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2 years ago

At what condition does a body becomes weightless at the equator?

Physics

1 answer:

ArbitrLikvidat [17]2 years ago

4 0

Answer:

The decrease is due to the bulge at the equator (putting more distance between the rest of the planet and the surface

Explanation:

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rt A For a given substance, do you expect the density of the substance in its liquid state to be closer to the density in the ga

Nat2105 [25]

Answer:

A substance in its liquid state is closer to the density of its solid phase than the density of its gaseous phase.

Explanation:

For a substance in its liquid state we can expect the density of the substance more closer to the density of its solid state than its gaseous state because the the inter-molecular space is much close near to incompressible in the liquid state and the the inter-molecular force of attraction is much higher as compared to gaseous state.

In contrast to the molecular properties in liquid state gases have almost negligible inter-molecular force of attraction and very huge inter-molecular spacing which makes it well compressible.

8 0

3 years ago

A 25 kg child is sitting at the top of a 4 m tall slide, what is his potential energy?

Ipatiy [6.2K]

100
U have to multiply
25x4

4 0

2 years ago

A single insulated duct flow experiment using air operating at steady-state is performed in a lab. One measurement location (Sta

weqwewe [10]

Answer:

a) -0.0934 kJ/kg. K

b) The direction of flow is from right to left.

Explanation:

A free flow diagram of the horizontal insulated duct is as shown below.

NOW,

Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:

$\frac{\sigma_{cv}}{m}= (s_2-s_1) \geq 0 \ \ \ -------> \ \ \ 1$

Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.

If the value is less than zero; then we conclude that the assumption is wrong.

Then, the flow is said to be in the opposite direction

Formula for the change in specific entropy can be calculated as:

$s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \ -------> \ \ \ 2$

where;

$s_1, s_2 , s^0(T_2), s^0(T1)$ are specific entropies

R = universal gas constant

$P_1$ = pressure at location 1

$P_2$ = pressure at location 2

We obtain the specific properties of air at temperature at $T_1$ = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)

$s^0(T1)$ = 1.8279 kJ/kg.K

We also obtain the specific properties of air at temperature $T_2$ = 22°C + 273) K = 295 K

From the table A- 22

$s^0(T_2)$ = 1.68515 kJ/kg . K

R = $\frac{8.314 kJ}{28.97 kg.K}$

$P_1 =$ 0.95 bar

$P_2 =$ 0.8 bar

Now replacing our values into equation (2) from above; we have;

$s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )$

$s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97} \ In (\frac{0.8}{0.95} )$

$s_2-s_1 = 1.68515 -1.8279+ 0.0493$

$s_2-s_1 =-0.0934 \ kJ/kg.K$

Equating our result to equation (1)

$s_2-s_1 \geq 0\\-0.0934 \leq 0$

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

We therefore conclude that the direction of flow is from right to left.

3 0

3 years ago

Two protons are released from rest, with only the electrostatic force acting. Which of the following statements must be true abo

ycow [4]

Answer:

(A)

Explanation:

We know , electric potential energy between two charge particles of charges "q" and "Q" respectively is given by kqQ/r where r is the distance between them.

Since the two charged particles are moving apart, the distance between them (r) increases and thus electrical potential energy decreases.

7 0

3 years ago

A heating element has a resistance of 180 Ω and draws a current of 1.60 A. What is its power?

iragen [17]

its either 288Watts or 112.5 Watts

5 0

2 years ago

Read 2 more answers