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3 years ago

What does a light year measure

Physics

2 answers:

SVEN [57.7K]3 years ago

8 0

It’s used in measuring distance from earth with other celestial object. Hope this helps mark brainest

IRINA_888 [86]3 years ago

5 0

Light years measure the distance in space

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Sphere 1 with radius R_1 has positive charge q, Sphere 2 with radius 4.50 R_1 is far from sphere 1 and initially uncharged. The

tia_tia [17]

Answer:

Explanation:

capacitance of sphere 2 will be 4.5 times sphere 1

a ) when spheres are in contact they will have same potential finally . So

V_1 / V_2 = 1

b )

Charge will be distributed in the ratio of their capacity

charge on sphere1 = q x 1 / ( 1 + 4.5 )

= q / 5.5

fraction = 1 / 5.5

c ) charge on sphere 2

= q x 4.5 / 5.5

fraction = 4.5 / 5.5

d ) surface charge density of sphere 1

= q /( 5.5 x A ) where A is surface area

surface charge density of sphere 2

= q x 4.5 /( 5.5 x 4.5² A ) where A is surface area

= q /( 5.5 x 4.5 A )

q_1/q_2 = 4.5

6 0

3 years ago

Calculate the total displacement if a toy car starts at 0 moves 5cm to the left then 8cm to the right then 3 cm to the left

never [62]

It’s a total of 16 cm because it starts at 0 moves then moves to left then the right then the left

6 0

3 years ago

a glass beaker has a mass of 50g. a liquid of density 1.8g/cm3 is poured into the beaker until it reaches the 200cm3 mark. calcu

xz_007 [3.2K]

<u>Answer:</u>

total mass = 410 g

<u>Explanation:</u>

density = 1.8 g/cm³

volume = 200 cm³

density = mass / volume

mass (of liquid) = density x volume

= 1.8 x 200

= 360 g

total mass (beaker + liquid) = 50 + 360 = 410 g [Ans]

Hope this helps!

5 0

3 years ago

Question 4 of 10

kiruha [24]

B hey what do u know i took that test to

5 0

2 years ago

A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the

lana [24]

(a) Let's convert the final speed of the car in m/s:
$v_f = 61 km/h = 16.9 m/s$
The kinetic energy of the car at t=19 s is
$K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J$

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
$P= \frac{W}{\Delta t}$
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
$P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W$

(c) The instantaneous power is given by
$P_i = Fv_f$
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
$a= \frac{v_f-v_i}{\Delta t}= \frac{16.9 m/s}{19 s}=0.89 m/s^2$
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
$P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W$

5 0

3 years ago