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2 years ago

For this graph to the right, what is the velocity of the object? Show your work.

Physics

1 answer:

mr Goodwill [35]2 years ago

7 0

I dont see a mark where it is, if there was a line it would be marking the constant speed of it. There is just time(s) which probably stands for speed, and position(m) which probably stands for miles. There is a measurement of 50, down to 10. Please give me more explanation to this question and I may be able to help!

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According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave lengt

scZoUnD [109]

Answer:1.55 times

Explanation:

Given

First wavelength $(\lambda _1)=450 nm$

Second wavelength $(\lambda _2)=700 nm$

According wien's diplacement law

$\lambda T=constant$

where $\lambda =wavelength$

T=Temperature

Let $T_1 and T_2$ be the temperatures corresponding to $\lambda _1 & \lambda _2$ respectively.

$\lambda _1\times T_1=\lambda _2\times T_2$

$\frac{T_1}{T_2}=\frac{\lambda _2}{\lambda _1}$

$\frac{T_1}{T_2}=\frac{700}{450}=1.55$

Thus object with $\lambda 450 nm$ is 1.55 times hotter than object with wavelength $\lambda =700 nm$

8 0

3 years ago

What is produced as the result of unequal warming of the earth's surface?

Soloha48 [4]

Unusual precipitation patterns

7 0

2 years ago

Death Star has a diameter of 160,000m and a mass of 5.1e17kg. Millennium Falcon has a mass of 1.36e6kg (data from Wookieepedia)

lions [1.4K]

Answer:

7229 N

Explanation:

The gravitational force between the Death Star and the Millenium Falcon is given by:

$F=G\frac{mM}{R^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$M=5.1\cdot 10^{17} kg$ is the mass of the Death Star

$m=1.36\cdot 10^6 kg$ is the mass of the Millennium Falcon

$R=\frac{160,000 m}{2}=80,000 m$ is the radius of the Death Star

Substituting numbers into the equation, we find the force

$F=(6.67\cdot 10^{-11})\frac{(1.36\cdot 10^6 kg)(5.1\cdot 10^{17} kg)}{(80,000 m)^2}=7229 N$

5 0

3 years ago

S A block of mass M is connected to a spring of mass m and oscillates in simple harmonic motion on a frictionless, horizontal tr

Elina [12.6K]

The period of oscillation is T = 2 * pi * sqrt ( ( m2/3 + m1) / k )

<h3>What is period of oscillation?</h3>

This is the time in seconds it takes to complete one oscillation. where an oscillation is a repetitive to and fro motion. period if the inverse of frequency and both are basic when calculation motion in simple harmonic motion.

The period of oscillation is given as T

T = 2 * pi * sqrt ( m / k )

where

m = mass on this case mass of the spring will be inclusive to the mass of the block such that we have:

m1 = mass of the block

m2 = mass pf the spring

k = force constant of the spring

including the two masses to the period gives

T = 2 * pi * sqrt ( ( m2/3 + m1) / k )

Read more on period of oscillation here: brainly.com/question/22499336

#SPJ4

7 0

1 year ago

Two buses are driving along parallel freeways that are 5mi apart, one heading east and the other heading west. Assuming that eac

Oksanka [162]

Answer:

101.54m/h

Explanation:

Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;

Let l be the be the distance further away at which they will meet from the current points;

$l=\sqrt{13^2-5^2}=12m\\\\\frac{dl}{dt}=-(55m/h+55m/h})\\\\=-110m/h$ #The speed toward each other.

$\frac{dh}{dt}=0, \ \ \ \ h=constant\\\\h^2+l^2=b^2\\\\2h\frac{dh}{dt}+2l\frac{dl}{dt}=2b\frac{db}{dt}\\\\2\times5\times0+2\times12\times(-110)=2\times13\frac{db}{dt}\\\\\frac{db}{dt}=-101.54m/h$

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h

4 0

2 years ago