All categories

3 years ago

How far will a freely falling object fall from rest in 5 seconds?

Physics

1 answer:

Travka [436]3 years ago

3 0

<h2>how far will a freely falling object fall from rest in 5 seconds?</h2>

If an object free falls from rest for 5 seconds, its speed will be about 50 m/s.

hope it helps

#carry on learning

You might be interested in

A person spins a tennis ball on a string in a horizontal circle (so that the axis of rotation is vertical). At the point indicat

Ilya [14]

Answer:

direction does the axis of rotation tilt toward after the blow is the z- direction

Explanation:

This is because

Due to the blow, there is a impulse imparted on the ball which gives a change in linear momentum () in the x direction

Thus When cross product of momentum in x direction is taking with r vector, we will get resultant in z direction. Hence change in angular momentum will be z direction.

5 0

3 years ago

Tammy leaves the office, drives 34 km due north, then turns onto a second highway and continues in a direction of 35◦ north of e

Natasha2012 [34]

Answer:

The total displacement is 102 km $51^o$ north of east.

Explanation:

We can treat this problem as a trigonometric one, so we need to calculate the total displacement on the north and east.

$d_n=34km+79km*sin(35^o)=79km$

and

$d_e=79*cos(35^o)=65km$

The total displacement is given by:

$D=\sqrt{(79km)^2+(65km)^2}=102km$

with an agle of:

$\alpha =arctg(\frac{79km}{65km})=51^o$

4 0

3 years ago

Suppose that a teacher driving a 1972 LeMans zooms out of a darkened tunnel at 34.5 m/s. He is momentarily blinded by the sunshi

valkas [14]

Answer:

489.19m

Explanation:

To find the minimum distance you first calculate the time in which the teacher stops:

$v=v_o-at\\\\t=\frac{v_o-v}{a}=\frac{34.5m/s-0m/s}{2.5m/s^2}=13.8s$

however, the reaction of the teacher is 0.31s later, then you use

t=13.8-0.31s=13.49s

during this time the camper has traveled a distance of:

$x=vt=(15.1m/s)(13.49s)=203.69m$ (1)

Next you calculate the distance that teacher has traveled for 13.6s:

$x=x_o+v_ot+\frac{1}{2}at^2\\\\x=0m+34.5m/s(13.49s)+\frac{1}{2}(2.5m/s^2)(13.49s)^2=692.88m$ (2)

The minimum distance between the driver and the camper will be the difference between (2) and (1):

$x_{min}=692.88m-203.69m=489.19m$

5 0

3 years ago

What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s 2 ) towards the Eart

Wittaler [7]

Answer:

The centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ is $1.020x10^{-3}m/s^{2}$

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude $71.9^{\circ}$ . Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

$\cos \theta = \frac{adjacent}{hypotenuse}$

$\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}}$ (3)

Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

$r_{71.9^{\circ}}$ can be isolated from equation 3:

$r_{71.9^{\circ}} = r_{e} \cos \theta$ (4)

$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

$r_{71.9^{\circ}} = 1.97x10^{6} m$

Then, equation 2 can be used

$v = \frac{2 \pi (1.97x10^{6} m)}{24h}$

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

$T = 24h . \frac{3600s}{1h}$ ⇒ $84600s$