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kiruha [24]
3 years ago
5

If the mass of an object is 200 kg and the applied force is 2600 N, calculate the acceleration

Physics
1 answer:
sweet [91]3 years ago
3 0

Answer: a = 13 m/s²

Explanation:

Mass of object = 200 Kg

Applied force = 2600 N

Acceleration  = ?

Solution:

Definition:

The acceleration is rate of change of velocity of an object with respect to time.

Formula:

a = Δv/Δt

a = acceleration

Δv = change in velocity

Δt = change in time

Units:

The unit of acceleration is m.s⁻².

Acceleration can also be determine through following formula,

F = m × a

a = F/m  (N = kgm/s²)

a = 2600 kgm/s² / 200 Kg

a = 13 m/s²

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A 275-g sample of nickel at 100.0°C is placed in 100.0 mL of water at 22.0°C. What is the final temperature of the water? Assume
Mademuasel [1]

Answer:

39.6138 °C

Explanation:

Heat gain by water = Heat lost by nickel

Thus,  

m_{water}\times C_{water}\times (T_f-T_i)=-m_{nickel}\times C_{nickel}\times (T_f-T_i)

Where, negative sign signifies heat loss

Or,  

m_{water}\times C_{water}\times (T_f-T_i)=m_{nickel}\times C_{nickel}\times (T_i-T_f)

For water:

Volume = 100.0 mL

Density of water= 1 g/mL

So, mass of the water:  

Mass\ of\ water=Density \times {Volume\ of\ water}  

Mass\ of\ water=1 g/mL \times {100.0\ mL}  

Mass of water  = 100 g

Initial temperature = 22.0 °C

Specific heat of water = 4.186 J/g°C

For nickel:

Mass = 275 g

Initial temperature = 100 °C

Specific heat of nickel = 0.444 J/gK = 0.444 J/g°C

So,  

100\times 4.186\times (T_f-22.0)=275\times 0.444\times (100-T_f)

418.6\times (T_f-22.0)=122.1\times (100-T_f)

418.6\times T_f+122.1\times T_f=21419.2

T_f=39.6138\ ^{0}C

Thus,  

The final temperature of the combined metals is 39.6138 °C

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The the figure shows a famous roller coaster ride. You can ignore friction. If the roller coaster leaves Point Q from rest, what
ch4aika [34]

Answer:

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3 years ago
Please do not round the numbers
slamgirl [31]

Answer:

1)   λ = 24.7 cm,  2) f = 13.88 Hz, 3)  L = 117.3 cm

Explanation:

1) This is a resonance process, that is, the wave going downwards will interfere with the wave going upwards.

This is a tube with one end closed and the other open, at the closed end there is a node and at the open end a belly, so the resonances are

       L = λ / 4

       λ = 4L                     1st harmonic

       λ = 4L / 3                third harmonic

       λ = 4L / 5                fifth harmonic

       λ = 4L / n ’              n’ odd number   n ’= (2n +1)

the wavelength is requested for the eighth resonance n = 8, the corresponding prime number is

          n ’= 2 8 +1

          n ’= 17

we substitute

     λ = 4 105/17

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2) the speed of the wave is related to the wavelength and frequency

          v =   λf

          f = v /λ

          f = 343 / 24.7

          f = 13.88 Hz

3) the next resonance occurs for n = 9, so the prime number is

         n ’= 2 9 +1

         n ’= 19

         L = n’ λ / 4

       

         L = 19 λ / 4  

We must suppose a value for the wavelength, if the wavelength is present in the tube and the length of the column increases, the resonance number increases

         L = 19 24.7/4

         L = 117.3 cm

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3 years ago
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