Belkis is pulling a toy by exerting a force of 1.5 newtons on a string attached to the toy.
2 answers:
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
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Answer:
Step-by-step explanation:
The current function of time is defined as follows:
where is the charge function.
For the given charge function of time we have the following current function:
In the problem it is proposed that .
Examining the expression of we obtained by deriving
with the expression proposed by the problem and comparing term by term:
We conclude that and
.
Part (a):
Before we begin, remember the difference between squares rule which is as follows:
a² - b² = (a+b)(a-b)
Now, for the given we have:
1 - r⁴
This can be rewritten as:
(1)² - (r²)²
We can apply the difference between squares as follows:
(1-r²)(1+r²)
Now, checking the result we reached, we can note that we can apply the difference between squares again on the first bracket.
Doing this, we will reach:
(1-r)(1+r)(1+r²) .............> This is the simplest factored form
Part (b):
The given expression is:
6x² + 7x - 49
This is a polynomial of second degree.
This means that we can use the quadratic formula to get the solutions. The quadratic formula is shown in the attached image
From the expression, we can note that:
a = 6
b = 7
c = -49
Substituting in the formula, we would find that:
either x = 7/3 ...........> This means that the first bracket is (3x-7)
or x = -7/2 .............> This means that the second bracket is (2x+7)
Based on the above, the simplest factored form of 6x² + 7x - 49 is:
(3x-7)(2x+7)
Hope this helps :)
Before we begin, remember the difference between squares rule which is as follows:
a² - b² = (a+b)(a-b)
Now, for the given we have:
1 - r⁴
This can be rewritten as:
(1)² - (r²)²
We can apply the difference between squares as follows:
(1-r²)(1+r²)
Now, checking the result we reached, we can note that we can apply the difference between squares again on the first bracket.
Doing this, we will reach:
(1-r)(1+r)(1+r²) .............> This is the simplest factored form
Part (b):
The given expression is:
6x² + 7x - 49
This is a polynomial of second degree.
This means that we can use the quadratic formula to get the solutions. The quadratic formula is shown in the attached image
From the expression, we can note that:
a = 6
b = 7
c = -49
Substituting in the formula, we would find that:
either x = 7/3 ...........> This means that the first bracket is (3x-7)
or x = -7/2 .............> This means that the second bracket is (2x+7)
Based on the above, the simplest factored form of 6x² + 7x - 49 is:
(3x-7)(2x+7)
Hope this helps :)