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3 years ago

Mercury is 0.39 AU from the sun. What is its distance from the sun in kilometers?

Physics

1 answer:

qaws [65]3 years ago

8 0

Answer:

0.39AU = 58344000 [km] or 58.500.000 [km]

Explanation:

This is a problem that consists simply in converting units:

AU = astronomic unit

$1 [AU] = 1.496*10^{8} [km]\\0.39 [AU] = x\\\\Therefore\\x = \frac{0.39*1.496*10^{8} }{1} \\x = 58344000[km]$

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In nuclear fission, a nucleus splits roughly in half. (a) What is the potential 2.00 × 10−14 m from a fragment that has 46 proto

Valentin [98]

Answer:

electric potential is 3.31 × $10^{6}$ V

potential energy is 152 MeV

Explanation:

given data

fragment charge Q = 46 protons = 46 × 1.6 × $10^{-19}$ C

to find out

electric potential and potential energy

solution

we know here distance from fragment d = 2 × $10^{-14}$ m

and constant for electric force k that is 9 × $10^{9}$ N-m²/C²

so that we can find electric potential = kQ/d

electric potential = 9 × $10^{9}[/tex ×46 × 1.6 × [tex]10^{-19}$ / ( 2 × $10^{-14}$ )

electric potential = 3.31 × $10^{6}$ V

and

we know relation between electric potential and potential

that is V = U/q

so U will be = qV

now put all value

we get potential energy U

potential energy = 46 × 3.31 × $10^{6}$

potential energy = 1.52 × $10^{8}$ eV

so potential energy = 152 MeV

4 0

3 years ago

how long does it take a 750watt heater operating at full rating to rais the temperature of 1kg of water From 40°C to 70°C {S.H.C

siniylev [52]

Answer:

168 seconds (2 min 48 s)

Explanation:

Find the heat absorbed by the water.

q = mCΔT

q = (1 kg) (4200 J/kg/K) (70°C − 40°C)

q = 126,000 J

Power is energy per time.

P = q / t

750 W = 126,000 J / t

t = 168 s

It takes 168 seconds (2 min 48 s).

6 0

3 years ago

A thin nonconducting rod with a uniform distribution of positive charge Q is bent into a complete circle of radius R. The centra

Dmitriy789 [7]

Answer:

(a). If z = 0, The electric field due to the rod is zero.

(b). If z = ∞, The electric field due to the rod is $E\propto\dfrac{1}{z^2}$ .

(c). The positive distance is $\dfrac{R}{\sqrt{2}}$

(d). The maximum magnitude of electric field is $1.54\times10^{4}\ N/C$

Explanation:

Given that,

Radius = 2.00 cm

Charge = 4.00 mC

(a). If the radius and charge are R and Q.

We need to calculate the electric field due to the rod

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}$

Where, Q = charge

z = distance

If z = 0,

Then, The electric field is

$E=0$

(b). If z = ∞, z>>R

So, R = 0

Then, the electric field is

$E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Q}{z^2}$

$E\propto\dfrac{1}{z^2}$

(c). In terms of R,

We need to calculate the positive distance

If $E\rightarrow E_{max}$

Then, $\dfrac{dE}{dz}=0$

$\dfrac{Q}{4\pi\epsilon_{0}}(\dfrac{(z^2+R^2)^\frac{3}{2}-\dfrac{3z}{2}(z^2+R^2)^\dfrac{1}{2}}{(z^2+R^2)^2})=0$

Taking only positive distance

$z=\dfrac{R}{\sqrt{2}}$

(d). If R = 2.00 and Q = 4.00 mC

We need to calculate the maximum magnitude of electric field

Using formula of electric field

$E_{max}=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}$

$E_{max}=9\times10^{9}\times\dfrac{4.0\times10^{-6}\times\dfrac{2.00}{\sqrt{2}}}{((\dfrac{2.00}{\sqrt{2}})^2+(2.00)^2)^{\frac{2}{3}}}$

$E_{max}=15418.7\ N/C$

$E_{max}=1.54\times10^{4}\ N/C$

Hence, (a). If z = 0, The electric field due to the rod is zero.

(b). If z = ∞, The electric field due to the rod is $E\propto\dfrac{1}{z^2}$ .

(c). The positive distance is $\dfrac{R}{\sqrt{2}}$

(d). The maximum magnitude of electric field is $1.54\times10^{4}\ N/C$

6 0

3 years ago

Calculate the amount of heat transferred when 710 grams of water warms from an initial temperature of 4.0 ºC to a final temperat

tigry1 [53]

Answer:

Q = 62383.44 Joules

Explanation:

Given that,

Mass of water, m = 710 gm

Initial temperature of water, $T_i=4^{\circ} C$

Final temperature of water, $T_f=25^{\circ} C$

The specific heat capacity of liquid water is, $c=4.184\ J/g\ ^oC$

Heat transferred is given by :

$Q=mc(T_f-T_i)$

$Q=710\times 4.184\times (25-4)$

Q = 62383.44 Joules

So, the amount of heat transferred is 62383.44 Joules. Hence, this is the required solution.

3 0

3 years ago

What’s the answer to this

ladessa [460]

Choices 1, 2, and 4 . . . . . Yes

Choices 3 and 5 . . . . . No

6 0

4 years ago