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Georgia [21]
3 years ago
11

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

. At 11:00 pm the storm is at 75 mi north. What's the direction of the average velocity of the storm? And the magnitude of the average velocity of the storm?
Physics
1 answer:
Rus_ich [418]3 years ago
4 0
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
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3 years ago
Hi! Can somebody please help?
dezoksy [38]

Answer:

Diagram A will reach the top first.

Explanation:

If it is going straight, it will go slower. The higher the movement speed the faster it is. Hope this helps!

7 0
3 years ago
A 12 oz can of soda is left in a car on a hot day. In the morning the soda temperature was 60oF with a gauge pressure of 40 psi.
Neko [114]
In this case, volume of the can remains constant. The relationship between pressure and temperature at constant volume is given by:

P/T = Constant

Then
\frac{ P_{1} }{ T_{1} } = \frac{ P_{2} }{ T_{2} }

Where
P1 = 40 psi
P2 = ?
T1 = 60°F ≈ 289 K
T2 = 90°F ≈ 305 K (note, 363 K is not right)

Substituting;
P_{2} = \frac{ P_{1}  T_{2} }{ T_{1} } = \frac{40*305}{289}  =42.21 psi
3 0
3 years ago
Kathy 82 kg performer standing on a diving board at the carnival dive straight down into a small pool of water. Just before stri
mixas84 [53]

Solution :

Given weight of Kathy = 82 kg

Her speed before striking the water, $V_o $ = 5.50 m/s

Her speed after entering the water, $V_f$= 1.1 m/s

Time = 1.65 s

Using equation of impulse,

$dP = F \times  dT$

Here, F =  the force ,

       dT =  time interval over which the force is applied for

            = 1.65 s

       dP  = change in momentum

dP = m x dV

    $= m \times [V_f - V_o] $

    = 82 x (1.1 - 5.5)

    = -360 kg

∴ the net force acting will be

$F=\frac{dP}{dT}$

$F=\frac{-360}{1.65}$

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8 0
3 years ago
a wire with mass per unit length 75 g/m runs horizontally at right angles to a uniform horizontal 0.12 T magnetic field. what am
Sindrei [870]

Answer:

The amount of current that must flow through the wire for it to be suspended against gravity by magnetic force = 6.125 A

Explanation:

Force on a wire carrying current in an electric field is given by

F = (B)(I)(L) sin θ

For this question,

The magnetic force must match the weight of the wire.

F = mg

mg = (B)(I)(L) sin θ

(m/L)g = (B)(I) sin θ

Mass per unit length = 75 g/m = 0.075 kg/m

B = magnetic field = 0.12 T

I = ?

g = acceleration due to gravity = 9.8 m/s

θ = angle between wire's current direction and magnetic field = 90°

0.075 × 9.8 = 0.12 × I sin 90°

I = 0.075 × 9.8/0.12 = 6.125 A

3 0
3 years ago
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