Ask question

Ask question

All categories

2 years ago

12

Jack and Jill are on two different floors of their high rise office building and looking out of their respective windows. Jack s

ees a flower pot go past his window ledge and Jill sees the same pot go past her window ledge a little while later. The time between the two observed events was 4.6 s. Assume air resistance is negligible.
a) If the speed of the pot as it passes Jill's window is 58.0 m/s, what was its speed when Jack saw it go by?
b) What is the height between the two window ledges?

1 answer:

Maru [420]2 years ago

5 0

Answer:

a) speed when Jack sees the pot : 12.92 meters per second

b) height difference 163.115 meters

Explanation:

First to calculate te initial speed we use the acceleration formula:

a= v1-v0/t

Acceleration being gravity's acceleration (9.8 m/s^2)

v1 being the speed when Jill sees the pot

v0 when Jack sees it

and t the time between

Solving for v0 it would be

v1 - a*t = v0

replacing

$58 m/s - 9.8 m/s^2 *4.6 s = v0 = 12.92 m/s$

For the second question we use the position formula setting y0 and t0 as the position and time when jack sees the pot. (and setting the positive axis downward I.E. one meter below jack would be 1m not -1m)

The formula is

$y0 + v0*t + 1/2 g *t^2 = yt$

replacing

$0m + 12.92m/s* 4.6 s + 1/2 * 9.8 m/s^2 * (4.6 s)^2 = 163.115 m$

You might be interested in

A person absentmindedly walks off the edge of a tall cliff. They will fall 50 m into either

kifflom [539]

The man can survive, and he lands 12.1 m from the base of the cliff (into the lake)

Explanation:

The motion of the person is equivalent to the motion of a projectile, which consists of two separate motions:

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

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

We start by analyzing the vertical motion, to find the time it take for the person to reach the ground level. We can use the following suvat equation:

$s=ut+\frac{1}{2}gt^2$

where

s = 50 m is the vertical distance covered (the height of the cliff)

u = 0 is the initial vertical velocity

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

t is the time of flight

Solving for t,

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(50)}{9.8}}=3.19 s$

The person moved horizontally at a constant speed of

$v_x = 3.8 m/s$

So, the horizontal distance covered by the man during his flight will be

$d_x = v_x t = (3.8)(3.19)=12.1 m$

So, the man will land on the lake (which starts from 12 m), so he can survive.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

8 0

2 years ago

An electron in an atom's orbital shell, labeled X in the model below, released enough energy to move to a different orbital shel

Delicious77 [7]

Answer:

Lower energy shell which will be nearer to the nucleus.

Explanation:

When electron move from one energy level to another, an electron must gain or lose just the right amount of energy.

When atoms releases energy, electrons move into lower energy levels. The electrons in the shells aways from the nucleus have more energy as compared to the electrons in the nearer shells.

Electrons with the lowest energy are found closest to the nucleus, where the attractive force of the positively charged nucleus is the greatest. Electrons that have higher energy are found further away

7 0

2 years ago

barxatty [35]

Answer:

NE DIYON INGILIZ MISIN SEN

7 0

2 years ago

A cylinder of radius R, length L, and mass M is released from rest on a slope inclined at angle θ. It is oriented to roll straig

inna [77]

Answer:

$\mu_s=\frac{1}{3}\tan \theta$

Explanation:

Let the minimum coefficient of static friction be $\mu_s$ .

Given:

Mass of the cylinder = $M$

Radius of the cylinder = $R$

Length of the cylinder = $L$

Angle of inclination = $\theta$

Initial velocity of the cylinder (Released from rest) = 0

Since, the cylinder is translating and rolling down the incline, it has both translational and rotational motion. So, we need to consider the effect of moment of Inertia also.

We know that, for a rolling object, torque acting on it is given as the product of moment of inertia and its angular acceleration. So,

$\tau =I\alpha$

Now, angular acceleration is given as:

$\alpha = \frac{a}{R}\\Where, a\rightarrow \textrm{linear acceleration of the cylinder}$

Also, moment of inertia for a cylinder is given as:

$I=\frac{MR^2}{2}$

Therefore, the torque acting on the cylinder can be rewritten as:

$\tau = \frac{MR^2}{2}\times \frac{a}{R}=\frac{MRa}{2}------ 1$

Consider the free body diagram of the cylinder on the incline. The forces acting along the incline are $mg\sin \theta\ and\ f$ . The net force acting along the incline is given as:

$F_{net}=Mg\sin \theta-f\\But,\ f=\mu_s N\\So, F_{net}=Mg\sin \theta -\mu_s N-------- 2$

Now, consider the forces acting perpendicular to the incline. As there is no motion in the perpendicular direction, net force is zero.

So, $N=Mg\cos \theta$

Plugging in $N=Mg\cos \theta$ in equation (2), we get

$F_{net}=Mg\sin \theta -\mu_s Mg\cos \theta\\F_{net}=Mg(\sin \theta-\mu_s \cos \theta)--------------3$

Now, as per Newton's second law,

$F_{net}=Ma\\Mg(\sin \theta-\mu_s \cos \theta)=Ma\\\therefore a=g(\sin \theta-\mu_s \cos \theta)------4$

Now, torque acting on the cylinder is provided by the frictional force and is given as the product of frictional force and radius of the cylinder.

$\tau=fR\\\frac{MRa}{2}=\mu_sMg\cos \theta\times R\\\\a=2\times \mu_sg\cos \theta\\\\But, a=g(\sin \theta-\mu_s \cos \theta)\\\\\therefore g(\sin \theta-\mu_s \cos \theta)=2\times \mu_sg\cos \theta\\\\\sin \theta-\mu_s \cos \theta=2\mu_s\cos \theta\\\\\sin \theta=2\mu_s\cos \theta+\mu_s\cos \theta\\\\\sin \theta=3\mu_s \cos \theta\\\\\mu_s=\frac{\sin \theta}{3\cos \theta}\\\\\mu_s=\frac{1}{3}\tan \theta............(\because \frac{\sin \theta}{\cos \theta}=\tan \theta)$

Therefore, the minimum coefficient of static friction needed for the cylinder to roll down without slipping is given as:

$\mu_s=\frac{1}{3}\tan \theta$

3 0

2 years ago

Read 2 more answers

How do you find average speed on a graph

coldgirl [10]

Answer:

speed equals distance over time 50 divided by 5.

5 0

2 years ago