When completely filled with water, the beaker and its contents have a total mass of 326.75 g. What volume does the beaker hold?
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Refer to the diagram shown below.
The net force acting on the vehicle is
F - R = 1060 -1010 = 50 N
The distance traveled is 21 m. Because the force is constant, the work done is
W = (50 N)*(21 m) = 1050 J
Assume that energy is not dissipated by air resistance or otherwise.
Conservation of energy requires that W = KE, where KE is the kinetic energy of the vehicle.
The KE is
KE = (1/2)*(2000 kg)*(v m/s)² = 1000v² J
Equate KE and W to obtain
1000v² = 1050
v² = 1.05
v = 1.025 m/s
Answer: 1.025 m/s
The net force acting on the vehicle is
F - R = 1060 -1010 = 50 N
The distance traveled is 21 m. Because the force is constant, the work done is
W = (50 N)*(21 m) = 1050 J
Assume that energy is not dissipated by air resistance or otherwise.
Conservation of energy requires that W = KE, where KE is the kinetic energy of the vehicle.
The KE is
KE = (1/2)*(2000 kg)*(v m/s)² = 1000v² J
Equate KE and W to obtain
1000v² = 1050
v² = 1.05
v = 1.025 m/s
Answer: 1.025 m/s
The speed of the particle to pass through the Selector is 2875 m/s
<u>Explanation:</u>
Given -
Electric field,(We can represent as E) = 460 N/C
Magnetic field, (We can represent as B) = 0.16 T
Speed,(We can represent as v) = ?
We know that the formula for finding the velocity ,
Therefore, speed of the particle to pass through the Selector is 2875 m/s
Answer:
Ein: 2.75*10^-3 N/C
Explanation:
The induced electric field can be calculated by using the following path integral:
Where:
dl: diferencial of circumference of the ring
circumference of the ring = 2πr = 2π(5.00/2)=15.70cm = 0.157 m
ФB: magnetic flux = AB (A: area of the loop = πr^2 = 1.96*10^-3 m^2)
The electric field is always parallel to the dl vector. Then you have:
Next, you take into account that the area of the ring is constant and that dB/dt = - 0.220T/s. Thus, you obtain:
hence, the induced electric field is 2.75*10^-3 N/C