Ples help qustion in the image

A ball falling through the air has what?

AysviL [449]

A ball falling through the air has a mass, a density, a volume...it is facing air resistance and is being acted on by gravity...it is accelerating and gaining velocity...and it is increasing in kinetic energy.
I suppose out of all those the biggest thing the ball has in this case is ENERGY. There are two main types to focus on...

Kinetic Energy - The further the ball fall the more KE it has...until terminal velocity is reach, then KE would become constant.
Potential Energy - Conversely to that of KE, the further the ball falls the less PE it will have.

<em>Heat/Thermal Energy is technically also present due to the friction from the air resistance, but the transfer of energy between the air and ball is quite complex and not necessary important for basic physics.
</em>
The question itself seem kind of vague and open ended, but I could just be viewing it the wrong way.
Comment if you need more help!

8 0

3 years ago

A transformer has a primary coil with 106 turns and a secondary coil of 340 turns. The AC voltage across the primary coil has a

UkoKoshka [18]

To solve this problem it is necessary to apply the concepts related to transformers, that is to say passive electrical device that transfers electrical energy from one electrical circuit to one or more circuits.

From the mathematical definition we have that the relationship between the voltage of the first coil and the second coil is proportional to the number of loops of the first and second loop, that is:

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$

Where

$V_p =$ input voltage on the primary coil.

$V_s=$ input voltage on the secondary coil.

$N_p=$ number of turns of wire on the primary coil.

$N_s =$ number of turns of wire on the secondary coil.

Replacing our values we have:

$V_p = 128V$

$N_p = 106$

$N_s = 340$

Replacing,

$\frac{V_s}{128} = \frac{340}{106}$

$V_s = 410.56V$

From the same relations of number of turns and the voltage of the first and second coil we also have the relation of electricity and voltage whereby:

$V_s I_s = V_p I_p$

Where

$I_p$ = Current Primary Coil

$I_s$ = Current secundary Coil

Therefore:

$I_s = \frac{V_p I_p}{V_s}$

$I_s = \frac{(128)(6)}{410.56}$

$I_s = 1.87 A$

Therefore the maximum values for the secondary coil of the voltage is 410.56V and Current is 1.87A

5 0

3 years ago

The maximum amount of pulling force a truck can apply when driving on

kupik [55]

C the correct but not sure?

5 0

3 years ago

An ideal parallel - plate capacitor consists of two parallel plates of area A separated by a distance d. This capacitor is conne

Svetradugi [14.3K]

Answer:

The capacitance is cut in half.

Explanation:

The capacitance of a plate capacitor is directly proportional to the area A of the plates and inversely proportional to the distance between the plates d. So if the distance was doubled we should expect that the capacitance would be cut in half. That can be verified by the following equation that is used to compute the capacitance in such cases:

C = (\epsilon)*(A/d)

Where \epsilon is a constant that represents the characteristics for the insulator between the plates. A is the area of the plates and d is the distance between them. When we double d we have a new capacitance, given by:

C_new = (\epsilon)*(A/2d)

C_new = (1/2)*[(\epsilon)*(A/d)]

Since C = (\epsilon)*(A/d)] we have:

C_new = (1/2)*C

4 0

3 years ago

A common method to measure thermal conductivity of a biomaterial is to insert a long metallic probe axially into the center of a

tia_tia [17]

Answer:

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

Explanation:

Let suppose that thermal conduction is uniform and one-dimensional, the conduction heat transfer ( $\dot Q$ ), measured in watts, in the hollow cylinder is:

$\dot Q = \frac{2\cdot k\cdot L}{\ln \left(\frac{D_{o}}{D_{i}} \right)}\cdot (T_{i}-T_{o})$

Where:

$k$ - Thermal conductivity, measured in watts per meter-Celsius.

$L$ - Length of the cylinder, measured in meters.

$D_{i}$ - Inner diameter, measured in meters.

$D_{o}$ - Outer diameter, measured in meters.

$T_{i}$ - Temperature at inner surface, measured in Celsius.

$T_{o}$ - Temperature at outer surface, measured in Celsius.

Now we clear the thermal conductivity in the equation:

$k = \frac{\dot Q}{2\cdot L\cdot (T_{i}-T_{o})}\cdot \ln\left(\frac{D_{o}}{D_{i}} \right)$

If we know that $\dot Q = 40.8\,W$ , $L = 0.6\,m$ , $T_{i} = 50\,^{\circ}C$ , $T_{o} = 20\,^{\circ}C$ , $D_{i} = 0.01\,m$ and $D_{o} = 0.04\,m$ , the thermal conductivity of the biomaterial is:

$k = \left[\frac{40.8\,W}{2\cdot (0.6\,m)\cdot (50\,^{\circ}C-20\,^{\circ}C)}\right]\cdot \ln \left(\frac{0.04\,m}{0.01\,m} \right)$

$k \approx 1.571\,\frac{W}{m\cdot ^{\circ}C}$

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

8 0

3 years ago