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Ymorist [56]
4 years ago
6

Why is the climate of nepalgunj and namchebazar different though they lie at the same. latitude​

Physics
1 answer:
svet-max [94.6K]4 years ago
4 0

Answer:

Though nepalgunj and taplegunj lie at the same latitude(28 degree) they greatly vary In climate..

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A powerful motorcycle can produce an acceleration of 3.00 m/s2 while traveling at 90.0 km/h. At that speed the forces resisting
nata0808 [166]

Answer:

1185 N

Explanation:

From Newton’s second law of motion,  

F=ma where m= mass of motorcycle, a is acceleration of the motorcycle and F=Force

Net force acting on motorcycle F_{net} is given by  F_{net}=F-f

Where F is force acting on motorcycle and f is frictional force

Substituting F-f for F_{net}

F_{net}=ma hence  ma= F- f Substituting a with 3, m with 245Kg and f with 450N as provided

245*3= F- 450

F=245*3 +450= 1185 N

6 0
4 years ago
Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d
Step2247 [10]
Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that F = k \frac{Q_1Q_2}{d^2} so if we double the charge on Q_1 and double the distance to (2d) we plug these into the equation to find

<span>F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}</span>

So we see the new force is exactly 1/2 of the old force so your answer should be \frac{1}{2}F if I can remember my physics correctly.

9 0
4 years ago
Read 2 more answers
Two parallel-plate capacitors, identical except that one has twice the plate separation of the other, are charged by the same vo
erik [133]

Answer:

The capacitor having less distance of separation has a stronger electric field.

Explanation:

The capacitors are identical and only difference between them is that one has twice the plate separation of the other. Therefore, capacitance of the given capacitors C1 and C2 is,

C1= Aε/d  and C2=Aε/2d

The charges Q1 and Q2 on the capacitors of capacitance C1 and C2 respectively, is then given by the equation,

Q1=VC1

Q1=VAε/d

Q2=VC2

Q2=VAε/2d

Therefore, the surface charge density σ1 and σ2 for the capacitors is,

σ1=Q1/A

σ1=VAε/(d*A)

σ1=Vε/d

Similarly,

σ2=Q2/A

σ2=Vε/2d

The electric field between the plates is directly proportional to the surface charge density. And so electric field is inversely proportional to the distance of separation. Therefore the capacitor whose distance of separation is less has a stronger electric field.

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8 0
2 years ago
Can any one here form light from lemon?​
sergey [27]

Answer:

I did not get it

Explanation:

what u r saying

8 0
3 years ago
Read 2 more answers
In a setup like that in Figure 27.7, a wavelength of 625 nm is used in a Young's double-slit experiment. The separation between
kap26 [50]

The separation between the slits is d = 8.96

What is fringe width?

  • Fringe width is the distance between two consecutive bright spots (maximas, where constructive interference take place)
  • Or two consecutive dark spots (minimas, where destructive interference take place).

Fringe width is given by β = λL/d

In the first case fringe width is β1 = λLA /d   = 625 x 10-9 x 0.36 / ( 1.4 x 10-5 )  = 0.016071428 m

The total width of the screen is 0.2 m . So, on one side of the central maximum, the width is 0.1 m

No. of fringes in this 0.1m = 0.1 / 0.016071428  = 6.222  

So, since there is a bright fringe after every fringe width, the number of bright fringes on one side of central maximum is 6.

In the second case fringe width is β1 = λLAB /d   = 625 x 10-9 x 0.25 / ( 1.4 x 10-5 )  = 0.011160714 m

The total width of the screen is 0.2 m . So, on one side of the central maximum, the width is 0.1 m

No. of fringes in this 0.1m = 0.1 / 0.011160714  = 8.96

So, since there is a bright fringe after every fringe width, the number of bright fringes on one side of central maximum is 8.  The ninth one will not be seen since the screen is less a little less in width.

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<u>The complete question is -</u>

In a setup like that in Figure 27.7, a wavelength of 625 nm is used in a Young's double-slit experiment. The separation between the slits is d = 1.4 × 10-3 m. The total width of the screen is 0.20 m. In one version of the setup, the separation between the double slit and the screen is LA = 0.36 m, whereas in another version it is LB = 0.25 m. On one side of the central bright fringe, how many bright fringes lie on the screen in the two versions of the setup? Do not include the central bright fringe in your counting. --Tm = 3 (Bright fringe) ++m = 0 (Bright fringe) -m = 3 (Bright fringe) Figure 27.7

5 0
1 year ago
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