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3 years ago

Why is it more helpful to know a tornadoes velocity rather than its speed

Physics

2 answers:

max2010maxim [7]3 years ago

6 0

So you know where the tornado is going

victus00 [196]3 years ago

5 0

So you not only know<span> how fast it is going, but where it is going.</span>

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What is convection?<br> what is radiation ?<br> what conduction?

stepan [7]

--ⓒᴄᴏɴᴠᴇᴄᴛɪᴏɴ ᴏᴄᴄᴜʀꜱ ᴡʜᴇɴ ᴘᴀʀᴛɪᴄʟᴇꜱ ᴡɪᴛʜ ᴀ ʟᴏᴛ ᴏꜰ ʜᴇᴀᴛ ᴇɴᴇʀɢʏ ɪɴ ᴀ ʟɪQᴜɪᴅ ᴏʀ ɢᴀꜱ ᴍᴏᴠᴇ ᴀɴᴅ ᴛᴀᴋᴇ ᴛʜᴇ ᴘʟᴀᴄᴇ ᴏꜰ ᴘᴀʀᴛɪᴄʟᴇꜱ ᴡɪᴛʜ ʟᴇꜱꜱ ʜᴇᴀᴛ ᴇɴᴇʀɢʏ. ... ʟɪQᴜɪᴅꜱ ᴀɴᴅ ɢᴀꜱᴇꜱ ᴇxᴘᴀɴᴅ ᴡʜᴇɴ ᴛʜᴇʏ ᴀʀᴇ ʜᴇᴀᴛᴇᴅ. ᴛʜɪꜱ ɪꜱ ʙᴇᴄᴀᴜꜱᴇ ᴛʜᴇ ᴘᴀʀᴛɪᴄʟᴇꜱ ɪɴ ʟɪQᴜɪᴅꜱ ᴀɴᴅ ɢᴀꜱᴇꜱ ᴍᴏᴠᴇ ꜰᴀꜱᴛᴇʀ ᴡʜᴇɴ ᴛʜᴇʏ ᴀʀᴇ ʜᴇᴀᴛᴇᴅ ᴛʜᴀɴ ᴛʜᴇʏ ᴅᴏ ᴡʜᴇɴ ᴛʜᴇʏ ᴀʀᴇ ᴄᴏʟᴅ.

--Ⓡʀᴀᴅɪᴀᴛɪᴏɴ ɪꜱ ᴇɴᴇʀɢʏ ᴛʜᴀᴛ ᴄᴏᴍᴇꜱ ꜰʀᴏᴍ ᴀ ꜱᴏᴜʀᴄᴇ ᴀɴᴅ ᴛʀᴀᴠᴇʟꜱ ᴛʜʀᴏᴜɢʜ ꜱᴘᴀᴄᴇ ᴀɴᴅ ᴍᴀʏ ʙᴇ ᴀʙʟᴇ ᴛᴏ ᴘᴇɴᴇᴛʀᴀᴛᴇ ᴠᴀʀɪᴏᴜꜱ ᴍᴀᴛᴇʀɪᴀʟꜱ. ... ᴛʜᴇ ᴋɪɴᴅꜱ ᴏꜰ ʀᴀᴅɪᴀᴛɪᴏɴ ᴀʀᴇ ᴇʟᴇᴄᴛʀᴏᴍᴀɢɴᴇᴛɪᴄ (ʟɪᴋᴇ ʟɪɢʜᴛ) ᴀɴᴅ ᴘᴀʀᴛɪᴄᴜʟᴀᴛᴇ (ɪ.ᴇ., ᴍᴀꜱꜱ ɢɪᴠᴇɴ ᴏꜰꜰ ᴡɪᴛʜ ᴛʜᴇ ᴇɴᴇʀɢʏ ᴏꜰ ᴍᴏᴛɪᴏɴ). ɢᴀᴍᴍᴀ ʀᴀᴅɪᴀᴛɪᴏɴ ᴀɴᴅ x ʀᴀʏꜱ ᴀʀᴇ ᴇxᴀᴍᴘʟᴇꜱ ᴏꜰ ᴇʟᴇᴄᴛʀᴏᴍᴀɢɴᴇᴛɪᴄ ʀᴀᴅɪᴀᴛɪᴏɴ

--Ⓒᴄᴏɴᴅᴜᴄᴛɪᴏɴ ɪꜱ ᴛʜᴇ ᴡᴀʏ ɪɴ ᴡʜɪᴄʜ ᴇɴᴇʀɢʏ ɪꜱ ᴛʀᴀɴꜱꜰᴇʀʀᴇᴅ (ᴛʜʀᴏᴜɢʜ ʜᴇᴀᴛɪɴɢ ʙʏ ᴄᴏɴᴛᴀᴄᴛ) ꜰʀᴏᴍ ᴀ ʜᴏᴛ ʙᴏᴅʏ ᴛᴏ ᴀ ᴄᴏᴏʟᴇʀ ᴏɴᴇ (ᴏʀ ꜰʀᴏᴍ ᴛʜᴇ ʜᴏᴛ ᴘᴀʀᴛ ᴏꜰ ᴀɴ ᴏʙᴊᴇᴄᴛ ᴛᴏ ᴀ ᴄᴏᴏʟᴇʀ ᴘᴀʀᴛ).

