Answer:
average value of the resulting force
Explanation:
The average module value of this resulting force is equivalent to 2.0. 10⁵ N.
The impulse of a force can be calculated by the product of the intensity of the force applied by the time interval in which it is applied -
I = F.Δt
Where,
F = Strength in Newtons
Δt = time interval in seconds
I = Impulse in N.s
The impulse of a force is equivalent to the variation of the amount of movement it causes in the body.
I = ΔQ
The amount of movement is a vector quantity that results from the multiplication of the mass of a body by its speed. Its direction and direction are the same as the velocity vector of the body.
Q = m-V
As the car goes to rest after the application of force, the amount of final movement of the car is equivalent to zero.
I = 0 - mV
F. Δt = - mV
F. 0,1 = - 1000. 20
F = - 20000/0,1
F = 200,000 N
F = 2,0. 10⁵ N
I'm not sure which two rocks you're talking about.
If the volumes of the two rocks are equal, then the rock with the greater density has more mass than the rock with the smaller density, and it also weighs more.
Answer:
The maximum speed of the ejected photoelectrons is 1.815 x 10⁶ m/s.
Explanation:
Given;
frequency of the light, f = 3.5 x 10¹⁵ Hz
work function of the metal, Φ = 5.11 eV
Φ = 5.11 x 1.602 x 10⁻¹⁹ J = 8.186 x 10⁻¹⁹ J
The energy of the incident light is given as sum of maximum kinetic energy and work function of the metal.
E = K.E + Φ
where;
E is the energy of the incident light, calculated as;
E = hf
E = (6.626 x 10⁻³⁴)(3.5 x 10¹⁵)
E = 2.319 x 10⁻¹⁸ J
The maximum kinetic energy of the photoelectrons is calculated as;
K.E = E - Φ
K.E = 2.319 x 10⁻¹⁸ J - 8.186 x 10⁻¹⁹ J
K.E = 2.319 x 10⁻¹⁸ J - 0.8186 x 10⁻¹⁸ J
K.E = 1.5004 x 10⁻¹⁸J
The maximum speed of the ejected photoelectrons in 10⁶ m/s is given as;
K.E = ¹/₂mv²
Therefore, the maximum speed of the ejected photon-electrons is 1.815 x 10⁶ m/s.
Quantum mechanics is all to do with space and different planets
Answer:
The difference between frictionless ramp and a regular ramp is that on a frictionless ramp the ball cannot roll it can only slide, but on a regular ramp the ball can roll without slipping.
We will use conversation of energy.
Note that initial potential energy is zero because the ball is on the bottom, and the final kinetic energy is zero because the ball reaches its maximum vertical distance and stops.
For the ball B;
The initial velocities of the balls are equal. Their maximum climbing point will be proportional to their final potential energy. Since their initial kinetic energies are equal, their final potential energies must be equal as well.
Hence, both balls climb the same point.
Explanation:
