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4 years ago

What factors affect attractive force

2 answers:

8 0

Two Factors That Affect How Much Gravity Is on an Object. Gravity is the force that gives weight to objects and causes them to fall to the ground when dropped. Two major factors, mass and distance, affect the strength of gravitational force on an object.

Dmitriy789 [7]4 years ago

5 0

attractive force only means force pulling two things together but factors affecting the force depend on the kind of force. gravity is an attractive force n the other helper gave the factors.

another attractive force is magnetic force and how strong the magnet will determine how strong the attractive force.

electrostatic force can also be attractive and the amount of electric charge will determine the force.

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2 Polnis

Marta_Voda [28]

Answer:

Oprion A

The length of time over which the conditions are measured

Explanation:

Weather captures the daily atmospheric conditions over a short duration hence it is short-term. Climate is the average of weather and covers a longer duration hence long-term. Therefore, what differentiatea these two is the length of time over which the conditions are measured.

7 0

3 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

4 years ago

what is the mechanical advantage of a crowbar when a worker uses 10N of force to pry open a window that has a resistance of 500N

Oksana_A [137]

Answer:

Explanation:

The mechanical advantage of a machine is given by

$MA=\frac{F_{out}}{F_{in}}$

where

$F_{out}$ is the output force

$F_{in}$ is the input force

For the crowbar in this problem,

$F_{in}=10 N$ is the force in input applied by the worker

$F_{out}=500 N$ is the force that the machine must apply in output to overcome the resistance of the window and to open it

Substituting into the equation, we find

$MA=\frac{500}{10}=50$

3 0

4 years ago

Read 2 more answers

What is 0 point energy times negative 0 point energy ????????

elixir [45]

1 point energy should be the answer

8 0

3 years ago

A permanent magnet has a magnetic dipole moment of 0.160 A · m^2. The magnet is in the presence of an external uniform magnetic

Elena L [17]

Answer:

the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm

the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J

Explanation:

The torque is given by :

$\bar {N} = \bar {m} * \bar {B}$

where ;

m = 0.160 A.m²

B = 0.0800 T

θ = 35°

So the magnitude of the torque N = mBsinθ

N = (0.160)(0.0800)(sin 35°)

N = 0.007341

N = 7.34×10⁻³ Nm

Hence, the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm

b) The potential energy $\bar{U} = \bar{-m} * \bar{B}$

U = -mBcosθ

U = (- 0.160)(0.0800)(cos 45)

U = -0.010485

U = -1.0485 ×10⁻² J

Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J

6 0

3 years ago