To solve this problem we will apply the concepts related to energy conservation. With this we will find the speed before the impact. Through the kinematic equations of linear motion we will find the velocity after the impact.
Since the momentum is given as the product between mass and velocity difference, we will proceed with the velocities found to calculate it.
Part A) Conservation of the energy
Part B) Kinematic equation of linear motion,
Here
v= 0 Because at 1.5m reaches highest point, so v=0
Therefore the velocity after the collision with the floor is 3.7m/s
PART C) Total change of impulse is given as,
Answer:
A) Zero newtos
Explanation:
To solve this exercise, let's use Newton's first and second law, in the first law it is established that every body is at rest or moving with constant speed if the external forces are zero. In the second law it is determined that the sum of the force is proportional to the acceleration of the body.
Let us apply our case, the body is at constant speed which implies that the sum of all external forces is zero. In addition, there is no friction, so we do not need a force that counteracts them.
Consequently, the force applied must be zero.
Answer:
I = 11.26 mA
Explanation:
given,
V = 0.7 V length = 80 m
diameter = 0.2 mm = 0.02 cm
radius = 0.01 × 10⁻² m
ρ for gold wire = 2.44 × 10⁻⁸ ohm-m at 20 °C
A = cross sectional area = π r² = π (0.01 × 10⁻² )²
= 31.4× 10⁻⁹ m²
R = 62.165 Ω
I = 11.26 mA
Answer:
<h3>The answer is 1600 kgm/s</h3>
Explanation:
The momentum of an object can be found by using the formula
<h3>momentum = mass × velocity</h3>
From the question
mass = 200 kg
velocity / speed = 8m/s
We have
momentum = 200 × 8
We have the final answer as
<h3>1600 kgm/s</h3>
Hope this helps you
Answer:
Where the electric potential is constant, the strength of the electric field is zero.
Explanation:
As a test charge moves in a given direction, the rate of change of the electric potential of the charge gives the potential gradient whose negative value is the same as the value of the electric field. In other words, the negative of the slope or gradient of electric potential (V) in a direction, say x, gives the electric field (Eₓ) in that direction. i.e
Eₓ = - dV / dx ----------(i)
From equation (i) above, if electric potential (V) is constant, then the differential (which is the electric field) gives zero.
<em>Therefore, a constant electric potential means that electric field is zero.</em>
