Which is the largest gas giant?

What is the difference between displacement and distance travelled

IrinaK [193]

Displacement in Space
It is the length of a body's real route. It is the shortest distance between the body's final and beginning positions.
It's a number with a scalar value. It's a quantity with a vector.
It can't possibly be negative. It might be a negative number, a zero number, or a positive number.

8 0

3 years ago

Read 2 more answers

A bird is flying with a speed of 18.0 m/s over water when it accidentally drops a 2.00 kg fish. If the altitude of the bird is 5

Alina [70]

Here we go.
My abbreviations; KE = Kinetic Energy; GPE = Gravitational Potential Energy.

So first off, we know the fish has KE right when the bird releases it. Why? Because it has horizontal velocity after released! So let’s calculate it:
KE = 1/2(m)(V)^2
KE = 1/2(2)(18)^2
KE = 324 J

Nice!
We also know that the fish has GPE at its maximum height before release:
GPE = mgh
GPE = (2)(9.81)(5.40)
GPE = 105.95 J

Now, based on the *queue dramatic voice* LAW OF CONSERVATION OF ENERGY, we know all of the initial energy of the fish will be equal to the amount of final energy. And since the only form of energy when it hits the water is KE, we can write:
KEi + GPEi = KEf
(Remember - we found the initial energies before!)
(324) + (105.95) = KEf
KEf = 429.95J
And that’s you’re final answer! Notice how this value is MORE than the initial KE from before (324 J) - this is because all of the initial GPE from before was transformed into more KE as the fish fell (h decreased) and sped up (V increased).

If this helped please like it and comment!

4 0

3 years ago

Analyze the given velocity time graph

a_sh-v [17]

Answer:

Where' the graph

Explanation:

5 0

3 years ago

A typical human lens has an index of refraction of 1.430 . The lens has a double convex shape, but its curvature can be varied b

rjkz [21]

Answer:

Maximum Power = 144.3 D

The associated focal length of the lens = $6.92*10^{-3} m$

Explanation:

According to the Lens maker's Formula:

$\frac{1}{f} = (n-1) (\frac{1}{R_1}-\frac{1}{R_2} )$

where;

$n_1$ = the refractive index of the medium

$R_1$ and $R_2$ = radius of curvature on each surface

For a convex lens, The radius of curvature in the front surface will be positive and that of the second surface will be negative . Therefore;

$\frac{1}{f} = (n-1) (\frac{1}{R_1}-\frac{1}{-R_2} ) \\ \\ \frac{1}{f} = (n-1) (\frac{1}{R_1}+\frac{1}{R_2} )$

At maximum power

$\frac{1}{f} = (1.430-1) (\frac{1}{6.50 \ mm}-\frac{1}{5.50 \ mm} )$

= $0.144 \ mm^{-1}$

This Implies

$f = 6.92 mm\\f = 6.92*10^{-3} \ m$

Therefore; the power is given by the formula:

$P_{max} = \frac{1}{f}$

$P_{max}= \frac{1}{6.92*10^{-3}}$

= 144.3 D

3 0

4 years ago

A particle executes simple harmonic motion with an amplitude of 3.00 cm. At what position does its speed equal half of is maximu

Mrac [35]

Answer:

x=±0.026m

Explanation:

In simple harmonic motion the maximum value of the magnitude of velocity

$v_{max}=wA=\sqrt{\frac{k}{m} }A$

The speed as a function of position for simple harmonic oscillator is given by

$v=w\sqrt{A^{2}-x^{2}$

where A is amplitude of motion

Given data

Amplitude A=3 cm =0.03 m

v=(1/2)Vmax

To find

We have asked to find position x does its speed equal half of is maximum speed

Solution

The speed of the particle the maximum speed as:

$v=\frac{V_{max} }{2}\\ w\sqrt{A^{2}-x^{2} }=\frac{wA}{2}\\ A^{2}-x^{2}=\frac{A^{2} }{4}\\ x^{2}=A^{2}- \frac{A^{2} }{4}\\ x^{2}=\frac{3A^{2}}{4} \\x=\sqrt{\frac{3A^{2}}{4}}$

x=±(√3(0.03)/2)

x=±0.026m

5 0

4 years ago