Fill in the blanks….

A meteorologist tracks the movement of a thunderstorm with Doppler radar. At 8:00pm the storm was 55 mi northeast of her station

Rus_ich [418]

At 8:00 pm, the velocity of the storm is 55 mi northeast. Assuming that the direction is exactly northeast, the angle is 45°
At 11:00 pm, the velocity is 75 mi north. The angle is 90°
In vector form
55 ∠ 45°
and
75 ∠ 90°
The magnitude and direction of the average velocity is
(55 ∠ 45° + 75 ∠ 90° ) / 3

4 0

2 years ago

The battery for a certain cell phone is rated at 3.70 V. According to the manufacturer it can produce math]3.15 \times 10^{4} J[

AysviL [449]

Answer:

The average current that this cell phone draws when turned on is 0.451 A.

Explanation:

Given;

voltage of the phone, V = 3.7 V

electrical energy of the phone battery, E = 3.15 x 10⁴ J

duration of battery energy, t = 5.25 h

The power the cell phone draws when turned on, is the rate of energy consumption, and this is calculated as follows;

$P = \frac{E}{t}$

where;

P is power in watts

E is energy in Joules

t is time in seconds

$P = \frac{3.15*10^4}{5.25*3600s} = 1.667 \ W$

The average current that this cell phone draws when turned on:

P = IV

$I = \frac{P}{V} =\frac{1.667}{3.7} = 0.451 \ A$

Therefore, the average current that this cell phone draws when turned on is 0.451 A.

5 0

3 years ago

3. The propeller of a World War II fighter plane is 2.30 m in diameter. (a) What is its angular velocity in radians per second i

Darya [45]

Answer:

(a) Angular velocity will be 125.6 rad/sec

(b) Linear velocity will be 144.44 m /sec

(c) Centripetal acceleration = 1849.3031 g

Explanation:

We have given diameter d = 2.30 m

So radius r = $\frac{d}{2}=\frac{2.30}{2}=1.15m$

(a) Speed is given as 1200 rev/min

We know that angular velocity is given by $\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 1200}{60}=125.6rad/sec$

(b) Linear speed is given by $v=\omega r=125.6\times 1.15=144.44m/sec$

(c) Centripetal acceleration is given by $a_c=\frac{v^2}{r}=\frac{144.44^2}{1.15}=18141.664m/sec^2$

We know that $g=9.81m/sec^2$

So $18141.66m/sec^2=\frac{18141.664}{9.81}=1849.3031g$

6 0

2 years ago

How do you find the voltage of a section of a parallel circuit?

sergiy2304 [10]

Answer: Voltage is the same across each component of the parallel circuit. The sum of the currents through each path is equal to the total current that flows from the source. You can find total resistance in a Parallel circuit with the following formula: 1/Rt = 1/R1 + 1/R2 + 1/R3 +.

Hope this helps!

3 0

3 years ago

spotlight on a boat is 2.5 m above the water, and the light strikes the water at a point that is 8.0 m horizontally displaced fr

ludmilkaskok [199]

Answer:

Explanation:

Let i be the angle of incidence and r be the angle of refraction .

From the figure

Tan ( 90 - i ) = 2.5 / 8

cot i = 2.5 / 8

Tan i = 8 / 2.5 = 3.2

i = 72.65°

From snell's law

sini / sin r = refractive index

sin 72.65 / sinr = 1.333

sin r = .9545 / 1.333

= .72

r = 46⁰

From the figure

Tan r = d / 4

Tan 46 = d /4

d = 4 x Tan 46

= 4 x 1.0355

=4.14 m .

3 0

3 years ago