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Ivan
2 years ago
9

Find the work done in moving a particle from p to q if the magnitude and direction

Physics
1 answer:
exis [7]2 years ago
3 0

<span>Work=32unit</span><span>...hopes thats helps</span>
You might be interested in
When it reaches an altitude of 3600 m, where the temperature is 5.0∘C and the pressure only 0.68 atm, how will its volume compar
dalvyx [7]

Answer:

The question was incomplete. Here is the complete question.

Explanation:

A helium-filled balloon escapes a child’s hand at sea level and 20.0C. When it reaches an altitude of 3600 m, where the temperature is 5.0∘C and the pressure only 0.68 atm, how will its volume compare to that at sea level?

The ideal gas equation:

PV = nRT

P = absolute pressure

V = Volume of a gas

n = no of moles of a gas

R = ideal gas constant

T = Absolute temperature of a gas

For initial and final states:

P_{i}V_{i} = nRT_{i}   \\P_{f}V_{f} = nRT_{f}   \\\frac{V_{f} }{V_{i} } = \frac{P_{i}T_{f} }{P_{f}T_{i} }

= 1.4

7 0
3 years ago
An overnight rainstorm has caused a major roadblock. Three massive rocks of mass m1=584 kg, m2=838 kg, and m3=322 kg have blocke
Elena-2011 [213]

Answer:

Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N

Explanation:

Total force required = Mass x Acceleration,

F = ma

Here we need to consider the system as combine, total mass need to be considered.

Total mass, a = m₁+m₂+m₃ = 584 + 838 + 322 = 1744 kg

We need to accelerate the group of rocks from the road at 0.250 m/s²

That is acceleration, a = 0.250 m/s²

Force required, F = ma = 1744 x 0.25 = 436 N

Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N

8 0
3 years ago
A 5kg goes from 4m/s to 20m/s determine the change in momentum
lukranit [14]

Answer:

i might know the answer to this, so can you help me with my question too?

Explanation:

5 0
3 years ago
. The inner and outer surfaces of a 4-m × 7-m brick wall of thickness 30 cm and thermal conductivity 0.69 W/m-K are maintained a
Anni [7]

Answer:

\frac{dQ}{dt} = 966 W

Explanation:

As we know that the rate of heat transfer due to temperature difference is given by the formula

\frac{dQ}{dt} = \frac{KA(\Delta T)}{L}

here we know that

K = 0.69 W/m-K

A = 4 m x 7 m

thickness = 30 cm

temperature difference is given as

\Delta T = 20 - 5 = 15 ^oC

now we have

\frac{dQ}{dt} = \frac{(0.69W/m-K)(28 m^2)(15)}{0.30}

\frac{dQ}{dt} = 966 W

4 0
3 years ago
A coin is dropped from a height of 421 m. calculate the velocity of the coin after 3 s​
Keith_Richards [23]

Answer:

29.4 m.s

Explanation:

Vf = vo + at       v o = original velocity = 0 in this case

Vf = at

   = 9.81 m/s^2 * 3 = 29.4 m/s

7 0
1 year ago
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