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kumpel [21]
3 years ago
15

How much work is done in accelerating a 2000 kg car from rest to a speed of 30 m/s?

Physics
1 answer:
Otrada [13]3 years ago
8 0
The work done is the same as the amount of energy increase. The formula for kinetic energy is 122
1
2
m
v
2
.

The initial KE of the car is 12(1000)×202=200,000
1
2
(
1000
)
×
20
2
=
200
,
000
joules.

The final KE of the car is 12(1000)×302=450,000
1
2
(
1000
)
×
30
2
=
450
,
000
joules.

The difference between these is the amount of work done: 450,000−200,000=250,000
450
,
000
−
200
,
000
=
250
,
000
joules.
14.2K viewsView upvotes

20



3
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Ishmael has an object that produces an electric field. He then places another charged object inside that electric field.
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The increase the magnitude of the electric force ;

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2) The distance of separation between the objects must be reduced.

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The increase the magnitude of the electric force ;

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2) The distance of separation between the objects must be reduced.

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3 0
2 years ago
Developments in technology have allowed scientists to conclude that Earth more than 4.5 billion years old. Why was the discovery
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4 0
4 years ago
An electron (mass 9 × 10-31 kg) is traveling at a speed of 0.92 in an electron accelerator. An electric force of 1.4 × 10-13 N i
pogonyaev

Given Information:

Mass of electron = m = 9x10⁻³¹ kg

initial speed of electron = v₁ = 0.92c

Force = F =  1.4x10⁻¹³ J

Distance = d = 3 m

Required Information:

Final speed of electron = v₂ = ?

Answer:

Final speed of electron = v₂ = 2.974x10⁸ m/s

Explanation:

As we know from the conservation of energy,

E₂ - E₁ = W

E₂ = E₁ + W

Where E₂ is the final energy of electron and E₁ is the initial energy of electron

The above equation can be written in the form of particle energy

γ₂mc² = γ₁mc² + W

where γ₁ and γ₂ are given by

γ₁ = 1/√1 - (v₁/c)²

γ₂ = 1/√1 - (v₂/c)²

First calculate γ₁

γ₁ = 1/√1 - (0.92c/c)²

γ₁ = 2.55 m

Now calculate γ₂

γ₂ = (γ₁mc² + W)/mc²

First we need to find the work done

W = F*d

W = 1.4x10⁻¹³*3

W = 4.2x10⁻¹³ J

so γ₂ is

γ₂ = (2.55*9x10⁻³¹*(3x10⁸)² + 4.2x10⁻¹³)/9x10⁻³¹*(3x10⁸)²

γ₂ = 7.73

Now we can find the new speed of the electron

γ₂ = 1/√1 - (v₂/c)²

Re-arranging the above equation results in

v₂ = c*√(1 - 1/γ₂²)

v₂ = 3x10⁸*√(1 - 1/7.73²)

v₂ = 2.974x10⁸ m/s

4 0
3 years ago
A parallel-plate, air-gap capacitor has a capacitance of 0.14 mu F. The plates are 0.5 mm apart, What is the area of each plate?
Marysya12 [62]

Answer:

7.9060 m²

8.57 Volts

5.142×10⁻⁶ Joule

1.2×10⁻⁶ Coulomb

Explanation:

C = Capacitance between plates = 0.14 μF = 0.14×10⁻⁶ F

d = Distance between plates = 0.5 mm = 0.5×10⁻³ m

Q = Charge = 1.2 μC = 1.2×10⁻⁶ C

ε₀ = Permittivity = 8.854×10⁻¹² F/m

Capacitance

C=\frac{\epsilon_{0}A}{d}\\\Rightarrow A=\frac{Cd}{\epsilon_{0}}\\\Rightarrow A=\frac{0.14\times 10^{-6}\times 0.5\times 10^{-3}}{8.854\times 10^{-12}}\\\Rightarrow A=7.9060\ m^2

∴ Area of each plate is 7.9060 m²

Voltage

V=\frac{Q}{C}\\\Rightarrow V=\frac{1.2\times 10^{-6}}{0.14\times 10^{-6}}\\\Rightarrow V=8.57\ Volts

∴ Potential difference between the plates if the capacitor is charged to 1.2 μC  is 8.57 Volts.

Energy stored

E=0.5CV²

⇒E = 0.5×0.14×10⁻⁶×8.57²

⇒E = 5.142×10⁻⁶ Joule

∴ Stored energy is 5.142×10⁻⁶ Joule

Charge

Q = CV

⇒Q = 0.14×10⁻⁶×8.57

⇒Q = 1.2×10⁻⁶ C

∴ Charge the capacitor carries before a spark occurs between the two plates is 1.2×10⁻⁶ Coulomb

6 0
3 years ago
The moons of Mars, Phobos (Fear) and Deimos (Terror), are very close to the planet compared to Earth's Moon. Their orbital radii
tankabanditka [31]

Answer:

0.2528

Explanation:

To calculate the period we need the formula:

T=\frac{2\pi r^{3/2}}{\sqrt{GM}}

Where r is the radius of the moon, G is the universal constant of gravitation and M is the mass of mars.

The period of Phobos:

T_{p}=\frac{2\pi r_{p}^{3/2}}{\sqrt{GM}}

The period of Deimos:

T_{D}=\frac{2\pi r_{D}^{3/2}}{\sqrt{GM}}

The ratio of the period of Phobos and Deimos:

\frac{T_{p}}{T_{D}}=\frac{\frac{2\pi r_{p}^{3/2}}{\sqrt{GM}}}{\frac{2\pi r_{D}^{3/2}}{\sqrt{GM}}}

\frac{T_{p}}{T_{D}}=\frac{\sqrt{GM}2\pi r_{p}^{3/2}}{\sqrt{GM}2\pi r_{D}^{3/2}}

Most terms get canceled and we have:

\frac{T_{p}}{T_{D}}=\frac{r_{p}^{3/2}}{r_{D}^{3/2}}

According to the problem

r_{p}=9,378km\\r_{D}=23,459km

so the ratio will be:

\frac{T_{p}}{T_{D}}=\frac{(9,378)^{3/2}}{(23,459)^{3/2}}=\frac{908166.22}{3593058.125}=0.25275 ≈ 0.2528

the ratio of the period of revolution of Phobos to that of Deimos is 0.2528

8 0
3 years ago
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