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2 years ago

Deduceți expresia relației Robert Mayer pentru căldurile specifice ale gazulu ideal! va r0g urgent

Physics

1 answer:

stepladder [879]2 years ago

3 0

Uh what..? English please.

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A horizontal wire of length 0.53 m. carrying a current of 7.5 A. is placed in a uniform external magnetic field. When the wire i

Lady_Fox [76]

Answer:

3.4 mT

Explanation:

L = 0.53 m

i = 7.5 A

Theta = 19 degree

F = 4.4 × 10^-3 N

Let B be the strength of magnetic field.

Force on a current carrying conductor placed in a magnetic field.

F = i × L × B × Sin theta

4.4 × 10^-3 = 7.5 × 0.53 × B × Sin 19

B = 3.4 × 10^-3 Tesla

B = 3.4 mT

6 0

3 years ago

Can someone help me with 1-2

ira [324]

Can't really plot a graph here for question 1.

2a) The car speeds up from A to B. The car travels at a constant speed from B to C. The car slows down to a stop from C to D.

b) From the graph, at 10 seconds, the car is moving at 20 m/s.

4 0

3 years ago

Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at

Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

$v = \dfrac{2\pi R}{T}$

$v = \dfrac{2\pi\times 8}{4.30}$

v = 11.69 m/s

now, Force does the ring push on her at the top

$- N - m g = \dfrac{-mv^2}{R}$

$N + m g = \dfrac{mv^2}{R}$

$N = \dfrac{mv^2}{R}- m g$

$N = m(\dfrac{v^2}{R}- g)$

$N = 58\times (\dfrac{11.69^2}{8}- 9.8)$

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

6 0

3 years ago

Why do scientist use different types of models to represent compounds

nexus9112 [7]

Because they are different they all show different traits.

5 0

3 years ago

Starting from rest, a 2.1x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts

nirvana33 [79]

Answer:

1.327363 m/s

0.00090243 m

Explanation:

u = Initial velocity

v = Final velocity

m = Mass of flea

Energy

$E=\frac{1}{2}m(v^2-u^2)\\\Rightarrow 3.7\times 10^{-4}=2.1\times 10^{-4}(v^2-0)\\\Rightarrow v=\sqrt{\frac{3.7\times 10^{-4}}{2.1\times 10^{-4}}}\\\Rightarrow v=1.32736\ m/s$

The velocity of the flea when leaving the ground is 1.327363 m/s

$W=F\times s\\\Rightarrow s=\frac{W}{F}\\\Rightarrow s=\frac{3.7\times 10^{-4}}{0.41}\\\Rightarrow s=0.00090243\ m$

The flea will travel 0.00090243 m upward

8 0

3 years ago