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3 years ago

What is the correct equation kinetic enery

Physics

1 answer:

Ivahew [28]3 years ago

4 0

Kinetic energy is directly proportional to the mass of the object and to the square of its velocity: K.E. = 1/2 m v2.

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How does the shape and size of the continents today help support the Theory of Continental Drift?

astraxan [27]

The size and shape of the continents help support the continental drift because they all most fit together like puzzle pieces

4 0

3 years ago

A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

4 0

2 years ago

According to a college survey, 22% of all students work full time. find the mean for the number of 3) students who work full tim

attashe74 [19]

Assuming that the students worldwide are being considered, because of the extremely large population, this can be considered as a binomial distribution. A normal distribution is used most usually as a fair approximation of the binomial. The mean is the expectation, therefore:<span>
E[x] = np = (16)(0.22) = 3.52
<span>μ = 3.52 </span></span>

3 0

3 years ago

If a mass of .343 kg moves down a 5m ramp in 5.8 seconds, what is the velocity and KE developed by the moving mass

Soloha48 [4]

Velocity 0.86m/s

0.13J

Explanation:

Given parameters:

Mass = 0.343kg

distance = 5m

time taken = 5.8s

Unknown: