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3 years ago

A cannon fires a 0.652 kg shell with initial

Physics

2 answers:

nataly862011 [7]3 years ago

8 0

Answer:

height= 6.429m

Explanation:

h is equall to Max height traveled by trajectory body.

Max height= (U²Sin²tita)/2g

Max height = (12²(sin61)²)/(2*9.8)

Max height =( 144*0.875)/19.6

Max height= 6.429 m

laila [671]3 years ago

6 0

Mass have no effect for the projectile motion and u want to know the height "h"
first,
find the vertical and horizontal components of velocity
vertical component of velocity = 12 sin 61
horizontal component of velocity = 12 cos 61
now for the vertical motion ;
S = ut + (1/2) at^2
where
s = h
u = initial vertical component of velocity
t = 0.473 s
a = gravitational deceleration (-g) = -9.8 m/s^2

h=[12×sin 610×0.473]+[−9.8×(0.473)2]

u can simplify this and u will get the answer

h=.5Gt2

H=1.09m

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What is a stable and unstable equilibrium

Citrus2011 [14]

Answer:

here

Explanation:

Equilibrium is a state of a system which does not change. ... An equilibrium is considered stable (for simplicity we will consider asymptotic stability only) if the system always returns to it after small disturbances. If the system moves away from the equilibrium after small disturbances, then the equilibrium is unstable.

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2 years ago

The (blank) of vibration of a wave is defined as that which has the lowest frequency

saveliy_v [14]

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3 years ago

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A merry go round (mass of 200kg and radius 5m) is rotating so that the outside edge is moving 10m/s. A giant bug of 6kd at a dis

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Answer:

2 rad/s

Explanation:

For a rotating object, the linear velocity is given by

$v=\omega r$

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$\omega=\dfrac{v}{r}$

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3 years ago

An example of when total internal reflection occurs is when all the light passing from a region of higher index of refraction to

Amiraneli [1.4K]

Answer:

is reflected back into the region of higher index

Explanation:

Total internal reflection is a phenomenon that occurs when all the light passing from a region of higher index of refraction to a region of lower index is reflected back into the region of higher index.

According to Snell's law, refraction of ligth is described by the equation

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

n1 is the refractive index of the first medium

n2 is the refractive index of the second medium

$\theta_1$ is the angle of incidence (in the first medium)

$\theta_2$ is the angle of refraction (in the second medium)

Let's now consider a situation in which

$n_1 > n_2$

so light is moving from a medium with higher index to a medium with lower index. We can re-write the equation as

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$

Where $\frac{n_1}{n_2}$ is a number greater than 1. This means that above a certain value of the angle of incidence $\theta_1$ , the term on the right can become greater than 1. So this would mean

$sin \theta_2 > 1$

But this is not possible (the sine cannot be larger than 1), so no refraction occurs in this case, and all the light is reflected back into the initial medium (total internal reflection). The value of the angle of incidence above which this phenomen occurs is called critical angle, and it is given by

$\theta_c =sin^{-1}(\frac{n_2}{n_1})$

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3 years ago

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