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2 years ago

Shadow is formed when an_____object comes in thewayof light.

Physics

2 answers:

Zanzabum2 years ago

7 0

Answer:

shadow is formed when an opaque object comes on path of light shadows are formed because light travels in a straight line and it can't pass through an opaque object

Ratling [72]2 years ago

5 0

Answer:

Opaque object

Explanation:

Shadow forms when an opaque object comes in the way of light. Because opaque objects does not allow light to pass through them and light travels in a straight line. Hence a retreating figure of the same opaque object ,forms and it's black in colour without any detail just outline.

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abruzzese [7]

Answer:

I think they use it at night because they want to avoid involving themselves in road accident while crossing the street. When there are reflective strips on their clothes, drivers do notice them even at far ends of the street since the strips will reflect the rays from the car lights to the driver,hence, the driver notifies that there is a pedestrian crossing and therefore slows down.

3 0

3 years ago

A block (mass = 61.2 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1/2MR2 kg · m2, wh

kolezko [41]

Answer:

The angular velocity is $w = 53.35 \ rounds /minute$

Explanation:

From the question we are told that

The mass of the block is $m = 61.2kg$

The of the pulley is $M = 14.2 kg$

The radius of the pulley is $R = 1.5m$

The radius of the cord around the pulley is $r = 1.5 m$

The distance of the block to the floor is $d = 8.0 m$

From the question we are told that the moment of inertia of the pulley is

$I = \frac{1}{2} MR^2 kg \cdot m^2$

Substituting value

$I = \frac{1}{2} * 14.2 * (1.5)^2$

$I = 15.975 kg \cdot m^2$

Using the Newtons law we can express the force acting on the vertical axis as

$ma = mg -T$

=> $T = mg -ma$

Now when the pulley is rotated that torque generated on the massless cord as a r result of the tension T and the radius of the cord around the pulley is mathematically represented as

$\tau = I \alpha$

Here $\alpha$ is the angular acceleration

Here $\tau$ is the torque which can be equivalent to

$\tau = T r$

Substituting this above

$Tr = I \alpha$

Substituting for T

$(mg - ma ) r = I\ r \alpha$

Here $a$ is the linear acceleration which is mathematically represented as

$a = r\alpha$

$(mg - m(r\alpha ) ) r = I\ r \alpha$

$mgr = I\alpha + m(r\alpha ) r$

$mgr = \alpha [ I + mr^2]$

making $\alpha$ the subject

$\alpha = \frac{mgr}{I -mr ^2}$

Substituting values

$\alpha = \frac{61.2 * 1.5 * 9.8}{15.975 + (61.2 ) * (1.5)^2}$

$\alpha =5.854 rad /s^2$

Now substituting into the equation above to obtain the acceleration

$a = 5.854 * 1.5$

$a=8.78 m/s^2$

This acceleration is $a = \frac{v}{t}$

and v is the linear velocity with is mathematically represented as

$v = \frac{d}{t}$

Substituting this into the formula acceleration

$a = \frac{d}{t^2}$

making t the subject

$t = \sqrt{\frac{d}{a} }$

substituting value

$t = \sqrt{\frac{8}{8.78}}$

$t = 0.9545 \ s$

Now the linear velocity is

$v = \frac{8}{0.9545}$

$v = 8.38 m/s$

The angular velocity is

$w = \frac{v}{r}$

$w = \frac{8.38}{1.5}$

$w = 5.59 rad/s$

Generally 1 radian is equal to 0.159155 rounds or turns

So 5.59 radian is equal to x

Now x is mathematically obtained as

$x = \frac{5.59 * 0.159155}{1}$

$= 0.8892 \ rounds$

Also

60 second = 1 minute

So 1 second = z

Now z is mathematically obtained as

$z = \frac{ 1}{60}$

z $= 0.01667 \ minute$

Therefore

$w = \frac{0.8892}{0.01667}$

$w = 53.35 \ rounds /minute$

8 0

3 years ago

On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wi

Lostsunrise [7]

Answer:

$0.06\Omega/m$

Explanation:

Firstly, when you measure the voltage across the battery, you get the emf,

E = 13.0 V

In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.

Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire ( $I=\frac{V}{R}$ ). But this is not the case.

Let the resistance of the ammeter be r

Hence, using Ohm's law we get the following 2 equations:

$\frac{13}{20z+r} =7.6$ .......(1)

$\frac{13}{40z+r} =4.5$ ......(2)

Substituting the value of r from (2) in (1), we have,

$13=152z+7.6\times\frac{13-180z}{4.5}$

which simplifying gives us, $z=0.0589\Omega/m\approx0.06\Omega/m$ (which is our required solution)

putting the value of z in either (1) or (2) gives us, r = 0.5325 $\Omega$

3 0

2 years ago

Are all elements naturally forming?

makkiz [27]

Answer:

1 through 92 occur naturally on Earth

3 0

3 years ago

A point charge q1 is held stationary at the origin. A second charge q2 is placed at point a, and the electric potential energy o

Brrunno [24]

The electric potential energy of the pair of charges when the second charge is at point b is 7.3 x 10⁻⁸ J.

<h3>Electric potential energy</h3>

When work is done on a positive test charge to move it from one location to another, potential energy increases and electric potential increases.

The electric potential energy between the charges when the second charge is at point b is calculated as follows;

ΔU = -w

Ui - Uf = w

Uf = Ui - w

where;

Uf is the final potential energy

Ui is the initial potential energy

w is the work done by the force

Uf = 5.4 x 10⁻⁸ J - (-1.9 x 10⁻⁸J)

Uf = 5.4 x 10⁻⁸ J + 1.9 x 10⁻⁸ J

Uf = 7.3 x 10⁻⁸ J

Thus, the electric potential energy of the pair of charges when the second charge is at point b is 7.3 x 10⁻⁸ J.

Learn more about electric potential energy here: brainly.com/question/14306881

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7 0

1 year ago

Shadow is formed when an_____object comes in thewayof light.​

Shadow is formed when an_____object comes in thewayof light.