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Maksim231197 [3]
2 years ago
7

Two pebbles, Pebble A and Pebble B, are thrown horizontally with the same force. Pebble A's mass is 3 times the

Physics
1 answer:
Hitman42 [59]2 years ago
5 0

Answer:

Pebble A has 1/3 the acceleration as pebble B.

Explanation:

F = m×a

mass of a = 3 × mass of b (m_a = 3 × m_b)

Same starting force, F

m_a = mass of a

m_b = mass of b

a_a = acceleration of a

a_b = acceleration of b

F = m_a × a_a = m_b × a_b

3 × m_b × a_a = m_b × a_b

3 × a_a = a_b

OR

a_a = a_b / 3

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2. A pebble is dropped down a well and hits the water 1.5 s later. Using the equations for motion with constant acceleration, de
kow [346]

Answer:

S = 11.025 m

Explanation:

Given,

The time taken by the pebble to hit the water surface is, t = 1.5 s

Acceleration due to gravity, g = 9.8 m/s²

Using the II equations of motion

                          S = ut + 1/2 gt²

Here u is the initial velocity of the pebble. Since it is free-fall, the initial velocity

                                u = 0

Therefore, the equation becomes

                            S = 1/2 gt²

Substituting the given values in the above equation

                              S = 0.5 x 9.8 x 1.5²

                                 = 11.025 m

Hence, the distance from the edge of the well to the water's surface is, S = 11.025 m

3 0
3 years ago
Helppp<br> In Static Electricity, The charges do not ____
Anna11 [10]

Answer:

basically I will tell you the definition

Explanation:

so when charges are unbalanced statistic energy is formed positive attract negative and negative attracts positive like repell while unlike attract .

5 0
3 years ago
A 1.40 mH inductor and a 1.00 µF capacitor are connected in series. The current in the circuit is described by I = 14.0 t, where
4vir4ik [10]

Answer:

Explanation:

Inductance L = 1.4 x 10⁻³ H

Capacitance C = 1 x 10⁻⁶ F

a )

current I = 14 .0 t

dI / dt  = 14

voltage across inductor

= L dI / dt

= 1.4 x 10⁻³ x 14

= 19.6 x 10⁻³ V

= 19.6 mV

It does not depend upon time because it is constant at 19.6 mV.

b )

Voltage across capacitor

V = ∫ dq / C

= 1 / C ∫ I dt  

= 1 / C ∫ 14 t dt

1 / C x 14 t² / 2

= 7 t² / C

= 7 t² / 1 x 10⁻⁶

c ) Let after time t energy stored in capacitor becomes equal the energy stored in capacitance

energy stored in inductor

= 1/2 L I²

energy stored in capacitor

= 1/2 CV²

After time t

1/2 L I² = 1/2 CV²

L I² =  CV²

L x ( 14 t )² = C x  ( 7 t² / C )²

L x 196 t² = 49 t⁴ / C

t² = CL x 196 / 49

t = 74.8 μ s

After 74.8 μ s energy stored in capacitor exceeds that of inductor.

7 0
3 years ago
76.8 kPa to mm of Hg
Bumek [7]

Answer:

The answer is 576.0473

Explanation:

Hope this helps.

Please mark my answer as brainliest?

3 0
3 years ago
What are some of the challenges of scuba diving?
EastWind [94]
Some of the challenges are the unpredictable fish and the risk of scratching againest coral or drowning for not focusing on your oxygen tank.
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