A boy with mass 50 kg running at 5m/s jumps on to a 20 kg trolley travelling in the same direction at 1.5 m/s. What is their com
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Answer:
6 teaspoons of baking powder required.
Explanation:
Given that
According to the recipe of pancake,
For every teaspoon of salt, 2 teaspoons of baking baking powder is required.
To find:
How much baking powder will be needed, if 1 teaspoon of salt was used ?
Solution:
This problem can be solved using ratio.
teaspoon of salt : 2 teaspoons of baking powder
Let us multiply the above ratio with 3.
teaspoon of salt : 2
3 teaspoons of baking powder
OR
1 teaspoon of salt : 6 teaspoons of baking powder
So, answer is <em>6 teaspoons </em>of baking powder required.
Also, we can use the unitary method:
teaspoon of salt needs = 2 teaspoons of baking powder
1 teaspoon of salt needs = teaspoons of baking powder
1 teaspoon of salt needs = 2 3 = <em>6</em> teaspoons of baking powder needed
So, the answer is:
<em>6 teaspoons of baking powder </em>required.
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 =
Where;
n = number of free electrons per cubic meter
q = electron charge
A = cross-sectional area of the wire
<em>First let's calculate the number of free electrons per cubic meter (n)</em>
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)
<em>Substitute the values of Nₐ, ρ and M into equation (ii) as follows;</em>
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
<em>Now let's calculate the drift electron</em>
Known values from question:
A = 5.261 mm² = 5.261 x 10⁻⁶m²
I = 23A
q = 1.6 x 10⁻¹⁹C
<em>Substitute these values into equation (i) as follows;</em>
v =
v =
v = 3.22 x 10⁻⁴ m/s
Therefore, the drift electron is 3.22 x 10⁻⁴ m/s