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3 years ago

An object appears red when viewed with white light. What color would it appear when viewed with blue light?

2 answers:

8 0

The colour of an object depends upon the light reflected and absorbed, so if you the object appears blue to you that means that the object reflects blue light and absorbs rest of the spectrum. That's why in black colour all light is absorbed and no light is emitted, thus absorbing all the radiation heat. Now, coming to the point, the red light incident on the object will contain light composed of many wavelength, some of which will be reflected by the body, so the body should reflected light of wavelength near to blue light thus giving a darkish red, with blueish tint. The resultant should be somewhat dark maroon.

zvonat [6]3 years ago

7 0

Answer:

Colour of any object is by the colour of light it reflects i.e if white light is incident on the object,it will reflect blue color.so it will appear blue.But if red light is incident on it,it will not reflect that and absorb it.so as it will not reflect any light it will appear black.

Explanation:

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in a football game, the kicker kicks a football a horizontal distance of 43 yards if the ball lands 3.9 seconds later, what is t

erica [24]

Answer:

10s

Explanation:

7 0

2 years ago

Please help quick please

Ede4ka [16]

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2 years ago

Two tiny particles having charges 20.0 μC and 8.00 μC are separated by a distance of 20.0 cm What are the magnitude and directio

Alecsey [184]

Answer:

The magnitude and direction of electric field midway between these two charges is $10.8\times10^{5}\ N/C$ along AB.

Explanation:

Given that,

First charge $q_{1}= 20\mu C$

second charge $q_{2}= 8\mu C$

Distance = 20 cm

We need to calculate the electric field

For first charge,

Using formula of electric field

$E_{1}= \dfrac{kq_{1}}{r^2}$

Put the valueinto the formula

$E_{1}=\dfrac{9\times10^{9}\times20\times10^{-6}}{10\times10^{-2}}$

$E_{1}=18\times10^{5}\ N/C$

Direction of electric field along AB

We need to calculate the electric field

For second charge,

Using formula of electric field

$E_{2}= \dfrac{kq_{2}}{r^2}$

Put the valueinto the formula

$E_{2}=\dfrac{9\times10^{9}\times8\times10^{-6}}{10\times10^{-2}}$

$E_{2}=7.2\times10^{5}\ N/C$

Direction of electric field along AO

We need to calculate the net electric field at midpoint

$E_{net}=E_{1}-E_{2}$

$E_{net}=(18-7.2)\times10^{5}\ N/C$

$E_{net}=10.8\times10^{5}\ N/C$

Direction of net electric field along AB

Hence, The magnitude and direction of electric field midway between these two charges is $10.8\times10^{5}\ N/C$ along AB.

8 0

3 years ago

Home Run

svp [43]

The force acting on the ball are unbalanced. Reactionary momentum force (that originated as a result of the swing of the bat) is the most powerful.

Yes friction is acting on the ball. In course of journey it would slow the ball down and make it trace a parabolic path rather than straight path as intended by hitter.

Explanation:

As the hitter hits the ball, momentum of the bat due to swing (mass of the bat*velocity provided by the batsman swinging action of bat) gets transferred on the ball on its impact with the bat.

Since ball’s mass is quite small as compared to the bat, the velocity of the ball increases by the same factor by which the ball’s mass is lower than the bat’s mass. This velocity causes forward motion of the ball (of course in the direction of bat’s motion, here the batsman intends to send the ball straight away hence the ball would move straight).

Various forces on ball is-

Reactionary momentum force -bat’s force (most powerful force)
The frictional force of the air (opposing the motion of the ball through the air)
Gravity force (pulling the ball down to the Earth)

As a combined effect of these force when all the force remains unbalanced, the ball moves away in the straight path under the impact of bats momentum which was most powerful of all.

Frictional force and Gravity force continue acting on the ball. While frictional forces decrease the ball velocity through the air, gravity force pulls it down thus deflecting its direction. Under the combined impact of declining bats momentum, friction force and gravity force, the ball traces a parabolic path (in accordance with the first law of motion from Newton)

8 0

3 years ago

A 700 g can of beans is dropped from a shelf that is 1.5 m high. What is the gravitational potential energy of this can? Round y

OLga [1]

<span>To find the gravitational potential energy of an object, we can use this equation: GPE = mgh m is the mass of the object in kg g = 9.80 m/s^2 h is the height of the object in meters GPE = mgh GPE = (0.700 kg) (9.80 m/s^2) (1.5 m) GPE = 10.3 J The gravitational potential energy of this can is 10.3 J</span>

4 0

3 years ago

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