Answer:
Explanation:
Momentum is a vector quantity that represents the "amount of motion" of an object.
Mathematically, the momentum of an object is given by
where
m is the mass of the object
v is the velocity
Since momentum is a vector, it also has a direction, which is the same as the velocity.
Therefore, if we have two objects, the total momentum of the two objects will be obtained from the vector sum of the individual momenta of the two objects.
In this problem we have:
is the momentum of object A
is the momentum of object B
Therefore, the total momentum of objects A and B can be obtained by adding each components of A to the corresponding component of B, so:
So the total initial momentum is
DaddyFed is right, it would be all of them.
Answer:
Radius, r = 0.00523 meters
Explanation:
It is given that,
Magnetic field, 
Current in the toroid, I = 9.6 A
Number of turns, N = 6
We need to find the radius of the toroid. The magnetic field at the center of the toroid is given by :
r = 0.00523 m
or
So, the radius of the toroid is 0.00523 meters. Hence, this is the required solution.
Answer:
Earth: 22.246 N
Moon: 3.71 N
Jupiter: 58.72 N
Explanation:
The mass of an object will remain constant in any location, its weight however, can fluctuate depending on its location. For example, a golf ball will weigh less on the moon, but its mass will not be different if it was on earth.
To calculate anything, we need to convert to standard measurements.
5.00 lbs = 2.27 kg
On earth, gravity is measured to be 9.8 m/s², so the weight in Newtons on Earth would be: (2.27 kg) x (9.8 m/s²) = 22.246 N
Repeated on the moon where gravity is (9.8 m/s²) x (1/6) = 1.633 m/s², so the weight in Newtons on the moon would be: (2.27 kg) x (1.633 m/s²) = 3.71 N
Repeated on Jupiter where gravity is (9.8 m/s²) x (2.64) = 25.87 m/s², so the wight in Newtons on Jupiter would be: (2.27 kg) x (25.87 m/s²) = 58.72 N
Answer:
Final temperature of the aluminum = 41.8 °C
Explanation:
We have the equation for energy
E = mcΔT
Here m = 55 g = 0.055 kg
ΔT = T - 27.5
Specific heat capacity of aluminum = 921.096 J/kg.K
E = 725 J
Substituting
E = mcΔT
725 = 0.055 x 921.096 x (T - 27.5)
T - 27.5 = 14.31
T = 41.81 ° C = 41.8 °C
Final temperature of the aluminum = 41.8 °C
