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4 years ago

As shown in the figure below, a seal and the penguin are playing a fun game of tag. The penguin first swims rightward

Physics

1 answer:

RSB [31]4 years ago

7 0

Answer:

Explanation:

Average velocity = Displacement/Time

ΔS = -10+6

ΔS = -4m

Given tme = 30s

Average velocity = -4/30

Average velocity = -0.13m/s

Average speed = Distance/Time

Total distance covered = 10m+6m

Total distance covered = 16m

Average speed = 16/30

Average speed = 0.53m/s

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How are heat and light waves produced on and in the sun?

enyata [817]

All of the electromagnetic energy radiated from the sun (and from
other stars) is the product of nuclear fusion in its core.

8 0

3 years ago

Read 2 more answers

The magnetic field at 8 cm distance from a long straight wire, carrying is 0.2x10^-5 T. How much is the electric current in the

FrozenT [24]

Answer:

The electric current in the wire is 0.8 A

Explanation:

We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:

$B=\frac{2\pi*a }{u*I}$

B= Magnetic field due to a straight and long wire that carries current

u= Free space permeability

I= Electrical current passing through the wire

a = Perpendicular distance from the wire to the point where the magnetic field is located

Magnetic Field Calculation

We cleared (I) of the formula (1):

$I=\frac{2\pi*a*B }{u}$ Formula(2)

$B=0.2*10^{-5} T = 0.2*10^{-5} \frac{weber}{m^{2} }$

a =8cm=0.08m

$u=4*\pi *10^{-7} \frac{Weber}{A*m}$

We replace the known information in the formula (2)

$I=\frac{2\pi*0.08*0.2*10^{-5} }{4\pi *10x^{-7} }$

I=0.8 A

Answer: The electric current in the wire is 0.8 A

4 0

3 years ago

Can someone solve this problem and explain to me how you got it

evablogger [386]

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒ $F\alpha\frac{q1*q2}{r^{2}}$