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avatar+1832 

I need help in these questions in details 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

physics
 Aug 22, 2014

Best Answer 

 #7
avatar+32822 
+10

xvxvxv said: 

"alan .. for Q 2 The answer in the model answer is A .. so the area in part D is positive !"

No! The question asks for net displacement not total distance travelled.  When the velocity is negative the player is moving in the opposite direction to when the velocity is positive, so he is reducing his displacement.  If the model answer is A then it is wrong!

(Total distance travelled would be 20m).

 Aug 23, 2014
 #1
avatar+121078 
+10

I'm not well-versed in Physics, but I thimk I can do 1 and 3 for you.

1)  The first part of the trip took 6mi/(30 m/h) = 1/5 hr. And the second part of the trip took 6 mi / (60m/h)= 1/10hr.   So the total time was   (1/5 + 1/10) hr  = 15/50 hr  = 3/10 hr. And the average rate = D / Total time = 12mi / (3/10)hr  = 120/3 mph = 40 mph.

3) We have the height at any time is given by

h = -4.9t^2 + 14t        ...... and we're assuming that the object's intial position is h = 0.....so we have

5 = -4.9t^2 + 14t    and rearranging, we have

-4.9t^2 + 14t -  5 = 0

And using the on-site calculator to solve this, we have

$${\mathtt{\,-\,}}{\mathtt{4.9}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{14}}{\mathtt{\,\times\,}}{\mathtt{t}}{\mathtt{\,-\,}}{\mathtt{5}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,-\,}}{\mathtt{10}}\right)}{{\mathtt{7}}}}\\
{\mathtt{t}} = {\frac{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{10}}\right)}{{\mathtt{7}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{0.418\: \!418\: \!884\: \!019\: \!217\: \!8}}\\
{\mathtt{t}} = {\mathtt{2.438\: \!723\: \!973\: \!123\: \!639\: \!3}}\\
\end{array} \right\}$$

The object is at the 5m level on the descent at about 2.4 sec.

 Aug 22, 2014
 #2
avatar+330 
+10

(Q-11)

This is a potential energy calculation. The energy is the same whether this is occurs in 4 seconds or 4 minutes. (Note: there is a difference in power because of this).

Here a mass is moved 0.8 meters vertically against a force of 9.81m/s2. Potential Energy = Mass x (Force of Gravity per Kg) x vertical distance 20Kg * (9.81m/s2) * 0.8M

$$E = m \times g \times h = 20 \textrm{ kg} \times 9.81 \frac{\textrm{N}}{\textrm{kg}} \times h$$

$$E = 20 \textrm{ kg} \times 9.81 \frac{\textrm{N}}{\textrm{kg}} \times 0.8m \ =\ 156.96J \\$$

---------------

(Q-21)

The answer is A.

------------

 

(Q-22) Which shown arrows represents correctly the electric field at point p.

The answer is A. Electric fields are conventionally shown as moving to a negative charge.

 

--------------

(Q-30)

Magnetic field = Permeability * Turn density * Current

Current = (Magnetic field) / (Permeability * Turn density)

The relative permeability here is assumed to be (1). (This is used as a multiplier of absolute permeability).  

Turn density = The number of “loops” of wire uniformly distributed over a length of 1 meter.

110turns/0.5Meters = 220turns/meter

Magnetic field in milliTesla’s : 2.5mT = 0.002500 T

The magnetic field inside a solenoid is proportional to the applied current and the number of turns per unit length. The field inside is considered constant and independent of the diameter of the solenoid.

The magnetic field only depends on the current (I = amps), the number of turns per unit length (N/L), and the permeability of the core (mu0). 

B= mu0(N/L) * I mu0 is  Permeability in a vacuum (or air)

I=B/(mu0(N/L)

 

$$I = \frac{B}{mu_0 \frac {N}{L}}$$

 

$$I = \frac{B}{220 *\mu_0} \ = \ 9.04 A\\$$

 

Where

 

$$\\ \mu_0 = \ 4\pi \* 10^-7} \ H M^-^1$$

 


Yields Tesla measure of the ratio between enclosed magnetic field and the current.

 

 0.002500 T corresponds to 9.04 Amps. (Answer D).

 

~~D~~

 Aug 23, 2014
 #3
avatar+32822 
+10

For question 2) you need to find the area under the curve as distance is the integral of velocity with respect to time. The areas of each section are easy to find as they are comprised of triangles and rectangles.  The only point to note is that the last section will contribute a negative area.  (You should find B to be the answer)

 Aug 23, 2014
 #4
avatar+1832 
0

alan .. for Q 2 The answer in the model answer is A .. so the area in part D is positive ! 

because we find area not integral ... right ! 

 Aug 23, 2014
 #5
avatar+1832 
0

CPhill ... for Q 1 the answer in the model answer is    A 35mph  !!  

 Aug 23, 2014
 #6
avatar+1832 
0

CPhill  ... for Q 3 

 

I answerd like this 

 

 

 

 Aug 23, 2014
 #7
avatar+32822 
+10
Best Answer

xvxvxv said: 

"alan .. for Q 2 The answer in the model answer is A .. so the area in part D is positive !"

No! The question asks for net displacement not total distance travelled.  When the velocity is negative the player is moving in the opposite direction to when the velocity is positive, so he is reducing his displacement.  If the model answer is A then it is wrong!

(Total distance travelled would be 20m).

Alan Aug 23, 2014
 #8
avatar+1832 
+5

Thanks for all ...  CPhill  ... DavidQD ... Alan 

 Aug 24, 2014

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