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vredina [299]
3 years ago
13

Amotor supplied by 240V requires 12A to lift a 2000 lb at a rate of 25 ft/min, me power input to the motor is-

Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
4 0

Answer:

Power input, P = 2880 watts

Explanation:

It is given that,

Voltage of the motor, V = 240 V

Current required, I = 12 A

Weight lifted, W = 2000 lb

It is lifting at a speed of 25 ft/min. We need to find the power input to the motor. The product of current and voltage is called power input of the motor.

P=I\times V

P=12\ A\times 240\ V

P = 2880 watts

So, the power input of the motor is 2880 watts. Hence, this is the required solution.

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Sharks are generally negatively buoyant; the upward buoyant force is less than the weight force. This is one reason sharks tend
Tresset [83]

Answer:

8.67807 N

34.7123 N

Explanation:

m = Mass of shark = 92 kg

\rho_{se} = Density of seawater = 1030 kg/m³

\rho_{f} = Density of freshwater = 1000 kg/m³

\rho_{sh} = Density of shark = 1040 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

Net force on the fin is (seawater)

F_n=mg-V_s\rho_{se}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{se}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1030\times 9.81\\\Rightarrow F_n=8.67807\ N

The lift force required in seawater is 8.67807 N

Net force on the fin is (freshwater)

F_n=mg-V_s\rho_{f}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{f}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1000\times 9.81\\\Rightarrow F_n=34.7123\ N

The lift force required in a river is 34.7123 N

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Iron has more mass than gold. Does this mean that the density of iron is greater
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What energy output objects work with the turbine
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Which of the following is not an example of a chemical reaction? Rust, Photosynthesis, Melting Ice, Heat from Fire?
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A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a
KatRina [158]

Answers:

a) 0.80 kg

b) 2.12 s

c) 1.093 m/s^{2}

Explanation:

We have the following data:

k=7 N/m is the spring constant

A=12.5 cm \frac{1 m}{100 cm}=0.125 m is the amplitude of oscillation

V=32 cm/s=0.32 m/s is the velocity of the block when x=\frac{A}{2}=0.0625 m

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

U_{o}+K_{o}=U_{f}+K_{f} (1)

Where:

U_{o}=k\frac{A^{2}}{2} is the initial potential energy

K_{o}=0  is the initial kinetic energy

U_{f}=k\frac{x^{2}}{2} is the final potential energy

K_{f}=\frac{1}{2} m V^{2} is the final kinetic energy

Then:

k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2} (2)

Isolating m:

m=\frac{k(A^{2}-x^{2})}{V^{2}} (3)

m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}} (4)

m=0.80 kg (5)

<h3>b) Period</h3>

The period T is given by:

T=2 \pi \sqrt{\frac{m}{k}} (6)

Substituting (5) in (6):

T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}} (7)

T=2.12 s (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration a_{max} is when the force is maximum F_{max}, as well :

F_{max}=m.a_{max}=k.x_{max} (9)

Being x_{max}=A

Hence:

m.a_{max}=kA (10)

Finding a_{max}:

a_{max}=\frac{kA}{m} (11)

a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg} (12)

Finally:

a_{max}=1.093 m/s^{2}

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