Answer:
Max height= 36000 meters
Total Time = 120 seconds
Explanation:
0 = U - at
U = at
U= 20*60
U= 1200 m/s
MAX altitude would be
(U²Sin²tita)/2g
Max height= 1200² *( SIN90)²/(2*20)
Time of FLIGHT
2 * 1200/20
2400/20
120 sec
onds
Atomic Number
or
Number of Protons
ΩΩΩΩΩΩΩΩΩΩ
Answer:
a) x = 4.33 m
, b) w = 2 rad / s
, f = 0.318 Hz
, c) a = - 17.31 cm / s²,
d) T = 3.15 s, e) A = 5.0 cm
Explanation:
In this exercise on simple harmonic motion we are given the expression for motion
x = 5 cos (2t + π / 6)
they ask us for t = 0
a) the position of the particle
x = 5 cos (π / 6)
x = 4.33 m
remember angles are in radians
b) The general form of the equation is
x = A cos (w t + Ф)
when comparing the two equations
w = 2 rad / s
angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 2 / 2pi
f = 0.318 Hz
c) the acceleration is defined by
a == d²x / dt²
a = - A w² cos (wt + Ф)
for t = 0
, we substitute
a = - 5,0 2² cos (π / 6)
a = - 17.31 cm / s²
d) El period is
T = 1/f
T= 1/0.318
T = 3.15 s
e) the amplitude
A = 5.0 cm
Explanation:
In the given question, the two metal spheres were hanged with the nylon thread.
When these two spheres were brought close together, they attracted each other. The attraction between these spheres is the result of the opposite charges between them.
The possible ways by which these two metal spheres can be charged are by induction that is touching the metal or by rubbing them.
During induction, the same charges are transferred to each sphere. In this case, either both the spheres will be negatively charged or positively charged.
It is not possible that after the sphere touch each other they will cling together because the same charge repels each other and during touching, if one sphere is neutral, then the charged one will transfer the same charge. And as we know that same charge repel each other therefore they will repel each other.
Answer: 2.49×10^-3 N/m
Explanation: The force per unit length that two wires exerts on each other is defined by the formula below
F/L = (u×i1×i2) / (2πr)
Where F/L = force per meter
u = permeability of free space = 1.256×10^-6 mkg/s^2A^2
i1 = current on first wire = 57A
i2 = current on second wire = 57 A
r = distance between both wires = 26cm = 0.26m
By substituting the parameters, we have that
Force per meter = (1.256×10^-6×57×57)/ 2×3.142 ×0.26
= 4080.744×10^-6/ 1.634
= 4.080×10^-3 / 1.634
= 2.49×10^-3 N/m
