Answer:
The new distance is d = 0.447 d₀
Explanation:
The electric out is given by Coulomb's Law
F = k q₁ q₂ / r²
This electric force is in balance with tension.
We reduce the charge of sphere B to 1/5 of its initial value (
=q₂ = q₂ / 5) than new distance (d = n d₀)
dat
q₁ = 
q₂ =
r = d₀
In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points
F = k q₁ q₂ / d₀²
F = k q₁ (q₂ / 5) / (n d₀)²
.k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)
5 n² = 1
n = √ 1/5
n = 0.447
The new distance is
d = 0.447 d₀
Answer:
The total amount of energy that would have been released if the asteroid hit earth = The kinetic energy of the asteroid = 1.29 × 10¹⁵ J = 1.29 PetaJoules = 1.29 PJ
1 PJ = 10¹⁵ J
Explanation:
Kinetic energy = mv²/2
velocity of the asteroid is given as 7.8 km/s = 7800 m/s
To obtain the mass, we get it from the specific gravity and diameter information given.
Density = specific gravity × 1000 = 3 × 1000 = 3000 kg/m³
But density = mass/volume
So, mass = density × volume.
Taking the informed assumption that the asteroid is a sphere,
Volume = 4πr³/3
Diameter = 30 m, r = D/2 = 15 m
Volume = 4π(15)³/3 = 14137.2 m³
Mass of the asteroid = density × volume = 3000 × 14137.2 = 42411501 kg = 4.24 × 10⁷ kg
Kinetic energy of the asteroid = mv²/2 = (4.24 × 10⁷)(7800²)/2 = 1.29 × 10¹⁵ J
Answer:
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Explanation:
Answer:
Explanation:
Expression for time period of a pendulum is as follows
T = 
l is length of pendulum from centre of bob and g is acceleration due to gravity
Given
Time period T = 1.583
g = 9.846
Substituting the values
1.583 = 
l = 
l = .6244 m
= 62.44 cm
Length of rod = length of pendulum - radius of bob
= 62.44 - 13.62
= 48.82 cm
= .488 m
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
