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1 year ago

14

Why are light-years more convenient than miles, kilometers, or astronomical units (au) for measuring the distances to stars and

galaxies? (select all that apply.)

1 answer:

Tomtit [17]1 year ago

5 0

Answer:

sEE BELOW

Explanation:

Well.....because the numbers are 'astronomical'....meaning VERY, VERY , VERY LARGE

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In your own words, explain what non-contact forces are.

katrin [286]

It’s a force which acts on an object without coming physically in contract with it.

4 0

2 years ago

) determine the density of a 32.5 g metal sample that displaces 8.39 ml of water.

sweet-ann [11.9K]

Density is the ratio of a substance's mass to its volume. On the other hand, according to Archimedes' principle, the volume of water displaced is equal to the volume of the object placed on the water. Thus, the density of the metal is equal to 8.39 mL. So, the density would be

Density = 32.5 g/8.39 mL = 3.87 g/mL

3 0

3 years ago

Read 2 more answers

A 100-g toy car is propelled by a compressed spring that starts it moving. the car follows a curved track. what is the final spe

wel

Answer:

0.687 m/s

Explanation:

Initial energy = final energy

1/2 mu² = mgh + 1/2 mv²

1/2 u² = gh + 1/2 v²

Given u = 2.00 m/s, g = 9.8 m/s², and h = 0.180 m:

1/2 (2.00 m/s)² = (9.8 m/s²) (0.180 m) + 1/2 v²

v = 0.687 m/s

5 0

3 years ago

Read 2 more answers

A sinusoidal transverse wave of amplitude ym = 8.4 cm and wavelength = 5.3 cm travels on a stretched cord. Find the ratio of the

Scilla [17]

Answer:

The ratio is 9.95

Solution:

As per the question:

Amplitude, $y_{m} = 8.4\ cm$

Wavelength, $\lambda = 5.3\ cm$

Now,

To calculate the ratio of the maximum particle speed to the speed of the wave:

For the maximum speed of the particle:

$v_{m} = y_{m}\times \omega$

where

$\omega = 2\pi f$ = angular speed of the particle

Thus

$v_{m} = 2\pi fy_{m}$

Now,

The wave speed is given by:

$v = f\lambda$

Now,

The ratio is given by:

$\frac{v_{m}}{v} = \frac{2\pi fy_{m}}{f\lambda}$

$\frac{v_{m}}{v} = \frac{2\pi \times 8.4}{5.3} = 9.95$

8 0

3 years ago

A balloon filled with helium gas has an average density of Q,-0.41 kg/m'. The density of the air is Qa-1.23 kg/m3. The volume of

Citrus2011 [14]

Answer:

a) $(Qa*g*Vb)-(Qh*Vb*g)=(Qh*Vb*a)\\where \\g=gravity [m/s^2]\\a=acceleration [m/s^2]$

b) a = 19.61[m/s^2]

Explanation:

The total mass of the balloon is:

$massball=densityheli*volumeheli\\\\massball=0.41 [kg/m^3]*0.048[m^3]\\massball=0.01968[kg]\\\\$

The buoyancy force acting on the balloon is:

$Fb=densityair*gravity*volumeball\\Fb=1.23[kg/m^3]*9.81[m/s^2]*0.048[m^3]\\Fb=0.579[N]$

Now we need to make a free body diagram where we can see the forces that are acting over the balloon and determinate the acceleration.

In the attached image we can see the free body diagram and the equation deducted by Newton's second law

6 0

2 years ago