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4 years ago

10) The closest star to the earth (other than

Physics

1 answer:

Reptile [31]4 years ago

8 0

Answer:

It takes a little, be patient.

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34.9x46x809 I need help again

Irina-Kira [14]

Answer:

hope that helps uh..☺

4 0

3 years ago

Read 2 more answers

A 350-g mass is attached to a spring whose spring constant is 64 N/m. Its maximum acceleration is 5.3 m/s2. What is the frequenc

max2010maxim [7]

The frequency of oscillation is 2.153 Hz

What is the frequency of spring?

Spring Frequency is the natural frequency of spring with a weight at the lower end. Spring is fixed from the upper end and the lower end is free.

For the mass-spring system in this problem,

The Frequency of spring is calculated with the equation:

$f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }$

Where,

f = frequency of spring

k = spring constant = 64 N/m

m = mass attached to spring = 350g = 0.350 kg

a = maximum acceleration = 5.3 m/s^2

Substituting the values in the equation,

$f = \frac{1}{2\pi } \sqrt{\frac{64}{0.350} }$

$f = \frac{1}{2\pi } ( 13.522)$

$f = 2.1535 Hz$

Hence,

The frequency of oscillation is 2.153 Hz

Learn more about frequency here:

brainly.com/question/13978015

#SPJ4

6 0

2 years ago

1) Name one thing that can cause global warming

max2010maxim [7]

Air pollution!! Can I have brainlyist:))

6 0

3 years ago

This problem has been solved!

lana66690 [7]

Answer:

Charge on each metal sphere will be $8\times 10^{8}C$

Explanation:

We have given number of electron added to metal sphere A $n=10^{12}electron$

As both the spheres are connected by rod so half -half electron will be distributed on both the spheres.

So electron on both the spheres $=\frac{10^{12}}{2}=5\times 10^{11}electron$

We know that charge on each electron $e=1.6\times 10^{-19}C$

So charge on both the spheres will be equal to $q=1.6\times 10^{-19}\times 5\times 10^{11}=8\times 10^{8}C$

So charge on each metal sphere will be equal to $8\times 10^{8}C$

6 0

3 years ago

B) A satellite with mass m orbits the Earth at a radius r. A second satellite also with mass m orbits the

Snezhnost [94]

Answer:

So, given the eqn Fg=G(m1+m2/r^2) where G is the gravitational constant, m is the mass of the satellite and m2 is the mass of the earth and r is the distance from earth to the satellite, the force of earths gravity should be quartered.

Cause (2r)^2 gets turned into (4r^2) where 4r^2 is compared to r^2

Explanation:

6 0

3 years ago