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Alexandra [31]
3 years ago
8

Help!!!! please!!! i’m not sure what the answer is for both questions please

Physics
1 answer:
n200080 [17]3 years ago
7 0

Answer:

a.) 600N

b.) 2 m/s^2

Explanation:

a.)

In order to solve for the needed force, you must use Newton's 2nd law which states that a force is equal to mass multiplied by acceleration.

Basically: Force = mass * acceleration ; F = ma

Since both mass and acceleration are already given, just plug them into the formula and solve for force:

Mass = 200kg

Acceleration = 3m/s^2

F = m * a = 200 * 3 = 600kg * m/s^2 = 600N

b.)

In order to find the acceleration, like the hint already says, you need to find the net force first and then use Newton's 2nd law.

To calculate net force:

Since both forces are parallel (affect the same axis) and in opposite directions, you can subtract the two in order to find resultant magnitude. The resultant direction is the direction of the force with the large magnitude.

Net Force: 8N - 4N = 4N

Net Force Direction: since 8N > 4N, the direction is in the same direciton as the 8N force = Left

Finally, you find acceleration using newton's 2nd law, F = ma.

Since you already know mass and net force, plug in to solve for acceleration:

4N = 2kg * a

a = 4 N / 2kg = 2m/s^2

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A uniform meter stick is pivoted at the 50.00 cm mark on the meter stick. A 400.0 gram object is hung at the 20.0 cm mark on the
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Answer:C

Explanation:

Given

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Taking ACW as positive thus

T_{net}=0.4\times g\times (0.5-0.2)-0.32\times g\times(0.75-0.50)

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THE WINNER WILL BE AWARDED THE BRAINLIEST PLEASE HELPP
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Physics question 28 plz help me
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Answer:

a. I = 30 A

b. E = 1080000 J = 1080 KJ

c. ΔT = 12.86°C

d. Cost = $ 4.32  

Explanation:

a.

The current in the coil is given by Ohm's Law:

V = IR\\I = \frac{V}{R}

where,

I = current = ?

V = Voltage = 120 V

R = Resistance = 4 Ω

Therefore,

I = \frac{120\ V}{4 \Omega}\\

<u>I = 30 A</u>

<u></u>

b.

The energy can be calculated as:

E = VIt\\E = (120\ V)(30\ A)(5\ min)(\frac{60\ s}{1\ min})\\

<u>E = 1080000 J = 1080 KJ</u>

<u></u>

c.

For the increase in the temperature of water:

E = mC\Delta T\\

where,

m = mass of water = 20 kg

C = specific heat of water = 4.2 KJ/kg.°C

Therefore,

1080\ KJ = (20\ kg)(4.2\ KJ/kg.^oC)\Delta T

<u>ΔT = 12.86°C</u>

<u></u>

d.

First, we will calculate the total energy consumed:

E=(Power)(Time)\\E=VI(Time)\\E = (120\ V)(30\ A)(0.5\ h/d)(30\ d)\\E = 54000\ Wh\\E = 54 KWh

Now, for the cost:

Cost = (Unit\ Cost)(Energy)\\Cost = (\$ 0.08\KWh)(54\ KWh)

<u>Cost = $ 4.32</u>

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