength λ=2.5 µm =2.5×10⁻⁶m
Diameter d=0.20 mm =0.20×10⁻³m
To find
Angle Θ in radians and degree
Solution
Circular apertures have first dark fringe at
Θ=(1.22λ)/d
Substitute the given values
So
Θ=[1.22(2.5×10⁻⁶m)]/0.20×10⁻³m
Θ=0.01525 rad
or
Θ=0.87°
Answer:
30m
Explanation:
Given parameters:
Speed of the goat = 1.5m/s
Time = 20s
Unknown:
Distance traveled = ?
Solution:
To solve this problem, we use;
Speed =
Distance = Speed x time
So;
Distance = 1.5m/s x 20s = 30m
a) Speed of the Moon: 1025 m/s
The speed of the moon is equal to the ratio between the circumference of its orbit and the orbital period:

where
is the radius of the orbit of the Moon
is the orbital period
Substituting into the formula, we find

b) Centripetal force: 
The centripetal force acting on the Moon is given by:

where
is the mass of the Moon
is its orbital speed
is the radius of the orbit
Substituting into the formula, we find

c) Gravitational force
The only relevant force that acts on the Moon, and that keeps the Moon in circular motion around the Earth, is the gravitational force exerted by the Earth on the Moon. In fact, this force "pulls" the Moon towards the Earth, so towards the centre of the orbit of the Moon, therefore it acts as source of centripetal force for the Moon.