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Vlada [557]
3 years ago
9

A baker uses a balance to weigh out some ingredients to make bread.

Physics
1 answer:
Natasha2012 [34]3 years ago
7 0
The Answer is C………
because it is.
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HELP ✋‼️‼️
Fed [463]

Answer:

wwadaww

Explanation:

6 0
3 years ago
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Electrically charged sunspot gases which escape the sun's chromosphere and enter the earth's atmosphere near the magnetic north
Kisachek [45]

Answer:

Northern Lights ( Aurora Borealis)

Explanation:

When the electricaly charged sunspot gases (they are named a solar wind) escape the sun's chromosphere and penetrates from the earth magnetic sheild which is called earth's magnetosphere then upon there interaction with atoms and molecules of our atmosphere there are little bursts of photons in the form of light which made up these northern lights.

6 0
3 years ago
A circular wire loop of radius LaTeX: RR lies in the xy-plane with the z-axis running through its center. There is initially no
frosja888 [35]

Answer:

Explanation:

Given a circular loop of radius R

r = R

Note: the radius lies in the xy plane

Area is given as

A = πr² = πR²

At t = 0, no magnetic field B=0

The magnetic field is given as a function of time

B = C•exp(t) •i + D•t² •k

Where C and D are constant

We want to find the magnitude of EMF in the circular loop.

EMF is given as

ε = - N•dΦ/dt

Where,

N is number of turn and in this case we will assume N = 1.

Φ is magnetic flux and it is given as

Φ = BA

ε = - N•d(BA)/dt

Where A is a constant, then we have

ε = - N•A•dB/dt

B = C•exp(t) •i + D•t² •k

dB/dt = C•exp(t) •i + 2D•t •k

Then,

ε = - N•A•dB/dt

ε = - 1•πR²•(C•exp(t) •i + 2D•t •k)

ε = -πR²•(C•exp(t) •i + 2D•t •k)

So, let find the magnitude of EMF

Generally finding magnitude of two vectors R = a•i + b•j

Then, |R| = √a² + b²

So, applying this we have,

ε = πR² (√(C²•exp(2t) + 4D²t²))

From the given magnetic field, we are given that,

B = 0 at t = 0

B = C•exp(t) •i + D•t² •k

B = 0 = C•exp(0) •i + D•0² •k

0 = C

Then, C = 0.

So, substituting this into the EMF.

ε = πR² (√(0²•exp(2t) + 4D²t²))

ε = πR² (√4D²t²)

ε = πR² × 2Dt

ε = 2πDR²t

So, the EMF is also a function of time

ε = 2πDR²t

4 0
4 years ago
1) a cathode-ray tube is a type of ______ tube.
galina1969 [7]
1) Vacuum tube
2) silicon 
3) integrated, I.C.s
4) this one is worded weird. Transistors are the main part of voltage regulators but they use (require) diodes and capacitors, too.  And they are all put onto a microchip in the power distribution section.  sorry for the long explanation.
5) away from
6) directly, inversely
4 0
3 years ago
A 40 N crate rests on a rough horizontal floor. A 12 N horizontal force is then applied to it. If the coecients of friction are
bixtya [17]

Answer:

Fr = 20 (N)

Explanation: See atached file ( free body diagram)

As for Newton´s low

∑ Fy  =  0

-mg + N = 0        ⇒  - 40 + N = 0       ⇒ N  = 40 [Newtons]

by definition :  Fr = μs * N        ⇒  Fr = 0,5 * 40      ⇒  Fr = 20 (N)

∑ Fx  =  0   body is at rest

Fe - Fr = 0

Fr > Fe

Fr > 12 (N)  the body is at rest

6 0
3 years ago
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