Is there a way to see the color of cosmic rays the human eye can't see

Q10. In an ideal situation without any air resistance, any free-falling body will fall at the same rate of acceleration, regardl

rosijanka [135]

Velocity of the object parameters would affect the fall if air drag is also considered.

<h3>What is velocity in physics with example?</h3>

Velocity can be defined as the rate at which that moves in a specific direction. as the velocity of a car driving north on a road or the pace at which a rocket takes off. Because the velocity vector is scalar, its integrand magnitude will always equal the motion's speed.

<h3>Can velocity be negative?</h3>

A moving object has a negative velocity when it is moving in the wrong direction. If an object is slowing down, its generate and evaluate is pointing away from the direction in which it is moving (in this case, a positive acceleration).

<h3>What is the principle of velocity?</h3>

An object must move at a constant speed and direction in order to have a constant velocity. The object can only travel in a straight line if the direction is constant. So, motion in a single direction at a constant speed is defined as having a constant velocity.

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7 0

1 year ago

A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are

Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of $x= 5t^3+1$

Coordinate of position of $y=3t^2+2$

Time = 2.00 s

We need to calculate the acceleration

$a = \dfrac{d^2x}{dt^2}$

For x coordinates

$x=5t^3+1$

On differentiate w.r.to t

$\dfrac{dx}{dt}=15t^2+0$

On differentiate again w.r.to t

$\dfrac{d^2x}{dt^2}=30t$

The acceleration in x axis at 2 sec

$a = 60i$

For y coordinates

$y=3t^2+2$

On differentiate w.r.to t

$\dfrac{dy}{dt}=6t+0$

On differentiate again w.r.to t

$\dfrac{d^2y}{dt^2}=6$

The acceleration in y axis at 2 sec

$a = 6j$

The acceleration is

$a=60i+6j$

We need to calculate the net force

$F = ma$

$F = 3.00\times(60i+6j)$

$F=180i+18j$

The magnitude of the force

$|F|=\sqrt{(180)^2+(18)^2}$

$|F|=180.89\ N$

Hence, The net force acting on this object is 180.89 N.

3 0

4 years ago

Calculate the wavelength (in nm) of light that produces its first minimum at an angle of 21.0° when falling on a single slit of

Cerrena [4.2K]

To solve this problem it is necessary to apply the related concepts to the principle of overlap, specifically to single slit diffraction experiment concept.

Mathematically this can be expressed as:

$dsin\theta = m\lambda$

Where,

d = Width of the slit

$\lambda =$ Wavelength

$\theta =$ Angle relative to the original direction of the light

m = Any integer which represent the order of the equation (number of repetition of the spectrum)

To solve the problem we need to rearrange the equation and find the wavelength

$\lambda = \frac{dsin\theta}{m}$

Our values are given as,

$d = 1.46\mu m = 1.46*10^{-6}m$

$\theta = 21\°$

$m = 1$

Replacing in our equation we have,

$\lambda = \frac{dsin\theta}{m}$

$\lambda = \frac{(1.46*10^{-6})sin(21)}{1}$

$\lambda = 5.232*10^{-7}m$

$\lambda = 523.2nm$

Therefore the wavelength is 523.2nm

7 0

4 years ago

The position of a car at time t is given by the function p(t)=t2 2t−4. What is the velocity when p(t)=11? assume t≥0

AysviL [449]

The velocity when function p(t)=11 is 8 .

According to the question

The position of a car at time t represented by function :

$p(t)=t^{2} +2t-4$

Now,

When function p(t) = 11 , t will be

$p(t)=t^{2} +2t-4$

11 = t²+2t-4

0 = t² + 2t - 15

or

t² +2t-15 = 0

t² +(5-3)t-15 = 0

t² +5t-3t-15 = 0

t(t+5)-3(t+5) = 0

(t-3)(t+5) = 0

t = 3 , -5

as t cannot be -ve as given ( t≥0)

so,

t = 3

Now,

the velocity when p(t)=11

As we know velocity = $\frac{position}{time}$

therefore to get the value of velocity from function p(t)

we have to differentiate the function with respect to time

$\frac{d(p(t))}{dt} =\frac{d}{dt} (t^{2} +2t-4)$

v(t) = 2t + 2

where v(t) = velocity at that time

as t = 3 for p(t)=11

so ,

v(t) = 2t + 2

v(t) = 2*3 + 2

v(t) = 8

Hence, the velocity when function p(t)=11 is 8 .

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4 0

2 years ago

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

3 years ago