All categories

2 years ago

Where would you feel the most motion?

Physics

1 answer:

Mrac [35]2 years ago

4 0

Answer:

Explanation:

since its the tip, you will feel it the most there.

You might be interested in

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]

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

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

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

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

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 roller coaster car rapidly picks up speed as it

TiliK225 [7]

Acceleration = (change of speed) / (time for the change)

Change in speed = (22 - 4) = 18 m/s.
Time for the change = 3 sec.

Acceleration = 18/3 = 6 m/s per second.

8 0

2 years ago

A converging-diverging nozzle has a throat area of 10 cm2 and an exit area of 28.96 cm2 . A normal shock stands in the exit when

Svetllana [295]

The tank pressure is 5.08 kPa and the mass flow rate is 2.6 kg/s.

The given parameters:

Throat area of the nozzle, $A^*$ = 10 cm² = 0.001 m²
The exit area of the nozzle, A = 28.96 cm² = 0.002896 m²
Air pressure at sea level = 101.325 kPa

The ratio of the areas of the converging-diverging nozzle is calculated as follows;

$= \frac{A}{A^*} \\\\= \frac{0.002896}{0.001} \\\\= 2.896$

From supersonic isentropic table, at $\frac{A}{A^*} = 2.896$ , we can determine the following;

$M_e = 2.6 \ kg/s\\\\\frac{P_o}{P_e} = 19.954$

The tank pressure is calculated as follows;

$\frac{P_o}{P_e} = 19.954 \\\\P_e = \frac{P_o}{19.954} \\\\P_e = \frac{101.325 \ kPa}{19.954} \\\\P_e = 5.08 \ kPa$

Thus, the tank pressure is 5.08 kPa and the mass flow rate is 2.6 kg/s.

Learn more about converging-diverging nozzle design here: brainly.com/question/13889483

8 0

2 years ago

Use the crisscross method to find the chemical formula for the ionic compound formed by strontium (Sr) and bromine (Br).

nignag [31]

Answer:it’s c SrBr2

Explanation:

8 0

2 years ago

Read 2 more answers

A ball is dropped from 8.5 meters above the ground. If it begins at rest, how long does it take to hit the ground?

Anna35 [415]

Answer:

Explanation:

Givens

d = 8.5 meters

vi = 0

a = 9.81

t = ?

Formula

d = vi * t + 1/2 a t^2

Solution

8.5 = 0 + 1/2 9.81 * t^2 multiply both sides by 2

8.5 = 4.095 t^2 Divide both sides by 4.095

8.5/4.095 = t^2

1.7329 = t^2 Take the square root of both sides

t = 1.316

It takes 1.316 seconds to hit the ground.

6 0

2 years ago

Where would you feel the most motion?​

Where would you feel the most motion?