All categories

4 years ago

In scientific notation 8700 is written

Physics

2 answers:

Alex73 [517]4 years ago

8 0

Answer:

8.7 x 10^3 is the correct answer.

Nookie1986 [14]4 years ago

6 0

Answer:

8.7 x 10^3

I have no explanation, I just know that it is written as this

You might be interested in

According to evidence presented in both the text and video, which statement is the most accurate regarding the relationship betw

andrey2020 [161]

Answer:

Stress does not cause illness, but restrains the immune system's ability to fight foreign invaders

Explanation:

It's correct :)

4 0

3 years ago

If the coefficient of friction is 0.3900 and the cylinder has a radius of 2.700 m, what is the minimum angular speed of the cyli

Aleks04 [339]

Answer:

w=3.05 rad/s or 29.88rpm

Explanation:

k = coefficient of friction = 0.3900

R = radius of the cylinder = 2.7m

V = linear speed of rotation of the cylinder

w = angular speed = V/R or to rewrite V = w*R

N = normal force to cylinder

N= $=\frac{m(V)^{2}}{R}=m*(w)^2*R$

$Friction force\\Ff = k*N\\Ff= k*m*w^2*R$

$Gravitational force \\Fg = m*g$

These must be balanced (the net force on the people will be 0) so set them equal to each other.

$Fg = Ff$

$m*g = k*m*w^2*R$

$g=k*w^{2}*R$

$w^2 =\frac{g}{k*R}$

$w=\sqrt{\frac{g}{k*R}} \\w =\sqrt{\frac{9.8\frac{m}{s^{2}}}{0.3900*2.7m}}\\ w=\sqrt{9.306}=3.05 \frac{rad}{s}$

There are 2*pi radians in 1 revolution so:

$RPM=\frac{w}{2\pi }*60\\RPM=\frac{3.05\frac{rad}{s}}{2\pi}*60\\RPM= 0.498*60\\RPM=29.88$

So you need about 30 RPM to keep people from falling out the bottom

7 0

4 years ago

A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr

VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

$rR:\\\\E=k\frac{Q}{r^2}$

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

$F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}$

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

$Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}$