Answer:
T = 27.92 N
Explanation:
For this exercise let's use Newton's second law
T - W = m a
The weight
W = mg
The acceleration can be found by derivatives
a = dv / dt
v = 2 t + 0.6 t²
a = 2 + 0.6 t
We replace
T - mg = m (2 + 0.6t)
T = m (g + 2 + 0.6 t) (1)
Let's look for the time for the speed of 15 m / s
15 = 2 t + 0.6 t²
0.6 t² + 2 t - 15 = 0
We solve the second degree equation
t = [-2 ±√(4 - 4 0.6 (-15))] / 2 0.6
t = [-2 ±√40] / 1.3 = [-2 ± 6.325] / 1.2
We take the positive time
t = 3.6 s
Let's calculate from equation 1
T = 2.00 (9.8 + 2 + 0. 6 3.6)
T = 27.92 N
Answer:
three times the original diameter
Explanation:
From the wire's resistance formula, we can calculate the relation between the diameter of the wire and its length:
Here, d is the wire's diameter, is the electrical resistivity of the material and R is the resistance of the wire. We have
Answer:
1.23×10⁸ m
Explanation:
Acceleration due to gravity is:
a = GM / r²
where G is the universal gravitational constant,
M is the mass of the planet,
and r is the distance from the center of the planet to the object.
When the object is on the surface of the Earth, a = g and r = R.
g = GM / R²
When the object is at height i above the surface, a = 1/410 g and r = i + R.
1/410 g = GM / (i + R)²
Divide the first equation by the second:
g / (1/410 g) = (GM / R²) / (GM / (i + R)²)
410 = (i + R)² / R²
410 R² = (i + R)²
410 R² = i² + 2iR + R²
0 = i² + 2iR − 409R²
Solve with quadratic formula:
i = [ -2R ± √((2R)² − 4(1)(-409R²)) ] / 2(1)
i = [ -2R ± √(1640R²) ] / 2
i = (-2R ± 2R√410) / 2
i = -R ± R√410
i = (-1 ± √410) R
Since i > 0:
i = (-1 + √410) R
R = 6.37×10⁶ m:
i ≈ 1.23×10⁸ m
So we want to know what do electric field lines show except strength of the electric field. Electric field lines are lines that show us the field strength and the DIRECTION of the electric field. If the lines point away from the charge that is producing the lines, the charge is positive, and if the lines point towards the charge that is producing them that charge is negative.
Answer: a) 6.67cm/s b) 1/2
Explanation:
According to law of conservation of momentum, the momentum of the bodies before collision is equal to the momentum of the bodies after collision. Since the second body was initially at rest this means the initial velocity of the body is "zero".
Let m1 and m2 be the masses of the bodies
u1 and u2 be their velocities respectively
m1 = 5.0g m2 = 10.0g u1 = 20.0cm/s u2 = 0cm/s
Since momentum = mass × velocity
The conservation of momentum of the body will be
m1u1 + m2u2 = (m1+m2)v
Note that the body will move with a common velocity (v) after collision which will serve as the velocity of each object after collision.
5(20) + 10(0) = (5+10)v
100 + 0 = 15v
v = 100/15
v = 6.67cm/s
Therefore the velocity of each object after the collision is 6.67cm/s
b) kinectic energy of the 10.0g object will be 1/2MV²
= 1/2×10×6.67²
= 222.44Joules
kinectic energy of the 5.0g object will be 1/2MV²
= 1/2×5×6.67²
= 222.44Joules
= 111.22Joules
Fraction of the initial kinetic transferred to the 10g object will be
111.22/222.44
= 1/2