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3 years ago

How old was martin king when he died

Physics

2 answers:

konstantin123 [22]3 years ago

7 0

He does at age 39
1929-1968

notka56 [123]3 years ago

5 0

He died April 4, 1968. He dies at age 39.

Hope this helped☺☺

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A 1.0-μm-diameter oil droplet (density 900 kg/m3) is negatively charged with the addition of 39 extra electrons. It is released

GREYUIT [131]

Answer:

$6.75\mu C/m^2$

Explanation:

We are given that

Diameter,d= $1\mu m=1\time 10^{-6} m$

$1\mu m=10^{-6} m$

Radius,r= $\frac{d}{2}=\frac{1}{2}\times 10^{-6}=0.5 \times 10^{-6} m$

Density, $\rho=900kg/m^3$

Total number of electrons,n=39

Charge on electron = $1.6\times 10^{-19} C$

Total charge= $q=ne=39\times 1.6\times 10^{-19}=62.4\times 10^{-19} C$

Distance,s=2mm= $2\times 10^{-3} m$

Mass = $density\times volume=900\times \frac{4}{3}\pi r^3=900\times \frac{4}{3}\pi(0.5\times 10^{-6})^3=4.7\times 10^{-16} kg$

Initial velocity,u=0

Final speed,v=4.5 m/s

$v^2-u^2=2as$

$(4.5)^2-0=2a(2\times 10^{-3})$

$20.25=4a\times 10^{-3}$

$a=\frac{20.25}{4\times 10^{-3}}=5062.5m/s^2$

Force,F=ma

$qE=ma$

$q(\frac{\sigma}{2\epsilon_0})=ma$

$\sigma=\frac{2\epsilon_0ma}{q}=\frac{2\times 8.85\times 10^{-12}\times 4.7\times 10^{-16}\times 5062.5}{62.4\times 10^{-19}}$

$\epsilon_0=8.85\times 10^{-12}$

$\sigma=6.75\times 10^{-6}C/m^2=6.75\mu C/m^2$

6 0

3 years ago

The Hubble Space Telescope has a mass of 1.16*10^ 4 kg and orbits the Earth at an altitude of 5.68 * 10 ^ 5 above Earth's surfac

andrezito [222]

Answer:

$E=8.13\times 10^{12}\ J$

Explanation:

Given that,

The mass of a Hubble Space Telescope, $m_1=1.16\times 10^4\ kg$

It orbits the Earth at an altitude of $5.68\times 10^5\ m$

We need to find the potential energy the telescope at this location. The formula for potential energy is given by :

$E=\dfrac{Gm_1m_e}{r}$

Where

$m_e$ is the mass of Earth

Put all the values,

$E=\dfrac{6.67\times 10^{-11}\times 1.16\times 10^4\times 5.97\times 10^{24}}{5.68\times 10^5}\\\\E=8.13\times 10^{12}\ J$

So, the potential energy of the telescope is $8.13\times 10^{12}\ J$ .

5 0

3 years ago

When a 12 N horizontal force is applied to a box on a horizontal tabletop, the box remains at rest. The force of friction acting

Mrrafil [7]

Answer:

12N

Explanation:

when a force is applied to a body but still stays at rest or moves at a constant speed , the frictional force is equal to the force applied

3 0

3 years ago

A 2.20-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 29.0 N is require

g100num [7]

Answer:

a. 145 N/m b. 1.29 Hz c. 1.62 m/s d. 0 m e. 13.2 m/s² f. ± 0.2 m g. 2.9 J h. 0.54 m/s i. 4.39 m/s²

Explanation:

a. The force constant of the spring

The spring force F = kx and k = F/x where k is the spring constant. F = 29.0 N and x = 0.200 m

k = 29.0 N/0.200 m = 145 N/m

b. The frequency of oscillations, f

f = 1/2π√(k/m) m = mass = 2.20 kg

f = 1/2π√(145 N/m/2.20 kg) = 1.29 Hz

c. maximum speed of the object

The maximum elastic potential energy of the spring = maximum kinetic energy of the object

1/2kx² = 1/2mv²

v = (√k/m)x where v is the maximum speed of the object

v = (√145/2.2)0.2 = 1.62 m/s

d Where does the maximum speed occur?

The maximum speed occurs at 0 m

e. The maximum acceleration

a = kx/m = 145 × 0.2/2.2 = 13.2 m/s²

f. The maximum acceleration occurs at x = ± 0.2 m

g. The total energy of the system is the maximum elestic potential energy of the system

E = 1/2kx² = 1/2 × 145 × 0.2² = 2.9 J

h. When x = x₀/3

1/2k(x₀/3)² = 1/2mv²

kx₀²/9 = mv²

v = 1/3(√k/m)x₀ = 1/3(√145/2.2)0.2 = 0.54 m/s

i When x = x₀/3

a = kx₀/3m = 145 × 0.2/(2.2 × 3)= 4.39 m/s²

8 0

3 years ago

Microphones and loudspeakers are used in an auditorium because the sound waves at the stage compared to the sound waves at the b

MaRussiya [10]

Acoustics (sounds)
Hope the this helps!

6 0

3 years ago