--ⓒ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).
Ok, assuming "mj" in the question is Megajoules MJ) you need a total amount of rotational kinetic energy in the fly wheel at the beginning of the trip that equals
(2.4e6 J/km)x(300 km)=7.2e8 J
The expression for rotational kinetic energy is
E = (1/2)Iω²
where I is the moment of inertia of the fly wheel and ω is the angular velocity.
So this comes down to finding the value of I that gives the required energy. We know the mass is 101kg. The formula for a solid cylinder's moment of inertia is
I = (1/2)mR²
We want (1/2)Iω² = 7.2e8 J and we know ω is limited to 470 revs/sec. However, ω must be in radians per second so multiply it by 2π to get
ω = 2953.1 rad/s
Now let's use this to solve the energy equation, E = (1/2)Iω², for I:
I = 2(7.2e8 J)/(2953.1 rad/s)² = 165.12 kg·m²
Now find the radius R,
165.12 kg·m² = (1/2)(101)R²,
√(2·165/101) = 1.807m
R = 1.807m
Answer:
35.3 N
Explanation:
U = 0, V = 0.61 m/s, s = 0.39 m
Let a be the acceleration.
Use third equation of motion
V^2 = u^2 + 2 as
0.61 × 0.61 = 0 + 2 × a × 0.39
a = 0.477 m/s^2
Force = mass × acceleration
F = 74 × 0.477 = 35.3 N
Answer:
D. only briefly while being connected or disconnected.
Explanation:
As we know that transformer works on the principle of mutual inductance
here we know that as per the principle of mutual inductance when flux linked with the primary coil charges then it will induce EMF in secondary coil
So here when AC source is connected with primary coil then it will give output across secondary coil because AC source will have change in flux with time.
Now when we connect DC source across primary coil then it will not induce any EMF across secondary coil because DC source is a constant voltage source in which flux will remain constant always
So here in DC source the EMF will only induce at the time of connection or disconnection when flux will change in it while rest of the time it will give ZERO output
so correct answer will be
D. only briefly while being connected or disconnected.
Answer : 6.022• 10^23 atoms of potassium
