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3 years ago

Which is the larger change in temperature 1 Kelvin or 1 degree Celsius

1 answer:

5 0

Those changes in temperature are equal. The absolute (Kelvin) scale and the Celsius scale use the same size degrees, (but their zeros are set at different temperatures).

Their 'degree' is 0.01 of the difference between the boiling and freezing temperatures of water.

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A dairy farmer notices that a circular water trough near the barn has become rusty and now has a hole near the base. The hole is

Crazy boy [7]

Answer:

$v=1.93m/s$

Explanation:

From the concept of fluids mechanics we know that if a tank has a hole at the bottom, the equation that we need to use is:

$v=\sqrt{2gh}$

Since we know gravity and its hight

$v=\sqrt{2*(9.81m/s^{2})(0.19m) }=1.93m/s$

5 0

3 years ago

A block of mass m = 4.8 kg slides from left to right across a frictionless surface with a speed Vi=7.3m/s. It collides with a bl

nirvana33 [79]

Answer:

The speed of the 11.5kg block after the collision is V≅4.1 m/s

Explanation:

ma= 4.8 kg

va1= 7.3 m/s

va2= - 2.5 m/s

mb= 11.5 kg

vb1= 0 m/s

vb2= ?

vb2= ( ma*va1 - ma*va2) / mb

vb2= 4.09 m/s ≅ 4.1 m/s

7 0

3 years ago

At a circus, a human cannonball is shot from a cannon at 15m/s at an angle of 40° above horizontal. She

Citrus2011 [14]

Answer:

The height is $H = 6.74 \ m$

The horizontal distance is $D = 23.74 \ m$

Explanation:

From the question we are told that

The speed is $v = 15 \ m/s$

The angle is $\theta = 40^o$

The height of the cannon from the ground is h = 2 m

The distance of the net from the ground is k = 1 m

Generally the maximum height she reaches is mathematically represented as

$H = \frac{v^2 sin^2 \theta }{2 * g } + h$

=> $H = \frac{(15)^2 [sin (40)]^2 }{2 * 9.8} + 2$

=> $H = 6.74 \ m$

Generally from kinematic equation

$s = ut + \frac{1}{2} at^2$

Here s is the displacement which is mathematically represented as

s = [-(h-k)]

=> s = -(2-1)

=> s = -1 m

There reason why s = -1 m is because upward motion canceled the downward motion remaining only the distance of the net from the ground which was covered during the first half but not covered during the second half

a = -g = -9.8

$u = v sin (\theta)$