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3 years ago

10

Magnet A and B are of equal magnetic strength and which position will magnets A&B have the greatest attraction for toward ea

ch other

2 answers:

Oksanka [162]3 years ago

8 0

Its D.

(sorry the picture got cut off)

77julia77 [94]3 years ago

3 0

When they are facing each other, almost touching

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2. If 66.2 J of thermal energy is removed from a 0.141 kg piece of iron, what is its change in temperature? The specific heat ca

yawa3891 [41]

Answer: I think is -5

Explanation:

3 0

2 years ago

AN airplane travels 4000 m in 20s on a heading 0f 35 degrees north west. Calculate average velocity .

alexandr402 [8]

Isn't velocity Distance over time? if the degree isn't adding resistance it should be 4000 ÷ 20 which gives you 200mps ("per second") which is the velocity without resistance.

6 0

3 years ago

Help with this please!

DochEvi [55]

Answer:

I have no clue.

Explanation:

3 0

3 years ago

A large plate carries a uniform charge density σ = 8. 85 × 10-9 c/m2. a pattern showing equipotential surfaces with a 5 v potent

vfiekz [6]

The potential difference comes out to be

$10 \times 10 {}^{ - 3} m$

Given:

σ = 8. 85 × 10-9 c/m2

we know,

$E = \frac{σ}{2ε0}$

$E = \frac{8.85 \times 10 {}^{ - 9} }{2ε0}$

$E = \frac{v}{d}$

given the potential difference between two equipotential surface=5v

E=∆v

∆d=∆v/E

$= \frac{5 \times 8.85 \times 10 { }^{ - 12} \times 2 }{8.85 \times 10 {}^{ - 9} }$

$Δ = 10 \times 10 {}^{ - 3} m$

Thus the potential difference is

$10 \times 10 {}^{ - 3} m$

Learn more about potential difference from here: brainly.com/question/28165869

#SPJ4

5 0

1 year ago

A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at lif

Serggg [28]

Answer:

Height of the rocket be one minute after liftoff is 40.1382 km.

Explanation:

$v(t)=-gt-v_e\times \ln \frac{m-rt}{m}$

v = velocity of rocket at time t

g = Acceleration due to gravity = $9.8 m/s^2$

$v_e$ = Constant velocity relative to the rocket = 2,900m/s.

m = Initial mass of the rocket at liftoff = 29000 kg

r = Rate at which fuel is consumed = 170 kg/s

Velocity of the rocket after 1 minute of the liftoff =v

t = 1 minute = 60 seconds'

Substituting all the given values in in the given equation:

$v(60)=-9.8 m/s^2\times 60 s-2,900m/s\times \ln (\frac{29,000 kg-170 kg/s\times 60 s}{2,9000 kg})$

$v(60) = 668.97 m/s$

Height of the rocket = h

$Velocity=\frac{Displacement}{time}$

$668.97 m/s=\frac{h}{60 s}$

$h=668.97 m/s\times 60 s=40,138.2 m = 40.1382 km$

Height of the rocket be one minute after liftoff is 40.1382 km.

4 0

3 years ago