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Katyanochek1 [597]
3 years ago
14

Does heat always give of heat

Physics
2 answers:
Marysya12 [62]3 years ago
6 0
If you meant doest heat always give off heat than yes

iren2701 [21]3 years ago
3 0
Yes it does (not to be  mean its kinda stupid for you to ask)
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A wire is formed into a circle having a diameter of 11.1 cm and is placed in a uniform magnetic field of 2.79 mT. The wire carri
Ber [7]

Answer:

Maximum torque on the wire is 1.34\times 10^{-4}\ N-m

Explanation:

It is given that,

Diameter of the wire, d = 11.1 cm = 0.111 m

Radius of wire, r = 0.0555 m

Magnetic field, B=2.79\ mT=0.00279\ T

Current, I = 5 A

We need to find the maximum torque on the wire. Torque is given by :

\tau=IAB\ sin\theta

Torque is maximum when, \theta=90

\tau=IAB

\tau=5\times \pi \times (0.0555)^2\times 0.00279

\tau=0.000134\ N-m

or

\tau=1.34\times 10^{-4}\ N-m

So, the maximum toque on the wire is 1.34\times 10^{-4}\ N-m. Hence, this is the required solution.

6 0
3 years ago
A crane lifts 1800kg mass through a vertical height of 6cm in 9 seconds, Taking (g) as 10N/kg. what is the cranes power output?​
valkas [14]

Explanation:

power output=(1800×10×0.06)/9=120watts

6 0
3 years ago
Convert 37.45 kilometers to meters.<br> A.3745 m<br> B. 37,450 m<br> C. 3.745 m<br> D. 0.03745 m
lbvjy [14]

Answer: 37,450

Explanation:

6 0
3 years ago
Read 2 more answers
If the merry-go-round makes one revolution in 10 seconds, what is the child’s linear speed?
Anton [14]
The child's linear speed is
              
    <em> (pi / 5) x (the child's distance from the center of the ride, in feet)</em>

                                                                                        feet per second.
5 0
3 years ago
A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi
slega [8]
We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is
dI =r^2*dm
where dm =M(b*dr)/(ab)
Now we find the moment of inertia by integrating from -a/2 to a/2
The moment of inertia is
I= \int\limits^{-a/2}_{a/2} {r^2*dm} = M \int\limits^{-a/2}_{a/2} r^2(b*dr)/(ab)=(M/a)(r^3/3) (from (-a/2) toI=(M/3a)(a^3/8 +a^3/8)=(Ma^2)/12 (a/2))



4 0
3 years ago
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