Answer: 4 ohms
Explanation:
For parallel connection we use this formula
R=equivalent resistance
R1=24 ohms
R2=8 ohms
R3=12 ohms
1/R=1/R1 + 1/R2 + 1/R3
1/R=1/24 + 1/8 + 1/12
1/R=(1x1+3x1+2x1) ➗ 24
Cross multiplying we get
24x1=(1x1+3x1+2x1) x R
24=(1+3+2) x R
24=6xR
Divide both sides by 6
24 ➗ 6=6xR ➗ 6
4=R
R=4 ohms
The correct answer would be cells
To solve this problem we must basically resort to the kinematic equations of movement. For which speed is defined as the distance traveled in a given time. Mathematically this can be expressed as
Where
d = Distance
t = time
For which clearing the time we will have the expression
Since we have two 'fluids' in which the sound travels at different speeds we will have that for the rock the time elapsed to feel the explosion will be:
In the case of the atmosphere -composite of air- the average speed of sound is 343m / s, therefore it will take
The total difference between the two times would be
Therefore 3.357s will pass between when they feel the explosion and when they hear it
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.
6.388 or 6.388468 cubic feet..
not sure
