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3 years ago

Why does the light from the touch reach upto only certain distance

1 answer:

6 0

It is also likely (but not certain) that the photons will be absorbed by atoms. ... Light particles( or photons) never”run out” or loose their energy, so they can go an infinite distance, or until it reaches an object, that reflects the light or obsorbs it. Ie, a planet, or a mirror.

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ASAP!! please What is the answer? MCQ. Thank you

sdas [7]

Answer:

Id say C

Explanation:

The hypotenuse for c is up and to the right so the unit vectors show that

8 0

3 years ago

Read 2 more answers

figure 2 shows a charged ball of mass m = 1.0 g and charage q = -24*10^-8 c suspended by massless string in the presence of a un

Vlad [161]

Answer:

E = 307667 N/C

Explanation:

Since the object's mass is 1 g, then its weight in newtons is 0.001 * 9.8 = 0.0098 N.

This weight should have the same magnitude of the vertical component of the tension T of the string (T * cos(37)) so we can find the magnitude of the tension T via:

0.0098 N = T * cos(37)

then T = 0.0098/cos(37) N = 0.01227 N

Knowing the tension's magnitude, we can find its horizontal component:

T * sin(37) = 0.007384 N

and now we can obtain the value of the electric field since we know the charge of the ball to be: -2.4 * 10^(-8) C:

0.007384 N = E * 2.4 * 10^(-8) C

Then E = 0.007384/2.4 * 10^(-8) N/C

E = 307667 N/C

8 0

3 years ago

car rides on four wheels that are connected to the body of the car by spring assemblies that let the wheels move up and down ove

Tomtit [17]

Answer:

78.4 KN/m

Explanation:

Given

mass of person 'm' =80 kg

car dips about i.e spring stretched 'x'= 1 cm => 0.01m

acceleration due to gravity 'g'= 9.8 m/s^2

as we know that,in order to find approximate spring constant we use Hooke's Law i.e F=kx

where,

F = the force needed

x= distance the spring is stretched or compressed beyond its natural length

k= constant of proportionality called the spring constant.

F=kx ---> (since f=mg)

mg=kx

k=(mg)/x

k=(80 x 9.8)/ 0.01

k=78.4x $10^3$

k=78.4 KN/m

3 0

3 years ago

Read 2 more answers

Two boats start together and race across a 48-km-wide lake and back. Boat A goes across at 48 km/h and returns at 48 km/h. Boat

nika2105 [10]

Answer:

Part 1)

Boat A will win the race

Part 2)

Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line

Part 3)

average velocity must be zero

Explanation:

As we know that the distance moved by the boat is given as

$d = 48 km$

now the time taken by the boat to move to and fro is given as

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

$t = \frac{48 + 48}{48}$

$t = 2 hrs$

Time taken by Boat B to cover the distance

$t = \frac{48}{24} + \frac{48}{72}$

$t = 2.66 h$

Part 1)

Boat A will win the race

Part 2)

Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line

Part 3)

Since the displacement of Boat A is zero

so average velocity must be zero

3 0

3 years ago

An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F

ololo11 [35]

Answer

given,

mass of jogger = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed = 1.3 m/s in 61 ° north of east.

jogger one

$P_1 = m_1 v_1 \hat{i}$

$P_1 = 67 \times 2.3\hat{i}$

$P_1 = 154.1 \hat{i}$

$P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}$

$P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}$

$P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}$

$P_2 = 44.12\hat{i} +79.59\hat{j}$

now

P = P₁ + P₂

$P = 198.22 \hat{i} +79.59 \hat{j}$

magnitude

$P = \sqrt{198.22^2 + 79.59^2}$

$P =213.60 kg.m/s$

$\theta = tan^{-1}\dfrac{79.59}{198.22}$

$\theta = 21.87$

the angle is $\theta = 21.87$ north of east

7 0

3 years ago