Are basins a collection of smaller watersheds

Heat is added to an open pan of water at 100.0°c, vaporizing the water. the expanding steam that results does 43.0 kj of work, a

vampirchik [111]

Heat = change in internal energy + Work done The internal energy of a system = heat added and mechanical work done by the system, i.e. U = Q + W rearranging the formula above, will give us: Q = deltaU + W
Q = U - W = 604 kJ - 43.0 kJ = 561,000 J would be the answer.

6 0

2 years ago

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Một xe đang di chuyển với tốc độ 36km/h thì gia tốc và sau 2s xe lên tới tốc độ 54km/h. Tính gia tốc của xe trong 2s ?

yulyashka [42]

Answer:

which language is these

Explanation:

we only use english , Hindi ,tamil , bengali and Malayalam

please tell us that it is which language. From that we will slove this

3 0

2 years ago

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Why can't speed ratio ever equal mechanical advantage?

OLga [1]

Efficiency of a simple machine is the ratio of mechanical advantage and speed ratio.

If mechanical advantage = speed ratio then efficiency = 1
Which is impossible for a practical machine due to presence of frictional force which wastes some energy and thus lowers the output

6 0

2 years ago

Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. At what radius (in nm) of sphe

Law Incorporation [45]

Answer:

3.85 nm

Explanation:

The volume of a sphere is:

V = 4/3 * π * r^3

The volume of a shell is the volume of the big sphere minus the volume of the small sphere

Vs = 4/3 * π * r2^3 - 4/3 * π * r1^3

Vs = 4/3 * π * (r2^3 - r1^3)

If the difference between the radii is 1 nm

r2 = r1 + 1 nm

Vs = 4/3 * π * ((r1 + 1)^3 - r1^3)

Vs = 4/3 * π * (r1^3 + 3*r1^2 + 3*r1 + 1 - r1^3)

Vs = 4/3 * π * (3*r1^2 + 3*r1 + 1)

The volme of the shell is equall to the volume of the inner shell:

4/3 * π * (r1^3) = 4/3 * π * (3*r1^2 + 3*r1 + 1)

r1^3 = 3*r1^2 + 3*r1 + 1

0 = -r1^3 + 3*r1^2 + 3*r1 + 1

Solving this equation electronically:

r1 = 3.85 nm

7 0

3 years ago

sp2606 [1]

Answer:

The new period is approximately 3.93 × 10⁻⁷ h

Explanation:

The radius of the planet = 5 × 10⁶ m

The acceleration due to gravity on the planet, a = 10 m/s²

The period of the planet = 25 h

The centripetal force, $F_c$ , is given by the following equation;

$F_c = \dfrac{m \cdot v^2}{r}$

Where;

v = The linear speed

r = The radius

Therefore, for the apparent weight, W, of an object to be zero, we have;

The weight of the object = The centripetal force of the object

W = Mass, m × Acceleration due to gravity, a

∴ W = $F_c$

Which gives;

$m \times a = \dfrac{m \cdot v^2}{r}$

$a = \dfrac{v^2}{r}$

∵ r = The radius of the planet

We have;

$10 = \dfrac{v^2}{5 \times 10^6}$

v² = 10 × 5 × 10⁶

v = √(10 × 5 × 10⁶) ≈ 7071.07 m/s

The new frequency = Radius of the planet/(Linear speed component of rotation)

∴ The new frequency = 5 × 10⁶/(7071.07) = 707.107 revolutions per second

The new frequency = 707.107 × 60 × 60 = 2545585.2 revolutions per second

The new period = 1/Frequency ≈ 3.93 × 10⁻⁷ hour.

3 0

2 years ago