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3 years ago

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A projectile falls beneath the straight-line path it would follow if there were no gravity. How many meters does it fall below t

his line if it has been traveling for 1 s?

1 answer:

maria [59]3 years ago

4 0

Answer:

Explanation:

Suppose v is the initial velocity and $\theta$ is the angle of inclination

distance traveled in vertical direction in t=1 s

When gravity is present

$y=vt+\frac{1}{2}at^2$

where $y=vertical\ distance$

$a=acceleration$

$t=time$

$v=initial\ velocity$

here initial velocity is v\sin \theta [/tex] so

$y=v\sin \theta \times 1-\frac{1}{2}gt^2$

$y=v\sin \theta -0.5g$

In absence of gravity

$y_2=v\sin \theta \times t$

$y_2=v\sin \theta \times 1$

$\Delta y=y_2-y=v\sin \theta -v\sin \theta +0.5 g=4.9\ m$

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(15 points) :^|

mihalych1998 [28]

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7 0

3 years ago

For a given velocity of projection in a projectile motion, the maximum horizontal distance is possible only at ө = 45°. Substant

myrzilka [38]

First let us imagine the projectile launched at initial velocity V and at angle θ relative to the horizontal. (ignore wind resistance)

Vertical component y:

The initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum height of h, the vertical velocity will be 0, therefore the time t taken to attain this maximum height is:

h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g

where g is acceleration due to gravity

Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g]
D = (2V²sinθ cosθ)/g
D = (V²sin2θ)/g

In order for D (horizontal distance) to be maximum, dD/dθ = 0
That is,

2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2 or θ = π/4

Therefore it is now<span> shown that the maximum horizontal travelled is attained when the launch angle is π/4 radians, or 45°.</span>

6 0

3 years ago

1. What is the kinetic energy of a 1.75 kg ball travelling at a speed of 54 m/s?

Over [174]

Answer:

We conclude that the kinetic energy of a 1.75 kg ball traveling at a speed of 54 m/s is 2551.5 J.

Explanation:

Given

Mass m = 1.75 kg

Velocity v = 54 m/s

To determine

Kinetic Energy (K.E) = ?

We know that a body can possess energy due to its movement — Kinetic Energy.

Kinetic Energy (K.E) can be determined using the formula

$K.E=\frac{1}{2}mv^2$

where

m is the mass (kg)

v is the velocity (m/s)

K.E is the Kinetic Energy (J)

now substituting m = 1.75, and v = 54 in the formula

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}\left(1.75\right)\left(54\right)^2$

$K.E=1458\times 1.75$

$K.E=2551.5$ J

Therefore, the kinetic energy of a 1.75 kg ball traveling at a speed of 54 m/s is 2551.5 J.

7 0

3 years ago

Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its

ruslelena [56]

Answer:

a

The orbital speed is $v= 2.6*10^{3} m/s$

b

The escape velocity of the rocket is $v_e= 3.72 *10^3 m/s$

Explanation:

Generally angular velocity is mathematically represented as

$w = \frac{2 \pi}{T}$

Where T is the period which is given as 1.6 days = $1.6 *24 *60*60 = 138240 sec$

Substituting the value

$w = \frac{2 \pi}{138240}$

$= 4.54*10^ {-5} rad /sec$

At the point when the rocket is on a circular orbit

The gravitational force = centripetal force and this can be mathematically represented as

$\frac{GMm}{r^2} = mr w^2$

Where G is the universal gravitational constant with a value $G = 6.67*10^{-11}$

M is the mass of the earth with a constant value of $M = 5.98*10^{24}kg$

r is the distance between earth and circular orbit where the rocke is found

Making r the subject

$r = \sqrt[3]{\frac{GM}{w^2} }$

$= \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }$

$= 5.78 *10^7 m$

The orbital speed is represented mathematically as

$v=wr$

Substituting value

$v= (5.78*10^7)(4.54*10^{-5})$

$v= 2.6*10^{3} m/s$

The escape velocity is mathematically represented as

$v_e = \sqrt{\frac{2GM}{r} }$

Substituting values

$= \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }$

$v_e= 3.72 *10^3 m/s$

7 0

3 years ago

A policeman in a stationary car measures the speed of approaching cars by means of an ultrasonic device that emits a sound with

beks73 [17]

Answer:

4.6 kHz

Explanation:

The formula for the Doppler effect allows us to find the frequency of the reflected wave:

$f'=(\frac{v}{v-v_s})f$

where

f is the original frequency of the sound

v is the speed of sound

vs is the speed of the wave source

In this problem, we have

f = 41.2 kHz

v = 330 m/s

vs = 33.0 m/s

Therefore, if we substitute in the equation we find the frequency of the reflected wave:

$f'=(\frac{330 m/s}{330 m/s-33.0 m/s})(41.2 kHz)=45.8 kHz$

And the frequency of the beats is equal to the difference between the frequency of the reflected wave and the original frequency:

$f_B = |f'-f|=|45.8 kHz-41.2 kHz|=4.6 kHz$

6 0

3 years ago

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