Which of these is closest to the age of our solar system?

Imagine that the ball on the left is given a nonzero initial velocity in the horizontal direction, while the ball on the right c

masya89 [10]

Answer:

vₓ = xg/2y

Explanation:

In this question, let us find the time it takes for the ball on the right that has zero initial velocity to reach the ground.

By newton equation of motion we know that

y = v₀ t - ½ g t²

t = 2y / g

This is the time it takes for the ball on the right to reach the ground; at this time the ball on the left travels a distance

vₓ = x/t

vₓ = xg/2y

Where we assume that x and y are known.

7 0

3 years ago

PLEASE HELP ASAP!!!!!

Igoryamba

While falling, both the sheet of paper and the paper ball experience air resistance. But the surface area of the sheet is much more than that of the spherical ball. And air resistance varies directly with surface area. Hence the sheet experiences more air resistance than the ball and it falls more slowly than the paper ball.

Hope that helps!

8 0

3 years ago

Nresistors, each having resistance equal to 1 2, are arranged in a circuit first in series and then in parallel. What is the rat

aleksandr82 [10.1K]

Answer:

option (b)

Explanation:

Let the resistance of each resistor is R.

In series combination,

The effective resistance is Rs.

rs = r + R + R + .... + n times = NR

Let V be the source of potential difference.

Power in series

Ps = v^2 / Rs = V^2 / NR ..... (1)

In parallel combination

the effective resistance is Rp

1 / Rp = 1 / R + 1 / R + .... + N times

1 / Rp = N / R

Rp = R / N

Power is parallel

Rp = v^2 / Rp = N V^2 / R ..... (2)

Divide equation (1) by equation (2) we get

Ps / Pp = 1 / N^2

5 0

3 years ago

Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm

Burka [1]

Answer:

The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

Explanation:

Given that,

Atmospheric pressure $P_{atm}= 1.013\times10^{5}\ Pa$

drop height h'= 27.1 mm

Density of mercury $\rho= 13.59 g/cm^3$

We need to calculate the height

Using formula of pressure

$p = \rho g h$

Put the value into the formula

$1.013\times10^{5}=13.59\times10^{3}\times9.8\times h$

$h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}$

$h=0.76\ m$

We need to calculate the new height

$h''=h - h'$

$h''=0.76-27.1\times10^{-3}$

$h''=0.76-0.027$

$h''=0.733\ m$

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

$P=\rho g h$

Put the value into the formula

$P= 13.59\times10^{3}\times9.8\times0.733$

$P=0.97622\times10^{5}\ Pa$

Hence, The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

7 0

3 years ago

Earth’s polar ice caps contain about 2.3 × 1019 kg of ice. This mass contributes essentially nothing to the moment of inertia of

sp2606 [1]

Answer:

Explanation:

Initial moment of inertia of the earth I₁ = 2/5 MR² , M is mss of the earth and R is the radius . If ice melts , it forms an equivalent shell of mass 2.3 x 10¹⁹ Kg

Final moment of inertia I₂ = 2/5 M R² + 2/3 x 2.3 x 10¹⁹ x R²

For change in period of rotation we shall apply conservation of angular momentum law

I₁ ω₁ = I₂ ω₂ , ω₁ and ω₂ are angular velocities initially and finally .

I₁ / I₂ = ω₂ / ω₁

I₁ / I₂ = T₁ / T₂ , T₁ , T₂ are time period initially and finally .

T₂ / T₁ = I₂ / I₁

(2/5 M R² + 2/3 x 2.3 x 10¹⁹ x R²) / 2/5 MR²

1 + 5 / 3 x 2.3 x 10¹⁹ / M

= 1 + 5 / 3 x 2.3 x 10¹⁹ / 5.97 x 10²⁴

= 1 + .0000064

T₂ = 24 (1 + .0000064)

= 24 hours + .55 s

change in length of the day = .55 s .

3 0

3 years ago