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natita [175]
3 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]3 years ago
8 0

Answer:

Angle B =50

Angle C =130

Explanation:

Nikitich [7]3 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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Which evidence taken from a crime scene would be the best evidence for a District Attorney to gain a conviction with DNA evidenc
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6 0
3 years ago
A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500
jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5  ×10 ⁻³ kg

initial velocity of stone (u_{stone}) = 0 m/s

Initial velocity of bullet (u_{bullet}) = (500 m/s)i

Speed of the bullet after collision (v_{bullet}) = (300 m/s) j

Suppose we represent (v_{stone}) to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}

(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}

(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j=  (0.100 \ kg)v_{stone}

v_{stone}= (1.25\  kg.m/s)i-(0.75\ kg m/s)j

v_{stone}= (12.5\  m/s)i-(7.5\ m/s)j

∴

The magnitude now is:

v_{stone}=\sqrt{ (12.5\  m/s)^2-(7.5\ m/s)^2}

\mathbf{v_{stone}= 14.6 \ m/s}

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

\theta = tan^{-1} (\dfrac{-7.5}{12.5})

\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}

5 0
3 years ago
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