It increases confidence because the more times you conduct the same experiment over and over should either prove your hypothesis right and wrong and eliminate any random occurrences that might affect your results.
It depends how you want it to work do you want to take a picture oh yea and to do that you must create a account btw
Magnitude of normal force acting on the block is 7 N
Explanation:
10N = 1.02kg
Mass of the block = m = 1.02 kg
Angle of incline Θ
= 30°
Normal force acting on the block = N
From the free body diagram,
N = mgCos Θ
N = (1.02)(9.81)Cos(30)
N = 8.66 N
Rounding off to nearest whole number,
N = 7 N
Magnitude of normal force acting on the block = 7 N
Answer:
Explanation:
given
T = 3months = 7.9 × 10⁶s
orbital speed = 88 × 10³m/s
V= 2πr÷T
∴ r = (V×T) ÷ 2π
r = (88km × 7.9 × 10⁶s) ÷ 2π
r = 1.10 × 10⁸km
using kepler's 3rd law
mass of both stars = (seperation diatance)³/(orbital speed)²
M₁ + M₂ = (2r)³/(
year)²
= (1.06 × 10²⁵)/(6.2×10¹³)
1.71×10¹²kg
since M₁ = M₂ =1.71×10¹²kg ÷ 2
M₁ = M₂ = 8.55×10¹¹kg
Answer:
16.1 m/s
Explanation:
We can solve the problem by using the law of conservation of energy.
At the beginning, the spring is compressed by x = 35 cm = 0.35 m, and it stores an elastic potential energy given by

where k = 316 N/m is the spring constant. Once the block is released, the spring returns to its natural length and all its elastic potential energy is converted into kinetic energy of the block (which starts moving). This kinetic energy is equal to

where m = 0.15 kg is the mass of the block and v is its speed.
Since the energy must be conserved, we can equate the initial elastic energy of the spring to the final kinetic energy of the block, and from the equation we obtain we can find the speed of the block:
