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valentinak56 [21]
3 years ago
6

lf an object's velocity changes from 25 meters per second to 15 meters per second in 2.0 seconds, the magnitude of the object's

acceleration is​
Physics
2 answers:
Mila [183]3 years ago
6 0

Answer:

-5 m/s^2

Explanation:

acceleration =  \frac{change \: in \: velocity}{time}  \\  \\  =  \frac{15 - 25}{2}  \\  \\  =  \frac{ - 10}{2}   \\  \\  =  -  5 \: m {s}^{ - 2}  \\  \\ negative \: sign \: represents \: that \:  \\ object \: is \: slowing \: down.

andreyandreev [35.5K]3 years ago
5 0

Explanation:

Acceleration is the change in velocity over time.

a = Δv / Δt

a = (15 m/s − 25 m/s) / 2.0 s

a = -5.0 m/s²

The magnitude of the acceleration is 5.0 m/s².

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The condition required to work to be done​
Dvinal [7]

Answer:

work=f(costheta)

Explanation:

work is done when a force acts on a body and displaces it on the direction of force

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The advantage of nuclear power include the low production of air pollution and the fact that is not produce a toxic waste materi
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Induced EMF and Current in a Shrinking LoopShrinking Loop. A circular loop of flexible iron wire has an initial circumference of
vodomira [7]

Answer:

Explanation:

Let c be the circumference and r be the radius

c = 2πr , r = c / 2π , area A = π r² = π (c/2π )²  = (1/4π) x c²

flux (ψ) = BA = 1 X 1/4π X c²

dψ/dt = 1/4π x 2c dc/dt =1/2π x c x dc/dt

at t = 8 s

c = 161 - 13 x 8 = 57 cm , dc/dt = 13 cm/s  

e = dψ/dt = (1 / 2π )x 57 x 13 x 10⁻⁴ = 118 x 10⁻⁴ V.

4 0
3 years ago
A circular parallel plate capacitor is constructed with a radius of 0.52 mm, a plate separation of 0.013 mm, and filled with an
olya-2409 [2.1K]

Answer:

289282

Explanation:

r = Radius of plate = 0.52 mm

d = Plate separation = 0.013 mm

A = Area = \pi r^2

V = Potential applied = 2 mV

k = Dielectric constant = 40

\epsilon_0 = Electric constant = 8.854\times 10^{-12}\ \text{F/m}

Capacitance is given by

C=\dfrac{k\epsilon_0A}{d}

Charge is given by

Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}

Number of electron is given by

n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}

The number of charge carriers that will accumulate on this capacitor is approximately 289282.

6 0
3 years ago
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