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Sophie [7]
3 years ago
6

Why is the value of g more in polar region than in equator ?​

Physics
1 answer:
LuckyWell [14K]3 years ago
3 0

likely due to the location.

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A stone is dropped into a well. The sound of the splash is heard 6.25 s later. What is the depth of the well? (Take the speed of
Naya [18.7K]

Answer:

depth of well is 163.30 m

Explanation:

Given data

speed of sound = 343 m/s

timer = 6.25 s

to find out

depth of well

solution

let us consider depth d

so equation will be

depth = 1/2 ×g ×t²    ..............1

and

depth = velocity of sound × time    .................2

here we have given time 6.25 that is sum of 2 time

when stone reach at bottom that time

another is sound reach us after stone strike on bottom

so time 1 + time 2 = 6.25 s

so from equation 1  and 2 we get

1/2 ×g ×t² = velocity of sound × time

1/2 ×9.8 × t1² = 343 × (6.25 - t1 )

t1 = 5.77376 sec

so height = 1/2 ×g ×t²

height = 1/2 ×9.8 × (5.773)²

height = 163.30 m

3 0
3 years ago
The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90
s2008m [1.1K]

Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

20.45 pC

Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

6 0
3 years ago
If someone throws a 3 gram fry accelerating at 5 meter/second2, what is the fry’s force?
Alex

Answer:

0.015 m/s2

Explanation:

Using Newtons 2nd law.

F = ma where F = Force applied, m = mass of the object and a = acceleration acquired.

So substitute the values in SI units.

m = \frac{3}{1000} kg

Therefore F = 0.003×5 = 0.015 m/s2

3 0
3 years ago
Define torque qualitatively and quantitatively.
abruzzese [7]
When you use a wrench to tighten or loosen a nut on a bolt, you are applying torque. It is measured in units of force times distance.  A force of F newtons pulling on a handle of L meters in length would supply a torque of F L newton-meters. More technically, torque is the vector cross product of force times perpendicular distance from the object, F x r = F r sin @
4 0
3 years ago
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shelly starts from rest on her bicycle at the top of a hill. After 6.0s she has reached a final velocity of 14m/s. what is shell
earnstyle [38]
Divide 14 by 6 and there is your answer with the unit of m
8 0
3 years ago
Read 2 more answers
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