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Rain falling vertically will make vertical streaks on a car's side window. However,if the car is moving, the streaks are slanted

Alla [95]

When car is at rest the steaks makes makes vertical lines

which means the rain is falling in vertically downward direction

Now when car is moving with some speed v

Now the steaks makes and an angle 45 degree

So here we can say that relative velocity of rain with car is 45 degree

Now this is the resultant speed of rain in car frame

$V_{rc} = V_r - V_c$

now if relative velocity makes 45 degree angle so this vector must have same components in vertical and horizontal direction

Since we know that relative velocity is resultant of rain velocity and car velocity so we can say here its two components are rain velocity and car velocity

So these two components must be of same magnitude

as it makes 45 degree

because when two vector are of same magnitude then the resultant vector always makes 45 degree with them if these two vectors are perpendicular to each other

car is moving at same speed as the speed of rain

7 0

3 years ago

Describe three events that you cannot explain.<br> About energy and matter

ZanzabumX [31]

Answer:

A teacher giving homework on fridays

A teacher giving tests on fridays

A teacher holding you back after class on fridays

Explanation:

because i luv friday

5 0

3 years ago

A particle of mass 4.0 kg is constrained to move along the x-axis under a single force F(x) = −cx3 , where c = 8.0 N/m3 . The pa

Pie

Answer:4.58 m/s

Explanation:

Given

mass of Particle $m=4 kg$

$F=-cx^3$

$a=\frac{F}{m}$

$a=-\frac{cx^3}{m}$

$a=-\frac{8x^3}{4}$

$a=-2x^3$

$v\frac{\mathrm{d} v}{\mathrm{d} x}=-2x^3$

$vdv=-2x^3dx$

integrating

$\int_{6}^{v_b}vdv=\int_{1}^{-2}-2x^3dx$

$\frac{v_b^2-6^2}{2}=-\frac{1}{2}\left [ \left ( -2\right )^4-\left ( 1\right )^4\right ]$

$\frac{v_b^2-36}{2}=-0.5\times 15$

$v_b^2=36-15$

$v_b=\sqrt{21}$

$v_b=4.58 m/s$

6 0

3 years ago

Consider two sizes of disk, both of mass M. One size of disk has radius R; the other has radius 4R. System A consists of two of

Harman [31]

Answer:

4 smaller disks

Explanation:

We are given;

Mass of smaller and larger disks = M

Radius of smaller disk = R

Radius of larger disk = 4R

Formula for moment of inertia about cylinder axis is:

I = ½MR²

Thus;

For small disk, I_small = ½MR²

For large disk, I_large = ½M(2R)² = 2MR²

We are told that moment of inertia of System A consists of two of the larger disks. Thus;

I_A = 2 × I_large = 2 × 2MR²

I_A = 4MR²

We are also told that System B consists of one of the larger disks and a number of the smaller disks. Thus;

I_B = I_large + n(I_small)

Where n is the number of smaller disks.

I_B = 2MR² + n(½MR²)

I_B = MR²(2 + n/2)

We are told that the moment of inertia for system A equals the moment of inertia for system B. Thus;

I_A = I_B

So;

4MR² = MR²(2 + n/2)

MR² will cancel out to give;

4 = 2 + n/2

Multiply through by 2 to give;

8 = 4 + n

n = 8 - 4

n = 4

5 0

3 years ago

Electrons in a particle beam each have a kinetic energy of 4.0 × 10−17 J. What is the magnitude of the electric field that will

denpristay [2]

Explanation:

Relation between work and change in kinetic energy is as follows.

$W_{net} = \Delta K$

Also, $\Delta K = K_{initial} - K_{final}$

= $(0 - 4.0 \times 10^{-17})$ J

= $-4.0 \times 10^{-17}$ J

Let us assume that electric force on the electron has a magnitude F. The electron moves at a distance of 0.3 m opposite to the direction of the force so that work done is as follows.

w = -Fd

$-4.0 \times 10^{-17} J = -F \times 0.3 m$