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maw [93]
3 years ago
9

Consider a projectile launched with an initial velocity of v0 = 120 ft/s, inclined at an angle, θ with the horizontal. Let us as

sume that the projectile lands on a spot that is at a height h = 10 feet, raised in elevation from its launch point. For this entire problem, you can neglect drag.
a) Determine an expression for the horizontal range R as a function of time. Here, range represents the total horizontal distance traveled from launch point to point of landing. Your answer will be in terms of initial velocity v0, time t, height h and acceleration due to gravity, g. At this point, do not substitute any of the numerical values yet.
b) Determine the time taken to achieve the maximum possible range, R. Think about maxima and minima from calculus when solving this part of the problem.
c) Once you have determined the time from part (b), substitute the numerical values for the terms and obtain the actual time in seconds. Then, determine the initial launch angle, θ (also called the angle of attack), such that the projectile achieves the maximum possible horizontal range.
d) What is the numerical value of this maximum R (based on your calculations from parts (b) and (d)).
e) Determine the velocity at the peak of the trajectory of the projectile.
f) Determine the time taken for the projectile to reach its peak height on its trajectory.
g) Determine the peak height reached by the projectile.
Physics
1 answer:
Natali [406]3 years ago
4 0

Answer:

How to find the maximum height of a projectile?

if α = 90°, then the formula simplifies to: hmax = h + V₀² / (2 * g) and the time of flight is the longest. ...

if α = 45°, then the equation may be written as: ...

if α = 0°, then vertical velocity is equal to 0 (Vy = 0), and that's the case of horizontal projectile motion.

Explanation:

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Spiderman, whose mass is 74.0 kg, is dangling on the free end of a 11.0-m-long rope, the other end of which is fixed to a tree l
Anestetic [448]

Answer:

W = -1844.513 J

Explanation:

GIVEN DATA:

mass of spider man is m  74 kg

vertical displacement if spider is 11 m

final displacement  =  11 cos 60.6 =  - 6.753 m

change in displacement is  = -6.753 - (-11) = 4.25 m

gravity force act on spiderman is f = mg = 74 × 9.8 = 725.2 N

work done by gravity is W = F \delta r cos\theta

W = 725.2 \times 4.25 \times cos 180

where 180 is the angle between spiderman weight and displacement

W = -1844.513 J

7 0
3 years ago
Which of these events is an example of the Doppler shift?
garik1379 [7]
The intensity of an electromagnetic wave increases with the field strength.
5 0
3 years ago
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A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with
Julli [10]

Explanation:

Given that,

Capacitor C=1.66\ \mu F

Resistor R=80.0\ \Omega

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

f_{c}=\dfrac{1}{2\pi R C}

Where, R = resistor

C = capacitor

Put the value into the formula

f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}

f_{c}=1198.45\ Hz

(B). We need to calculate the V_{R} when f = \dfrac{1}{2f_{c}}

Using formula of  V_{R}

V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})

Put the value into the formula

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=2.280\ Volt

(C). We need to calculate the V_{R} when f = f_{c}

Using formula of  V_{R}

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=3.606\ Volt

(D). We need to calculate the V_{R} when f = 2f_{c}

Using formula of  V_{R}

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=4.561\ Volt

Hence, This is the required solution.

8 0
3 years ago
A tin can has a volume of 1100 cm³ and a mass of 80 g. Approximately how many grams of lead shot can it carry without sinking in
Kruka [31]

Answer:

1020g

Explanation:

Volume of can=1100cm^3=1100\times 10^{-6}m^3

1cm^3=10^{-6}m^3

Mass of can=80g=\frac{80}{1000}=0.08kg

1Kg=1000g

Density of lead=11.4g/cm^3=11.4\times 10^{3}=11400kg/m^3

By using 1g/cm^3=10^3kg/m^3

We have to find the mass of lead which shot can it carry without sinking in water.

Before sinking the can  and lead inside it they are floating in the water.

Buoyancy force =F_b=Weight of can+weight of lead

\rho_wV_cg=m_cg+m_lg

Where \rho_w=10^3kg/m^3=Density of water

m_c=Mass of can

m_l=Mass of lead

V_c=Volume of can

Substitute the values then we get

1000\times 1100\times 10^{-6}=0.08+m_l

1.1-0.08=m_l

m_l=1.02 kg=1.02\times 1000=1020g

1 kg=1000g

Hence, 1020 grams of lead shot can it carry without sinking water.

4 0
3 years ago
A very narrow beam of white light is incident at 40.80° onto the top surface of a rectangular block of flint glass 11.6 cm thick
DerKrebs [107]
Dispersion angle = 0.3875 degrees. 
Width at bottom of block = 0.09297 cm 
Thickness of rainbow = 0.07038 cm 
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 n1*sin(θ1) = n2*sin(θ2)
 where
 n1, n2 = indexes of refraction for the different mediums
 Î¸1, θ2 = angle of incident rays as measured from the normal to the surface. 
 Solving for θ2, we get
 n1*sin(θ1) = n2*sin(θ2)
 n1*sin(θ1)/n2 = sin(θ2)
 asin(n1*sin(θ1)/n2) = θ2 
 The index of refraction for air is 1.00029, So let's first calculate the angles of the red and violet rays.
 Red:
 asin(n1*sin(θ1)/n2) = θ2
 asin(1.00029*sin(40.80)/1.641) = θ2
 asin(1.00029*0.653420604/1.641) = θ2
 asin(0.398299876) = θ2
 23.47193844 = θ2 
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 asin(n1*sin(θ1)/n2) = θ2
 asin(1.00029*sin(40.80)/1.667) = θ2
 asin(1.00029*0.653420604/1.667) = θ2
 asin(0.39208764) = θ2
 23.08446098 = θ2 
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 23.47193844 - 23.08446098 = 0.38747746 degrees. 
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 tan(θ) = X/11.6
 11.6*tan(θ) = X
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 11.6*tan(θ) = X
 11.6*tan(23.47193844) = X
 11.6*0.434230136 = X
 5.037069579 = X 
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 11.6*tan(θ) = X
 11.6*tan(23.08446098) = X
 11.6*0.426215635 = X
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 So the width as measured from the bottom of the block is: 5.037069579 cm - 4.944101361 cm = 0.092968218 cm 
 The actual width of the beam after it exits the flint glass block will be thinner. The beam will exit at an angle of 40.80 degrees and we need to calculate the length of the sides of a 40.80/49.20/90 right triangle. If you draw the beams, you'll realize that:
 cos(θ) = X/0.092968218
 0.092968218*cos(θ) = X 
 0.092968218*cos(40.80) = X
 0.092968218*0.756995056 = X
 0.070376481 = X 
 So the distance between the red and violet rays is 0.07038 cm.
7 0
3 years ago
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