Which of the following is NOT an example of a girl on a bike who is accelerating?

What are the magnitude and direction of the acceleration of an electron at a point where the electric field has magnitude 7400 N

bezimeni [28]

Answer: a = 1.32 * 10^18m/s² due north

Explanation: The magnitude of the force required to move the electron is given as

F = ma

The force exerted on the charge by the electric field of intensity (E) is given by

F = Eq

Thus

Eq = ma

a = E * q/ m

Where a = acceleration of charge

E = strength of electric field = 7400N/c

q = magnitude of electronic charge = 1.609 * 10^-6c

m = mass of an electronic charge = 9.109 * 10^-31kg

a = 7400 * 1.609 * 10^-16/ 9.109 * 10^-31

a = 11906.6 * 10^-16 / 9.019 * 10^-31

a = 1.19 * 10^-12 / 9.019 * 10^-31

a = 0.132 * 10^19

a = 1.32 * 10^18m/s²

As stated in the question, the direction of the electric field is due north hence, the direction of it force will also be north thus making the electron experience a force due north ( according to Newton second law of motion)

5 0

3 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

During the first 50 s a truck traveled at constant speed of 25 m/s. Find the distance that it is traveled.

Allushta [10]

Time=50s
speed=25m/s

Distance = speed×time
=25×50
=1250m

DISTANCE TRAVELLED IS =1250m

6 0

3 years ago

How is velocity different from speed? Based on distance and/or direction

Mnenie [13.5K]

Answer:

Speed is solved with time and distance but has no direction

Average velocity is solved with Δx/Δt and has a direction

6 0

2 years ago

A red train travelling at 72 km/h and a green train travelling at 144 km/h are headed toward each

abruzzese [7]

Answer:

Collision will occur.

Speed of red train when they collide = 0 m/s.

Speed of green train when they collide = 10 m/s.

Explanation:

Speed of red train = 72 km/h = 20 m/s

Speed of green train = 144 km/h = 40 m/s.

Deceleration of trains = 1 m/s²

For red train:-

Equation of motion v = u + at

u = 20 m/s

v = 0 m/s

a = -1 m/s²

Substituting

0 = 20 - 1 x t

t = 20 s.

Equation of motion s = ut + 0.5at²

u = 20 m/s

t = 20 s

a = -1 m/s²

Substituting

s = 20 x 20 - 0.5 x 1 x 20² = 200 m

So red train travel 200 m before coming to stop.

For green train:-

Equation of motion v = u + at

u = 40 m/s

v = 0 m/s

a = -1 m/s²

Substituting

0 = 40 - 1 x t

t = 40 s.

Equation of motion s = ut + 0.5at²

u = 40 m/s

t = 40 s

a = -1 m/s²

Substituting

s = 40 x 40 - 0.5 x 1 x 40² = 800 m

So green train travel 800 m before coming to stop.

Total distance traveled = 800 + 200 = 1000 m>950 m.

So both trains collide.

Distance traveled by green train when red train stops(t=20s)

Equation of motion s = ut + 0.5at²

u = 40 m/s

t = 20 s

a = -1 m/s²

Substituting

s = 40 x 20 - 0.5 x 1 x 20² = 600 m

Total distance after 20 s = 600 + 200 = 800 m< 950m . So they collide after red train stops.

Speed of red train when they collide = 0 m/s.

Distance traveled by green train when they collide = 950 - 200 = 750 m

Equation of motion v² = u² + 2as

u = 40 m/s

s= 750 m

a = -1 m/s²

Substituting

v² = 40² - 2 x 1 x 750 = 100

v = 10 m/s

Speed of green train when they collide = 10 m/s.

6 0

3 years ago