Ask question

Ask question

All categories

sleet_krkn [62]

2 years ago

15

What happens to friction if the velocity is constant? What will it equal to and what formula can be used to find it? Is it Fn =

Ff?

1 answer:

dimaraw [331]2 years ago

7 0

Answer:

If the velocity is constant there is no friction it will equal to zero. It is Fn=Ff

Explanation:

You might be interested in

"A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two firs

Fantom [35]

Answer:

A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if

the dispersion is great

6 0

3 years ago

Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive

Setler [38]

Answer:

The mass of the massive object at the center of the Milky Way galaxy is $3.44\times10^{37}\ Kg$

Explanation:

Given that,

Diameter = 10 light year

Orbital speed = 180 km/s

Suppose determine the mass of the massive object at the center of the Milky Way galaxy.

Take the distance of one light year to be 9.461×10¹⁵ m. I was able to get this it is 4.26×10³⁷ kg.

We need to calculate the radius of the orbit

Using formula of radius

$r=\dfrac{d}{2}$

$r=\dfrac{15\times9.461\times10^{15}}{2}$

$r=7.09\times10^{16}\ m$

We need to calculate the mass of the massive object at the center of the Milky Way galaxy

Using formula of mass

$M=\dfrac{v^2r}{G}$

Put the value into the formula

$M=\dfrac{(180\times10^3)^2\times7.09\times10^{16}}{6.67\times10^{-11}}$

$M=3.44\times10^{37}\ Kg$

Hence, The mass of the massive object at the center of the Milky Way galaxy is $3.44\times10^{37}\ Kg$

5 0

3 years ago

A racecar experiences a centripetal acceleration of 36.0 m/s2 as it travels at a constant speed of 27.0 m/s along a circular arc

alexdok [17]

Answer:

20.25 m

Explanation:

<u>Centripetal acceleration </u>is given by; the square of the velocity, divided by the radius of the circular path.

That is;

<em><u>ac = v²/r</u></em>

<em> </em><em><u> Where; ac = acceleration, centripetal, m/s², v is the velocity, m/s and r is the radius, m</u></em>

Therefore;

r = v²/ac

= 27²/36

= 20.25 m

Hence the radius is 20.25 meters

5 0

2 years ago

Read 2 more answers

A charge q1 of -5.00 x 10^-9 C and a charge q2 of -2.00x 10^-9 C are separated by a distance of 40.0 cm. Find the equilibrium po

Blababa [14]

The magnitude of charge on a proton and electron is the same, 1.602 x 10-19 C. Protons are +, and electrons -.

5 0

3 years ago

A spherical helium filled balloon (B) with a hanging passenger cage being held by a single vertical cable (C) attached to Earth

Gre4nikov [31]

Answer:

The tension is $T = 4326.7 \ N$

Explanation:

From the question we are told that

The total mass is $m = 200 \ kg$

The radius is $r = 5 \ m$

The density of air is $\rho_a = 1.225 \ kg/m^3$

Generally the upward force acting on the balloon is mathematically represented as

$F_N = T + mg$

=> $(\rho_a * V * g ) = T + mg$

=> $T = (\rho_a * V * g ) - mg$

Here V is the volume of the spherical helium filled balloon which is mathematically represented as

$V = \frac{4}{3} * \pi r^3$

=> $V = \frac{4}{3} * 3.142 *(5)^3$

=> $V = 523.67\ m^3$

So

$T = (1.225 * 523.67* 9.8 ) - 200 * 9.8$

$T = 4326.7 \ N$

5 0

3 years ago