A particle (q = +5.0 µC) is released from rest when it is 2.0 m from a charged particle which is held at

Bryan Allen pedaled a human-powered aircraft across the English channel from the cliffs of Dover to Cap Gris-Nez on June 12, 197

tigry1 [53]

Answer:

The answer is " $5.53 \ \frac{m}{s}$ "

Explanation:

apply the formula for calculating the average velocity to the relative air

$V_{PG} =V_{PA}+V_{AG}$

$\Rightarrow V_{PA} = V_{PG} -V_{AG}$

Given value:

$V_{AG} = -2 \ \frac{m}{s}$

$V_{PG} =3.53$

$\Rightarrow V_{PA} = 3.53 - (-2) \\\\\Rightarrow V_{PA} = 3.53 +2 \\\\\Rightarrow V_{PA} = 5.53 \\\\$

The final answer is " $5.53 \ \frac{m}{s}$ " in the south-east direction.

4 0

2 years ago

you are piloting a small plane and you want to reach an airport 450 km due south in 3.0 h a wind is blowing from the west 50.0 k

alex41 [277]

Answer:

You should choose airspeed 158.11 km/h at 18.4° west of south

Explanation:

The distance to the air port is 450 km due to south

You should to reach the airport in 3 hours

→ Velocity = distance ÷ time

→ Distance = 450 km , time = 3 hours

→ The velocity of your plane = 450 ÷ 3 = 150 km/h due to south

A wind is blowing from west 50 km/h

We need to know what heading and airspeed you should choose to

reach your destination

At first we must find the resultant velocity of your plane and the wind

The south and west are perpendicular, then the resultant velocity is

→ $v_{R}=\sqrt{(v_{p})^{2}+(v_{w})^{2}}$

→ $v_{p}=150$ km/h , $v_{w}=50$ km/h

→ $v_{R}=\sqrt{(150)^{2}+(50)^{2}}=158.11$ km/h

To cancel the velocity of the wind, the pilot should maintain the velocity

of the plane at 158.11 km/h

The direction of the velocity is the angle between the resultant velocity

and the vertical (south)

→ The direction of the velocity is $tan^{-1}\frac{50}{150}=18.4$ °

The direction of the velocity is 18.4° west of south

<em>You should choose airspeed 158.11 km/h at 18.4° west of south</em>

8 0

3 years ago

Formula for percentage error

GarryVolchara [31]

Answer:

PE = (|accepted value – experimental value| \ accepted value) x 100%

Explanation:

7 0

2 years ago

A ship anchored at sea is rocked by waves that have crests 14 m apart the waves travel at 7.0 m/s how often do the wave crest re

xeze [42]

Answer:

I think its 2 seconds

Explanation:

14/7

3 0

3 years ago

An electron with a speed of 0.95c is emitted by a supernova, where cc is the speed of light. What is the magnitude of the moment

krok68 [10]

Answer:

2.59×10¯²² Kgm/s

Explanation:

Data obtained from the question include:

Velocity of electron = 0.95c

Momentum =?

Next, we shall determine the velocity of the electron. This can be obtained as follow:

Velocity of electron = 0.95c

Velocity of Light (c) = 3×10⁸ m/s

Velocity of electron = 0.95c

Velocity of electron = 0.95 × 3×10⁸

Velocity of electron = 2.85×10⁸ m/s

Finally, we shall determine the mometum of the electron.

Momentum is simply defined as the product of mass and velocity. Mathematically, it is expressed as:

Momentum = mass x Velocity

Thus, with the above formula, we calculate the momentum of the electron as follow:

Mass of electron = 9.1×10¯³¹ Kg

Velocity of electron = 2.85×10⁸ m/s

Momentum of electron =?

Momentum = mass x Velocity

Momentum = 9.1×10¯³¹ × 2.85×10⁸

Momentum = 2.59×10¯²² Kgm/s

Therefore, the momentum of the electron is 2.59×10¯²² Kgm/s

3 0

3 years ago