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zlopas [31]
3 years ago
11

Determine the accelerations that result when a 45 N net force is applied to 3kg object

Physics
2 answers:
drek231 [11]3 years ago
6 0

Answer:

18

Explanation:

svp [43]3 years ago
4 0

Answer:

acceleration is 18

Explanation:

45N/3kg=18

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A composite load consists of three loads connected in parallel. One draws 100 W at a PF of 0.92 lagging, another takes 250 W at
fenix001 [56]

Answer:

a) I_{RMS} = 4.79 A

b) PF = 0.908

Explanation:

Get the reactive powers for each of the loads:

Reactive power = Real Power * tanθ

For load 1

Active power, P₁ = 100 W

Power factor, cos \theta_{1} = 0.92

\theta_{1} = cos^{-1} 0.92\\\theta_{1} = 23.074

Q_{1}= P_{1} tan \theta_{1} \\Q_{1}= 100tan 23.074\\Q_{1}= 42.60 W

For load 2

Active power, P₂ = 250 W

Power factor, cos \theta_{2} = 0.8

\theta_{2} = cos^{-1} 0.8\\\theta_{2} = 36.87

Q_{2}= P_{1} tan \theta_{2} \\Q_{2}= 250tan 36.87\\Q_{2}= 187.5 W

For load 3

Active power, P₃ = 250 W

Power factor, cos \theta_{3} = 1

\theta_{3} = cos^{-1} 1\\\theta_{3} =0

Q_{2}= P_{1} tan \theta_{3} \\Q_{3}= 150tan 0\\Q_{3}= 0 W

Calculate the total reactive power, Q_{net} = 42.6 + 187.5 + 0

Q_{net} = 230.1 W

Calculate the total active power, P_{net} = 100 + 250 + 150 = 500 W

S_{net} = P_{net} + Q_{net} \\S_{net} = 500 + j230.1

P_{net} = IVcos \theta_{net}

\theta_{net} = tan^{-1} \frac{230.1}{500} \\\theta_{net} = 24.712

V = 115 V_{rms}

500 = I_{RMS}  * 115 cos 24.712\\I_{RMS} = 500/104.47\\ I_{RMS} = 4.79 A

b) Power factor of the composite load is cos\theta_{net}

\theta_{net}  = 24.712\\PF = cos 24.712\\PF = 0.908

4 0
3 years ago
A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turn
Cerrena [4.2K]

Answer:

e. Not enough information to determine

Explanation:

This question is incomplete. Here is the complete question with my solution afterwards;

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turntable (initially at rest) begins to rotate with its rate of rotation constantly increasing.

What is the first event that will occur?(Assume non-zero frictional force and the same coefficients of friction for both bugs.)

a. The ladybug begins to slide

b. The gentleman bug begins to slide

c. Both bugs begin to slide at the same time

d. Nothing ever happens

e. Not enough information to determine

The centripetal force acting on a rotating body or bugs can be written as,

F=mrw^2

m= mass of the corresponding bugs

r= corresponding radial distance of each bug

w= angular speed of the turntable

The centripetal force tries to slide the bugs in an outward direction and it is directly proportional to the products of its mass and radial distance from the axis of rotation of the turntable

F ∝ mr

Since the radial distance from the axis of rotation of the turntable for each bug is given, but the mass is not given, the given information is therefore not enough to determine which bugs will slide first.

Option "e" is correct.

7 0
3 years ago
Read 2 more answers
What is the measure of the force that gravity applies to an object
GarryVolchara [31]

Answer:

Gravity, or gravitation, is a natural phenomenon by which all things with mass or energy-including planets,stars,galaxies, and even light-are brought toward one another.

Explanation:

4 0
3 years ago
Help sdsfdfdsasfsdfdsaf
WITCHER [35]

Answer:

sorry i don't no !

i am also in trouble now

nobody help me !

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4 0
3 years ago
A closed system contains 30g of gas. How much heat(in joules) is added to or rejected by the system to produce 5000 N-m of work
finlep [7]

Answer : The heat rejected by the system is 1000 J

Explanation :

As per first law of thermodynamic,

q=\Delta U+w

where,

\Delta U = internal energy  of the system

q = heat  added or rejected by the system

w = work done of the system

First we have to determine the internal energy for 30 grams of gas.

As, 1 gram of gas has internal energy = 200 J

So, 30 grams of gas has internal energy = 200 × 30 = 6000 J

Now we have to determine the heat of the system.

q=\Delta U+w

\Delta  = -6000 J

w = 5000 N.m = 5000 J

Now put all the given values in the above formula, we get:

q=-6000J+5000J

q=-1000J

The negative sign indicate that the heat rejected by the system.

Hence, the heat rejected by the system is 1000 J

8 0
3 years ago
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