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2 years ago

The amount of water vapor in the air is known as the _______.

Physics

2 answers:

storchak [24]2 years ago

5 0

It is C the humidity

Bond [772]2 years ago

5 0

<span>The amount of water vapor in the air is known as the _______.

c. Humidity

</span>

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A constant torque of 3 Nm is applied to an unloaded motor at rest at time t = 0. The motor reaches a speed of 1,393 rpm in 4 s.

irakobra [83]

Answer:

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

Explanation:

From Newton's Laws of Motion and Principle of Motion of D'Alembert, the net torque of a system ( $\tau$ ), measured in Newton-meters, is:

$\tau = I\cdot \alpha$ (1)

Where:

$I$ - Moment of inertia, measured in Newton-meter-square seconds.

$\alpha$ - Angular acceleration, measured in radians per square second.

If motor have an uniform acceleration, then we can calculate acceleration by this formula:

$\alpha = \frac{\omega - \omega_{o}}{t}$ (2)

Where:

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\omega$ - Final angular speed, measured in radians per second.

$t$ - Time, measured in seconds.

If we know that $\tau = 3\,N\cdot m$ , $\omega_{o} = 0\,\frac{rad}{s }$ , $\omega = 145.875\,\frac{rad}{s}$ and $t = 4\,s$ , then the moment of inertia of the motor is:

$\alpha = \frac{145.875\,\frac{rad}{s}-0\,\frac{rad}{s}}{4\,s}$

$\alpha = 36.469\,\frac{rad}{s^{2}}$

$I = \frac{\tau}{\alpha}$

$I = \frac{3\,N\cdot m}{36.469\,\frac{rad}{s^{2}} }$

$I = 0.0823\,N\cdot m\cdot s^{2}$

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

5 0

2 years ago

You place an ice cube of mass 7.50×10−3kg and temperature 0.00∘C on top of a copper cube of mass 0.540 kg. All of the ice melts,

lbvjy [14]

Answer:

The value is $T_c = 12 .1 ^oC$

Explanation:

From the question we are told that

The mass of the ice cube is $m_i = 7.50 *10^{-3} \ kg$

The temperature of the ice cube is $T_i = 0^o C$

The mass of the copper cube is $m_c = 0.540 \ kg$

The final temperature of both substance is $T_f = 0^oC$

Generally form the law of thermal energy conservation,

The heat lost by the copper cube = heat gained by the ice cube

Generally the heat lost by the copper cube is mathematically represented as

$Q = m_c * c_c * [T_c - T_f ]$

The specific heat of copper is $c_c = 385J/kg \cdot ^oC$

Generally the heat gained by the ice cube is mathematically represented as

$Q_1 = m_i * L$

Here L is the latent heat of fusion of the ice with value $L = 3.34 * 10^{5} J/kg$

$Q_1 = 7.50 *10^{-3} * 3.34 * 10^{5}$

=> $Q_1 = 2505 \ J$

$2505 = 0.540 * 385 * [T_c - 0 ]$

=> $T_c = 12 .1 ^oC$

4 0

2 years ago

Lori’s family is on a road trip. They split their drive into the five legs listed in the table. Find the average velocity for ea

NARA [144]

It's not possible to answer the question exactly the way it's written.
That's because we don't know anything about the direction they
drive at any time during the trip.

You see, "velocity" is not just a word that you use for 'speed' when
you want to sound smart and technical, like this question is doing.
"Velocity" is a quantity that's made up of speed AND THE DIRECTION
of the motion. If you don't know the direction of the motion, then you
CAN'T tell the velocity, only the speed.

Here are the average speeds that Lori's family drove on each leg
of their trip:

Speed = (distance covered) / (time to cover the distance) .

Leg-A:
Speed = 15km/10min = 1.5 km/min

Leg-B:
Speed = 20km/15min = (1 and 1/3) km/min

Leg-C
Speed = 24km/12min = 2 km/min

Leg-D:
Speed = 36km/9min = 4 km/min

Leg-E:
Speed = 14km/14min = 1 km/min

From lowest speed to highest speed, they line up like this:

[Leg-E] ==> [Leg-B] ==> [Leg-A] ==> [Leg-C] ==> [Leg-D]
1.0 . . . . . . . . 1.3 . . . . . . . 1.5 . . . . . . . 2.0 . . . . . . . 4.0 . . . . km/minute

Whoever drove Leg-D should have been roundly chastised
and then abandoned by the rest of the family.  36 km in 9 minutes
(4 km per minute) is just about 149 miles per hour !

4 0

2 years ago

Read 2 more answers

A small sphere with mass mcarries a positive chargeqand is attached to one end of a silk fiber of lengthL. The other end of the

Aleksandr-060686 [28]

Answer:

(a): The magnitude of the electric force on the small sphere = $\dfrac{q\sigma}{2\epsilon_o}.$

(b): Shown below.

Explanation:

<u>Given:</u>

m = mass of the small sphere.
q = charge on the small sphere.
L = length of the silk fiber.
$\sigma$ = surface charge density of the large vertical insulating sheet.

When the dimensions of the sheet is much larger than the distance between the charge and the sheet, then, according to Gauss' law of electrostatics, the electric field experienced by the particle due to the sheet is given as:

$\rm E = \dfrac{\sigma}{2\epsilon_o}.$

<em>where,</em>

$\epsilon_o$ is the electrical permittivity of the free space.

The electric field at a point is defined as the amount of electric force experienced by a unit positive test charge, placed at that point. The magnitude electric field at a point and the magnitude of the electric force on a charge q placed at that point are related as:

$\rm F_e=qE.$

Thus, the magnitude of the electric force on the small sphere is given by

$\rm F_e = q\times \dfrac{\sigma }{2\epsilon_o}=\dfrac{q\sigma}{2\epsilon_o}.$

The sheet and the small sphere both are positively charged, therefore, the electric force between these two is repulsive, which means, the direction of the electric force on the sphere is away from the sheet along the line which is perepndicular to the sheet and joining the sphere.

When the sphere is in equilibrium, the tension in the fiber is given by the resultant of the weight of the sphere and the electric force experienced by it as shown in the figure attached below.

According to the fig.,

$\rm \tan \theta = \dfrac{F_e}{W}.$

<em>where,</em>

$\rm F_e$ = electric force on the sphere, acting along left.
$\rm W$ = weight of the sphere, acting vertically downwards.

<em />

$\rm F_e = \dfrac{q\sigma}{2\epsilon_o}\\\\W=mg\\\\Therefore,\\\\\tan\theta = \dfrac{\dfrac{q\sigma}{2\epsilon_o}}{mg}=\dfrac{q\sigma}{2mg\epsilon_o}.\\\Rightarrow \theta=\tan^{-1}\left ( \dfrac{q\sigma}{2mg\epsilon_o}\right ) .$

g is the acceleration due to gravity.

6 0

2 years ago

A stone is thrown upward with an initial upward velocity of 15 m/s from a

tensa zangetsu [6.8K]

Answer

for every meters it will go up 15 so if it was 2 secoonds it woudl be 30 and if it was 3 seconds it would be 45 meters

Explanation:

7 0

2 years ago