Which of the following are true for an object in projectile motion?
1 answer:
The correct options are: B, C and D.
In fact:
- B is true, because the motion of the projectile can be decomposed in two independent motions on the x and y axis. On the x-axis, there is no acceleration; while on the y-axis, the projectile is accelerating towards the ground with
(acceleration of gravity)
- C is true, because the two motions on the horizontal and vertical direction are independent. In particular, the horizontal motion is a uniform motion (constant velocity), while the vertical motion is a uniformly accelerated motion (because of the gravitational acceleration g acting on the projectile)
- D is true, in fact (if we neglect air resistance) gravity is the only force acting on the projectile.
In fact:
- B is true, because the motion of the projectile can be decomposed in two independent motions on the x and y axis. On the x-axis, there is no acceleration; while on the y-axis, the projectile is accelerating towards the ground with
- C is true, because the two motions on the horizontal and vertical direction are independent. In particular, the horizontal motion is a uniform motion (constant velocity), while the vertical motion is a uniformly accelerated motion (because of the gravitational acceleration g acting on the projectile)
- D is true, in fact (if we neglect air resistance) gravity is the only force acting on the projectile.
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Answer:
the penny loses contact at the piston's highest point.
f = 2.5 Hz
Explanation:
Concepts and Principles
1- Newton's Second Law: The net force F on a body with mass m is related to the body's acceleration a by
∑F = ma (1)
2- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed w and the amplitude A as follows:
a_max = -w^2A (2)
3- The angular frequency w of a wave is related to the frequency f by:
w = 2π f (3)
Given Data
- The amplitude of the piston is: A = (4.0 cm) ( 1/ 100 cm)= 0.04 m.
- The frequency of oscillation of the piston is steadily increased.
Required Data
<em>In part (a), we are asked to determine the point at which the penny first loses contact with the piston. </em>
<em>In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle. </em>
Solution
(a)
The free-body diagram in Figure 1 shows the forces acting on the penny; mi is the gravitational force exerted by the Earth on the penny andrt is the normal contact force exerted by the piston on the penny.
figure 1 is attached
Apply Newton's second law from Equation (1) in the vertical direction to the penny:
∑F_y -mg= ma
Solve for n=m(g+a) The penny loses contact with the surface of the oscillating piston when the normal force n exerted by the piston is zero. So
0 = m(g + a)
a = —g
Therefore, the penny loses contact with the piston when the piston starts accelerating downwards. The piston first acceleratesdownward at its highest point and hence the penny loses contact at the piston's highest point.
(b)
The maximum acceleration of the penny at the highest point of the piston is found from Equation (2):
a = —w^2A
where a = —g at the highest point. So
g = w^2A
Solve for w:
w =√g/A
Substitute for w from Equation (3):
2πf = √g/A
Solve for f :
f = 1/2π√g/A
Substitute numerical values:
f = 1/2π√9.8 m/s^2/0.04
f = 2.5 Hz
The velocity at point B is 10 m/s with a pressure of 262500 Pa
<h3>Bernoulli equation</h3>
According to the continuity equation:
A₁V₁ = A₂V₂
Where A is the area and V is velocity, ρ = density of water = 1000 kg/m³
Hence:
4(5) = 2(V₂)
V₂ = 10 m/s
Using Bernoulli equation:
The velocity at point B is 10 m/s with a pressure of 262500 Pa
Find out more on Bernoulli equation at: brainly.com/question/14082066
Answer:
The intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰
Explanation:
Given;
rock concert sound intensity level, β₁ = 120 dB
whisper sound intensity level, β₂ = 20 dB
The sound intensity level is given as;
where;
I₀ is the threshold sound intensity of hearing = 10⁻¹² W/m²
I is the sound intensity
Intensity of sound at rock concert ;
The intensity of sound of a whisper;
Determine the factor by which the intensity of sound at a rock concert louder than that of a whisper
Therefore, the intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