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Simora [160]
3 years ago
8

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

mon velocity?
Physics
1 answer:
beks73 [17]3 years ago
8 0

Answer:

Explanation:

Boy information

Mass = 50kg

Velocity = 5m/s

Trolley information

Mass = 20 kg

Velocity = 1.5 m/s

so using conservation law of momentum

MV + mv = ( M + m ) v'

50*5 + 20*1.5 = ( 50 + 20 ) v'

250 + 30 = 70 v'

280 = 70 v'

so v' = 4m/s which is their common velocity.

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lcompute the increase in length of 500 m of copper wire when it's temperature changes from 12 degrees celsius to 32 degrees Cels
marshall27 [118]

Explanation:

step by step explanation

7 0
2 years ago
The Carson family's pancake recipe uses 2 teaspoons of baking powder for every 1/3 of a teaspoon of salt. How much baking powder
melomori [17]

Answer:

6 teaspoons of baking powder required.

Explanation:

Given that

According to the recipe of pancake,

For every \frac{1}{3} 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.

\frac{1}3 teaspoon of salt : 2 teaspoons of baking powder

Let us multiply the above ratio with 3.

\frac{1}{3}\times 3 teaspoon of salt : 2 \times 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:

\frac{1}3 teaspoon of salt needs =  2 teaspoons of baking powder

1  teaspoon of salt needs =  \frac{2}{\frac{1}3} teaspoons of baking powder

1  teaspoon of salt needs = 2 \times 3 = <em>6</em>  teaspoons of baking powder needed

So, the answer is:

<em>6 teaspoons of baking powder </em>required.

5 0
3 years ago
Read 2 more answers
A kicker kicks a ball off the ground at 29.5 mi/hr and at 42.5 degrees.
deff fn [24]

Answer:

20.3m

Explanation:

the formula used was

s=(u^2sin^2∆)÷2g

4 0
3 years ago
I have a pure breed homozygous long claw monster (CC), Claude and
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8 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

<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 = \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
3 years ago
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