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4 years ago

Who was the scientist that discovered Pluto wasnt a planet

Physics

1 answer:

Romashka-Z-Leto [24]4 years ago

3 0

The scientist that contributed to the declassification of Pluto as a planet is Mike Brown.

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Lizard: A lizard is running in a straight line according to the following: xx(tt) = tt3⁄3 − tt2 + tt He starts at tt = 0. (a) De

insens350 [35]

Answer:

Explanation:

a ) x ( t ) = t³ / 3 - t² + t

v = dx / dt = 3 t² / 3 - 2 t + 1 = t² - 2 t + 1

b ) lizard is at rest , v( t ) = 0

t² - 2 t + 1 = 0

( t - 1 )² = 0

t = 1

c )

velocity is positive when

t² - 2 t + 1 > 0

( t - 1 ) ² > 0

Here we see that LHS is a square so it is always positive whatever be the value of t

So velocity is always positive or lizard is always moving in positive x direction .

d ) It never moves in negative x direction .

e )

a ( t ) = dv / dt = 2t - 2

t = 1

so it has zero acceleration at t = 0 .

7 0

3 years ago

How does sound travel through a medium?

Papessa [141]

Particles vibrate parallel to the direction the sound travels. It's a longitudinal wave.

4 0

3 years ago

Read 2 more answers

A bowling ball with a mass of 9 kg is moving at a speed of 4 m/s. What is its

Ilia_Sergeevich [38]

I’m pretty sure it’s B

4 0

3 years ago

A -4.00 nC point charge is at the origin, and a second -5.50 nC point charge is on the x-axis at x = 0.800 m.

mafiozo [28]

Answer:

a. $f=1.22*10^{-15} N$

b. $f=53.6*10^{-17} N$

Explanation:

The force existing between two charges is given as

$f=\frac{kq_{1}q_{2}}{r^{2}}$

where q= charge,

k=constant

r= distance between the two charges

Note: this force can either be repulsive or attractive force depending on the charges involve. it is repulsive if they are similar charge and it is attractive if it is opposite charges.

Also the charge of an electron is

$-1.602*10^{-19}$

A. we first determine the magnitude force between the -4nC and the electron

$f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{0.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{0.04}\\f_{21}=1.44*10^{-15}Ni$

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis

for the -5.50nC the distance between them is 0.600m as can be seen in the diagram the magnitude of the force is

$f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.6^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.36}\\f_{23}=-(0.22*10^{-15})N i$

this this force will be repulsive force and it points away from the electron i.e points towards the -ve x-axis.

The total net force on the electron is thus

$f=f_{21}+f_{23}\\ f=1.44*10^{-15}-0.22*10^{-15}\\ f=1.22*10^{-15} N$

b. at distance of x=1.20m, this is shown on the diagram below (attachment 2)

we first determine the magnitude force between the -4nC and the electron

$f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{1.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{1.44}\\f_{21}=4.0*10^{-17}Ni$

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

for the -5.50nC the distance between them is 0.4m as can be seen in the diagram the magnitude of the force is

$f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.4^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.16}\\f_{23}=49.6*10^{-17}Ni$

this this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

The total net force on the electron is thus

$f=f_{21}+f_{23}\\ f=4.0*10^{-15}+49.6*10^{-17}\\ f=53.6*10^{-17} N$

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3 years ago

What happens when a roller coaster car moves down from the top of a hill?

VLD [36.1K]

When the roller coaster moves down from the top of the hill, all of its stored potential energy is converted into kinetic energy to move it and when it goes back up the hill it turns kinetic into potential.

3 0

3 years ago

Read 2 more answers