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3 years ago

Which describes how light travels?

Physics

2 answers:

trasher [3.6K]3 years ago

4 0

Answer:

the Answer Would be D

Explanation:

From what I know light travels 300,000 km/second , travels at fast speeds and travels in a straight line

Romashka-Z-Leto [24]3 years ago

4 0

D. All of the above

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The amount of fluid displaced by a submerged object depends on its

Daniel [21]

The amount of fluid displaced by a submerged object depends on its volume.

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3 years ago

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A bicyclist is travelling at 25 m/s when he begins to decelerate at -4m/s2 . How fast is travelling after 5 seconds

Umnica [9.8K]

Initial velocity (Vi) = 25 m/s

acceleration (a) = $-4 m/s^{2}$

time interval (t) = 5 sec

let us assume that final velocity after 5 sec be Vf

As acceleration is constant, we can apply the the equation of motion with constant acceleration i.e. $V_{f} = V_{i} + at$

Hence, $V_{f} = 25 +(-4)(5) = 25 -20 = 5 m/s$

so, the velocity of bicyclist will be 5 m/s after 5 sec

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3 years ago

Please help, all about Mirrors and Lenses

romanna [79]

Answer: I think it’s 20cm.

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2 years ago

Scientists have organized all forms of energy into two types—kinetic and potential. Kinetic energy is the energy of motion, and

mihalych1998 [28]

Answer:

Potential energy only

Explanation:

at the top of its swing the pendulum stops moving , (therefore it has no KINETIC energy) thus all of the energy is stored as potential energy.

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2 years ago

Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge

NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is $V=\frac{1}{4\pi\epsilon} \frac{q}{d}$

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as- $E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}$

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining two charges where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

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3 years ago