3 0

2 years ago

A long solenoid has 103 turns/cm and carries current i. An electron moves within the solenoid in a circle of radius 2.60 cm perp

Aneli [31]

Answer:

0.23348 A

Explanation:

B = Magnetic field

v = Velocity of electron = $1.38\times 10^7\ m/s$

q = Charge of electron = $1.6\times 10^{-19}\ C$

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

r = Radius of circle = 0.026 m

N = Number of turns = 103 turns/cm = $103\times 100\ turns/m$

I = Current

The magnetic and centripetal force will be balanced

$Bqv=m\frac{v^2}{r}\\\Rightarrow B=\frac{mv}{qr}$

The magnetic field in solenoid is given by

$B=N\mu_0 I\\\Rightarrow I=\frac{B}{N\mu_0}$

From the first equation

$I=\frac{\frac{mv}{qr}}{N\mu_0}\\\Rightarrow I=\frac{mv}{N\mu_0qr}\\\Rightarrow I=\frac{9.11\times 10^{-31}\times 1.38\times 10^7}{103\times 100\times 4\pi \times 10^{-7}\times 1.6\times 10^{-19}\times 0.026}\\\Rightarrow I=0.23348\ A$

The current in the solenoid is 0.23348 A

8 0

2 years ago

A 50 g copper calorimeter contains 250 g of water at 20 C. How much steam be condensed into the water to make the final temperat

Nostrana [21]

Answer:

Approximately $13\; \rm g$ of steam at $100\; \rm ^\circ C$ (assuming that the boiling point of water in this experiment is $100\; \rm ^\circ C\!$ .)

Explanation:

Latent heat of condensation/evaporation of water: $2260\; \rm J \cdot g^{-1}$ .

Both mass values in this question are given in grams. Hence, convert the specific heat values from this question to $\rm J \cdot g^{-1}$ .

Specific heat of water: $4.2\; \rm J \cdot g^{-1}\cdot \rm K^{-1}$ .

Specific heat of copper: $0.39\; \rm J \cdot g^{-1}\cdot K^{-1}$ .

The temperature of this calorimeter and the $250\; \rm g$ of water that it initially contains increased from $20\; \rm ^\circ C$ to $50\; \rm ^\circ C$ . Calculate the amount of energy that would be absorbed:

$\begin{aligned}& Q(\text{copper}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 0.39\; \rm J \cdot g^{-1}\cdot K^{-1} \times 50\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 585\; \rm J \end{aligned}$ .

$\begin{aligned}& Q(\text{cool water}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; \rm J \cdot g^{-1}\cdot K^{-1} \times 250\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 31500\; \rm J \end{aligned}$ .

Hence, it would take an extra $585\; \rm J + 31500\; \rm J = 32085\; \rm J$ of energy to increase the temperature of the calorimeter and the $250\; \rm g$ of water that it initially contains from $20\; \rm ^\circ C$ to $50\; \rm ^\circ C$ .

Assume that it would take $x$ grams of steam at $100\; \rm ^\circ C$ ensure that the equilibrium temperature of the system is $50\; \rm ^\circ C$ .

In other words, $x\; \rm g$ of steam at $100\; \rm ^\circ C$ would need to release $32085\; \rm J$ as it condenses (releases latent heat) and cools down to $50\; \rm ^\circ C$ .

Latent heat of condensation from $x\; \rm g$ of steam: $2260\; {\rm J \cdot g^{-1}} \times (x\; {\rm g}) = (2260\, x)\; \rm J$ .

Energy released when that $x\; {\rm g}$ of water from the steam cools down from $100\; \rm ^\circ C$ to $50\; \rm ^\circ C$ :

$\begin{aligned}Q = \;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; {\rm J \cdot g^{-1}\cdot K^{-1}} \times (x\; \rm g) \times (100\;{\rm ^\circ C} - 50\;{\rm ^\circ C}) \\ =\; & (210\, x)\; \rm J \end{aligned}$ .

These two parts of energy should add up to $32085\; \rm J$ . That would be exactly what it would take to raise the temperature of the calorimeter and the water that it initially contains from $20\; \rm ^\circ C$ to $50\; \rm ^\circ C$ .

$(2260\, x)\; {\rm J} + (210\, x)\; {\rm J} = 32085\; \rm J$ .

Solve for $x$ :

$x \approx 13$ .

Hence, it would take approximately $13\; \rm g$ of steam at $100\; \rm ^\circ C$ for the equilibrium temperature of the system to be $50\; \rm ^\circ C$ .

4 0

2 years ago

Novosadov [1.4K]

Explanation:

The are the same but difference

8 0

2 years ago

Explain how a generator creates electricity.

vlada-n [284]

Generators don't actually create electricity. Instead, they convert mechanical or chemical energy into electrical energy. They do this by capturing the power of motion and turning it into electrical energy by forcing electrons from the external source through an electrical circuit. hope it helps man

5 0

3 years ago

Read 2 more answers