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Afina-wow [57]
3 years ago
8

Water cannot be used in barometer.why.​

Physics
2 answers:
Natasha2012 [34]3 years ago
8 0

Answer:

water cannot be used as barometric liquid because it's density is lower than Mercury . water's density is 1000 gram per cubic meter. hence it requires a barometer whose height is around 11 meter

photoshop1234 [79]3 years ago
3 0

Answer:

the above answer is correct hope it's help you have a great day ☺ ☺ keep smiling be happy stay safe

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How can i open leaderboards column in brainly
natta225 [31]

Answer:

Go to the answers column there you will see on the top, it is writen Meet Our Leaders then click on it and then leaderboards column will appear.

Explanation:

6 0
2 years ago
Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's
mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

T^2\alpha R^3\\T^2=kR^3.......................(1)

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

\frac{T^2}{R^3}=k.......................(2)

Let the orbital period of the earth be T_e and its mean distance of from the sun be R_e.

Also let the orbital period of the planet be T_p and its mean distance from the sun be R_p.

Equation (2) therefore implies the following;

\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)

We make the period of the planet T_p the subject of formula as follows;

T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

R_p=16R_e...............(5)

Substituting equation (5) into (4), we obtain the following;

T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}

R_e^3 cancels out and we are left with the following;

T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)

Recall that the orbital period of the earth is about 365.25 days, hence;

T_p=64*365.25\\T_p=23376days

4 0
3 years ago
A ball rolls across a floor with an acceleration of 0.100 m/s2 in a direction opposite to its velocity. The ball has a velocity
Westkost [7]

Answer:

4.15 m/s

Explanation:

Its given that acceleration is 0.1 m/s² with a direction opposite to the velocity. Since, the direction of acceleration is opposite to the velocity, this gives us a hint that the velocity is decreasing and so acceleration would be negative.

i.e.

acceleration = a = - 0.1 m/s²

Distance covered = S = 6m

Velocity after covering 6 meters = Final velocity = v_{f} = 4 m/s

We need to find the initial speed, which will be the same as the magnitude of initial velocity.

Initial velocity = v_{i} =  ?

3rd equation of motion relates the acceleration, distance, final velocity and initial velocity as:

2aS = (v_{f})^{2}-(v_{i})^{2}

Using the known values in the formula, we get:

2(-0.1)(6)=(4)^{2}- (v_{i})^{2}\\\\ (v_{i})^{2}=16+1.2\\\\ (v_{i})^{2}=17.2\\\\ v_{i}=4.15

Thus, the initial speed of the ball was 4.15 m/s

3 0
3 years ago
Cells that remain as sources of other cells are called
inna [77]
They are called stem cells. This cells are undifferentiated which means it can specialize in other types when it receives the right stimuli. They can divide through mitoses and become more stem cell or become a bone, muscle, blood cell, etc.

They can have 2 origins: embryos or some human tissue; their function is to regenerate or substitute damaged cells
8 0
3 years ago
A bolt of lightning discharges 9.7 C in 8.9 x 10^-5 s. What is the average current during the discharge?
Anastaziya [24]

Answer: 1.089\times 10^5\ A

Explanation:

Given

Charge discharged Q=9.7\C

time taken t=8.9\times 10^{-5}\ s

Current is given as rate of change of discharge i.e.

\Rightarrow I=\dfrac{Q}{t}\\\\\Rightarrow I=\dfrac{9.7}{8.9\times 10^{-5}}\\\\\Rightarrow I=1.089\times 10^5\ A

Therefore, the average current is 1.089\times 10^5\ A

3 0
2 years ago
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