Answer:
<h2><u>Constant</u></h2>
Explanation:
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<h2>Thanks</h2>
Answer:
The velocity of the student has after throwing the book is 0.0345 m/s.
Explanation:
Given that,
Mass of book =1.25 kg
Combined mass = 112 kg
Velocity of book = 3.61 m/s
Angle = 31°
We need to calculate the magnitude of the velocity of the student has after throwing the book
Using conservation of momentum along horizontal direction
Put the value into the formula
Hence, The velocity of the student has after throwing the book is 0.0345 m/s.
Answer: One quarter of the force
Explanation:
According to Newton's law of Gravitation, the force 
exerted between two bodies of masses 
and 
and separated by a distance 
is equal to the product of their masses and inversely proportional to the square of the distance:
(1)
Where 
is the gravitational constant
This means that the gravity force decreases when the distance between these two bodies increases.
In this context, if the distance between the capsule and the Earth increases twice, the new distance will be 
.
Substituting this distance in (1):
(2)
<u>Finally:</u>
>>>This means the force toward Earth becomes one quarter "weaker"
Use formula for Echo which is Velocity=2(Distance)/time so 343=2(150)/T 343T=300..T=300/343=0.9 seconds
Answer:
speed of the charge electric is v = - (Eo q/m) cos t
Explanation:
The electric charge has a very small mass so it follows the oscillations of the electric field. We force ourselves on the load,
F = q Eo sint
a) To find the velocity of the particle, let's use Newton's second law to find the acceleration and of this by integration the velocity
F = ma
q Eo sint = ma
a = Eo q / m sint
a = dv / dt
dv = adt
∫ dv = ∫ a dt
v-vo = I (Eoq / m) sin t dt
v- vo = Eo q / m (-cos t)
We evaluate the integral from the initial point, as the particle starts from rest Vo = 0, for t = 0
v = - (Eo q / m) cos t
b) Kinetic energy
K = ½ m v2
K = ½ m (Eoq / m)²2 (sint)²
K = ¹/₂ Eo² q² / m sin² t
c) The average kinetic energy over a period
K = ½ m v2
<v2> = (Eoq / m) 2 <cos2 t>
The average of cos2 t = ½, substitute and calculate
K = ½ m (Eoq / m)² ½
K = ¼ Eo² q² / m
