you throw an object straight up at 15m/s. how fast is it going when it gets back to the height from which you threw it?
1 answer:
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Answer:
The final components of velocity of the composite object is 3.33 m/s.
Explanation:
Given;
mass of the first object, m₁ = 3.35 kg
initial velocity of the first object, u₁ = 4.90 m/s in positive x-direction
mass of the second object, m₂ = 1.88 kg
initial velocity of the second object, u₂ = 3.12 m/s in negative y-direction
initial momentum of the first object, P₁ = 3.35 x 4.9 = 16.415 kgm/s
initial momentum of the second object, P₂ = 1.88 x 3.12 = 5.8656 kgm/s
The resultant velocity of the two objects is given by;
R² = 16.415² + 5.8656²
R² = 303.858
R = √303.858
R = 17.432 kgm/s
Apply the principle of conservation of linear momentum for inelastic collision;
total initial momentum before = total final momentum after collision
P₁(x) + P₂(y) = Pf
R = Pf
R = v(m₁ + m₂)
17.432 = v(m₁ + m₂)
where;
v is the final components of velocity of the composite object
Therefore, the final components of velocity of the composite object is 3.33 m/s.
1) See three Kepler laws below
2a) Acceleration is
2b) Tension in the string: 27.4 N
3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position
3b) The kinetic energy of the object is 2.25 J
Explanation:
1)
There are three Kepler's law of planetary motion:
- 1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
- 2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
- 3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically,
, where T is the period of revolution and r is the semi-major axis of the orbit
2a)
To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.
For the block on the right, we have:
(1)
where
is the component of the weight of the block parallel to the incline, with
M = 8.0 kg (mass of the block)
(acceleration of gravity)
T = tension in the string
a = acceleration of the block
For the block on the left, we have similarly
(2)
where
m = 3.5 kg (mass of the block)
From (2) we get
Substituting into (1),
Solving for a,
2b)
The tension in the string can be calculated using the equation
where
m = 3.5 kg (mass of lighter block)
(acceleration found in part 2)
Substituting,
3a)
The kinetic energy of an object is the energy due to its motion. It is calculated as
where
m is the mass of the object
v is its speed
The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by
where
m is the mass of the object
g is the strength of the gravitational field
h is the heigth of the object relative to the ground
3b)
The kinetic energy of an object is given by
where
m is the mass of the object
v is its speed
For the object in this problem,
m = 500 g = 0.5 kg
v = 3 m/s
Substituting, we find its kinetic energy:
Learn more about acceleration and forces:
And about kinetic energy:
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