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marshall27 [118]
3 years ago
11

A boulder of mass 4 kg rolls over a cliff and reaches the beach below with a velocity of 20 metres per second minus 1.

Physics
1 answer:
Ksenya-84 [330]3 years ago
5 0

Given:-

Mass of boulder = 10kg

Final Velocity = 15m/s

Acceleration due to gravity = +10m/s

To Find:-

The kinetic energy of the boulder as it lands

The potential energy of the boulder when it was at the top of the cliff

The height of the cliff

Formulae used:-

Kinetic energy = mv²/2

Potential energy = mgh

Now,

We will first find the Kinetic enrgy,

→ K. E = ½mv²

→ K. E = ½ × 10 × (15)²

→ K. E = 5 × 225

→ K. E =1125J

But we know that when the body starts moving the Potential enrgy of the body starts Converting into Kinetic enrgy. Before reaching to the last spot it has equal kinetic energy and Potential energy.

Therefore,

→ mgh = 1125J

The Potential enrgy of the boulder when it was on top is 1125J

Now,

→ mgh = 1125J

→ 10 × 10 × h = 1125J

→ 100h = 1125J

→ h = 1125/100

→ h = 11.25m

Hence, The Height of the Cliff is 11.25m

Thanks!

Mark me brainliest!

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xz_007 [3.2K]

Answer:

Explanation:its hard

And ok

8 0
3 years ago
People jump from much greater heights than this, explain how this is possible
dusya [7]

Answer:

a

Explanation:

When people jump from greater height,  energy gets absorbed into the muscles as potential. And when the person flexes her legs on landing energy is trans formed not only into elastic energy in the bones, but also into elastic energy in other tissues and into thermal energy.

air resistance is very small hence, negligible. So, the statement a is correct and rest are wrong.

5 0
4 years ago
Electro-magnets are created by...
mojhsa [17]
Wrapping in insulated wire around a medal with ferromagnetic properties and applying an electric current
6 0
3 years ago
Unpolarized light whose intensity is 1.46 W/m2 is incident on the polarizer in the drawing. (a) What is the intensity of the lig
Serhud [2]

Answer:

0.73 W/m²

0.2522 W/m²

Explanation:

I_0 = Unpolarized light = 1.46 W/m²

\theta = Analyzer angle = 54°

Light through first filter

I_1=\frac{I_0}{2}\\\Rightarrow I_1=\frac{1.46}{2}\\\Rightarrow I_1=0.73\ W/m^2

The intensity of the light leaving the polarizer is 0.73 W/m²

After passing through analyzer

I=I_1cos^2\theta\\\Rightarrow I=0.73\times cos^2(54)\\\Rightarrow I=0.2522\ W/m^2

The intensity of the light that reaches the photocell is 0.2522 W/m²

6 0
3 years ago
Se usa una bombona de He para inflar globos, de volumen 15L con una presión de 200 atm. Si los globos tienen un volumen de 1.5L
Bingel [31]

Answer:

La cantidad de globos llenos es de aproximadamente 1.974 globos

Explanation:

Los parámetros dados son;

El gas utilizado para inflar el globo = Él

El volumen del gas He, V₁ = 15 L

La presión del gas He, P₁ = 200 atm = 20,265,000 Pa

El volumen de cada globo = 1,5 L

La presión del He dentro del globo, P₂ = 770 mmHg = 102,641 Pa

Sea 'V₂', que represente el volumen total que el gas ocupa en todos los globos, y 'x', que represente el número de globos;

Por lo tanto, tenemos;

1,5 · x = V₂

Según la ley de Boyle, tenemos;

P₁ · V₁ = P₂ · V₂

∴ V₂ = P₁ · V₁ / (P₂)

Al introducir los valores conocidos, obtenemos;

V₂ = 20,265,000 Pa × 15 L / (102,641 Pa) = 2,961.53548 L

1,5 L × x = V₂

∴ 1,5 L × x = 2,961,53548 L

x = 2,961.53548 L / (1.5 L) = 1,974.35699 redondeamos hacia abajo para obtener un número entero de globos de la siguiente manera;

El número de globos, x ≈ 1,974 globos.

8 0
3 years ago
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