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2 years ago

Which three factors are used to calculate gravitational potential energy?

Physics

1 answer:

NeX [460]2 years ago

7 0

Answer:

Height, mass, acceleration.

Explanation:

I hope it helps u dear! ^_^

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To dry the brakes, drive slowly in __________ and apply light pressure on your brake pedal.

ira [324]

<span>The answer is low gear. To dry the brakes of a vehicle, one must driv slowly in low gear and apply light pressure on the vehicle's brake pedal. Dry brakes are less expensive and are cheap when it needs replacement, yet under pressure they may cause more damage to the system than wet brakes, which are better in terms of emergency.</span>

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3 years ago

What are the names of neptunes moons

alexandr402 [8]

Answer:

1. naiad

2.Thalassa

3 Despina

4 Galatea

5 Larissa

6 Hippocamp

7 Proteus ˈ

8 Triton

9 Nereid

10 Halimede

11 Sao

12 Laomedeia

13 Psamathe

14 Neso

Explanation:

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5 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

How many newtons of force are represented by the following amount: 3 kg·m/sec^2?

Jet001 [13]

1N=1kg•m/s^2 so the answer is 3N

8 0

4 years ago

A very long insulating cylinder has radius R and carries positive charge distributed throughout its volume. The charge distribut

blsea [12.9K]

Answer:

1. $E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. $E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3.The results from part 1 and 2 agree when r = R.

Explanation:

The volume charge density is given as

$\rho (r) = \alpha (1-\frac{r}{R})$

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.

1. Since the cylinder is very long, Gauss’ Law can be applied.

$\int {\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

$\int\, da = 2\pi r h$

where ‘h’ is the length of the imaginary Gaussian surface.

$Q_{enc} = \int\limits^r_0 {\rho(r)h} \, dr = \alpha h \int\limits^r_0 {(1-r/R)} \, dr = \alpha h (r - \frac{r^2}{2R})\left \{ {{r=r} \atop {r=0}} \right. = \alpha h (\frac{2Rr - r^2}{2R})\\E2\pi rh = \alpha h \frac{2Rr - r^2}{2R\epsilon_0}\\E(r) = \alpha \frac{2R - r}{4\pi \epsilon_0 R}\\E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

$Q_{enc} = \int\limits^R_0 {\rho(r)h} \, dr = \alpha \int\limits^R_0 {(1-r/R)h} \, dr = \alpha h(r - \frac{r^2}{2R})\left \{ {{r=R} \atop {r=0}} \right. = \alpha h(R - \frac{R^2}{2R}) = \alpha h\frac{R}{2} \\E2\pi rh = \frac{\alpha Rh}{2\epsilon_0}\\E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3. At the boundary where r = R:

$E(r=R) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R}) = \frac{\alpha}{4\pi \epsilon_0}\\E(r=R) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r} = \frac{\alpha}{4\pi \epsilon_0}$

As can be seen from above, two E-field values are equal as predicted.

4 0

3 years ago