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2 years ago

Find the force on an object which has the mass of 20kg and an acceleration of 10m/s

Physics

1 answer:

Elis [28]2 years ago

4 0

Answer:

200 Newtons

Explanation:

20 kg x 10 m/s squared

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Two containers (A and B) are in thermal contact with an environment at temperature T = 280 K. The two containers are connected b

zmey [24]

Answer:

The maximum amount of work is $W = 1563.289 \ J$

Explanation:

From the question we are told that

The temperature of the environment is $T = 280\ K$

The volume of container A is $V_A = 2 m^3$

Initially the number of moles is $n = 1.2 \ moles$

The volume of container B is $V_B = 3.5 \ m^3$

At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as

$W = P_A V_A ln[ \frac{V_B}{V_A} ]$

Now from the Ideal gas law

$P_A V_A = nRT$

So substituting for $P_A V_A$ in the equation above

$W = nRT ln [\frac{V_B}{V_A} ]$

Where R is the gas constant with a values of $R = 8.314 \ J/mol$

Substituting values we have that

$W = 1.2 * (8.314) * (280) * ln [\frac{3.5}{2} ]$

$W = 1563.289 \ J$

8 0

2 years ago

A large sheet of charge has a uniform charge density of 9  μCm2. What is the electric field due to this charge at a point just

Alex73 [517]

Answer:

Explanation:

Surface charge density, σ = 9 μC/m² = 9 x 10^-6 C/m²

According to the Gauss theorem,

Electric field due to the sheet is given by

$E = \frac {\sigma }{2\epsilon _{0}}$

$E = \frac{9\times 10^{-6}}{2\times 8.854\times 10^{-12}}$

E = 5.08 x 10^5 N/C

7 0

2 years ago

Why can't fossils form in igneous rocks Magma flows too quickly for it to settle in The extreme heat destroys all of the remains

AleksandrR [38]

Answer:

you are right

Explanation:

7 0

2 years ago

Read 2 more answers

PLEASE I NEED THIS NOW ITS EASY I JUST DIDNT STUDY

dem82 [27]

Answer:

Explanation:

Frequency*speed

= 0.5*2.5

= 1.25m

3 0

2 years ago

An engine absorbs 1749 J from a hot reservoir and expels 539 J to a cold reservoir in each cycle. a. What is the engine’s effici

Masja [62]

Answer:

a)η = 69.18 %

b)W= 1210 J

c)P=3967.21 W

Explanation:

Given that

Q₁ = 1749 J

Q₂ = 539 J

From first law of thermodynamics

Q₁ = Q₂ +W

W=Work out put

Q₂=Heat rejected to the cold reservoir

Q₁ =heat absorb by hot reservoir

W= Q₁- Q₂

W= 1210 J

The efficiency given as

$\eta=\dfrac{W}{Q_1}$

$\eta=\dfrac{1210}{1749}$

$\eta=0.6918$

η = 69.18 %

We know that rate of work done is known as power

$P=\dfrac{W}{t}$

$P=\dfrac{1210}{0.305}\ W$

P=3967.21 W

8 0

2 years ago