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3 years ago

5

IS IT B!! if not pls helpp!

1 answer:

natita [175]3 years ago

6 0

Answer:

I believe you are correct but we just started this unit

Explanation:

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Discuss why the article says the fact that we’re in our universe complicates our understanding of the expansion of the universe.

MArishka [77]

<span>Because of our perception of the universe from inside the universe, we are unable to see how and towards what the universe is expanding. Also, our understanding of it is further complicated because we are moving as part of the expansion, thus distorting our perception of it.</span>

4 0

3 years ago

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6.If a 250. gram cart moving to the right with a velocity of .31 m/s collides inelastically with a 500. gram cart traveling to t

spin [16.1K]

Answer:

The total momentum of the system before the collision is 0.0325 kg-m/s due east direction.

Explanation:

Given that,

Mass of the cart, m = 250 g = 0.25 kg

Initial velocity of the cart, u = 0.31 m/s (due right)

Mass of another cart, m' = 500 g = 0.5 kg

Initial velocity of the another cart u' = -0.22 m/s (due left)

Let p is the total momentum of the system before the collision. It is given by :

$p=mu+m'u'\\\\p=0.25\times 0.31+0.5\times (-0.22)\\\\p=-0.0325\ kg-m/s$

So, the total momentum of the system before the collision is 0.0325 kg-m/s due east direction.

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3 years ago

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Two ladders of uniform density and equal mass m are propped up against

-BARSIC- [3]

Answer:

i need some more coins LMAI

7 0

3 years ago

calculate the efficiency of a light bulb that had an input of 400 j transfers 100j as a light and 300j as heat.

beks73 [17]

Efficiency = useful energy out / total energy in x 100
= 100/400 x 100
=0.25 x 100
= 25%

25%

6 0

3 years ago

An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m/s to 4.0 × 107 m/s over a distance

jeka94

Answer:

$1.8\times 105 N/C$

Explanation:

We are given that

$u=2\times 10^7 m/s$

$v=4\times 10^7 m/s$

d=1.9 cm= $\frac{1.9}{100}=0.019 m$

Using 1m=100 cm

We have to find the electric field strength.

$v^2-u^2=2as$

Using the formula

$(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)$

$16\times 10^{14}-4\times 10^{14}=0.038a$

$0.038a=12\times 10^{14}$

$a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2$

$q=1.6\times 10^{-19} C$

Mass of electron,m $=9.1\times 10^{-31} kg$

$E=\frac{ma}{q}$

Substitute the values

$E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}$

$E=1.8\times 105 N/C$

7 0

4 years ago