Ask question

Ask question

All categories

2 years ago

7

For a flourish at the end of her act, a juggler tosses a single ball high in the air. She catches the ball 3.3 s later at the sa

me height from which it was thrown.
Part A What was the initial upward speed of the ball?
Express your answer to two significant figures and include appropriate units.

1 answer:

lora16 [44]2 years ago

8 0

Answer:

Initial speed of the ball, u = 16 m/s

Explanation:

It is given that, a juggler tosses a single ball high in the air. She catches the ball 3.3 s later at the same height from which it was thrown. Let u is the initial upward speed of the ball.

At the highest point, the final speed of the ball, v = 0

Using equation of motion as :

$v=u+at$

a = -g

$0=u-gt$

Let $t_a\ and\ t_d$ are time of ascent and descent respectively.

So, total time, $t=\dfrac{2u}{g}$

$3.3=\dfrac{2u}{9.8}$

u = 16.17 m/s

or

u = 16 m/s

So, the initial upward speed of the ball is 16 m/s. Hence, this is the required solution.

You might be interested in

Unlike acceleration and velocity, speed does not need to specify

frosja888 [35]

Unlike acceleration and velocity, speed does not need to specify the direction of motion. Speed is a scalar quality.

4 0

1 year ago

Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i

SCORPION-xisa [38]

Answer:

The value is $r = 5.077 \ m$

Explanation:

From the question we are told that

The Coulomb constant is $k = 9.0 *10^{9} \ N\cdot m^2 /C^2$

The charge on the electron/proton is $e = 1.6*10^{-19} \ C$

The mass of proton $m_{proton} = 1.67*10^{-27} \ kg$

The mass of electron is $m_{electron } = 9.11 *10^{-31} \ kg$

Generally for the electron to be held up by the force gravity

Then

Electric force on the electron = The gravitational Force

i.e

$m_{electron} * g = \frac{ k * e^2 }{r^2 }$

$\frac{9*10^9 * (1.60 *10^{-19})^2 }{r^2 } = 9.11 *10^{-31 } * 9.81$

$r = \sqrt{25.78}$

$r = 5.077 \ m$

7 0

2 years ago

Steam undergoes an adiabatic expansion in a piston–cylinder assembly from 100 bar, 360°C to 1 bar, 160°C. What is work in kJ per

vfiekz [6]

Answer:

work is 130.5 kJ/kg

entropy change is 1.655 kJ/kg-k

maximum theoretical work is 689.4 kJ/kg

Explanation:

piston cylinder assembly

100 bar, 360°C to 1 bar, 160°C

to find out

work and amount of entropy and magnitude

solution

first we calculate work i.e heat transfer - work = specific internal energy @1 bar, 160°C - specific internal energy @ 100 bar, 360°C .................1

so first we get some value from steam table with the help of 100 bar @360°C and 1 bar @ 160°C

specific volume = 0.0233 m³/kg

specific enthalpy = 2961 kJ/kg

specific internal energy = 2728 kJ/kg

specific entropy = 6.004 kJ/kg-k

and respectively

specific volume = 1.9838 m³/kg

specific enthalpy = 2795.8 kJ/kg

specific internal energy = 2597.5 kJ/kg

specific entropy = 7.659 kJ/kg-k

now from equation 1 we know heat transfer q = 0

so - w = specific internal energy @1 bar, 160°C - specific internal energy @ 100 bar, 360°C

work = 2728 - 2597.5

work is 130.5 kJ/kg

and entropy change formula is i.e.

entropy change = specific entropy ( 100 bar @360°C) - specific entropy ( 1 bar @160°C )

put these value we get

entropy change = 7.659 - 6.004

entropy change is 1.655 kJ/kg-k

and we know maximum theoretical work = isentropic work

from steam table we know specific internal energy is 2038.3 kJ/kg

maximum theoretical work = specific internal energy - 2038.3

maximum theoretical work = 2728 - 2038.3

maximum theoretical work is 689.4 kJ/kg

3 0

2 years ago

A plastic rod is rubbed against a wool shirt, thereby acquiring a charge of −4.9 µc. how many electrons are transferred from the

creativ13 [48]

To find the number of electrons transferred, we should divide the total charge acquired by the rod $Q=-4.9 \mu C=-4.9 \cdot 10^{-6}C$ by the charge of a single electron ( $e=-1.6 \cdot 10^{-19}C$ ), and we find:
$N= \frac{Q}{e}= \frac{-4.9 \cdot 10^{-6}C}{-1.6 \cdot 10^{-19}C} =3.1 \cdot 10^{13}$

5 0

2 years ago

A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro

yanalaym [24]

Answer:

a) $v_{o} =16m/s$

b) $v=9.8m/s$

c) $\beta =-35.46º$

Explanation:

From the exercise we know that the ball strikes the building 16m away and its final height is 8m more than the initial

Being said that, we can calculate the initial velocity of the ball

a) First we analyze its horizontal motion

$x=v_{ox}t$

$x=v_{o}cos(60)t$

$v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)}$ (1)

That would be our first equation

Now, we need to analyze its vertical motion

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

$y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2$

Knowing $v_{o}$ in our first equation (1)

$8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2$

$\frac{1}{2}(9.8)t^2=16tan(60)-8$

Solving for t

$t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s$

So, the ball takes to seconds to get to the other building. Now we can calculate its <u>initial velocity</u>

$v_{o}=\frac{16m}{(2s)cos(60)}=16m/s$

b) To find the <u>magnitude of the ball just before it strikes the building</u> we need to calculate its x and y components

$v_{x}=v_{ox}+at=16cos(60)=8m/s$

$v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s$

So, the magnitude of the velocity is:

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s$

c) The <u><em>direction of the ball</em></u> is:

$\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º$

4 0

2 years ago