Find the constant of variation k for the direct variation. Then write an equation for the inverse variation. Y= 4.5 when x= 3
1 answer:
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According to the United States Mint, the mass of 1 nickel is 5.000 g.
There are 15 nickles in 75 cents worth of nickels.
15 * 5.000 g = 75.000 g
Answer: 75.000 grams
There are 15 nickles in 75 cents worth of nickels.
15 * 5.000 g = 75.000 g
Answer: 75.000 grams
From what I imagine in the problem description, the toy is located on the ground. Then, Belkis is pulling on the string to keep the toy moving just like how you put a leash on a dog. My illustration is shown in the picture. The F arrow is where Belkis pulls it. Its component vectors are Fx and Fy, as shown.
From the figure, we can see that it forms a right triangle. So, it makes it easy because the pythagorean theorems are applicable and we can easily use trigonometric functions. First, let's determine Fy which is located opposite to the angle. Since we know the hypotenuse F to be 1.5 N, so we use the sine function:
sin 52° = Fy/F = Fy/1.5
Fy = 1.182 N
For the horizontal component, Fx is parallel to the adjacent side with respect to the angle. Thus, we use the cosine function.
cos 52° = Fx/F = Fx/1.5
Fx = 0.923 N
From the figure, we can see that it forms a right triangle. So, it makes it easy because the pythagorean theorems are applicable and we can easily use trigonometric functions. First, let's determine Fy which is located opposite to the angle. Since we know the hypotenuse F to be 1.5 N, so we use the sine function:
sin 52° = Fy/F = Fy/1.5
Fy = 1.182 N
For the horizontal component, Fx is parallel to the adjacent side with respect to the angle. Thus, we use the cosine function.
cos 52° = Fx/F = Fx/1.5
Fx = 0.923 N
Answer:
a) j(t) = -16t^2 +20t +6
b) 1 second
c) less time
Step-by-step explanation:
The general equation for vertical motion is ...
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a) j(t) = -16t^2 +20t +6 . . . . . filling in the given initial values, with g=32 ft/s^2
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b) We want to find the larger of the solutions to ...
j(t) = 10 = -16t^2 +20t +6
Subtracting 6 and dividing by -4, we have ...
-1 = 4t^2 -5t
4t^2 -5t +1 = 0 . . . . add 1, put in standard form
(4t -1)(t -1) = 0 . . . . . factor
t = 1/4 or 1
The ball is in the air 1 second before reaching the hoop.
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c) In generic terms, the equation of motion for a given shooting height at the same upward velocity will be ...
10 = -16t^2 +20t + h0
-16t^2 +20t +(h0 -10) = 0
Using the quadratic formula, we find the larger solution to be ...
t = (-(20) -√(20² -4(-16)(h0 -10)))/(2(-16))
t = (5/8) + √(64h0 -240)/32 = (5 +√(4h0 -15))/8
This value increases as h0 increases, so the ball will be in the air less time when Raymond shoots.
