The y-component of the stone's velocity when it is 8 m below the hand is 14.86 m / s
v² = u² + 2 a s
s = Displacement
u = Initial velocity
a = Acceleration
u = 8 m / s
s = 8 m
v² = 8² + 2 * 9.8 * 8
v² = 64 + 156.8
v = √ 220.8
v = 14.86 m / s
The equation used to solve the problem is an equation of motion. These equations are designed to locate an object in motion using components such as velocity, displacement, acceleration and time.
Therefore, the y-component of the stone's velocity is 14.86 m / s
To know more about Equations of motion
brainly.com/question/5955789
#SPJ1
Answer:
<h2>3000 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 1000 × 3
We have the final answer as
<h3>3000 N</h3>
Hope this helps you
They relate because the further up you go, the colder it gets, and the air pressure decreases the further up you go. The altitude temperature and the air pressure both decrease, and that is their relationship. Altitude temperature decreases, the higher you go, and air pressure also decreases, the higher up you go. Therefore, the lower down you go, the higher the air pressure, and the higher the altitude temperature.
Hope this helps and have a nice day:)
Answer:
ΔΦ = -3.39*10^-6
Explanation:
Given:-
- The given magnetic field strength B = 0.50 gauss
- The angle between earth magnetic field and garage floor ∅ = 20 °
- The loop is rotated by 90 degree.
- The radius of the coil r = 19 cm
Find:
calculate the change in the magnetic flux δφb, in wb, through one of the loops of the coil during the rotation.
Solution:
- The change on flux ΔΦ occurs due to change in angle θ of earth's magnetic field B and the normal to circular coil.
- The strength of magnetic field B and the are of the loop A remains constant. So we have:
Φ = B*A*cos(θ)
ΔΦ = B*A*( cos(θ_1) - cos(θ_2) )
- The initial angle θ_1 between the normal to the coil and B was:
θ_1 = 90° - ∅
θ_1 = 90° - 20° = 70°
The angle θ_2 after rotation between the normal to the coil and B was:
θ_2 = ∅
θ_2 = 20°
- Hence, the change in flux can be calculated:
ΔΦ = 0.5*10^-4*π*0.19*( cos(70) - cos(20) )
ΔΦ = -3.39*10^-6
