The only thing acting as the roller coaster gets to the bottom is gravity thus
acceleration is 9.81
so u use the formula
vf^2 = vi^2 + 2abetaD
thus
distance = vf^2 - vi^2 divided by 2a
so 26^2 - 0^2 divided by 2(9.81m/s)
equals 676 divided by 2(9.81m/s)
= 34.45463812m
then use the appropriate 3 sig digs
so 34.5m
hopes this helps
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Answer:
864 mT
Explanation:
The magnetic field due to a long straight wire B = μ₀i/2πR where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, i = current in wire, and R = distance from center of wire to point of magnetic field.
The magnitude of magnetic field due to the first wire carrying current i = 2.70 A at distance R which is mid-point between the wires is B = μ₀i/2πR.
Since the other wire also carries the same current at distance R, the magnitude of the magnetic field is B = μ₀i/2πR.
The resultant magnetic field at B is B' = B + B = 2B = 2(μ₀i/2πR) = μ₀i/πR
Now R = 2.50 cm/2 = 1.25 cm = 1.25 × 10⁻² m and i = 2.70 A.
Substituting these into B' = μ₀i/πR, we have
B' = 4π × 10⁻⁷ H/m × 2.70 A/π(1.25 × 10⁻² m)
B = 10.8/1.25 × 10⁻⁵ T
B = 8.64 × 10⁻⁵ T
B = 864 × 10⁻³ T
B = 864 mT
Answer:
The mechanical energy of the ball-Earth-floor system the instant the ball left the floor is 7 Joules.
Explanation:
It is given that,
Initial gravitational potential energy of the ball-Earth-floor system is 10 J.
The ball then bounces back up to a height where the gravitational potential energy is 7 J.
Let U is the mechanical energy of the ball-Earth-floor system the instant the ball left the floor. Due to the conservation of energy, the mechanical energy is equal to difference between initial gravitational potential energy and the after bouncing back up to a height.
Initial mechanical energy is 10 + 0 = 10 J
Mechanical energy just before the collision is 0 + 10 = 10 J
Final mechanical energy, 7 + 0 = 7 J
Hence, the mechanical energy of the ball-Earth-floor system the instant the ball left the floor is 7 Joules.
False. solar comes from the sun while wind energy comes from the wind. hope this helps!
