A circular dartboard has a radius of 2 meters and a red circle in the center. Assume you hit the target at a random point. For w
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Answer:
(a) the final velocity of the cart is 0.857 m/s
(b) the impulse experienced by the package is 21.43 kg.m/s
(c) the fraction of the initial energy lost is 0.71
Explanation:
Given;
mass of the package, m₁ = 10 kg
mass of the cart, m₂ = 25 kg
initial velocity of the package, u₁ = 3 m/s
initial velocity of the cart, u₂ = 0
let the final velocity of the cart = v
(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
10 x 3 + 25 x 0 = v(10 + 25)
30 = 35v
v = 30 / 35
v = 0.857 m/s
(b) the impulse experienced by the package;
The impulse = change in momentum of the package
J = ΔP = m₁v - m₁u₁
J = m₁(v - u₁)
J = 10(0.857 - 3)
J = -21.43 kg.m/s
the magnitude of the impulse experienced by the package = 21.43 kg.m/s
(c)
the initial kinetic energy of the package is calculated as;
the final kinetic energy of the package;
the fraction of the initial energy lost;
Answer:
= 391.67 Hz
Explanation:
The sound of lowest frequency which is produced by a vibrating sting is called its fundamental frequency ().
The For a vibrating string, the fundamental frequency () can be determined by:
=
Where v is the speed of waves of the string, and L is the length of the string.
L = 42.0 cm = 0.42 m
v = 329 m/s
=
=
= 391.6667 Hz
The fundamental frequency of the string is 391.67 Hz.
Answer:
1) When <
(hence
< f ) and they are both in front of the mirror (positive), the image will be larger and inverted
2) When >
(and
< f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted
3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror
Explanation:
The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation
Where:
f = Focal length of the mirror = R/2
= Image distance from the mirror
= Object distance from the mirror
= Image height
= Object height
is positive for an object placed in front of the mirror and negative for an object placed behind the mirror
is positive for an image formed in front of the mirror and negative for an image formed behind the mirror
m is positive when the orientation of the image and the object is the same
m is negative when the orientation of the image and the object is inverted
f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)
∴ When <
(hence
< f ) and they are both in front of the mirror (positive), the image will be larger and inverted
When >
(and
< f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted
When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.