Answer:
0.187 m
Explanation:
We'll begin by calculating the acceleration of the ball. This can be obtained as follow:
Mass (m) = 0.450 Kg
Force (F) = 38 N
Acceleration (a) =?
F = m × a
38 = 0.450 × a
Divide both side by 0.450
a = 38 / 0.450
a = 84.44 m/s²
Finally, we shall determine the distance. This can be obtained as follow:
Initial velocity (u) = 2.20 m/s.
Final velocity (v) = 6 m/s
Acceleration (a) = 84.44 m/s²
Distance (s) =?
v² = u² + 2as
6² = 2.2² + (2 × 84.44 × s)
36 = 4.4 + 168.88s
Collect like terms
36 – 4.84 = 168.88s
31.52 = 168.88s
Divide both side by 168.88
s = 31.52 / 168.88
s = 0.187 m
Thus, the distance is 0.187 m
A. Is very attractive. If it's sublimation directly from water vapor in the air to ice on the glass, then yes. But from liquid water mist to water ice, no. Ice is less dense than water. That's why cubes float in your soda. Better leave 'A' alone. . . . D. Ice pellets turn to liquid. That one's good.
The figure is showing a volume of 2.4 mL becuase it's feel 4 little segments.
Therefore, the answer is 2.4 mL.
Answer:
(a) 1462.38 m/s
(b) 2068.13 m/s
Explanation:
(a)
The Kinetic energy of the atom can be given as:
K.E = (3/2)KT
where,
K = Boltzman's Constant = 1.38 x 10⁻²³ J/k
K.E = Kinetic Energy of atoms = 343 K
T = absolute temperature of atoms
The K.E is also given as:
K.E = (1/2)mv²
Comparing both equations:
(1/2)mv² = (3/2)KT
v² = 3KT/m
v = √[3KT/m]
where,
m = mass of Helium = (4 A.M.U)(1.66 X 10⁻²⁷ kg/ A.M.U) = 6.64 x 10⁻²⁷ kg
v = RMS Speed of Helium Atoms = ?
Therefore,
v = √[(3)(1.38 x 10⁻²³ J/K)(343 K)/(6.64 x 10⁻²⁷ kg)]
<u>v = 1462.38 m/s</u>
(b)
For double temperature:
T = 2 x 343 K = 686 K
all other data remains same:
v = √[(3)(1.38 x 10⁻²³ J/K)(686 K)/(6.64 x 10⁻²⁷ kg)]
<u>v = 2068.13 m/s</u>
Answer:
question5: F=74312.5N
question6: charge at the end of antenna=0.37N
Explanation:
Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.
⇒
∴
where 
is the force of attraction or repulsion
is Coulumb's constant=
and 
are the magnitude of the charges
is the distance between two charges
The force between the two charges is attractive if they are of different polarity
The force between the two charges is repulsive if they are of same polarity
Question5:
Given: q1=0.041 C, q2=0.029 C, r=12 m
therefore by Coulumb's law,
Question6:
Given: q1=
, r=5 m, F=
therefore by Coulumb's law,
⇒
