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3 years ago

9

What is always true of a liquid's shape and volume?

1 answer:

Natalija [7]3 years ago

4 0

Molecules of liquid are far away from each other but less compared to gas. Liquid has the ability to flow. Liquid molecules can move. Liquid occupies the shape of the container. Thus, they do not have fixed shape. Liquid has fixed volume because they are not compressible(easily)

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A car of mass m accelerates from rest along a level straight track, not at constant acceleration but with constant engine power,

lisabon 2012 [21]

<span>Because P = W ÷ t, and W = F*t, you can substitute (W) for (F*t). Then substitute (F) for (m*a). This will leave you with P = (m*a*d)/t. Since you need velocity, youd want to solve for a so you can use v = a*t. a = (P*t)/(m*d) therefore, substituting a in v = a*t, v = (P*t*t)/(m*d)</span>

4 0

2 years ago

Which statement about elements and atoms is true?

Vika [28.1K]

Atoms are the smallest unit of an element

8 0

3 years ago

[High Dive) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (Air r

blsea [12.9K]

Answer:

(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b) it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained

Explanation:

(a)

To find the final velocity $V_{f}$ for an object traveling distance h taking the initial vertical component of velocity as $V_{i}$ the kinematics equation is written as

$V_{f}^{2}=V_{i}^{2}+2ah$ where a is acceleration

Substituting g for a where g is gravitational force value taken as 9.81

$V_{f}^{2}=V_{i}^{2}+2gh$

Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h

$V_{f}=\sqrt {(2gh)}= V_{f}=\sqrt {(2*9.81*21.3)}$ = 20.44275

Therefore, the divers enter with a speed of 20.4 m/s

The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)

The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation

$V_{f}^{2}=V_{i}^{2}+2gh$

Since we have final velocity of 25 m/s

$V_{i}^{2}=2gh-V_{f}^{2}$

$V_{i}=\sqrt{(V_{f}^{2}-2gh)}$

$V_{i}=\sqrt{(25^{2}-2*9.81*21.3)}$ = 14.390761 m/s

Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s

In conclusion, the upward initial velocity can’t be physically attained

3 0

3 years ago

Calculate the approximate volume of a uranium nucleus, 23592u. (you can ignore the mass defect in this calculation and simply ta

tester [92]

Radius of nuclei is given by formula

$R = R_oA^{1/3}$

now we can say volume of the nuclei is given as

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

now the density is given as

density = mass / volume

mass of nuclei = mass of neutron + mass of protons

$m = z*m_p + (A- z)*m_n$

$m_p = m_n = 1.008u$

$m = A*1.008u$

Now density is given as

$\rho = \frac{A*1.008u}{\frac{4}{3}\pi R_0^3* A}$

here we know that

$R_0$ = 1.2 fm

$\rho = \frac{1.008u}{\frac{4}{3}\pi*(1.2*10^{-15})^3}$

$\rho = 2.31 * 10^{17} kg/m^3$

So from above we can say that density of all nuclei is almost same.

5 0

3 years ago

Two spherical objects have masses of 6.2 x 105 kg and 13 x 103 kg. The gravitational attraction between them is 130 N. How far a

Neporo4naja [7]

Answer:

r = 6.4 cm

Explanation:

F = GMm/r²

r = √(GMm/F)

r = √((6.674e-11)(6.2e5)(13e3)/130)

r = 0.06432... m

Those are some high density materials!

8 0

3 years ago