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2 years ago

How many neutrons are contained in 2 kg? Mass of one neutron is 1.67x10-27 kg.

Physics

1 answer:

liraira [26]2 years ago

4 0

Answer:

1.2 × 10^27 neutrons

Explanation:

If one neutron = 1.67 × 10^-27 kg

then in 2kg...the number of neutrons

; 2 ÷ 1.67 × 10^-27

There are.... 1.2 × 10^27 neutrons

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2. A pebble is dropped down a well and hits the water 1.5 s later. Using the equations for motion with constant acceleration, de

kow [346]

Answer:

S = 11.025 m

Explanation:

Given,

The time taken by the pebble to hit the water surface is, t = 1.5 s

Acceleration due to gravity, g = 9.8 m/s²

Using the II equations of motion

S = ut + 1/2 gt²

Here u is the initial velocity of the pebble. Since it is free-fall, the initial velocity

u = 0

Therefore, the equation becomes

S = 1/2 gt²

Substituting the given values in the above equation

S = 0.5 x 9.8 x 1.5²

= 11.025 m

Hence, the distance from the edge of the well to the water's surface is, S = 11.025 m

3 0

2 years ago

We know that the earth's axis is tilted 23 ½ degrees. On or about June 21 or 22 each year, the summer solstice occurs for those

Dima020 [189]

D) The sun's rays will never be directly overhead. The latitude of 23 ½ degrees north is known as the Tropic of Cancer. Above this imaginary line the sun's rays hit earth with decreased angles.

8 0

2 years ago

Read 2 more answers

The planet Neptune orbits the Sun. Its orbital radius is 30.130.130, point, 1 astronomical units (\text{AU})(AU)left parenthesis

lord [1]

Answer:

The distance the planet Neptune travels in a single orbit around the Sun is 60.2π AU.

Explanation:

As it is given that the Neptune's orbit is circular, the formula that we have to use is the circumference of a circle in order to find the distance it travels in a single orbit around the Sun. In other words, you can say that the circumference of the circle is equivalent to the distance it travels around the Sun in a single orbit.

The circumference of the circle = Distance Travelled (in a single orbit) = 2*π*R ---- (A)

Where,

R = Orbital radius (in this case) = 30.1 AU

Plug the value of R in the equation (A):

(A) => The circumference of the circle = 2*π*(30.1)

The circumference of the circle = 60.2π

Therefore, the distance the planet Neptune travels in a single orbit around the Sun is 60.2π AU.

5 0

2 years ago

On a ring road, 12 trams are spaced at regular intervals and travel at a constant speed. How many trams need to be added to the

Sphinxa [80]

3 trams must be added

Explanation:

In this problem, there are 12 trams along the ring road, spaced at regular intervals.

Calling L the length of the ring road, this means that the space between two consecutive trams is

$d=\frac{L}{12}$ (1)

In this problem, we want to add n trams such that the interval between the trams will decrease by 1/5; therefore the distance will become

$d'=(1-\frac{1}{5})d=\frac{4}{5}d$

And the number of trams will become

$12+n$

So eq.(1) will become

$\frac{4}{5}d=\frac{L}{n+12}$ (2)

And substituting eq.(1) into eq.(2), we find:

$\frac{4}{5}(\frac{L}{12})=\frac{L}{n+12}\\\rightarrow n+12=15\\\rightarrow n = 3$

Learn more about distance and speed:

brainly.com/question/8893949

#LearnwithBrainly

4 0

2 years ago

1. A2 .7-kg copper block is given an initial speed of 4.0m/s on a rough horizontal surface. Because of friction, the block final

tatyana61 [14]

Answer:

A. Increase in temperature is 0.0176 degree Celsius. b. the remaining energy will be lost.

Explanation:

The mass of copper block = 7kg

Initial speed = 4.0 m/s

Specific heat of copper = 0.385 j/g degree Celcius.

a. The increase in temperature is calculated below:

$\text{Change in kinetic energy} = \frac{1}{2}mv^{2} \\= \frac{1}{2} \times 2.7 \times 4^{2} = 21.6 J \\$

85% of energy is converted into internal energy.

$mc\DeltaT = 21.6 \times 0.85 \\2.7 \times 385 \times \Delta T = 21.6 \times 0.85 \\\Delta T = 0.0176 degree \ celsius$

b. The remaining 15 per cent of kinetic energy will be lost and it will be changed into other forms.

7 0

2 years ago