a lemon with mass 0.3 kg falls out of a tree from a height of 1.8 m. How much mechanical energy does the lemon have just before
2 answers:
Answer:
5.292 Joules
Explanation:
Hello
applying the law of the conservation of energy, and assuming that no energy is lost due to friction with the air, we can affirm that the energy that has just before the impact is the same energy that it has just before being released, that energy is due to height, mass and gravity and is defined by:
Let
pute the values into the equation
Have a nice day.
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Answer: E =
Explanation: The formulae for intensity of an electric field of a solid metal sphere relative to a point is given below
r
where ,
, r = 0.1m r = is the position vector of the charge.
it has been stated in the question that the charge is placed at the center thus it has no position vector.
Answer:
x = A cos (w \sqrt{2y_{o}/g})
a) maximun Ф= \sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }
b) minimun Ф = - \sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }
Explanation:
For this exercise let's use kinematics to find the time it takes for the mass to reach the floor
y = y₀ + v₀ t - ½ g t²
as the mass is released from rest, its initial velocity is zero (vo = 0) and its height upon reaching the ground is zero (y = 0)
0 = y₀ - ½ g t²
t =
The bucket-spring system has a simple harmonic motion, which is described by
x = A cos wt
in this expression we assumed that the phase constant (Ф) is zero
let's replace the time
x = A cos (w \sqrt{2y_{o}/g})
this is the distance where the system must be for the mass to fall into it.
a) The new system has a total mass of m ’= 3.0 kg, so its angular velocity changes
w =
In the initial state
w = \sqrt{k/2}
When the mass changes
w ’= \sqrt{k/3}
the displacement in each case is
x = A cos (wt)
for the new case
x ’= A cos (w’t + Ф)
the phase constant is included to take into account possible changes due to the collision of the mass.
we see that this maximum expressions when the cosine is maximum
cos (w´t + Ф) = 1
w’t + Ф = 0
Ф = -w ’t
Ф = -
\sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }
b) the function is minimun if
cos (w’t + fi) = 0
w’t + Ф = π / 2
Ф = π / 2 - w ’t
Ф = - \sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }
Answer:
(a) Acceleration of positron is 6.03 x 10¹³ m/s²
(b) Speed of positron after 8.70 x 10⁻⁹ s is 5.24 x 10⁵ m/s
Explanation:
Given :
Constant electric field, E = 343 N/C
Mass of positron, m = 9.1 x 10⁻³¹ kg
Charge of positron, q = +e = 1.6 x 10⁻¹⁹ C
(a) Coulomb force on the positron is determine by the relation :
F = q x E ....(1)
But, force is also equals to product of mass and acceleration. So,
F = ma .....(2)
Here a is acceleration.
From equation (1) and (2).
m x a = q x E
Substitute the values of q, E and m in the above equation.
a = 6.03 x 10¹³ m/s²
(b) Initially, the positron is at rest, so its initial speed is zero.
The equation of motion for positron is :
v = u + at
Here v is final speed, u is initial speed and t is time.
Since, u is zero, so the equation becomes :
v = at
Substitute 8.70 x 10⁻⁹ s for t and 6.03 x 10¹³ m/s² for a in the above equation.
v = 6.03 x 10¹³ x 8.70 x 10⁻⁹ m/s
v = 5.24 x 10⁵ m/s