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3 years ago

During a ______ Moon phase, the side of the Moon facing Earth isn't reflecting any sunlight from the Sun.

Physics

1 answer:

attashe74 [19]3 years ago

4 0

Answer:

B) New

During a new Moon phase, the side of the Moon facing Earth isn't reflecting any sunlight from the Sun.

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A charge per unit length given by l(x) = bx, where b = 12 nc/m2, is distributed along the x axis from x = +9.0 cm to x = +16 cm.

Pavel [41]

Hopefully this will help you.

4 0

3 years ago

The length of second, s pendulum at a place where gravitational acceleration ]g] is 9.8 m/s

amm1812

O.99 m long .simple pendulum time period is 2s for second formula then use formula T=2pi.rt(lenght/gravity)

6 0

3 years ago

Write the first equation of motion. Under what condition(s) is this equation valid?

Zepler [3.9K]

Explanation:

The first equation of motion in kinematics is given by :

$v=u+at$ .....(1)

u is initial speed

a is acceleration

v is final speed

t is time

Equation (1) is valid when the object is moving with constant acceleration. This equation gives relation between velocity and time.

3 0

3 years ago

Bill leaves his 60 W desk lamp on every day, including weekends, for eight hours. After one month (30 days), how much total ener

maxonik [38]

' W ' is the symbol for 'Watt' ... the unit of power equal to 1 joule/second.

That's all the physics we need to know to answer this question.
The rest is just arithmetic.

(60 joules/sec) · (30 days) · (8 hours/day) · (3600 sec/hour)

= (60 · 30 · 8 · 3600) (joule · day · hour · sec) / (sec · day · hour)

= 51,840,000 joules
__________________________________

Wait a minute ! Hold up ! Hee haw ! Whoa !
Excuse me. That will never do.
I see they want the answer in units of kilowatt-hours (kWh).
In that case, it's

(60 watts) · (30 days) · (8 hours/day) · (1 kW/1,000 watts)

= (60 · 30 · 8 · 1 / 1,000) (watt · day · hour · kW / day · watt)

= 14.4 kW·hour

Rounded to the nearest whole number:

14 kWh

7 0

3 years ago

Dragsters can actually reach a top speed of 145.0 m/s in only 4.45 s. (a) Calculate the average acceleration for such a dragster

astraxan [27]

Answer:

a) 32.58 m/s²

b) 161.84 m/s

Explanation:

Initial velocity = u = 0

Final velocity = v = 145 m/s

Time taken = t = 4.45 s

s = Displacement of dragster = 402 m

a = Acceleration

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{145-0}{4.45}\\\Rightarrow a=32.58\ m/s^2$

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as-u^2}\\\Rightarrow v=\sqrt{2\times 32.58\times 402-0^2}\\\Rightarrow v=161.84\ m/s$

The final velocity is greater than the velocity used to find the average acceleration due to the gear changes. The first gear in a dragster has the most amount of toque which means the acceleration will be maximum. The final gears have less torque which means the acceleration is lower here. The final gears have less acceleration but can spin faster which makes the dragster able to reach higher speeds but slowly.

7 0

3 years ago