What is the formula for silver nitrate

Calculating Orbital Period The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. If HST has a tangential speed o

arsen [322]

The answer is, 5630 seconds.

5 0

2 years ago

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A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

1.

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

2.

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago

A 5.0 kg chunk of putty moving at 10m/s collides and sticks to a 7.0 kg bowling ball that is initially at rest.What is the total

Flura [38]

Answer:

Total momentum = 50kgm/s

Explanation:

Given the following data;

Mass, M1 = 5kg

Mass, M2 = 7kg

Velocity, V1 = 10m/s

Velocity, V2 = 0m/s (since it's at rest).

To find the total momentum;

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

$Momentum = Mass * Velocity$

The law of conservation of momentum states that the total linear momentum of any closed system would always remain constant with respect to time.

Total momentum = M1V1 + M2V2

Substituting into the equation, we have;

Total momentum = 5*10 + 7*0

Total momentum = 50 + 0

Total momentum = 50 kgm/s

Therefore, the total momentum of the bowling ball and the putty after they collide is 50 kgm/s.

8 0

2 years ago

Sam receives the kicked football on the 3 yd line and runs straight ahead toward the goal line before cutting to the right at th

Pie

Answer:

Distance: 21 yd, displacement: 15 yd, gain in the play: 12 yd

Explanation:

The distance travelled by Sam is just the sum of the length of each part of Sam's motion, regardless of the direction. Initially, Sam run from the 3 yd line to the 15 yd line, so (15-3)=12 yd. Then, he run also 9 yd to the right. Therefore, the total distance is

d = 12 + 9 = 21 yd

The displacement instead is a vector connecting the starting point with the final point of the motion. Sam run first 12 yd straight ahead and then 9 yd to the right; these two motions are perpendicular to each other, so we can find the displacement simply by using Pythagorean's theorem:

$d=\sqrt{12^2+9^2}=15 yd$

Finally, the yards gained by Sam in the play are simply given by the distance covered along the forward-backward direction only. Since Sam only run from the 3 yd line to the 15 yd line along this direction, then the gain in this play was

d = 15 - 3 = 12 yd

7 0

3 years ago

Tarzan, in one tree, sees Jane in another tree. He grabs the end of a vine with length 20m that makes an angle of 45° with the v

nadezda [96]

Answer:

yes

Explanation:

3 0

3 years ago