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3 years ago

How to solve arctan[tan(7pi /4)] ...?

1 answer:

7 0

Well,
arctan is a bijection from R into (-pi/2 , pi/2)*and
pi is a period of tangent function:
so
as we have : tan(7pi/4) = tan(pi - pi/4) = - tan(pi/4)
we finally get :

arctan(tan(7pi/4)) = artan(tan(- pi/4)) = - pi/4


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Describe how kinetic and potential energy change as a diver climbs up to a diving board and then dives into the water below.

Dima020 [189]

A diver having mass m climbs up the diving board.
We know that Gravitational potential energy is given as PEG=mgΔh

What changes is his Gravitational potential energy due to change of height Δh with reference to the ground/water level.
While standing on the diving board his velocity is zero. As such kinetic energy is also zero.

Once he jumps off the springboard we see he gets additional energy from the springboard and falls down under action of gravity g. Due to decrease of height above the ground level Gravitational potential energy decreases and gets converted in to his kinetic energy. 1/2mv2.

While in air he encounters air resistance. Some of his energy is spent in overcoming this resistance. Gets converted in to kinetic and thermal energy of surrounding air and his body.

Once diver reaches the water, we see water splashing and hear noise of splash. Thereafter the diver comes to rest. Now his potential energy becomes zero. And converted kinetic energy has been converted in to kinetic energy, heat energy and sound energy of water.

As such energy transformation equation looks like

Gravitational PE+Elastic PE of springboard→Kinetic energy of air and water+Sound energy of splash+thermal energy

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3 years ago

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The first one is B, the second is C, the third is B, the last is C

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A bullet is fired with a velocity of 100 m/s from the ground at an angle of 60° with the horizontal. Calculate the horizontal ra

seraphim [82]

1) The horizontal range of the bullet is 884 m

2) The maximum height attained by the bullet is 383 m

Explanation:

The motion of the bullet is a projectile motion, which consists of two separate motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

From the equation of motions along the x- and y- directions, it is possible to find an expression for the horizontal range covered by a projectile, and it is:

$d=\frac{u^2 sin 2\theta}{g}$

where

u is the initial speed of the projectile

$\theta$ is the angle of projection

$g=9.8 m/s^2$ is the acceleration of gravity

For the bullet in the problem, we have

u = 100 m/s (initial speed)

$\theta=60^{\circ}$ (angle)

Solving the equation, we find the horizontal range:

$d=\frac{(100)^2sin(2\cdot 60^{\circ})}{9.8}=884 m$

To find the maximum height, we have to analyze the vertical motion of the bullet. We can do it by using the following suvat equation:

$v_y^2 - u_y^2 = 2as$

where

$v_y$ is the vertical velocity of the bullet after having covered a vertical displacement of $s$

$u_y$ is the initial vertical velocity

$a =-g=$ is the acceleration (negative, since it points downward)

The vertical component of the initial velocity is given by

$u_y = u sin\theta$

Also, the maximum height s is reached when the vertical velocity becomes zero,

$v_y =0$

Substituting into the equation and re-arranging for s, we find the maximum height:

$s=\frac{u^2 sin^2 \theta}{2g}=\frac{(100)^2(sin 60^{\circ})^2}{2(9.8)}=383 m$

Learn more about projectile motion:

brainly.com/question/8751410

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A 0.80-kg soccer ball experinces an impulse of 25 N x s . Determine the momentum change of the soccer ball.

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If the impulse is 25 N-s, then so is the change in momentum.
The mass of the ball is extra, unneeded information.

Just to make sure, we can check out the units:

Momentum = (mass) x (speed) = kg-meter / sec

Impulse = (force) x (time) = (kg-meter / sec²) x (sec) = kg-meter / sec

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