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natita [175]
4 years ago
9

Let mAngleA = 40°. If AngleB is a complement of AngleA, and AngleC is a supplement of AngleB, find these measures.

Physics
2 answers:
VMariaS [17]4 years ago
8 0

Answer:

Angle B =50

Angle C =130

Explanation:

Nikitich [7]4 years ago
7 0

Answer:

mAngleB= 50

mAngleC= 130

Explanation:

Complementary angles = 90

So If angle A is 40 subtract that to 90 to find angle B which is 50

and 50+40= 90 = Complementary

Supplementary angles = 180

So if angle C is supplementary to angle B you would need to subtract 50 from 180 to find angle C which is 130

130+20= 180 = supplementary  

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A 82-kg fisherman in a 112-kg boat throws a package of mass m = 15 kg horizontally toward the right with a speed of vi = 4.8 m/s
abruzzese [7]

Answer:

0.37 m/s to the left

Explanation:

Momentum is conserved.  Initial momentum = final momentum.

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

Initially, both the fisherman/boat and the package are at rest.

0 = m₁ v₁ + m₂ v₂

Plugging in values and solving:

0 = (82 kg + 112 kg) v + (15 kg) (4.8 m/s)

v = -0.37 m/s

The boat's velocity is 0.37 m/s to the left.

8 0
3 years ago
Instrumento que se utiliza para medir fuerzas
Klio2033 [76]

Answer:

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Explanation:

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8 0
3 years ago
A sealed container holding 0.0255 L of an ideal gas at 0.981 atm and 65 ∘ C is placed into a refrigerator and cooled to 41 ∘ C w
user100 [1]

Answer:

0.911 atm

Explanation:

In this problem, there is no change in volume of the gas, since the container is sealed.

Therefore, we can apply Gay-Lussac's law, which states that:

"For a fixed mass of an ideal gas kept at constant volume, the pressure of the gas is proportional to its absolute temperature"

Mathematically:

p\propto T

where

p is the gas pressure

T is the absolute temperature

For a gas undergoing a transformation, the law can be rewritten as:

\frac{p_1}{T_1}=\frac{p_2}{T_2}

where in this problem:

p_1=0.981 atm is the initial pressure of the gas

T_1=65^{\circ}+273=338 K is the initial absolute temperature of the gas

T_2=41^{\circ}+273=314 K is the final temperature of the gas

Solving for p2, we find the final pressure of the gas:

p_2=\frac{p_1 T_2}{T_1}=\frac{(0.981)(314)}{338}=0.911 atm

3 0
4 years ago
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lilavasa [31]

Answer:

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Explanation:

For this to balance, the moments around the fulcrum must sum to zero.

On the left you have   .21   ( is that down? I will assume it is)

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        .21 * 40     +  1.0 * 20    

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        .5 * 20     +     F * 45

these moments must equal each other

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7 0
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Zarrin [17]

Answer:

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because 1 m =100m so, 3000x100=300000

3 0
4 years ago
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