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2 years ago

If the x-component of a force vector is 5.17 newtons and its y-component is 8.80 newtons, then what is its magnitude?

Physics

1 answer:

olya-2409 [2.1K]2 years ago

3 0

Answer:

10.21 N

Explanation:

As the force is a vector, it can be decomposed in two components perpendicular each other, so there is no projection of one component in the direction of the other.

When divided in this way, the magnitude of the resultant vector can be found simply applying trigonometry, as follows:

F² = Fx² + Fy² ⇒ F = √(Fx)²+(Fy)²

Replacing by Fx= 5.17 N and Fy = 8.8 N, we get:

F = √(5.17)²+(8.8)² =10.21 N

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f the magnitude of the acceleration of a propeller blade's tip exceeds a certain value amaxamax, the blade tip will fracture. If

Luden [163]

Answer:

The angular velocity is $w= \sqrt[4]{\frac{a_{max}^2}{r^2} - \alpha ^2}$

Explanation:

Generally the acceleration experienced by the propeller blade's is broken down into

The Radial acceleration which is mathematically represented as

$a_r = \frac{v^2}{r} = w^2r$

And the Tangential acceleration which is mathematically represented as

$a_r = \alpha r$

The net acceleration is evaluated as

$a = \sqrt{a_r^2 + a_t^2}$

Now since angular speed varies directly with angular acceleration so when acceleration is maximum the angular velocity is maximum also and this point if the propeller blade's tip exceeds it the blade would fracture

So at maximum angular acceleration we a have

$a_{max} = \sqrt{a_r^2 + a_t^2}$

$a_{max}^2 = a_r^2 + a_t^2$

$a_{max}^2 = (w^2r)^2 + (\alpha r)^2$

$a_{max}^2 = r^2 w^4 + r^2 \alpha ^2$

$a_{max}^2 = r^2 (w^4 + \alpha^2 )$

$w^4 +\alpha ^2 = \frac{a_{max}^2}{r^2}$

$w^4 = \frac{a_{max}^2}{r^2} - \alpha ^2$

$w= \sqrt[4]{\frac{a_{max}^2}{r^2} - \alpha ^2}$

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Which adaptation is likely to increase the chances of survival of an animal in a rainforest?

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A block of mass 200g is oscillating on the end of a horizontal spring of spring constant 100 N/m and natural length 12 cm. When

malfutka [58]

In order to determine the acceleration of the block, use the following formula:

$F=ma$

Moreover, remind that for an object attached to a spring the magnitude of the force acting over a mass is given by:

$F=kx$

Then, you have:

$ma=kx$

by solving for a, you obtain:

$a=\frac{kx}{m}$

In this case, you have:

k: spring constant = 100N/m

m: mass of the block = 200g = 0.2kg

x: distance related to the equilibrium position = 14cm - 12cm = 2cm = 0.02m

Replace the previous values of the parameters into the expression for a:

$a=\frac{(\frac{100N}{m})(0.02m)}{0.2\operatorname{kg}}=10\frac{m}{s^2}$

Hence, the acceleration of the block is 10 m/s^2

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