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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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A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,00
Aleksandr-060686 [28]

Answer:

Explanation:

We have the following relation between power, P and intensity, I

I = P/(4*pi*r^2)

= 10^3/(4*pi*(35000*10^3))

= 6.5*10^-14 W/M^2

We also have the following relationship between electric field and electromagnetic radiation thus

I = (ceE^2)/2

Hence E = \sqrt{2I/ce}

substituting the values of I, c and e, we have

7*10^-6 V/m

3 0
3 years ago
What happens to the temperature of a substance when it is changing state of matter?
noname [10]
Solid substances have molecules held tightly and close together
Liquid substances have molecules moving loosely
Gaseous molecules are moving completely freely

As moleclues get further apart, i.e. As a substance changes state from solid to liquid to gas, molecules gain kinetic energy and vibrate/move more. This means they gain heat energy (the averge energy a substance has) so the temperature increases

Substances exist in different states at different temperatures and different substances will exist in different states at the same temperature. This is to do with the forces between molecules and how much heat (energy) is required to break them
5 0
3 years ago
A length of copper wire has a resistance 44 Ω. The wire is cut into three pieces of equal length, which are then connected as pa
WINSTONCH [101]

Answer:

\frac{R}{1} = \frac{44}{9}\ohm

Explanation:

Let us imagine that there are three wire of length equal length having equal resistances each of 44/3 Ω

Now connect these wires in parallel to so that their equivalent resistance is R.

then

\frac{1}{R} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}

\frac{1}{R} = \frac{3}{44}+\frac{3}{44}+\frac{3}{44}

\frac{1}{R} = \frac{9}{44}

⇒\frac{R}{1} = \frac{44}{9}\ohm

5 0
3 years ago
Read 2 more answers
The free-fall acceleration on Mars is 3.7 m/s^2. What length of pendulum has a period of 1.0 s on Earth? What length of pendulum
NemiM [27]

Answer:

Explanation:

Given

Free fall acceleration on mars g_{m}=3.7\ m/s^2

Time Period of pendulum on earth T=1\ s

Time period of Pendulum is given by

T=2\pi \sqrt{\frac{L}{g}}

for earth

1=2\pi\cdot \sqrt{\frac{L}{9.8}}

L=\frac{9.8}{(2\pi )^2}

L=0.498\ m

(b)For same time period on mars length is given by

L'=\frac{g_m}{(2\pi )^2}

L'=\frac{3.7}{39.48}

L'=0.0936\ m

L'=9.36\ cm                            

3 0
4 years ago
En un m.A.S. La amplitud tiene un valor de 10 centimetros y el periodo es de 2 segundos calcular el valor de la velocidad de 0.8
Ivanshal [37]

Answer:

v1=18.46m/s

v2=29.8cm/s

Explanation:

We know that

A=10cm\\T=2s

the equation of the motion is

x=Acos(\omega t)\\

we can calculate w by using

\omega=\frac{2\pi}{T}=\frac{2\pi}{2}=\pi

Hence, we have that

x=10cm*cos(\pi t)\\

the speed will be

v=-\omega*Asin(\omega t)\\|v(0.8)|=|\pi*10cm*sin(\pi *0.8)|=18.46\frac{cm}{s}\\|v(1.4)|=|\pi*10cm*sin(\pi *1.4)|=29.8\frac{cm}{s}

hope this helps!

6 0
3 years ago
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