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Alex Ar [27]
2 years ago
5

Assuming a roughly spherical shape and a density of 4000 kg/m3, estimate the diameter of an asteroid having the average mass of

1017 kg.
Physics
1 answer:
vredina [299]2 years ago
8 0

Explanation:

It is given that,

Density of asteroid, \rho=4000\ kg/m^3

Mass of asteroid, m=10^{17}\ kg

We need to find the diameter of the asteroid. The formula of density is given by:

\rho=\dfrac{m}{V}

V is the volume of spherical shaped asteroid, V=\dfrac{4}{3}\pi r^3

r^3=\dfrac{3M}{4\pi \rho}

r^3=\dfrac{3\times 10^{17}\ kg}{4\pi \times 4000\ kg/m^3}

r^3=\sqrt{5.96\times 10^{12}}

r = 2441311.12 m

Diameter = 2 × radius

d = 4882622.24 m

or

d=4.88\times 10^6\ m

Hence, this is the required solution.

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8 0
3 years ago
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What is the wavelength of a 6.00*10^2 Hz sound wave in air at 20C
Alex Ar [27]

Answer:

Solution

λ=v/n

Here, v=344 m s−1

n=22 MHz =22×106 Hz

λ=344/22×106=15.64×10−6m=15.64μm.

5 0
2 years ago
three condensers are connected in series across a 150 volt supply. The voltages across them are 40,50 and 60 volts respectively,
ioda

Explanation:

Given that,

The voltages across them are 40,50 and 60 volts respectively, and the charge on each condenser is 6×10⁻⁸ C.

(a) Capacitance of capacitor 1,

C_1=\dfrac{Q}{V_1}\\\\C_1=\dfrac{6\times 10^{-8}}{40}\\\\C_1=1.5\times 10^{-9}\ F\\\\C_1=1.5\ nF

Capacitance of capacitor 2,

C_2=\dfrac{Q}{V_2}\\\\C_2=\dfrac{6\times 10^{-8}}{50}\\\\C_2=1.2\times 10^{-9}\ F\\\\C_2=1.2\ nF

Capacitance of capacitor 3,

C_3=\dfrac{Q}{V_3}\\\\C_3=\dfrac{6\times 10^{-8}}{60}\\\\C_3=1\times 10^{-9}\ F\\\\C_3=1\ nF

(b) The equivalent capacitance in series combination is :

\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}\\\\\dfrac{1}{C}=\dfrac{1}{1.5}+\dfrac{1}{1.2}+\dfrac{1}{1}\\\\C=0.4\ nF

Hence, this is the required solution.

5 0
3 years ago
To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T
vladimir1956 [14]

Answer:

Explanation:

kinetic energy required = 1.80 MeV

= 1.8 x 10⁶ x 1.6 x 10⁻¹⁹ J

= 2.88 x 10⁻¹³ J

If v be the velocity of proton

1/2 x mass of proton x v² = 2.88 x 10⁻¹³

= .5 x 1.67 x 10⁻²⁷ x v² = 2.88 x 10⁻¹³

v² = 3.45 x 10¹⁴

v = 1.86 x 10⁷ m /s

If V be the potential difference required

V x e = kinetic energy . where e is charge on proton .

V x 1.6 x 10⁻¹⁹ = 2.88 x 10⁻¹³

V = 1.8 x 10⁶ volt .

3 0
2 years ago
05. The time required to complete one lap around a perfectly circular track having a radius of 1,835 meters is 86
likoan [24]

Answer:

v = 134.06 m/s

Explanation:

Given that,

Radius of a circular track is 1,835 m

Time required to complete one lap around a perfectly circular track is 86 seconds

We need to find the car's velocity. Velocity is equal to,

v=d/t

On circular path,

v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 1835}{86}\\\\v=134.06\ m/s

So, car's velocity is 134.06 m/s.

7 0
3 years ago
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