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3 years ago

An object falls for 7 seconds. How far does it fall?

Physics

1 answer:

aliina [53]3 years ago

5 0

Answer:

It depends on what the object is and what planet it is on and what the conditions are.

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The particles of a GAS within a closed container will collide with the container walls, exerting a FORCE. The force per unit of

Katyanochek1 [597]

Answer:

Pressure

Explanation:

One of the theories propounded by the Kinetic molecular theory, and which also provides an explanation of the several gas laws, is the statement that the gas molecules in a container, travel in straight lines and are in constant collision with themselves and the walls of the container, thus exerting force. This force is the pressure which is defined as the force per unit area.

There is no loss of energy in the collisions involving the gas molecules and that is why their movement can be described as elastic. The descriptions of the behavior of gas molecules in the Kinetic Molecular Theory, give rise to Charles law, Boyle's Law, Avogadro's Laws, Dalton's Law, and Amonton's Law.

6 0

3 years ago

Read 2 more answers

The function x = (5.2 m) cos[(5πrad/s)t + π/5 rad] gives the simple harmonic motion of a body. At t = 5.3 s, what are the (a) di

gulaghasi [49]

Answer:

(a) Displacement = - 3.0576 m

(b) Velocity $=-66.48$ m/s

(c)Acceleration = -753.39 m²/s

(d)The phase motion is 26.7 $\pi$ .

(e)Frequency =2.5 Hz.

(f)Time period =0.4 s

Explanation:

Given function is

$x= (5.2 m)cos[ (5\pi \ rad/s)t+ \frac\pi5]$

(a)

The displacement includes the parameter t, so,at time t=5.3 s

$x|_{t=5.3}= (5.2 m)cos[ (5\pi \ rad/s)5.3+ \frac\pi5]$

$= (5.2 m)cos[ 26.5\pi+ \frac\pi5]$

=(5.2)(-0.588)m

= - 3.0576 m

(b)

$x= (5.2 m)cos[ (5\pi \ rad/s)t+ \frac\pi5]$

To find the velocity of simple harmonic motion, we need to find out the first order derivative of the function.

$v=\frac{dx}{dt}$

$=\frac{d}{dt} (5.2 m)cos[ (5\pi \ rad/s)t+ \frac\pi5]$

$= (5.2 m)(-5\pi)sin[ (5\pi \ rad/s)t+ \frac\pi5]$

$= -26\pi sin[ (5\pi \ rad/s)t+ \frac\pi5]$

Now we can plug our value t=5.3 into the above equation

$v= -26\pi sin[ (5\pi \ rad/s)5.3\ s+ \frac\pi5]$

$=-66.48$ m/s

(c)

To find the acceleration of simple harmonic motion, we need to find out the second order derivative of the function.

$v= -26\pi sin[ (5\pi \ rad/s)t+ \frac\pi5]$

$a=\frac{d^2x}{dt^2}$

$=\frac{dv}{dt}$

$=\frac{d}{dt}( -26\pi sin[ (5\pi \ rad/s)t+ \frac\pi5])$

$= -26\pi (5\pi)cos[ (5\pi \ rad/s)t+ \frac\pi5]$

$= -130\pi^2cos[ (5\pi \ rad/s)t+ \frac\pi5]$

Now we can plug our value t=5.3 into the above equation

$a= -130\pi^2cos[ (5\pi \ rad/s)5.3 \ s+ \frac\pi5]$

= -753.39 m²/s

(d)

The general equation of SHM is

$x=x_mcos(\omega t+\phi)$

$x_m$ is amplitude of the displacement, $(\omega t+\phi)$ is phase of motion, $\phi$ is phase constant.

So,

$(\omega t+\phi)=5\pi t+\frac\pi5$

Now plugging t=5.3s

$(\omega t+\phi)=5\pi \times 5.3+\frac\pi5$

=26.7 $\pi$

The phase motion is 26.7 $\pi$ .

