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2 years ago

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The formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle invlove and use of which fun

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1 answer:

Anastaziya [24]2 years ago

3 0

Answer:

Explanation:

The formula to analyze motion in the vertical (or y) dimension is

$A_y=Asin\theta$ which says that the motion in the y-dimension is equal to the magnitude of the A vector times the sin of the angle.

The formula to analyze motion in the horizontal (or x) dimension is

$A_x=Acos\theta$ which says that the motion in the x-dimension is equal t the mignitude of the A vector times the cosine of the angle.

Many times this is used to find the upwards velocity and the horizontal velocity when an overall velocity is given with an angle of inclination.

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A weight lifter lifts a 345 N set of weights from ground level to a position over his head, a vertical distance of 1.89 m. How m

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Answer:652.05 J

Explanation:

Given

Weight of lifter $W=345 N$

vertical distance move $h=1.89 m$

Work done in lifting the weight is equal change in Potential Energy of weight

Change in Potential Energy $=m g h$

$\Delta PE=345 \times 1.89=652.05 J$

therefore work done is equal to $652.05 J$

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2 years ago

What waves travel through a medium? Transverse, longitudinal, surface, electromagnetic, and mechanical are the choices

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2 years ago

An airplane is flying in a horizontal circle at a speed of 480 km/h. If its wings are tilted 40° to the horizontal, what is the

guajiro [1.7K]

Answer:

r = 2161.9 m

Explanation:

Aerodynamic lift(L) is perpendicular to the wing, which is tilted 40 degrees to the horizontal.

Since the plane is moving in a horizontal circle, the vertical component of the lift must cancel the weight W of the airplane, but the horizontal component is the centripetal force that keeps it in a circle.

L is perpendicular to wing at angle θ with respect to horizontal

Thus,

Vertical component of lift is:

L cosθ = W = mg

Thus, m = L cosθ / g - - - - (eq1)

Horizontal component of lift is:

L sinθ = centripetal force = mv² / r - - - - (eq2)

Combining equations 1 and 2,we have;

L sinθ = (L cosθ / g)(v² / r)

L cancels out on both sides to give;

tanθ = v²/ rg

r = v² / (g tanθ)

We are given;

velocity; v = 480 km/hr = 480 x 10/36 = 133.33 m/s

r = 133.33²/[(9.8) tan(40)] = 2161.9 m

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2 years ago

A baseball pitcher throws the ball towards the batter at 90 mph. His bat connects with the ball for a line drive, after which th

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Answer:

$F=-18412.9N$ , where the minus indicates the direction is opposite to that of the throw.

Explanation:

a)

Since MKS stands for meter-kilogram-second and we know that:

$1\ hour = 3600\ seconds$

$1\ mile = 1600\ meters$

$1000g = 1kg$

We can write that:

$\frac{1\ hour}{3600\ seconds}=1$

$\frac{1600\ meters}{1\ mile}=1$

$\frac{1kg}{1000g}=1$

These are conversion factors, equal to 1, so multiplying our results by them won't change their value, only their units.

So we have that:

$90 mph=90 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=40m/s$

$110 mph=110 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=48.89m/s$

$145 g=145 g(\frac{1kg}{1000g})=0.145kg$

b)

Newton's 2nd Law tells us that F=ma, and the definition of acceleration is $a=\frac{\Delta v}{\Delta t}$ , so we have:

$F=m\frac{\Delta v}{\Delta t}=m\frac{v_f-v_i}{t}$

Taking the throw direction as the positive one, for our values we have:

$F=m\frac{v_f-v_i}{t}=(0.145kg)\frac{(-48.89m/s)-(+40m/s)}{0.0007s}=-18412.9N$

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vova2212 [387]

Idk I would go back to pre-k

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