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3 years ago

How are speed and velocity different?

Physics

2 answers:

liubo4ka [24]3 years ago

4 0

Velocity tells you how fast and in what direction. Speed only tells how fast.

dlinn [17]3 years ago

3 0

Velocity tells you how fast and gives a direction. Speed only tells how fast.

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what moon phase will occur about 14 days after a first quarter?

garik1379 [7]

Answer:

waxing gibbous is the answer.

3 0

3 years ago

Water is being boiled in an open kettle that has a 0.52-cm-thick circular aluminum bottom with a radius of 12.0 cm. If the water

tangare [24]

Answer:

$T_b=107.3784\ ^{\circ}C$

Explanation:

Given:

thickness of the base of the kettle, $dx=0.52\ cm=5.2\times 10^{-3}\ m$

radius of the base of the kettle, $r=0.12\ m$

temperature of the top surface of the kettle base, $T_t=100^{\circ}C$

rate of heat transfer through the kettle to boil water, $\dot Q=0.409\ kg.min^{-1}$

We have the latent heat vaporization of water, $L=2260\times 10^3\ J.kg^{-1}$

and thermal conductivity of aluminium, $k=240\ W.m^{-1}.K^{-1}$

$\dot Q=\frac{0.409\times 2260000}{60}$

$\dot Q=15405.67\ W$

<u>From the Fourier's law of conduction we have:</u>

$\dot Q=k.A.\frac{dT}{dx}$

$\dot Q=k\times \pi.r^2\times \frac{T_b-T_t}{5.2\times 10^{-3}}$

where:

$A=$ area of the surface through which conduction occurs

$T_b=$ temperature of the bottom surface

$15405.67=240\times \pi\times 0.12^2\times \frac{T_b-100}{5.2\times 10^{-3}}$

$T_b=107.3784\ ^{\circ}C$ is the temperature of the bottom of the base surface of the kettle.

6 0

3 years ago

the refractive index of cooking oil is 1.47, and the refractive index of water is 1.33. A thick layer of cooking oil is floating

VARVARA [1.3K]

When light travels from a medium with higher refractive index to a medium with lower refractive index, there is a critical angle after which all the light is reflected (so, there is no refraction).

The value of this critical angle can be derived by Snell's law, and it is equal to
$\theta_C = \arcsin ( \frac{n_2}{n_1} )$
where n2 is the refractive index of the second medium and n1 is the refractive index of the first medium.

In our problem, n1=1.47 and n2=1.33, so the critical angle is
$\theta_C = \arcsin( \frac{1.33}{1.47} )=\arcsin (0.91)=65^{\circ}$

4 0

3 years ago

Energy is released during the fission of pu-239 atoms as a result of the

pav-90 [236]

<h2>Answer: process of converting matter into energy</h2><h2></h2>

Nuclear fission consists of dividing a heavy nucleus into two or more lighter or smaller nuclei, by means of the bombardment with neutrons to make it unstable. In this process that takes place in the atomic nucleus, neutrons, gamma rays and <u>large amounts of energy are emitted. </u>

Then, with this division a great release of energy occurs and the emission of two or three neutrons, other particles and gamma rays.

This means fission is a process in which energy is released by the separation of the components of the nucleous of the atom.

In other words:

<h2>Matter is converted to energy .</h2>

6 0

3 years ago

If a plane is flying level at 910 km/h and the banking angle is not to exceed 50 ∘, what's the minimum curvature radius for the

hoa [83]

Answer:

5.5 km

Explanation:

First, we convert the distance from km/h to m/s

910 * 1000/3600

= 252.78 m/s

Now, we use the formula v²/r = gtanθ to get our needed radius

making r the subject of the formula, we have

r = v²/gtanθ, where

r = radius of curvature needed

g = acceleration due to gravity

θ = angle of banking

r = 252.78² / (9.8 * tan 50)

r = 63897.73 / (9.8 * 1.19)

r = 63897.73 / 11.662

r = 5479 m or 5.5 km

Thus, we conclude that the minimum curvature radius needed for the turn is 5.5 km

4 0

3 years ago