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Rina8888 [55]
2 years ago
5

Specialized businesses that help coordinate dealings between other companies are called:

Physics
1 answer:
sineoko [7]2 years ago
3 0

Answer:

The answer would be Agencies. This is grade 2 stuff.

Explanation:

Please mark me Brainliest.

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The timeline below shows some major discoveries in biology.
34kurt

Answer:

The cell theory

Explanation:

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27. The traffic officer issued violation tickets to traffic
Evgen [1.6K]

Answer:

27: 85

28:75%

Explanation:

27:68=80

?=100 hence (68×100)÷80

=85

28:<em>1</em><em>8</em><em>/</em><em>2</em><em>4</em><em>×</em><em> </em><em>1</em><em>0</em><em>0</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>7</em><em>5</em><em>%</em>

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3 years ago
you are hiking along a river and see a tall tree on thhe opposite bank. You measure the angle of elevation of the top of the tre
Sidana [21]

Answer:

Explanation:

Let the height of the tree is y and the distance of tree from point B is x.

According to the diagram

tan61 = \frac{y}{x}

x = 0.55 y ..... (1)

tan49.5 = \frac{y}{50+x}

(50 + 0.55y) 1.17 = y ..... from equation (1)

58.5 + 0.644 y = y

0.356 y = 58.5

y = 164.3 ft

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3 years ago
How much space is taken up by a package with a length of 45 cm, a width of 10 cm, and a height of 8 cm?
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The answer is C. 3,600cm cubed

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What is the potential difference between a point 0.48 mm from a charge of 2.9 nc and a point at infinity?
Nuetrik [128]
The potential at a distance r from a charge Q is given by
V(r) = k_e  \frac{Q}{r}
where ke is the Coulomb's constant.

The charge in our problem is Q=2.9 nC=2.9 \cdot 10^{-9} C; for the point at r=0.48 mm=0.48 \cdot 10^{-3} m, the potential is
V_1 = k_e  \frac{Q}{r}= (8.99 \cdot 10^9 Nm^2 C^{-2}) \frac{2.9 \cdot 10^{-9} C}{0.48 \cdot 10^{-3} m}=  5.43 \cdot 10^4 V

For the point at infinity, we immediately see that the potential is zero, because r= \infty and so V_2 = 0.

Therefore, the potential difference between the two points is
\Delta V = V_1 - V_2 = V_1 = 5.43 \cdot 10^4 V
6 0
3 years ago
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