The speed change : Δv = 0.41 m/s
<h3>Further explanation</h3>
Given
mass = 5.5 kg
Force = 15 N
time = 0.15 s
Required
the speed change
Solution
Newton 2nd's law
Impulse and momentum
F = m.a
F = m . Δv/t
F.t = m.Δv
Input the value :
15 N x 0.15 s = 5.5 kg x Δv
Δv = 0.41 m/s
Work in general is given by W=F·d where F is the force vector and d is the displacement vector. The dot symbol is the dot product which is a measure of how parallel two vectors are. It can be replaced by the cosine of the angle between the two vectors and the vectors replaced by their magnitudes. If F and d are parallel then the angle is zero and the cosine is unity. So in this case work can be defined as the product of the magnitudes of the force and distance:
W=Fd
Answer:
125.83672 seconds
Explanation:
P = Power of the horse = 1 hp = 746 W (as it is not given we have assumed the horse has the power of 1 hp)
m = Mass of professor = 103 kg
g = Acceleration due to gravity = 9.8 m/s²
h = Height of professor = 93 m
Work done would be equal to the potential energy
Power is given by
The time taken by the horse to pull the professor is 125.83672 seconds
Answer:
Explanation:
You pull a sled exerting a 50 N force on it , sled also exerts a force on you . These forces are action and reaction force , as per third law of Newton . These two forces are equal and opposite . But they do not act on the same object so they do not cancel each other . They act on different objects , one on the sledge and the other on you . Due to force on sledge , sledge moves in the direction of force or towards you . You will start moving in opposite direction if frictional force of ground is nil or less .
Answer:
the charge per unit area on the plastic sheet is - 3.23 x 10⁻⁷ C/m²
Explanation:
given information:
styrofoam mass, m = 16 g = 0.016 kg
net charge, q = - 8.6 μC
to calculate the charge per unit area on the plastic sheet, we can use the following equation:
where
the force between the electric field
m = mass
g = gravitational force
where
q = charge
E = electric field
and
E = σ/2ε₀
where
ε₀ = permitivity
thus
mg = qσ/2ε₀
σ = (2mg ε₀)/q
= 2 (0.016) (9.8) (8.85 x 10⁻¹²)/( - 8.6 x 10⁻⁶)
= - 3.23 x 10⁻⁷ C/m²
