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alexandr1967 [171]

3 years ago

12

A constant current source supplies a electric current of 200 mA to a load of 2kΩ. When the load changed to 50Ω, the load current

will be
8A
9mA
700mA
12A

1 answer:

serious [3.7K]3 years ago

4 0

Answer:

700ma

Explanation:

all u do is have to add them up.

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In a two-source circuit, one source acting alone produces 10 ma through a given branch. the other source acting alone produces 8

pashok25 [27]

Refer to the figure below.
R = resistance.

Case 1:
The voltage source is V₁ and the current is 10 mA. Therefore
V₁ = (10 mA)R

Case 2:
The voltage source is V₂ and the current is 8 mA. Therefore
V₂ = (8 mA)R

Case 3:
The voltage across the resistance is V₁ - V₂. Therefore the current I is given by
V₁ - V₂ = IR
10R - 8R = (I mA)R
2 = I
The current is 2 mA.

Answer: 2 mA

6 0

3 years ago

A 4 kg car with frictionless wheels is on a ramp that makes a 25° angle with the horizontal. The car is connected to a 1 kg mass

stellarik [79]

Refer to the diagram shown below.

W₁ = (4 kg)*(9.8 m/s²) = 39.2 N
W₂ = (1 kg)*(9.8 m/s²) = 9.8 N

The normal reaction on the 4-kg mass is
N = (39.2 N)*cos(25°) = 35.5273 N
The force acting down the inclined plane due to the weight is
F = (39.2 N)*sin(25°) = 16.5666 N

The net force that accelerates the 4-kg mass at a m/²s down the plane is
F - W₂ = (4 kg)*(a m/s²)
4a = 16.5666 - 9.8
a = 1.6917 m/s²

Answer: 1.69 m/s² (nearest hundredth)

7 0

4 years ago

Motion maps for two objects, Y & Z, are shown.

serg [7]

Answer: it’s 3 seconds

Explanation:

5 0

2 years ago

A mechanic needs to replace the motor for a merry-go-round. The merry-go-round should accelerate from rest to 1.5 rad/s in 6.0s

pashok25 [27]

Answer:

109656.25 Nm

Explanation:

$\omega_f$ = Final angular velocity = 1.5 rad/s

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

t = Time taken = 6 s

m = Mass of disk = 29000 kg

r = Radius = 5.5 m

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{1.5-0}{6}\\\Rightarrow \alpha=0.25\ rad/s^2$

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=\dfrac{1}{2}mr^2\alpha\\\Rightarrow \tau=\dfrac{1}{2}29000\times 5.5^2\times 0.25\\\Rightarrow \tau=109656.25\ Nm$

The torque specifications must be 109656.25 Nm

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At what point on the hill will the car have zero gravitational potential energy?

Mila [183]

At the bottom of the hill

8 0

3 years ago

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