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12345 [234]
3 years ago
14

During a chemical reaction, what remains constant? composition or tempature or or energy or bonds

Physics
1 answer:
Luda [366]3 years ago
6 0
Energy. Energy cannot be created or destroyed 
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A coil of wire with 50 turns lies in the plane of the page and has an initial area of 0.180 m2. The coil is now stretched to hav
butalik [34]

Answer:

Explanation:

Given that,

Number of turns of coil

N = 50 turns

Initial area of plane

A1 = 0.18 m²

The coil it stretch to a no area in time t = 0.1s

No area implies that the final area is 0, A2 = 0 m²

Constant magnetic field strength

B = 1.51 T

EMF?

EMF is given as

Using far away Lenz law

ε = —N• dΦ/dt

Where Φ = BA

Then,

ε = —N• d(BA)/dt

Since B is constant,

ε = —N•B dA/dt

ε = —N•B (∆A/∆t)

ε = —N•B(A2—A1)/(t2-t1)

ε = —50 × 1.51 (0—0.18)/(0.1—0)

ε =—75.5 × —0.18 / 0.1

ε = 135.9 V

The induced EMF is 135.9V

Fleming’s left hand rule stated that if the index finger points toward magnetic flux, the thumb towards the motion of the conductor, then the middle finger points towards the induced emf.

Since the area lines in the plane, then the induced emf will be out of the page

5 0
4 years ago
Fix any punctuation or capitalization errors below. Click "Submit Answer" if there
asambeis [7]

Answer:slightly broken

Explanation:

8 0
3 years ago
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Which is not an example of how an object gains elastic potential energy by stretching?
lora16 [44]
C because you don’t put a force of which you choose and the force wouldn’t be strong enough
4 0
4 years ago
A 1200-kg car pushes a 2100-kg truck that has a dead battery to the right. When the driver steps on the accelerator, the drive w
Tresset [83]
2,900 N is the answer if you need an explanation tell me and i will comment the explanation
8 0
4 years ago
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The electric field 1.5 cm from a very small charged object points toward the object with a magnitude of 180,000 N/C. What is the
Ray Of Light [21]

Answer:

q = 4.5 nC

Explanation:

given,

electric field of small charged object, E = 180000 N/C

distance between them, r = 1.5 cm = 0.015 m

using equation of electric field

E = \dfrac{kq}{r^2}

k = 9 x 10⁹ N.m²/C²

q is the charge of the object

q= \dfrac{Er^2}{k}

now,

q= \dfrac{180000\times 0.015^2}{9\times 10^9}

      q = 4.5 x 10⁻⁹ C

      q = 4.5 nC

the charge on the object is equal to 4.5 nC

8 0
3 years ago
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