Answer:
a 15.22 m/s
b 45.65 m
Explanation:
Using the same formula,
x = vt, where
x is now 45.65, and
t is 3 s, then
45.65 = 3v
v = 45.65/3
v = 15.22 m/s
See the attachment for the part b. We used the distance gotten in part B, to find question A
Answer:
E.true only when no charge is enclosed within the Gaussian surface.
Explanation:
Because Gauss’s law states that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge.
The magnitude of the change in momentum of the stone is about 18.4 kg.m/s

<h3>Further explanation</h3>
Let's recall Impulse formula as follows:

<em>where:</em>
<em>I = impulse on the object ( kg m/s )</em>
<em>∑F = net force acting on object ( kg m /s² = Newton )</em>
<em>t = elapsed time ( s )</em>
Let us now tackle the problem!

<u>Given:</u>
mass of ball = m = 0.500 kg
initial speed of ball = vo = 20.0 m/s
final kinetic energy = Ek = 70% Eko
<u>Asked:</u>
magnitude of the change of momentum of the stone = Δp = ?
<u>Solution:</u>
<em>Firstly, we will calculate the final speed of the ball as follows:</em>



→ <em>negative sign due to ball rebounds</em>


<em>Next, we could find the magnitude of the change of momentum of the stone as follows:</em>

![\Delta p_{stone} = - [ mv - mv_o ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%20-%20%5B%20mv%20-%20mv_o%20%5D)
![\Delta p_{stone} = m[ v_o - v ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%20m%5B%20v_o%20-%20v%20%5D)
![\Delta p_{stone} = m[ v_o + v_o\sqrt{0.7} ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%20m%5B%20v_o%20%2B%20v_o%5Csqrt%7B0.7%7D%20%5D)
![\Delta p_{stone} = mv_o [ 1 + \sqrt{0.7} ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%20mv_o%20%5B%201%20%2B%20%5Csqrt%7B0.7%7D%20%5D)
![\Delta p_{stone} = 0.500 ( 20.0 ) [ 1 + \sqrt{0.7} ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%200.500%20%28%2020.0%20%29%20%5B%201%20%2B%20%5Csqrt%7B0.7%7D%20%5D)


<h3>Learn more</h3>

<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Dynamics
Answer:
Option D
The air pressure inside the car is greater than the pressure outside.
Explanation:
When considering airflow over and around a surface, from Bernoulli's equation, air flow regions with higher velocity have a lower pressure, and regions with lower velocity have a higher pressure.
The air outside the convertible is moving faster than the air inside the convertible. This leads to a higher pressure zone just below the surface of the roof (inside the car) causing the roof of the convertible to bulge upwards
Answer:
a)106.48 x 10⁵ kg.m²
b)144.97 x 10⁵ kgm² s⁻¹
Explanation:
a)Given
m = 5500 kg
l = 44 m
Moment of inertia of one blade
= 1/3 x m l²
where m is mass of the blade
l is length of each blade.
Putting all the required values, moment of inertia of one blade will be
= 1/3 x 5500 x 44²
= 35.49 x 10⁵ kg.m²
Moment of inertia of 3 blades
= 3 x 35.49 x 10⁵ kg.m²
= 106.48 x 10⁵ kg.m²
b) Angular momentum 'L' is given by
L =
x ω
where,
= moment of inertia of turbine i.e 106.48 x 10⁵ kg.m²
ω=angular velocity =2π f
f is frequency of rotation of blade i.e 13 rpm
f = 13 rpm=>= 13 / 60 revolution per second
ω = 2π f => 2π x 13 / 60 rad / s
L=
x ω =>106.48 x 10⁵ x 2π x 13 / 60
= 144.97 x 10⁵ kgm² s⁻¹