I)

A crane raises a crate with a mass of 150 kg to a height of 20 m. Given that

Virty [35]

Answer:

$\boxed {\boxed {\sf 29,400 \ Joules}}$

Explanation:

Gravitational potential energy is the energy an object possesses due to its position. It is the product of mass, height, and acceleration due to gravity.

$E_P= m \times g \times h$

The object has a mass of 150 kilograms and is raised to a height of 20 meters. Since this is on Earth, the acceleration due to gravity is 9.8 meters per square second.

m= 150 kg
g= 9.8 m/s²
h= 20 m

Substitute the values into the formula.

$E_p= 150 \ kg \times 9.8 \ m/s^2 \times 20 \ m$

Multiply the three numbers and their units together.

$E_p=1470 \ kg*m/s^2 \times 20 m$

$E_p=29400 \ kg*m^2/s^2$

Convert the units.

1 kilogram meter square per second squared (1 kg *m²/s²) is equal to 1 Joule (J). Our answer of 29,400 kg*m²/s² is equal to 29,400 Joules.

$E_p= 29,400 \ J$

The crate has 29,400 Joules of potential energy.

7 0

3 years ago

Which statement bet explains the relationship between the electric force between two charged objects and the distance between th

Nataliya [291]

Coulomb's Law: Force = k x q1x q2 divided distance square
where k=9x10^9 , q1 and q2 are the charge
So if you distance is halved, your force is stronger by 4 times
and if you distance is doubled, your force is 1/4
Ask me again if you aren't clear :)

4 0

3 years ago

A fireman standing on a 15 m high ladder operates a water hose with a round nozzle of diameter 2.02 inch. The lower end of the h

Nataly_w [17]

Answer:

v₁ = 1,606 10⁴ m / s

Explanation:

For this exercise we must use Bernoulli's equation, let's use index 1 for the nozzle on the stairs and index 2 the pump on the street

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² +ρ g y₂

The pressure when the water comes out is the atmospheric pressure

P₁ = P_atm

The difference in height between the street and the nozzle on the stairs is

y₂-y₁ = 15 m

Now let's use the continuity equation

v₁ A₁ = v₂ A₁

The area of a circle is

A = π r² = π (d/2)²

v₁ π d₁²/ 4 = v₂ π d₂²/ 4

v₂ = v₁ d₁² / d₂²

Let's replace

P₂-P_atm + ½ ρ [ (v₁ d₁² / d₂²)²- v₁² ] + ρ g (y2-y1) = 0

P₂- P_Atm + ρ g (y₂-y₁) = ½ ρ v₁² [1- (d₁/d₂)⁴]

v₁² [1- (d₁/d₂)⁴] = (P₂-P_atm) ρ / 2 + g (y₂-y₁) / 2

Let's reduce the magnitudes to the SI system

d₂ = 3.37 in (2.54 10⁻² m / 1 in) = 8.56 10⁻² m

d₁ = 2.02 in = 5.13 10⁻² m

Let's calculate

v₁² [1- (5.13 / 8.56) 4] = 449.538 10³ 10³/2 -9.8 15/2

v₁² [0.8710] = 2.2477 10⁸ - 73.5 = 2.2477 10⁸

v₁ = √ 2.2477 10⁸ /0.8710

v₁ = 1,606 10⁴ m / s

4 0

2 years ago

A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 38

oksano4ka [1.4K]

Answer:

the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

Explanation:

Given

$Initial Temperature T_1=387 KFinal Temperature T_2=774 K$

The RMS velocity of molecules in a gas is given by

$V_{rms}=\sqrt{\dfrac{3k_bT}{m}}$

where T=temperature

$k_b=constant$

For T = 387K

$V_1=\sqrt{\frac{3k_b\cdot 387}{m}}----1$

For T = 774

$V_2=\sqrt{\frac{3k_b\cdot 774}{m}}----(2)$

dividing eqn 1 and eqn 2

$\frac{V_2}{V_1}=\sqrt{\frac{774}{387}}$

$\frac{V_2}{V_1}=\sqrt{2}$

Thus,the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

5 0

3 years ago

Read 2 more answers

What is the final temperature in degrees centigrade of 100g of water at 30 c if it is mixed with 50 g of water at 0?

natka813 [3]

Answer:

The final temperature = 293 K or 20 °C

Explanation:

Heat gained by water at 0°C = heat lost by water at 30°C

c₁m₁(T₃ - T₁) = c₂m₂(T₂-T₃).......................... Equation 1

Making T₃ the subject of the equation,

T₃ = (c₂m₂T₂ + c₁m₁T₁)/(c₁m₁+c₂m₂)............. Equation 2

Where m₁ =mass of water at 0°C, c₁ = specific heat capacity of water at 0°C , c₂ = specific heat capacity of water at 30°C, m₂ = mass of water at 30°C, T₁ = initial temperature of water at 0°C, T₂ = initial temperature of water at 30°C, T₃ = final temperature.

Given: m₁ = 50 g = (50/1000) kg = 0.05 kg, m₂ = 100 g = (100/1000) kg = 0.1 kg., T₁ = 0°C = 273 K, T₂ = 30°C = (30+273 )= 203 K

Constants: c₁ = c₂ = 4200 J/kg.K

Substituting these values into equation 2,

T₃ = (4200×0.1×303 + 4200×0.05×273)/(4200×0.1 + 4200×0.05)

T₃ = (127260 + 57330)/(420+210)

T₃ = 184590/630

T₃ = 293 K.

Therefore the final temperature = 293 K or 20 °C

3 0

3 years ago