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marishachu [46]
3 years ago
7

What is the net force acting on the object above?

Physics
1 answer:
photoshop1234 [79]3 years ago
6 0
The answer would be C (the one you picked)

Hope this helps

Have a great day/night
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A car accelerates while trying to merge onto the freeway. Its speed goes from 0 km/h to 70km/h in 10 seconds. What is its accele
atroni [7]

Answer:

1.944m_s²

Explanation:

First convert speed from km/h to m/s the use the formula a=v-u

t

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Help me someone ?...
Burka [1]
Gradpoint? The answer is B
8 0
3 years ago
Arrange the core steps of the scientific method in sequential order.
andriy [413]
<span>the picture bellow gives you all the answers, this picture was given to my class by my science teacher a few years ago so it is accurate

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4 0
4 years ago
Read 2 more answers
If a 2kg ball has and initial velocity of
lakkis [162]

Answer:

\boxed {\boxed {\sf 12 \ Newtons }}

Explanation:

Force is equal to the product of mass and acceleration.

F=m*a

We know the mass, but not the acceleration. Therefore, we must calculate it before we can calculate force.

1. Calculate Acceleration

Acceleration is the change in velocity over the change in time.

a=\frac{V_f-V_i}{t}

The final velocity is 10 meters per second and the initial velocity is 4 meters per second. The time is 1 second.

V_f=10 \ m/s \\V_i= 4 \ m/s \\t= 1 \ s

Substitute the values into the formula.

a=\frac{10 \ m/s-4 \ m/s }{1 \ s}

Solve the numerator.

a=\frac{6 \ m/s}{1 \ s }

Divide.

a= 6 \ m/s/s=6 \ m/s^2

2. Calculate Force

Now we know the acceleration and the mass.

m= 2 \ kg \\a= 6 \ m/s^2

Substitute the values into the fore formula.

F= 2 \ kg * 6  \ m/s^2

Multiply.

F= 12 \ kg*m/s^2

  • 1 kilogram meter per square second is equal to 1 Newton.
  • Our answer of 12 kg*m/s² is equal to 12 Newtons

F= 12 \  N

The force applies to the ball was <u>12 Newtons.</u>

8 0
3 years ago
A 1 m3tank containing air at 10oC and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35oC and 15
Sveta_85 [38]

Answer:

- the volume of the second tank is 1.77 m³

- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

Explanation:

Given that;

V_{A} = 1 m³

T_{A} = 10°C = 283 K

P_{A} = 350 kPa

m_{B} = 3 kg

T_{B} = 35°C = 308 K

P_{B} = 150 kPa

Now, lets apply the ideal gas equation;

P_{B} V_{B} = m_{B}RT_{B}

V_{B} = m_{B}RT_{B} / P_{B}

The gas constant of air R = 0.287 kPa⋅m³/kg⋅K

we substitute

V_{B} = ( 3 × 0.287 × 308) / 150

V_{B} = 265.188 / 150  

V_{B} = 1.77 m³

Therefore, the volume of the second tank is 1.77 m³

Also, m_{A} =  P_{A}V_{A} / RT_{A} = (350 × 1)/(0.287 × 283) = 350 / 81.221

m_{A}  = 4.309 kg

Total mass, m_{f} = m_{A} + m_{B} = 4.309 + 3 = 7.309 kg

Total volume V_{f} = V_{A} + V_{B}  = 1 + 1.77 = 2.77 m³

Now, from ideal gas equation;

P_{f} =  m_{f}RT_{f} / V_{f}

given that; final temperature T_{f} = 20°C = 293 K

we substitute

P_{f} =  ( 7.309 × 0.287 × 293)  / 2.77

P_{f} =  614.6211119 / 2.77

P_{f} =  221.88 kPa ≈ 222 kPa

Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

6 0
3 years ago
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