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Paladinen [302]

2 years ago

14

. A child drops a ball from a window. The ball strikes the ground in 3.0 seconds. What is the. A child drops a ball from a windo

w. The ball strikes the ground in 3.0 seconds. What is the
velocity of the ball the instant before it hits the ground?

1 answer:

Vsevolod [243]2 years ago

3 0

Answer:

30m/s

Explanation:

Use formula for height calculation: Time = √2h/√g
From here calculate height which comes out to be: 45m(if g is taken 10)
Then use formula for velocity calculation: v = √2gh

The velocity comes out to be 30m/s

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What is the acceleration of a 10 kg mass pushed by a 5 N force?

yanalaym [24]

$g-\ gravitational \ acceleration \\ g= \frac{G}{m} = \frac{5N}{10kg}= \\ g=0,5 \frac{N}{kg}$

8 0

3 years ago

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2. Four resistors of 2 ohm each are connected first in series and then in parallel with a battery.Find the ratio of electric cur

Serjik [45]

Answer:

Ratio of series current to parallel

= 1 : 8

Explanation:

Total resistance Rt

For series, Rt = 2+2+2+2 = 4ohms

For parallel, 1/Rt = 1/2 + 1/2 + 1/2 + 1/2

1/Rt = 4/2, Rt = 2/4 ohms.

If we use a 1V battery, then,

I = V/Rt

I = 1/4 = 0.25 ampere for series arrangement.

I = 1/0.5 = 2 ohms.

Ratio of current of series to parallel = 0.25 : 2

= 1 : 8

4 0

3 years ago

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Please please help

dsp73

Answer:

true

Explanation:

The law of conservation of charge states that whenever electrons are transferred between objects, the total charge remains the same.

3 0

3 years ago

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates

34kurt

The given question is incomplete. The complete question is as follows.

A parallel-plate capacitor has capacitance $C_{0}$ = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed $3.00 \times 10^{4}$ V/m?

Explanation:

It is known that relation between electric field and the voltage is as follows.

V = Ed

Now,

Q = CV

or, Q = $C \times Ed$

Therefore, substitute the values into the above formula as follows.

Q = $C \times Ed$

= $8.50 pF \times (\frac{10^{-12} F}{1 pF})(3 \times 10^{4} m/s)(1 mm)(\frac{10^{-3} m}{1 mm})$

= $2.55 \times 10^{-10} C$

Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is $2.55 \times 10^{-10} C$ .

3 0

3 years ago

Can someone help me?

Setler [38]

Answer: (1, 30), (2,10), (3,40), (4,20)

Explanation:

4 0

2 years ago