Answer: C
high; large
Explanation:
The wave energy is related to its amplitude and frequency.
The wave energy is proportional to the amplitude of the wave. So, wave with the most energy will have high amplitude.
Also, frequency is related to wave energy. The larger the frequency, the more the energy of the wave.
Therefore, The waves with the MOST energy have high amplitudes and large
frequencies.
To solve this problem we will apply the concepts related to the conservation of momentum. That is, the final momentum must be the same final momentum. And in each state, the momentum will be the sum of the product between the mass and the velocity of each object, then
Here,
= Mass of each object
= Initial velocity of each object
= Final velocity of each object
When they position the final velocities of the bodies it is the same and the car is stationary then,
Rearranging to find the final velocity
The expression for the impulse received by the first car is
Replacing,
The negative sign show the opposite direction.
An interesting problem, and thanks to the precise heading you put for the question.
We will assume zero air resistance.
We further assume that the angle with vertical is t=53.13 degrees, corresponding to sin(t)=0.8, and therefore cos(t)=0.6.
Given:
angle with vertical, t = 53.13 degrees
sin(t)=0.8; cos(t)=0.6;
air-borne time, T = 20 seconds
initial height, y0 = 800 m
Assume g = -9.81 m/s^2
initial velocity, v m/s (to be determined)
Solution:
(i) Determine initial velocity, v.
initial vertical velocity, vy = vsin(t)=0.8v
Using kinematics equation,
S(T)=800+(vy)T+(1/2)aT^2 ....(1)
Where S is height measured from ground.
substitute values in (1): S(20)=800+(0.8v)T+(-9.81)T^2 =>
v=((1/2)9.81(20^2)-800)/(0.8(20))=72.625 m/s for T=20 s
(ii) maximum height attained by the bomb
Differentiate (1) with respect to T, and equate to zero to find maximum
dS/dt=(vy)+aT=0 =>
Tmax=-(vy)/a = -0.8*72.625/(-9.81)= 5.9225 s
Maximum height,
Smax
=S(5.9225)
=800+(0.8*122.625)*(5.9225)+(1/2)(-9.81)(5.9225^2)
= 972.0494 m
(iii) Horizontal distance travelled by the bomb while air-borne
Horizontal velocity = vx = vcos(t) = 0.6v = 43.575 m/s
Horizontal distace travelled, Sx = (vx)T = 43.575*20 = 871.5 m
(iv) Velocity of the bomb when it strikes ground
vertical velocity with respect to time
V(T) =vy+aT...................(2)
Substitute values, vy=58.1 m/s, a=-9.81 m/s^2
V(T) = 58.130 + (-9.81)T =>
V(20)=58.130-(9.81)(20) = -138.1 m/s (vertical velocity at strike)
vx = 43.575 m/s (horizontal at strike)
resultant velocity = sqrt(43.575^2+(-138.1)^2) = 144.812 m/s (magnitude)
in direction theta = atan(43.575,138.1)
= 17.5 degrees with the vertical, downward and forward. (direction)
From the principle of energy conservation, the kinetic energy of the pendulum at 0.5 m is 14.7 J.
<h3>What is a pendulum?</h3>
A pendulum swings back and forth and can be used to show the change of potential energy to kinetic energy and vice versa.
Given that the kinetic energy is converted to the potential energy; the potential energy at 0.5 m is 3 * 9.8 * 0.5 = 14.7 J.
Following the principle of energy conservation, the kinetic energy of the pendulum at 0.5 m is 14.7 J.
Learn more about pendulum:brainly.com/question/14759840
#SPJ1
Answer:
The answer is below
Explanation:
Centripetal acceleration is the acceleration due to the movement of an object in a uniform circular motion. The acceleration is directed towards the center of the circle.
Centripetal acceleration is given by the formula:
a = v² / r; where v is the speed of the object and r is the radius of curvature (distance from object ot the center of circle).
Let us assume the car has a velocity of v m/s. For the curve with radius of curvature r:
a₁ = v² / r
For the curve with radius of curvature r = 2r:
a₂ = v² / 2r = (1/2)a₁
Therefore the centripetal acceleration is greater in the curve with radius pf curvature r and smaller in the curve with twice the radius of curvature of the other.
