To start with solving this problem, let us assume a launch angle of 45 degrees since that gives out the maximum range for given initial speed. Also assuming that it was launched at ground level since no initial height was given. Using g = 9.8 m/s^2, the initial velocity is calculated using the formula:

(v sinθ)^2 = (v0 sinθ)^2 – 2 g d

where v is final velocity = 0 at the peak, v0 is the initial velocity, d is distance = 11 m

Rearranging to find for v0: 
v0 = sqrt (d * g/ sin(2 θ))

v0 = 10.383 m/s

8 0

4 years ago

According to the Big Bang theory, the universe was once very __________ and is now __________.

Ne4ueva [31]

dense; expanding is the one

8 0

4 years ago

Read 2 more answers

ariana is accelerating her car at 4.6 m/s^2 it took her 10 seconds to reach this acceleration her starting lost the day was 0 me

arlik [135]

Acceleration of the car driven by Ariana = 4.6 m/s^2
Time taken to reach this acceleration = 10 seconds
Let us assume the final velocity of the car (V2) = x m/s
Initial velocity of the car driven by Ariana (V1) = 0 m/s
Then
Acceleration = (V2 - V1)/ Time taken
4.6 = (x - 0)/10
4.6 = x/10
x = 4.6 * 10 m/s
= 46 m/s
So the final velocity of the car Ariana was driving was 46 meters per second.

5 0

3 years ago

A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri

Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = $\frac{I}{nqA}$

Where;

n = number of free electrons per cubic meter

q = electron charge

A = cross-sectional area of the wire

First let's calculate the number of free electrons per cubic meter (n)

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M) -------------(ii)

Substitute the values of Nₐ, ρ and M into equation (ii) as follows;

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

Now let's calculate the drift electron

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

Substitute these values into equation (i) as follows;

v = $\frac{I}{nqA}$

v = $\frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}$

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

6 0

4 years ago