One of the formulas for calculating electrical power is
Power = (voltage)² / (resistance)
Since this problem wants us to find the voltage, let's
re-arrange the formula.
Multiply each side by (resistance):
(Voltage)² = (power) x (resistance)
Take the square root of each side:
Voltage = √ (power x resistance)
THERE's the formula we can use to find the voltage for this problem.
Voltage = √ (2,083 W x 25.4 Ω)
Voltage = √ (52,908.2 W-Ω )
Voltage = 230 volts
Answer:
The horizontal component of the 70 N force = 35·√3 N
The vertical component of the 70 N force = 35 N
Explanation:
The magnitude of the given force, F = 70 N
The angle of inclination of the given force to the horizontal, θ = 30°
By the resolution of forces, we can resolve the given force, F, into its horizontal component, Fₓ, and vertical components,
, as follows;
The horizontal component of the given force = Fₓ = F × cos(θ)
Substituting the values, we have;
The horizontal component of the 70 N force, Fₓ = 70 × cos(30°) = 35·√3 N
The vertical component of the given force =
= F × sin(θ)
Substituting the values, we have;
The vertical component of the 70 N force,
= 70 × sin(30°) = 35 N.
1) The distance travelled by the rocket can be found by using the basic relationship between speed (v), time (t) and distance (S):

Rearranging the equation, we can write

In this problem, v=14000 m/s and t=150 s, so the distance travelled by the rocket is

2) We can solve the second part of the problem by using the same formula we used previously. This time, t=300 s, so we have:

Answer:
0.3m
Explanation:
Given parameters;
Elastic energy = 20J
Spring constant = 445N/m
Unknown
Final extension of the spring = ?
Solution:
The elastic potential energy of a stretched spring can be determined using the expression below;
EP =
k e²
k is the spring constant
e is the extension
Insert the parameters and solve;
20 =
x 445 x e²
multiply those sides by 2;
2(20) = 445e²
40 = 445e²
e² =
= 0.09
e = √0.09 = 0.3m
The work done by the electric field when moving the charge from a to b is 0.0378 J.
<h3>
Distance between a and b</h3>
r = √[(a₂ - a₁)² + (b₂ - b₁)²]
r = √[(3 - 6)² + (4 - 0)²]
r = 5 m
<h3>Work done in moving the charge</h3>
W = Fr
W = kq₁q₂/r
W = (9 x 10⁹ x 2.1 x 10⁻⁵ x 1 x 10⁻⁶)/(5)
W = 0.0378 J
Thus, the work done by the electric field when moving the charge from a to b is 0.0378 J.
Learn more about work done here: brainly.com/question/8119756
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