No,
one of the 3 sides need to be 90
Using vector concepts, it is found that:
The component form is of approximately (-9.58, 7,22). It means that the ship is about 9.58 miles to the west and about 7.22 miles to the north of where the ship left the port.
<h3>How can a vector be represented in component notation?</h3>
Given a magnitude M and angle 
, then a vector V can be represented as follows in component notation:
In this problem, the magnitude and the angle are given, respectively, by:
Hence:
V = [12cos(143º), 12sin(143º)] = (-9.58, 7,22).
Which means a displacement of 9.58 miles to the west(negative x = west) and 7.22 miles to the north(positive y = north).
The component form is of approximately (-9.58, 7,22). It means that the ship is about 9.58 miles to the west and about 7.22 miles to the north of where the ship left the port.
More can be learned about vectors at brainly.com/question/24606590
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Answer:
1/4
Step-by-step explanation:
The chance of it landing heads up once would be 1/2
The chance of it landing heads up twice would be 1/4
The chance of it landing heads up three times would be 1/8
The chance of it landing heads up four times would be 1/16
See the pattern?
Pattern: <em><u>Divide the denominator by 2.</u></em>
<h2>
<em>Please give brainliest!!! :) </em></h2>
Best Answer: 2 LiCl = 2 Li + Cl2
mass Li = 56.8 mL x 0.534 g/mL=30.3 g
moles Li = 30.3 g / 6.941 g/mol=4.37
the ratio between Li and LiCl is 2 : 2 ( or 1 : 1)
moles LiCl required = 4.37
mass LiCl = 4.37 mol x 42.394 g/mol=185.3 g
Cu + 2 AgNO3 = Cu/NO3)2 + 2 Ag
the ratio between Cu and AgNO3 is 1 : 2
moles AgNO3 required = 4.2 x 2 = 8.4 : but we have only 6.3 moles of AgNO3 so AgNO3 is the limiting reactant
moles Cu reacted = 6.3 / 2 = 3.15
moles Cu in excess = 4.2 - 3.15 =1.05
N2 + 3 H2 = 2 NH3
moles N2 = 42.5 g / 28.0134 g/mol=1.52
the ratio between N2 and H2 is 1 : 3
moles H2 required = 1.52 x 3 =4.56
actual moles H2 = 10.1 g / 2.016 g/mol= 5.00 so H2 is in excess and N2 is the limiting reactant
moles NH3 = 1.52 x 2 = 3.04
mass NH3 = 3.04 x 17.0337 g/mol=51.8 g
moles H2 in excess = 5.00 - 4.56 =0.44
mass H2 in excess = 0.44 mol x 2.016 g/mol=0.887 g
