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3 years ago

A crane raises a crate with a mass of 75 kg to a height of 10 m. Given that the

Physics

1 answer:

Crazy boy [7]3 years ago

7 0

Answer: C. 7,350 J

Explanation:

Just did it in Apex

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A 2.0-kg cart collides with a 1.0-kg cart that is initially at rest on a low-friction track. After the collision, the 1.0-kg car

nikitadnepr [17]

To solve this problem we will apply the concepts related to the conservation of momentum. The momentum can be defined as the product between the mass of the object and its velocity, and the conservation of the momentum as the equality between the change of the initial momentum versus the final momentum. Mathematically, this relationship can be described as

$m_1u_1+m_2u_2 = m_1v_2+m_2v_2$

Here,

$m_{1,2}$ = Mass of each object

$u_{1,2}$ = Initial velocity of each object

$v_{1,2}$ = Final velocity of each object

According to the statement one of the bodies does not have initial velocity, therefore said term would be zero. And the equation could be rewritten as,

$m_1u_1= m_1v_2+m_2v_2$

Replacing the values respectively (The mass of your body with its respective speed we would have)

$2kg(u_1) = 2kg(0.3m/s)+1kg(0.5m/s)$

$u_1 = \frac{2kg(0.3m/s)+1kg(0.5m/s)}{2kg}$

$u_1 = 0.55m/s$

Therefore the initial velocity of the 2kg cart is 0.55m/s

4 0

2 years ago

A 12 cm diameter piston-cylinder device contains air at a pressure of 100 kPa at 24oC. The piston is initially 20 cm from the ba

lina2011 [118]

Answer:

ΔQ = 0.1 kJ

$\mathbf{v_f = 1.445*10^{-3} m^3}$

$\mathbf{P_f = 156.5 \ kPa}$

ΔS = -0.337 J/K

The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.

Explanation:

GIven that:

Diameter of the piston-cylinder = 12 cm

Pressure of the piston-cylinder = 100 kPa

Temperature =24 °C

Length of the piston = 20 cm

Boundary work ΔW = 0.1 kJ

The gas is compressed and The temperature of the gas remains constant during this process.

We are to find ;

a. How much heat was transferred to/from the gas?

According to the first law of thermodynamics ;

ΔQ = ΔU + ΔW

Given that the temperature of the gas remains constant during this process; the isothermal process at this condition ΔU = 0.

Now

ΔQ = ΔU + ΔW

ΔQ = 0 + 0.1 kJ

ΔQ = 0.1 kJ

Thus; the amount of heat that was transferred to/from the gas is : 0.1 kJ

b. What is the final volume and pressure in the cylinder?

In an isothermal process;

Workdone W = $\int dW$

$W = \int pdV \\ \\ \\W = \int \dfrac{nRT}{V}dv \\ \\ \\ W = nRt \int \dfrac{dv}{V} \\ \\ \\ W = nRT In V |^{V_f} __{V_i}} \\ \\ \\ W = nRT \ In \dfrac{V_f}{V_i}$

Since the gas is compressed ; then $v_f< v_i$

However;

$W =- nRT \ In \dfrac{V_f}{V_i}$

$W =- P_1V_1 \ In \dfrac{V_f}{V_i}$

The initial volume for the cylinder is calculated as ;

$v_1 = \pi r^2 h \\ \\ v_1 = \pi r^2 L \\ \\ v_1 = 3.14*(6*10^{-2})^2*(20*10^{-2}) \\ \\ v_1 = 2.261*10^{-3} \ m^3$

Replacing over values into the above equation; we have :

$100 = - ( 100*10^3 *2.261*10^{_3}) In (\dfrac{v_f}{v_i}) \\ \\ - In (\dfrac{v_f}{v_i})= \dfrac{100}{(100*10^3*2.261*10^{-3})} \\ \\ - In \ v_f + In \ v_i = \dfrac{100}{226.1} \\ \\ - In \ v_f = - In \ v_i + \dfrac{100}{226.1} \\ \\ - In \ v_f = - In (2.261*10^{-3} + \dfrac{100}{226.1 } \\ \\ - In \ v_f = 6.1 + 0.44 \\ \\ - In \ v_f = 6.54 \\ \\ - In \ v_f = -6.54 \\ \\ v_f = e^{-6.54} \\ \\ \mathbf{v_f = 1.445*10^{-3} m^3}$

The final pressure can be calculated by using :

$P_1V_1 = P_2V_2 \\ \\ P_iV_i = P_fV_f$

$P_f =\dfrac{P_iV_i}{V_f}$

$P_f =\dfrac{100*2.261*10^{-3}}{1.445*10^{-3}}$

$P_f = 1.565*10^2 \ kPa$

$\mathbf{P_f = 156.5 \ kPa}$

c. Find the change in entropy of the gas. Why is this value negative if entropy always increases in actual processes?

The change in entropy of the gas is given by the formula:

$\Delta S=\dfrac{\Delta Q}{T}$

where

T = 24 °C = (24+273)K

T = 297 K

$\Delta S=\dfrac{-100 \ J}{297 \ K}$

ΔS = -0.337 J/K

The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.

4 0

2 years ago

A ball was kicked upward at a speed of 64.2 m/s. How fast was the ball going 1.5 seconds later

Kobotan [32]

The ball's velocity $v$ at time $t$ is

$v=v_0-gt$

where $g=9.81\,\frac{\mathrm m}{\mathrm s^2}$ is the acceleration due to gravity. So after 1.5 seconds, the ball's velocity is

$v=64.2\,\dfrac{\mathrm m}{\mathrm s}-\left(9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)(1.5\,\mathrm s)$

$\implies v=49.5\,\dfrac{\mathrm m}{\mathrm s}$

so the ball's speed after 1.5 seconds is also 49.5 m/s.

3 0

3 years ago

22. A 2.0 kg ball, traveling at 1.0 m/s, rolls off of a cliff of height 3.0 m. If the

RoseWind [281]

GOOD MORNING HAVE A NICE DAY

3 0

2 years ago

The direct sunlight at Earth's surface is about 1050 W/m2 . Compute mass lost by Sun in a thousand years as a fraction of Earth

dimaraw [331]

Answer:

Explanation:

Energy falling on 1 m² surface of earth per second = 1050

Energy in one million years on 1 m²

= 1050 x 60 x 60 x 24 x 365 x 10⁶ = 3.311 x 10¹⁶ J

In order to calculate total energy coming out of the surface of the sun , we shall have to sum up this energy for the while spherical surface of imaginary sphere having radius equal to distance between sun and earth.

Area of this surface = 4π R² = 4 X 3.14 X (149.6 X 10⁹ )²

= 2.8 X 10²³ m²

So total energy coming out of the sun = 2.8 x 10²³ x 3.311 x 10¹⁶

= 9.271 x 10³⁹ J

From the formula

E = mc² { energy mass equivalence formula }

m = E / c² = $\frac{9.271 \times10^{39}}{9 \times 10^{16}}$

1.03 x 10²³ kg

mass of earth = 5.972 x 10²⁴

Answer in percentage of mass of earth

= $\frac{1.03\times10^23}{5.972\times10^{24}}\times100$

= 1.72 %

3 0

2 years ago