Answer:
W = 8.01 × 10^(-17) [J]
Explanation:
To solve this problem we need to know the electron is a subatomic particle with a negative elementary electrical charge (-1,602 × 10-19 C), The expression to calculate the work is given by:
W = q*V
where:
q = charge = 1,602 × 10^(-19) [C]
V = voltage = 500 [V]
W = work [J]
W = 1,602 × 10^(-19) * 500
W = 8.01 × 10^(-17) [J]
Pluto revolves around a star instead of our SUN. So, it's an "Exo-planet" rather than a Planet. It has declared in 2006 by the scientists.
Hope this helps!
Answer:
Acceleration of the crate is 0.362 m/s^2.
Explanation:
Given:
Mass of the box, m = 40 kg
Applied force, F = 15 N
Angle at which the force is applied, 
= 15°
We have to find the magnitude of the acceleration.
Let the acceleration be "a".
FBD is attached with where we can see the horizontal and vertical component of force.
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and ⇒ 
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⇒ Applying concept of forces.
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⇒ 
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<em> ...Newtons second law Fnet = ma</em>
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⇒ Plugging the values.
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<em>...f is the friction which is zero here.</em>
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Magnitude of the acceleration of the crate is 0.362 m/s^2.
Answer: Socratic
Explanation: i’m not sure but you can use Socratic it’s a good app that helps
The net speed due west is = distance traveled in west / time taken = 120/0.5 = 240 km/h.
so airspeed due west is = net speed - speed of plane = 240-220= 20 km/h.
airspeed due south is = distance traveled in west / time taken= 20/0.5= 40 km/h.
the magnitude of the wind velocity = √[(airspeed due south )² + (airspeed due west)²] = √ ( 40^2 + 20^2 ) = 44.72 km/h
the angle of airspeed south of west is tan⁻¹ ( airspeed due south / airspeed due west )= tan⁻¹(40/20)=63.43 degrees.
if wind velocity is 40 km/h due south, her velocity should have 20 km/h component in north.
so component west = sqrt ( 220^2 - 40^2 ) = 216.33 km/h.
the angle north of west is arctan( 40/216.33 ) = 10.47 degrees.
