If I move 15ft foward, 15 ft backwards, 15 ft to the right, 15ft to the left where am I?

Does mars has a bulge near its equator ?

Keith_Richards [23]

Yessir it sure does

7 0

2 years ago

An inkjet printer's speed is measured in ____. select one:

Vladimir [108]

<span>Back in the day, one measured a printer's speed in CPM, which stands for characters per minute. Most of the modern printers that exist today, including the inkjet printer measure their speed in PPM, which is also known as pages per minute.</span>

5 0

3 years ago

During a clinic visit a4 year old girl suddenly yells don't sit on erin the parents wispers that erin is an imaginary friend wha

igor_vitrenko [27]

Try and calm the child and say that you won't sit on Erin. If you tell the young girl that Erin isn't real at that age, they may have social issues later.

6 0

3 years ago

A velocity selector, in which charged particles of a specific speed pass through undeflected while those of greater or lesser sp

uranmaximum [27]

Answer:

a) 351351.35m/s

b) 1.044*10^{-8}kg/C

Explanation:

a) Electric force and magnetic force over the charge must have the same magnitude. From there we can compute the seep of the charge.

$F_E=F_B\\\\qE=qvB\\\\v=\frac{E}{B}=\frac{1.95*10^{5}V/m}{0.555T}=351351.35\frac{m}{s}$

b) the mass-charge ratio is given by:

$\frac{m}{q}=\frac{rB}{v}=\frac{(6.61*10^{-3}m)(0.555T)}{351351.35m/s}=1.044*10^{-8}\frac{kg}{C}$

hope this helps!!

5 0

3 years ago

A couple of astronauts agree to rendezvous in space after hours. Their plan is to let gravity bring them together. She has a mas

Oduvanchick [21]

Answer:

(a) Acceleration of female astronaut = 9.33*10^-12 m/s^2; Acceleration of male astronaut = 7.56*10^-12 m/s^2

(b) 27.013 days

(c) No, their accelerations would not be constant.

Explanation:

In the given question, we have:

Mass of female astronaut $M_{1}$ = 60.0 kg

Mass of male astronaut $M_{2}$ = 74.0 kg

Distance between them (S) = 23.0 m

(a) The free-body diagram is shown in the attached figure.

Using the equations below, the initial accelerations of the two astronauts can be calculated:

Force of gravity (F) = $\frac{G*M_{1}*M_{2}}{S^{2} }$

G = 6.67*10^-11 $\frac{m^{3} }{kg*s^{2} }$

F = (6.67*10^-11 *60*74)/23^2 = 5.598*10^-10 N

For the female astronaut, her initial acceleration = F/ $M_{1}$ = 5.598*10^-10/60 = 9.33*10^-12 m/s^2

For the male astronaut, his acceleration = F/ $M_{2}$ = 5.598*10^-10/74 = 7.56*10^-12 m/s^2

(b) Since the different between their mass is not much, we can deduce that:

$a_{average} = \frac{a_{1}+a_{2}}{2}$ = (9.33*10^-12 + 7.56*10^-12)/2 = 8.445*10^-12 m/s^2

Using the equation below, we can calculate the the time:

S = ut + 1/2 (at^2) where u = 0

23 = 1/2 (8.445*10^-12)*t^2

t^2 = 5.447*10^12

t = 2333883.044 s = 27.013 days

(c) No, their accelerations will not be constant. It will increase because their radii would be decreasing.

8 0

2 years ago