Answer:
7327 kg or 7.3 tons
Explanation:
We have Newton formula for attraction force between 2 objects with mass and a distance between them:

where
is the gravitational constant on Earth.
is the masses of the 2 objects. and R = 2.6m is the distance between them.
is the attraction force.



or 7.3 tons
In a closed system, the loss of momentum of one object is same as________ the gain in momentum of another object
according to law of conservation of momentum, total momentum before and after collision in a closed system in absence of any net external force, remains conserved . that is
total momentum before collision = total momentum after collision
P₁ + P₂ = P'₁ + P'₂
where P₁ and P₂ are momentum before collision for object 1 and object 2 respectively.
P'₁ - P₁ = - (P'₂ - P₂)
so clearly gain in momentum of one object is same as the loss of momentum of other object
Answer: d = 4750n/3.1+95n
Explanation:
Using the principle of moment to solve the question.
Sum of clockwise moments = sum of anti clockwise moments
Since there are n identical coins with mass 3.1g placed at point 0cm, 1 coin will have mass of 3.1/n grams
Taking moment about the pivot,
Mass 3.1/n grams will move anti-clockwisely while the mass 95g will move in the clockwise direction.
Since its a meter rule (100cm) the distance from the center mass(95g) to the pivot will be 50-d (check attachment for diagram).
To get 'd'
We have 3.1/n × d = 95 × (50-d)
3.1d/n = 4750-95d
3.1d = 4750n-95dn
3.1d+95dn=4750n
d(3.1+95n) = 4750n
d = 4750n/3.1+95n
Answer:
7.1 Hz
Explanation:
In a generator, the maximum induced emf is given by

where
N is the number of turns in the coil
A is the area of the coil
B is the magnetic field strength
f is the frequency
In this problem, we have
N = 200


B = 0.030 T
So we can re-arrange the equation to find the frequency of the generator:

The mass of the puck is
m = 0.15 kg.
The diameter of the puck is 0.076 m, therefore its radius is
r = 0.076/2 = 0.038 m
The sliding speed is
v = 0.5 m/s
The angular velocity is
ω = 8.4 rad/s
The rotational moment of inertia of the puck is
I = (mr²)/2
= 0.5*(0.15 kg)*(0.038 m)²
= 1.083 x 10⁻⁴ kg-m²
The kinetic energy of the puck is the sum of the translational and rotational kinetic energy.
The translational KE is
KE₁ = (1/2)*m*v²
= 0.5*(0.15 kg)*(0.5 m/s)²
= 0.0187 j
The rotational KE is
KE₂ = (1/2)*I*ω²
= 0.5*(1.083 x 10⁻⁴ kg-m²)*(8.4 rad/s)²
= 0.0038 J
The total KE is
KE = 0.0187 + 0.0038 = 0.0226 J
Answer: 0.0226 J