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3 years ago

The gravitational potential energy of an object is due to

Physics

1 answer:

ExtremeBDS [4]3 years ago

8 0

d. all of these

look at the picture to know from where we get this law with application on it

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An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.

Anna35 [415]

Answer

Given,

Energy absorbed, $Q_H = 1.69\ kJ$

Energy expels, $Q_C = 1.25\ kJ$

Temperature of cold reservoir, T = 27°C

a) Efficiency of engine

$\eta = \dfrac{Q_H - Q_C}{Q_H}\times 100$

$\eta = \dfrac{1.69 - 1.25}{1.69}\times 100$

$\eta =26.03 %$

b) Work done by the engine

$W = Q_H- Q_C$

$W =1.69 - 1.25$

$W = 0.44\ kJ$

c) Power output

t = 0.296 s

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

$P = \dfrac{0.44}{0.296}$

$P = 1.486\ kW$

8 0

3 years ago

Horseshoe bats (genus Rhinolophus) emit sounds from their nostrils, then listen to the frequency of the sound reflected from the

Triss [41]

Answer:

Check the explanation

Explanation:

This is the step by step explanation to the above question:

$v_i = v [ f_L *(v - v_b) - f_s*(v + v_b)] / [f_L * (v - v_b) + f_s*(v +v_b)]$

= v * (83.1 * (v-4.3) - 80.7 ( v+4.3))/ [83.1 *(v - 4.3) + 80.7*(v + 4.3)]

v = 344 m/s

vi = 344 * ( 83.1* (344-4.3) - 80.7*(344+4.3) ) / (83.1 *(344 - 4.3) + 80.7*(344 + 4.3))

= 0.74 m/s

8 0

3 years ago

One mole of water is equivalent to 18 grams of water. A glass of water has a mass of 200 g. How many moles of water is this?

Black_prince [1.1K]

200g*1 mole/ 18g=11.1 moles There are 11.1 moles of water.

4 0

3 years ago

Read 2 more answers

Hat is the approximate pressure of air at sea level?

Tanya [424]

c) 101kPa

Hope I helped! ( Smiles )

5 0

4 years ago

Read 2 more answers

if a stone is projected at an angle of 50 degrees to the horizontal with an initial velocity of 50m/s, what is the vertical comp

Vilka [71]

Answer:

38.3 m/s

Explanation:

To find vertical component of initial velocity, you'd have to use sine ratio:

$\displaystyle{\sin \theta = \dfrac{u_y}{u}}$

$\displaystyle{u_y}$ is vertical component of initial velocity and $\displaystyle{u}$ is initial velocity given which is 50 m/s.

A stone is projected at an angle of 50 degrees so $\displaystyle{\theta}$ = 50°. Substitute in the formula:

$\displaystyle{\sin 50^{\circ} = \dfrac{u_y}{50}}\\\\\displaystyle{50 \sin 50^{\circ} = u_y}\\\\\displaystyle{u_y = 38.3 \ \, \sf{m/s}}$

Therefore, the vertical component of initial velocity is approximately 38.3 m/s

(The picture is also attached for visual reference!)

3 0

2 years ago