A ____(concave, convex) mirror is used to collect light in a reflecting telescope

Electrical current is the flow of ____________

Amanda [17]

Solution :

A) Electric current is the flow of electrons .

B) We know, by ohm's law :

V = I × R

Putting given values in above equation, we get :

V = 0.56 × 72 V

V = 40.32 V

Hence, this is the required solution.

3 0

3 years ago

A driver of a car traveling at -15m/s applies the brakes, causing a uniform acceleration of +2.0m/s2. If the brakes are applied

klemol [59]

Given :

Initial velocity, u = -15 m/s.

Acceleration , a = 2 m/s².

Time taken to applied brake, t = 2.5 s.

To Find :

The velocity of the car at the end of the braking period.

How far has the car moved during the braking period.

Solution :

By equation :

$v = u+at\\\\v=-15 + 2\times 2.5\\\\v=-10 \ m/s$

Now, distance covered by car is :

$s=ut+\dfrac{at^2}{2}\\\\s=(-15)(2.5)+\dfrac{2(2.5)^2}{2}\\\\s=-31.25\ m$

Hence, this is the required solution.

3 0

4 years ago

If the moon orbited twice as far from earth how far would it "fall" each second?

Kamila [148]

I can't imagine that this is going to do you much good, but
I'm sure going to enjoy solving it.
-------------------------
Skip this whole first section.
It was an attempt to master a bunch of trees, while
the forest was right there in front of me all the time.
Drop down below the double line.
-------------------------

Kepler's 3rd law says:

 (square of the orbital period) / (cube of the orbital radius) = constant

 T₀² = K R₀³

I put the zero subscripts in there, because you doubled 'R'
and I need to know how that affected 'T'.

 new-T² = K(2 R₀)³

 new-T² = 8 K (R₀)³ = 8 old-T₀²

 new-T = √8 old-T <=== that's what I was after

I just teased out the Moon's new orbital period if it's distance were doubled.
Instead of 1 month, it's now √8 months.

To put a somewhat sharper point on it, the moon's period of revolution
changes from 27.322 days to 27.322√8 = 77.278 days (rounded) .

Using 385,000 km for the moon's current average distance, the current orbital speed is
 (2π x 385,000 km) / (27.322 days) = 1,024.7 m/s
(One online source says 1.023 km, so we're not doing too badly so far.)

================================================

I'm such a dummy. I don't need to go through all of that.

If the moon were twice as far from Earth as it really is, then it would
average 770,000 km instead of the present 385,000 km.

That's 120.86 times the Earth's radius of 6,371 km.

So the acceleration of gravity out there would be

 (1 / 120.86)² of the (9.807 m/s²) that it is here on the surface.

 new-G = 0.000671 m/s²

Distance a dropped object falls = 1/2 g t²

 In the first second, that's 1/2 g (1)² = 1/2 g

For an orbiting object, every second is the "first"second, because ...
as we often explain orbital motion qualitatively ... the Earth "falls away"
just as fast as the curved orbit falls.

Distance an object falls in the 1st second =

 1/2 G = 0.000336 m/s = 0.336 millimeter per second

I estimate the probability of a mistake somewhere during this process
at approx 99.99% . But I don't have anything better right now, and I've
wasted too much time on it already, so I'll stick with it.

5 0

4 years ago

An unsuspecting bird is coasting along in an easterly direction at 1.00 mph1.00 mph when a strong wind from the south imparts a

grin007 [14]

Answer:

2.375 m, 54°

Explanation:

I will be doing this applying the vector format of i and j. I hope that's not confusing for you. Where, i = East and j = North.

First, we convert the velocity from miles per hour to meters per second. Remember that 1 mile = 1609 meters

u = 1 mph East = 1.00i + 0j mph

u = (1.00i + 0j) mph * (1609 m / 1 mile) * ( 1 hr / 3600 s) = 0.45i + 0j m/s

a = 0.4 m/s² FROM THE SOUTH

a = 0.4 m/s² North = 0i + 0.4j m/s²

Next, we proceed to use the equation of motion to solve

sf = si + ut + 1/2at²

Let's assume that si = 0, this means that

sf = ut + 1/2at², now we substitute for u, t and a to get

sf = (0.45i + 0j m/s) * 3.1 s + 1/2 * (0i + 0.4j m/s²) * (3.1s)²

sf = (1.395i + 0j m) + (0+1.922j m)

sf = 1.395i + 1.922j m

This means that after 3.1 s, the birds displacement from the start is 1.395 m in the East direction and 1.922 m in the North direction.

Finding the resultant now will be

|sf| = √(1.395² + 1.922²) m

|sf| = √(1.946 + 3.694) m

|sf| = √5.64

|sf| = 2.375 m

direction = Tanθ = (j/i)

θ = Tan^-1 (j/i)

θ = Tan^-1(1.922/1.395)

θ = Tan^-1(1.378)

θ = 54° North East

4 0

3 years ago

"Several light bulbs each of resistance 1.25 Ω are connected in series across a 115 V power supply. If the current through the c

Bumek [7]

Answer:

The number of bulbs = 46 bulbs

Explanation:

Ohm's Law: Ohm's law states that the current flowing through a conductor is directly proportional to its potential difference across its ends. At constant temperature and pressure. It can be expressed mathematically as

V = IR............. Equation 1

making R the subject of formula in equation 1

R = V/I.............. Equation 2

Where R = combined resistance of the several bulbs, V = potential difference, I = current through the circuit.

Given: V= 115 V. I = 2 A.

Substituting these values into equation 2,

R = 115/2

R = 57.5 Ω

Since the bulbs are connected in series,

Number of bulbs = combined resistance/resistance of each bulb

Number of bulbs = R/Rₙ...................... Equation 3

Where Rₙ = resistance of each bulb, R =combined resistance of the bulb.

Also given: Rₙ = 1.25 Ω, and R = 57.5 Ω

Substituting these values into equation 3

Number of bulbs = 57.5/1.25

Number of bulbs = 46 bulbs.

Therefore the number of bulbs = 46 bulbs

8 0

3 years ago