The angular frequency $\omega = 5\pi$

(e)

The relation between angular frequency and frequency is

$\omega =2\pi f$

$\therefore f=\frac{\omega}{2\pi}$

$=\frac{5\pi}{2\pi}$

$=\frac52$

= 2.5 Hz

Frequency =2.5 Hz.

(f)

The relation between frequency and time period is

$T=\frac1 f$

$=\frac1{2.5}$

=0.4 s

Time period =0.4 s

6 0

3 years ago

A stone with a weight of 5.30 N is launched vertically from ground level with an initial speed of 23.0 m/s, and the air drag on

frozen [14]

Answer:

a) $h=25.7\ m$

b) $v'=21.8733\ m.s^{-1}$

Explanation:

Given:

weight of the stone, $w=5.3\ N$

initial velocity of vertical projection, $u=23\ m.s^{-1}$

air drag acting opposite to the motion of the stone, $D=0.266\ N$

The mass of the stone:

$m=\frac{w}{g}$

$m=\frac{5.3}{9.8}$

$m=0.5408\ kg$

Now the acceleration of the stone opposite of the motion:

$D=m.d$

where:

d = deceleration

$0.266=0.5408\times d$

$d=0.4918\ m.s^{-2}$

<u>In course of going up the net acceleration on the stone will be:</u>

$g'=g+d$

$g'=9.8+0.4918$

$g'=10.2918\ m.s^{-2}$

Now using the equation of motion:

$v^2=u^2-2 g'.h$

where:

$v=$ final velocity when the stone reaches at the top of the projectile = 0

h = height attained by the stone before starting to fall down

$0^2=23^2-2\times 10.2918\times h$

$h=25.7\ m$

during the course of descend from the top height of the projectile:

initial velocity, $v=0\ m.s^{-1}$

The acceleration will be:

$g"=g-d$

$g"=9.8-0.4918$

$g"=9.3082\ m.s^{-2}$

here the gravity still acts downwards but the drag acceleration acts in the direction opposite to the motion of the stone, now the stone is falling down hence the drag acts upwards.

Using equation of motion:

$v'^2=v^2+2g".h$ (+ve acceleration because it acts in the direction of motion)

$v'^2=0^2+2\times 9.3082\times 25.7$

$v'=21.8733\ m.s^{-1}$

8 0

3 years ago

How can you double the frequency of a wave if you have control over both the wavelength and the wave velocity?

sammy [17]

Answer:

Change the wavelength to half while keeping velocity constant or change the wave velocity to 2 times while keeping wavelength constant.

Explanation:

The frequency of a wave can be defined as the rate of change of wave speed with respect to the wavelength of the wave.

Mathematically,

$f=\frac{v}{\lambda}$

Here, v is the velocity, lambda is the wavelength, and f is the frequency of the wave.

If the observer want to double the frequency of the wave, then he should increase the wave velocity two times and take wavelength constant for this case or either he should half the wavelength of the wave to take the velocity of the wave constant.

5 0

3 years ago

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A 98-kg fullback is running along at 8.6 m / s when a 76-kg defensive back running in the same direction at 9.8 m / s jumps on h

algol13

The total momentum before and after the collision must be conserved.

The total momentum before the collision is:
$p_i = m_1 v_1 + m_2 v_2$
where m1 and m2 are the masses of the two players, and $v_1$ and $v_2$ their initial velocities. Both are considered with positive sign, because the two players are running toward the same direction.

The final momentum is instead
$p_f = (m_1+m_2)v_f$
because now the two players are moving together with a total mass of (m1+m2) and final speed vf.

By requiring that the momentum is conserved
$p_i=p_f$
we can calculate vf, the post-collision speed:
$m_1 v_1 + m_2 v_2 = (m_1+m_2)v_f$
$v_f = \frac{m_1 v_1 + m_2 v_2}{m_1 +m_2}= \frac{(98 kg)(8.6 m/s)+(76 kg)(9.8m/s)}{98 kg+76 kg}=9.1 m/s$
and the direction is the same as the direction of the players before the collision.

6 0

3 years ago