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3 years ago

What is the density of the object that has a mass 13.87g and a volume of 31.75cm3?

Physics

1 answer:

damaskus [11]3 years ago

3 0

Answer:

d=0.4368 g/cm³

Explanation:

Density of an object is defined as the ratio of its total mass to the mass volume.

So that it is

d= m/V

d= (13.87g) / (31.75cm³)

d= 0.4368g/cm³

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Cheetah cubs are in danger from predators like lions and hyenas which can track their prey by scent and so the mother and her cubs leave an area when their scent is too strong so that they are not hunted and the cubs survive.

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3 years ago

A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

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3 years ago

A tetrahedron has an equilateral triangle base with 25.0-cm-long edges and three equilateral triangle sides. The base is paralle

djverab [1.8K]

Answer:

a. 7.046 Nm²/C

b. 2.348 Nm²/C

Explanation:

Data given:

Base of equilateral triangle = 25.0 cm = 0.25 m

Strength of electric field = 260 N/C

In order to find the electric flux we first have to find out the area of triangle.

Area of triangle = $\frac{\sqrt{3} }{4} a^{2}$

= $\frac{\sqrt{3} }{4} (0.25)^{2}$

= 0.0271 m³

Lets find electric flux,

Electric Flux = E. A

= 260×0.0271

= 7.046 Nm²/C

Now we can find the electric flux through each of the three sides.

Electric flux through three sides = $\frac{7.046}{3}$

= 2.348 N m²/C

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3 years ago

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3 years ago

What is the density of the object that has a mass 13.87g and a volume of 31.75cm3?​

What is the density of the object that has a mass 13.87g and a volume of 31.75cm3?