Answer:
The work done by the applied force is 259.22 J.
Explanation:
The work done by the applied force is given by:
Where:
F: is the applied horizontal force = 108.915 N
d: is the distance = 2.38 m
Hence, the work is:
Therefore, the work done by the applied force is 259.22 J.
I hope it helps you!
Answer:
v = 10 [m/s]
Explanation:
The largest mass is that of 4 [kg], in this way the momentum can be calculated by means of the product of the mass by velocity.
where:
P = momentum [kg*m/s]
m = mass = 4 [kg]
v = velocity = 5 [m/s]
Now the momentum:
![P=4*5\\P=20[kg*m/s]](https://tex.z-dn.net/?f=P%3D4%2A5%5C%5CP%3D20%5Bkg%2Am%2Fs%5D)
This same momentum is equal for the other mass, in this way we can find the velocity.
![P=m*v\\20=2*v\\v=10[m/s]](https://tex.z-dn.net/?f=P%3Dm%2Av%5C%5C20%3D2%2Av%5C%5Cv%3D10%5Bm%2Fs%5D)
The angle that the cart rolls with the horizontal. The closer the ramp gets to 90 degrees the faster the cart will accelerate.
Answer:
F_Balance = 46.6 N ,m' = 4,755 kg
Explanation:
In this exercise, when the sphere is placed on the balance, it indicates the weight of the sphere, when another sphere of opposite charge is placed, they are attracted so that the balance reading decreases, resulting in
∑ F = 0
Fe –W + F_Balance = 0
F_Balance = - Fe + W
The electric force is given by Coulomb's law
Fe = k q₁ q₂ / r₂
The weight is
W = mg
Let's replace
F_Balance = mg - k q₁q₂ / r₂
Let's reduce the magnitudes to the SI system
q₁ = + 8 μC = +8 10⁻⁶ C
q₂ = - 3 μC = - 3 10⁻⁶ C
r = 0.3 m = 0.3 m
Let's calculate
F_Balance = 5 9.8 - 8.99 10⁹ 8 10⁻⁶ 3 10⁻⁶ / (0.3)²
F_Balance = 49 - 2,397
F_Balance = 46.6 N
This is the balance reading, if it is calibrated in kg, it must be divided by the value of the gravity acceleration.
Mass reading is
m' = F_Balance / g
m' = 46.6 /9.8
m' = 4,755 kg
