70x3=210 mph... its that easy
The answer is the option c. generator. Generator is a machine that
converts mechanical energy into electricity. The way it is made is
because the mechanical movement may be transformed in rotation of
magnets placed inside a coiled wire, and this induces the movement of
electrons through the coiled wire. As you know electricity is the flow
of electrons.
Answer:9.75 m/s
Explanation:
Given
Length of ladder 
Foot the ladder is moving away with speed of 
From diagram
------1
at 
Now differentiating equation 1 w.r.t time
negative indicates distance is decreasing with time
Answer:
a. A = 0.735 m
b. T = 0.73 s
c. ΔE = 120 J decrease
d. The missing energy has turned into interned energy in the completely inelastic collision
Explanation:
a.
4 kg * 10 m /s + 6 kg * 0 m/s = 10 kg* vmax
vmax = 4.0 m/s
¹/₂ * m * v²max = ¹/₂ * k * A²
m * v² = k * A² ⇒ 10 kg * 4 m/s = 100 N/m * A²
A = √1.6 m ² = 1.26 m
At = 2.0 m - 1.26 m = 0.735 m
b.
T = 2π * √m / k ⇒ T = 2π * √4.0 kg / 100 N/m = 1.26 s
T = 2π *√ 10 / 100 *s² = 1.99 s
T = 1.99 s -1.26 s = 0.73 s
c.
E = ¹/₂ * m * v²max =
E₁ = ¹/₂ * 4.0 kg * 10² m/s = 200 J
E₂ = ¹/₂ * 10 * 4² = 80 J
200 J - 80 J = 120 J decrease
d.
The missing energy has turned into interned energy in the completely inelastic collision
Answer:
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Explanation:
For this problem let's use Newton's second law applied to each body
Body A
X axis
T = m_A a
Axis y
N- W_A = 0
Body B
Vertical axis
W_B - T = m_B a
In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension
We write the equations
T = m_A a
W_B –T = M_B a
We solve this system of equations
m_B g = (m_A + m_B) a
a = m_B / (m_A + m_B) g
In this initial case
m_A = M
m_B = M
a = M / (1 + 1) M g
a = ½ g
Let's find the tension
T = m_A a
T = M ½ g
T = ½ M g
Now we change the mass of the second block
m_B = 2M
a = 2M / (1 + 2) M g
a = 2/3 g
We seek tension for this case
T’= m_A a
T’= M 2/3 g
Let's look for the relationship between the tensions of the two cases
T’/ T = 2/3 M g / (½ M g)
T’/ T = 4/3
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
