If mass1 of a student is 70 kg and mass of the Jupiter is 1.901 x 10^27 kg 1

why can scientists ignore the forces of attraction between particles in a gas under ordinary conditions?

Anna35 [415]

<span>The particles in a gas are apart and moving fast, so the forces of attraction are too weak to have a noticeable effect.</span>

5 0

3 years ago

Read 2 more answers

How many test tubes of blood equal a pint?

irina1246 [14]

It would take 57 test tubes to equal a pint

4 0

3 years ago

Early cameras were little more than a box with a pinhole on the side opposite the film. (a) What angular resolution would you ex

ollegr [7]

Answer:

angular resolution = 0.07270° = 1.269 × $10^{-3}$ rad

greatest distance from the camera = 118.20 m = 0.118 km

Explanation:

given data

diameter = 0.50 mm = 0.5 × $10^{-3}$ m

distance apart = 15 cm = 15× $10^{-2}$ m

wavelength λ = 520 nm = 520 × $10^{-9}$ m

to find out

angular resolution and greatest distance from the camera

solution

first we expression here angular resolution that is

sin θ = $\frac{1.22* \lambda }{D}$ .......................1

put here value λ is wavelength and d is diameter

we get

sin θ = $\frac{1.22*520*10^{-9}}{0.5*10^{-3}}$

θ = 0.07270° = 1.269 × $10^{-3}$ rad

and

distance from camera is calculate here as

θ = $\frac{I}{r}$ .................2

I = $\frac{15*10^{-2}}{1.269*10^{-3}}$

I = 118.20 m = 0.118 km

3 0

3 years ago

What occurs as a ray of light passes from<br> al inilo water?

iVinArrow [24]

Answer:

this may be wrong but I am not sure

3 0

3 years ago

A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 280 N/m. The

lina2011 [118]

Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s

Explanation:

1) We can find the speed of the object by conservation of energy:

$E_{i} = E_{f}$

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s$

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s$

3) When the object is at the equilibrium position, the speed of the object is:

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s$

I hope it helps you!

8 0

3 years ago