Why would you want a gold ring rather than a copper ring? (I know gold is more valuable)

A bowling ball has an initial momentum of +30 kg m/s and hits a stationary bowling pin. After the collision, the bowling ball le

dangina [55]

Momentum should be conserved. The momentum of both objects must balance with their initial and final momentum.

Let m1 and v1 be the mass and velocity of the bowling ball

And m2 and v2 be the mass and velocity of the bowling pin

(m1v1)i + (m2v2)i = (m1v1)f + (m2v2)f

30 kg m/s + (1.5 kg)(0 m/s) = 13kg m/s + 1.5v2f

V2f = 11.33 m/s

So the momentum = 1.5 kg(11.33 m/s) = 17 kg m/s

8 0

3 years ago

How close would you have to bring 1 C of positive chargeand 1 C of negative charge for them to exert forces of 1 N onone another

Harman [31]

Answer:

94,800 m

Explanation:

F = kq1 q2/r^2

1 = 9 x 10^9 x 1 / r^2

4 0

2 years ago

Identify the prefix that would be used to express 2,000,000,000 bytes of computer memory?

Reika [66]

Answer:

It would be 2 GB

4 0

3 years ago

Read 2 more answers

A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet

lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

Current flowing in the wire, I = 4.00 mA
Initial diameter of the wire, d₁ = 4 mm = 0.004 m
Final diameter of the wire, d₂ = 1 mm = 0.001 m
Length of wire, L = 2.00 m
Density of electron in the copper, n = 8.5 x 10²⁸ /m³

The initial area of the copper wire;

$A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2$

The final area of the copper wire;

$A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2$

The initial drift velocity of the electrons is calculated as;

$v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s$

The final drift velocity of the electrons is calculated as;

$v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7} \ m/s$

The change in the mean drift velocity is calculated as;

$\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s$

The time of motion of electrons for the initial wire diameter is calculated as;

$t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s$

The time of motion of electrons for the final wire diameter is calculated as;

$t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s$

The average acceleration of the electrons is calculated as;

$a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2$

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0

3 years ago

EARTH SCIENCE PLEASE ANSWER

Rzqust [24]

The Earth Science answers are shown below.

Explanation:

1. The movement of the sun will change the angle it has on the sky in 30 minutes, it is always moving from the east to the west, so in 30 minutes it would move more west, no matter at what time you make the experiment. From Earth, the Sun looks like it moves across the sky in the daytime and appears to disappear at night. This is because the Earth is spinning towards the east. The Earth spins about its axis, an imaginary line that runs through the middle of the Earth between the North and South poles

2. No, both marks are the same distance from the ground. the amount of stick above the mark will not affect the distance that the shadow of the mark moves at all. The Sun's clockwise motion is an apparent motion caused by the rotation of the Earth. The counterclockwise rotation of the Earth in the Sun's light causes the shadow of the gnomon to move clockwise. As the Sun appears to move higher above the horizon before solar noon, the shadow grows shorter and shorter.

3. In the summer the shadows are shorter, and in the winter the shadows are longer. In the morning your shadow will point west and in the afternoon it will point east. If your shadow is long, it is near sunrise or sunset. Your shadow is shortest around noon.

4. If the sun rises in the east and sets in the west, then the Earth should rotate in the opposite direction from west to east (anti-clockwise). Earth's spin (or rotation) on its axis. Earth rotates or spins toward the east, and that's why the Sun, Moon, planets, and stars all rise in the east and make their way westward across the sky.

4 0

4 years ago