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2 years ago

The density of a substance is its mass per unit volume.

Physics

1 answer:

Over [174]2 years ago

7 0

Answers:

a) $kg/m^{3}$

b) derived

Explanation:

Density $D$ is a characteristic property of substances and materials and is defined as the relationship between the mass $m$ and volume $V$ of a specific substance or material, as shown below:

$D=\frac{m}{V}$

This means density is inversely proportional to volume.

In addition, density is a derived unit, because is based on two basic units of the International Sistem of Units: Mass (kilograms) and Volume (cubic meters)

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3 years ago

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2 years ago

How do land and sea breeze work?

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3 years ago

A 120-kg roller coaster cart is being tested on a new track, and a crash-test dummy is loaded into it. The roller coaster starts

avanturin [10]

Answer:

vb = 22.13 m/s

So, the only thing that was measured here was the height of point A relative to point B. And the Law of Conservation of Energy was used.

Explanation:

In order to find the speed of roller coaster at Point B, we will use the law of conservation of Energy. In this situation, the law of conservation of energy states that:

K.E at A + P.E at A = K.E at B + P.E at B

(1/2)mvₐ² + mghₐ = (1/2)m(vb)² + mg(hb)

(1/2)vₙ² + ghₐ = (1/2)(vb)² + g(hb)

where,

vₙ = velocity of roller coaster at point a = 0 m/s

hₙ = height of roller coaster at point a = 25 m

g = 9.8 m/s²

vb = velocity of roller coaster at point B = ?

hb = Height of Point B = 0 m (since, point is the reference point)

Therefore,

(1/2)(0 m/s)² + (9.8 m/s²)(25 m) = (1/2)(vb)² + (9.8 m/s²)(0 m)

245 m²/s² * 2 = vb²

vb = √(490 m²/s²)

<u>So, the only thing that was measured here was the height of point A relative to point B. And the Law of Conservation of Energy was used.</u>

5 0

2 years ago

A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh

Elanso [62]

Answer:

The value is $v = 2.3359 *10^{4} \ m/s$

Explanation:

From the question we are told that

The initial speed is $u = 2.05 *10^{4} \ m/s$

Generally the total energy possessed by the space probe when on earth is mathematically represented as

$T__{E}} = KE__{i}} + KE__{e}}$

Here $KE_i$ is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

$KE_i = \frac{1}{2} * m * u^2$

=> $KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2$

=> $KE_i = 2.101 *10^{8} \ \ m \ \ J$

And $KE_e$ is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

$KE_e = \frac{1}{2} * m * v_e^2$

Here $v_e$ is the escape velocity from earth which has a value $v_e = 11.2 *10^{3} \ m/s$

=> $KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2$

=> $KE_e = 6.272 *10^{7} \ \ m \ \ J$

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

$KE_p = \frac{1}{2} * m * v^2$

Generally from the law energy conservation we have that

$T__{E}} = KE_p$

$2.101 *10^{8} m + 6.272 *10^{7} m = \frac{1}{2} * m * v^2$

=> $5.4564 *10^{8} = v^2$

=> $v = \sqrt{5.4564 *10^{8}}$

=> $v = 2.3359 *10^{4} \ m/s$

4 0

2 years ago