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Sav [38]
3 years ago
12

Eclipse of the sun occurs(A).When the moon is between the sun and the earth.(B).

Physics
1 answer:
lina2011 [118]3 years ago
6 0

Answer:

option a

Explanation:

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If you were trying to describe the difference between power and work you could say:
avanturin [10]
Power is the energy transferred or "WORK DONE" in one second
6 0
3 years ago
Read 2 more answers
A parcel of land is in the shape of an isosceles triangle. The base has a length of 425 ft.; the other sides, which are of equal
kogti [31]

Answer:

The answer to your question is 636.6 ft    

Explanation:

Data

base = 425 ft

angle = 39°

See the picture below

1.- Divide the triangle to get two right triangles.

    Now the superior angle will measure 19.5° and the opposite side will measure 212.5 ft

2.- Use the trigonometric function sine to find the hypotenuse

     sin 19.5 = 212.5/hyp

solve for hyp

    hyp = 212.5 / sin 19.5

Result

    hyp = 212.5/ 0.333

    hyp = 636.6 ft    

4 0
3 years ago
A 1500 kg car traveling 5.0 m/s collides head on with a 3000 kg truck traveling
Setler [38]
  • m1=1500kg
  • m_2=3000kg
  • v_1=5m/s
  • v_2=7m/s

Using law of conservation of momentum

\\ \sf\Rrightarrow m_1v_1-m_2v_2=(m1+m2)v_3

\\ \sf\Rrightarrow 1500(5)-3000(7)=(1500+3000)v_3

\\ \sf\Rrightarrow 7500-21000=4500v_3

\\ \sf\Rrightarrow -13500=4500v_3

\\ \sf\Rrightarrow v_3=-3m/s

6 0
3 years ago
Determine a formula for the acceleration of the system in terms of mA, mB, θ, and g. Ignore the mass of the cord and pulley. Exp
jekas [21]

Answer:

a=\frac{mBg-mAgSin\theta}{mA+mB}

Explanation:

Given two mass on an incline code mA and mB and an angle of inclination \theta. g. Assume that mA is the weight being pulled up and mB the hanging weight.

-The equations of motion from Newton's Second Law are:

mBg-T=mBa where a is the acceleration.

#Substituting for T (tension) gives:

mBg-mAsin\theta-mAa=mBa

#and solving for a:

a=\frac{mBg-mAgSin\theta}{mA+mB} which is the system's acceleration.

8 0
3 years ago
A square coil ℓ = 2cm on a side with 30 turns rotates in a uniform magnetic field, B~ = B0zˆ = 0.1Tˆz, such that the normal of t
kow [346]

Answer:

a) 1.2*10^{-3}cos(1.25t)

b) 0.49mV

Explanation:

a) The coil rotates periodically with period T. Hence, we can write the variation of the magnetic flux with a sinusoidal function, and with max flux NAB. Thus, we have that:

\Phi_B(t)=NABcos(\omega t)\\\\\omega=\frac{2\pi}{T}=1.25\frac{rad}{s}\\\\A=l^2=(0.02m)^2=4*10^{-4}m^2\\\\B=0.1T\\\\\Phi_B(t)=1.2*10^{-3}cos(1.25 t) W

where we have used the values given by the information of the problem for N B and A.

b)

the emf is given by:

emf=-\frac{d\Phi_B}{dt}=-NBA\omega sin(\omega t)\\\\emf(t=12.5s)=-(30)(0.1T)(4*10^{-4})(1.25\frac{rad}{s})sin(1.25*12.5)=1.49*10^{-4}V=0.49mV

hope this helps!!

5 0
3 years ago
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