All categories

3 years ago

HELPPP ME PLEASE!!!!

Physics

2 answers:

Leya [2.2K]3 years ago

6 0

The answer is c my drilla

andriy [413]3 years ago

3 0

Answer:

Explanation:

You might be interested in

Three features describe the motion of an object. they are...

jasenka [17]

I would think B, Position, direction and speed. I tried looking it up and B seems to be the closest answer that resembles my results as I found that energy didn’t describe the motion of an object.

6 0

3 years ago

Read 2 more answers

In an early attempt to understand atomic structure, Niels Bohr modeled the hydrogen atom as an electron in uniform circular moti

AleksandrR [38]

Answer:

The answer is below

Explanation:

Using Coulomb's law of electric field which is:

$F=k\frac{q_1q_1}{r^2}\\\\ k =constant=9*10^9\ Nm^2/C^2,q_1=q_2=1.6*10^{-19}C,r=5.29*10^{-11}m\\\\Substituting\ gives:\\\\F=9*10^9*\frac{(1.6*10^{-19})*(1.6*10^{-19})}{(5.29*10^{-11})^2} =8.22*10^{-8}N\\\\Both\ centripetal\ force\ is\ given\ by:\\\\F=m\frac{v^2}{r} \\\\m = mass\ of \ electron=9.11*10^{-31}g,v=speed\ of\ electron\\\\F=m\frac{v^2}{r} \\\\v=\sqrt{\frac{F*r}{m} } \\\\subsituting:\\\\v=\sqrt{\frac{8.22*10^{-8}*5.29*10^{-11}}{9.11*10^{-31}} } \\\\v=2.18*10^6\ m/s\\\\$

$But\ \omega=\frac{v}{r}=\frac{2.18*10^6}{5.29*10^{-11}} =4.13*10^{17}\\\\\omega=2\pi f; f=frequency\\\\f=\frac{\omega}{2\pi} =\frac{4.13*10^{17}}{2\pi} \\\\f=6.57*10^{15}\ Hz$

6 0

3 years ago

(a) In the baggage claim area of an airport, a particular baggage carousel is shaped like a section of a large cone, steadily ro

Tom [10]

Answer:

Part a)

$F_f = 107.8 N$

Part b)

$\mu = 0.415$

Explanation:

Part a)

Time period of one revolution is given as

T = 42 s

now the angular speed of the belt is given as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{42}$

$\omega = 0.15 rad/s$

Now at rest position the force along the surface of carousel must be constant

so we will have

$mgsin\theta + m\omega^2 r cos\theta = F_f$

$30(9.81)sin20.5 + 30(0.15^2)(7.46)cos20.5 = F_f$

$F_f = 107.8 N$

Part b)

New Time period of one revolution is given as

T = 30 s

now the angular speed of the belt is given as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{30}$

$\omega = 0.21 rad/s$

Now at rest position the force along the surface of carousel must be constant

so we will have

$mgsin\theta + m\omega^2 r cos\theta = F_f$

$30(9.81)sin20.5 + 30(0.21^2)(7.94)cos20.5 = F_f$

$F_f = 112.9 N$

Also we know that in perpendicular direction also force is balanced

$F_n + m\omega^2 r sin\theta = mgcos\theta$

$F_n = mgcos\theta - m\omega^2 r sin\theta$

$F_n = 30(9.81)cos20.5 - 30(0.21)^2(7.94)sin20.5$

$F_n = 272 N$

now for friction coefficient we will have

$F_f = \mu F_n$

$112.9 = \mu 272$

$\mu = 0.415$

8 0

3 years ago

A ball rolls onto the path of your car as you drive down a quiet neighborhood street. To avoid hitting the child that runs to re

ehidna [41]

Answer:

F = 7.143 kN

Explanation:

given,

time taken to apply break = 1.05 s

car slows down from 15 m/s to 9 m/s

mass of the car = 1250 Kg

force is equal to the change in momentum with respect to time.

$F = \dfrac{\Delta P}{t}$

$F = \dfrac{m(v_f - v_i)}{t}$

$F = \dfrac{1250\times (9 - 15)}{1.05}$

$F = \dfrac{-7500}{1.05}$

F = -7142.85 N

F = - 7.143 kN

Force is acting opposite direction of velocity of car i.e. the sign is negative.

7 0

3 years ago

In the data table , distance is measured in meters and time is in seconds. Calculate the mans average velocity using the equatio

Artist 52 [7]

Answer:

3.626 m/s

Explanation:

v=d/t

1. -0.02/0 = 0 m/s

2. 0.86/0.2 = 4.3 m/s

3. 1.71/0.4 = 4.275 m/s

4. 2.54/0.6 = 4.23 m/s

5. 3.32/0.8 = 4.15 m/s

6. 4.08/1.0 = 4.08 m/s

7. 4.79/1.2 = 3.99 m/s

8. 5.48/1.4 = 3.91 m/s

9. 6.15/1.6 = 3.84 m/s

10. 6.76/1.8 = 3.76 m/s

11. 7.37/2.0 = 3.66 m/s

12. 7.92/2.2 = 3.6 m/s

13. 8.45/2.4 = 3.52 m/s

14. 8.96/2.6 = 3.45 m/s

the mean of these numbers is 3.626

his average velocity ks 3.626 m/s

6 0

3 years ago