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2 years ago

10

Determine the scalar components Ra and Rb of the force R along the nonrectangular axes a and b. Also determine the orthogonal pr

ojection Pa of R onto axi

1 answer:

Liula [17]2 years ago

6 0

Answer: Hello your question is incomplete attached below is the complete question

Ra = 1132 N

Rb = 522.6 N

Pa = 679.7 N

Explanation:

To determine the scalar components Ra

$\frac{Ra}{Sin 120^o} = \frac{750}{sin 35^o}$

therefore : Ra = $\frac{sin120^o * 750}{sin 35^o}$ = 1132 N

To determine the scalar component Rb

$\frac{Rb}{sin 25^o} = \frac{750}{sin 35^o}$

therefore : Rb = $\frac{sin 25^o * 750}{sin 35^o}$ = 522.6 N

To determine the orthogonal projection Pa of R onto

Pa = 750 cos $25^o$ = 679.7 N

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Answer:

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t = Time taken

u = Initial velocity

v = Final velocity

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a = Acceleration due to gravity = 9.81 m/s²

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When s = Db, t = 2t

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2 years ago

14 gauge copper wire has a diameter of 1.6 mm. what length of this wire has a resistance of 4.8ω?

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The relationship between resistance R and resistivity $\rho$ is
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2 years ago

A stationary particle of charge q = 2.1 × 10-8 c is placed in a laser beam (an electromagnetic wave) whose intensity is 2.9 × 10

alisha [4.7K]

(a) The intensity of the electromagnetic wave is related to the amplitude of the electric field by
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2 years ago

An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s afte

aniked [119]

Answer:

35,79 meters

Explanation:

So, we got an archer, and we got a target. Lets call the distance between this two d.

Now, the archer fires the arrow, that, in a time $t_{arrow}$ travels the distance d with a speed $v_{arrow}$ of 40 m/s and hits the target. We can see that the equation will be:

$v_{arrow} * t_{arrow} = d\\ \\40 \frac{m}{s} * t_{arrow} = d$

Immediately after this, the arrow produces a muffled sound, which will travel the distance d at 340 m/s in a time $t_{sound}$ . Obtaining :

$v_{sound} * t_{sound} = d\\ \\340 \frac{m}{s} * t_{sound} = d$ .

Finally, the sound reaches the archer, exactly 1 second after he fired the bow, so:

$t_{arrow} + t _{sound} = 1 s$ .

This equation allows us to write:

$t _{sound} = 1 s - t_{arrow}$ .

Plugging this relationship in the distance equation for the sound:

$340 \frac{m}{s} * t_{sound} = d \\ \\ 340 \frac{m}{s} * (1 s- t_{arrow}) = d$ .

Now, we can replace d from the first equation, and obtain:

$40 \frac{m}{s} * t_{arrow} = d \\ 40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * (1 s- t_{arrow})$ .

Now, we can just work a little bit:

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Now, we can just plug this value into the first equation:

$40 \frac{m}{s} * t_{arrow} = d$

$40 \frac{m}{s} * 340/380 s = 35,79 s = d$

6 0

2 years ago