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ArbitrLikvidat [17]

4 years ago

8

The launching velocity of a missile is 20.0 m/s, and it is shot at 53º above the horizontal. What is the vertical component of t

he velocity at launch?

1 answer:

Lyrx [107]4 years ago

6 0

Given that

Velocity of missile (v) = 20 m/s ,

Angle of missile (Θ) = 53°

Determine , Vertical component = v sin Θ

= 20 sin 53°

= 15.97 m/s

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Zepler [3.9K]

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3 years ago

Part A: Explain why x = 5 makes 4x − 1 ≤ 19 true but not 4x − 1 < 19. (5 points) Part B: What value from the set {6, 7, 8, 9,

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Answer:

A: ≤ means less than OR equal to. < only means less than

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3 years ago

1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal.

harkovskaia [24]

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(a) The net force is 80.394 N

The acceleration of the crate is 0.804 m/s²

(b) the final velocity of the crate is 5.02 m/s

Explanation:

Given;

mass of the crate, m = 100 kg

applied force, F = 250 N

angle of inclination, θ = 45°

coefficient of friction, μ = 0.12

Applied force in y-direction, $F_y = Fsin \theta = 250sin45 = 176.78 \ N$

Applied force in x-direction, $F_x = Fcos \theta = 250cos45 = 176.78 \ N$

The normal force is calculated as;

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$F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N$

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F = ma

$a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2$

(b) the velocity of the crate after 5.0 s

$F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s$

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