V = I · R
Voltage = (current) · (Resistance)
Voltage = (250 A) · (2.09 x 10⁴)
Voltage = 5,225,000 volts .
I may be out of line here, but I'm pretty sure
that the resistance is 2.09 x 10⁻⁴ .
Then
Voltage = 0.05225 volt (not 5 million and something)
Answer:
v = 1 i ^ m / s, v = 1 m / s
Explanation:
In this problem we show the difference between vectors and scalars.
The average speed is
v = Δx / Δt
The bold are vectors
Δx = 20-10 = 10 i^ m
v = 10/10
v = 1 i ^ m / s
the unit vector indicates that the velocity is on the x-axis
The average speed is
v =Δx / Δt
v = 10/10
v = 1 m / s
in this case we have a scalar
A descriptive observation may very well be a mixture of both quantitative and qualitative as it can utilize elements of both types. Qualitative deals with the kinds of observations that cannot be measured in numerical form. Quantitative data is just that.
Answer:
block K = 29.39 J and spring #1 Ke = 360 J
Explanation:
In this problem we have that the elastic energy of the spring becomes part kinetic energy and the part in work against the force of friction, so, to use the law of conservation of energy, the decrease in energy is the rubbing force work
= Ef - E₀
Let's look for the energies
Initial
E₀ = Ke = ½ k₁ x₁²
Final, this is just before starting to compress the spring
Ef = Ke = ½ m v²
The work of the rubbing force is
= -fr x
Let's write Newton's second law the y axis
N-W = 0
N = W
fr = μ N
fr = μ mg
Let's replace
-μ mg x = ½ m v² - ½ k₁ x₁²
v² = 2/m (½ k₁ x1₁² -μ mg x)
v² = 2/6 (½ 2000 0.6²2 - 0.5 6 9.8 1) = 1/3 (360 - 29.4)
v = 3.13 m / s
With this value we calculate the energy of the block
K = ½ m v²
K = ½ 6 3.13²
K = 29.39 J
Calculate eenrgy of the spring ke 1
Ke = ½ k₁ x₁²
Ke = ½ 2000 0.60²
Ke = 360 J
Answer:
5
Explanation:
Hope this helps
PLEASE MARK AS BRAINLIEST !!!