The magnitude of the normal force acting on the space vehicle will be 5070 newtons. So option (4) is correct.
Given that,
The mass of the space vehicle is 1300 kilograms
It travels at a speed of 4.8 meters per second along the level surface of mars.
The g' (gravitational field strength) on the surface of mars is 3.9 newtons per kilogram
We need to find the normal force acting on the space vehicle
We know, N = mg' (where N is normal force)
N = (1300 × 3.9) Newton
N = 5070 Newton
So, the magnitude of the normal force acting on the space vehicle that is traveling at 4.8 meters per second along the level surface of mars will be 5070 newtons.
The normal force is the force that surfaces exert to prevent solid objects from passing through each other. The normal force is one type of ground reaction force.
Learn more about the normal force here:
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Answer:
L = ¼ k g / m
Explanation:
This is an interesting exercise, in the first case the spring bounces under its own weight and in the second it oscillates under its own weight.
The first case angular velocity, spring mass system is
w₁² = k / m
The second case the angular velocity is
w₂² = L / g
They tell us
w₂ = ½ w₁
Let's replace and calculate
√ (L / g) = ½ √ (k / m)
L / g = ¼ k / m
L = ¼ k g / m
The compass magnet aligns itself with the earths magnetic field.
The magnitude of the E-field decreases as the square of the distance from the charge, just like gravity.
Location ' x ' is √(2² + 3²) = √13 m from the charge.
Location ' y ' is √ [ (-3)² + (-2)² ] = √13 m from the charge.
The magnitude of the E-field is the same at both locations.
The direction is also the same at both locations ... it points toward the origin.
Answer:
Explanation:
As we know that
also we know that
it is given as
also we can find the magnitude of two vectors as
similarly we have
now we know the formula of dot product as
