The range of the piece of paper is C) 1.4 m
Explanation:
The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:
- A uniform motion along the horizontal direction, with constant velocity
- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity, 
)
From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:
where
u is the initial velocity
is the angle of projection
g is the acceleration of gravity
For the piece of paper in this problem,
u = 4.3 m/s
Substituting,
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Answer: 996m/s
Explanation:
Formula for calculating velocity of wave in a stretched string is
V = √T/M where;
V is the velocity of wave
T is tension
M is the mass per unit length of the wire(m/L)
Since the second wire is twice as far apart as the first, it will be L2 = 2L1
Let V1 and V2 be the speed of the shorter and longer wire respectively
V1 = √T/M1... 1
V2 = √T/M2... 2
Since V1 = 249m/s, M1 = m/L1 M2 = m/L2 = m/2L1
The equations will now become
249 = √T/(m/L1) ... 3
V2 = √T/(m/2L1)... 4
From 3,
249² = TL1/m...5
From 4,
V2²= 2TL1/m... 6
Dividing equation 5 by 6 we have;
249²/V2² = TL1/m×m/2TL1
{249/V2}² = 1/2
249/V2 = (1/2)²
249/V2 = 1/4
V2 = 249×4
V2 = 996m/s
Therefore the speed of the wave on the longer wire is 996m/s
Answer:
Explanation:
a )
We shall apply the concept of impulse .
Impulse = force x time = change in momentum
= 5 x 4 = 2 ( V - 3 ) , where V is final velocity of the object
20 = 2V - 6
V = 13 m /s
b )
Impulse applied = - 7 x 4 = - 28 kg m/s ( negative as direction of force is opposite motion )
If v be the final velocity
2 x 3 - 28 = 2 v ( initial momentum - change in momentum = final momentum )
- 22 = 2v
v = - 11 m /s
object will move with 11 m /s in opposite direction .
On a similar problem wherein instead of 480 g, a 650 gram of bar is used:
Angular momentum L = Iω, where
<span>I = the moment of inertia about the axis of rotation, which for a long thin uniform rod rotating about its center as depicted in the diagram would be 1/12mℓ², where m is the mass of the rod and ℓ is its length. The mass of this particular rod is not given but the length of 2 meters is. The moment of inertia is therefore </span>
<span>I = 1/12m*2² = 1/3m kg*m² </span>
<span>The angular momentum ω = 2πf, where f is the frequency of rotation. If the angular momentum is to be in SI units, this frequency must be in revolutions per second. 120 rpm is 2 rev/s, so </span>
<span>ω = 2π * 2 rev/s = 4π s^(-1) </span>
<span>The angular momentum would therefore be </span>
<span>L = Iω </span>
<span>= 1/3m * 4π </span>
<span>= 4/3πm kg*m²/s, where m is the rod's mass in kg. </span>
<span>The direction of the angular momentum vector - pseudovector, actually - would be straight out of the diagram toward the viewer. </span>
<span>Edit: 650 g = 0.650 kg, so </span>
<span>L = 4/3π(0.650) kg*m²/s </span>
<span>≈ 2.72 kg*m²/s</span>
