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4 years ago

What can you infer about the density of two objects if they object remains suspended in the liquid?

Physics

2 answers:

ZanzabumX [31]4 years ago

5 0

Answer:

Equal Densities

Explanation:

if the density of the object was greater than that of the liquid, it would sink to the bottom. if the density od the object was lesser than the liquid, it would float :)

4vir4ik [10]4 years ago

5 0

Answer:

The densities of the objects are equal

Explanation:

higher density objects sink to the bottom while lower density objects rise to the top

since the object remains suspended in the liquid, it has neither higher or lower density than the water because it is neither sinking nor floating

therefore, the object has equal densities with the liquid

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If an 800 kg roller coaster is at the top of its 50 m high track, it will have a potential energy pf 392,000 J and a kinetic ene

Viefleur [7K]

I just woke him crying crying laughing crying and laughing I’m so mad he got me blocked him lol I got a hold on her phone the answer is 0 m

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3 years ago

3. What do we call the ONLY part of the electromagnetic spectrum that we can

otez555 [7]

Answer:

Visible Light

wavelength = 4000 - 7000 Angstroms = 400 - 700 milli-microns

1 A unit = 10^-10 m

1 mμ = 10^-9 m

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2 years ago

The horizontal surface on which the block (mass 2.0 kg) slides is frictionless. The speed of the block before it touches the spr

ch4aika [34]

Answer:3.67 m/s

Explanation:

mass of block(m)=2 kg

Velocity of block=6 m/s

spring constant(k)=2 KN/m

Spring compression x=15 cm

Conserving Energy

energy lost by block =Gain in potential energy in spring

$\frac{m(v_1^2-v_2^2)}{2}=\frac{kx^2}{2}$

$2\left [ 6^2-v_2^2\right ]=2\times 10^3\times \left [ 0.15\right ]^2$

$v_2=3.67 m/s$

7 0

4 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin

Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is $14,000 + 10,000x − 26,000x^{2}$ , where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

W = $\int_{0}^{0.54} F dx$

= $\int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx$

= $[14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}$

= 7560 + 1458 - 1364.69

= 7653.31 J

or, = 7.65 kJ (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

5 0

3 years ago

Someone help please by providing work and answers please :)

Nastasia [14]

First we gotta use an equation of motion:

$d = ut + \frac{1}{2} a {t}^{2}$

Our vertical distance d= 100 m, initial vertical speed u = 0 m/s (because velocity is fully horizontal), and vertical acceleration a = 9.8 m/s2 because of gravity. Let's plug it all in!

$100 = 0 + \frac{1}{2} (9.8) {t}^{2}$

Now we just need to solve for t:

${t}^{2} = \frac{2(100)}{9.8} \\ \\ t = \sqrt{\frac{2(100)}{9.8}}$

Hit the calculators, and you'll get 4.5 seconds!

5 0

3 years ago