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OLEGan [10]
3 years ago
11

If a rock climber accidentally drops a 52.5-g piton from a height of 325 meters, what would its speed be just before striking th

e ground? ignore the effects of air resistance.
Physics
1 answer:
alex41 [277]3 years ago
6 0
Ignoring air resistance, the Kinetic energy before hitting the ground will be equal to the potential energy of the Piton at the top of the rock.  
So we have 1/2 MV^2 = MGH 
V^2 = 2GH 
V = âš2GH 
V = âš( 2 * 9.8 * 325)  
V = âš 6370
 V = 79.81 m/s
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The laws of thermodynamics state that, in a heat engine, _____.
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3 years ago
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Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic colli
serious [3.7K]

Answer:

pf = 198.8 kg*m/s

θ = 46.8º N of E.

Explanation:

  • Since total momentum is conserved, and momentum is a vector, the components of the momentum along two axes perpendicular each other must be conserved too.
  • If we call the positive x- axis to the W-E direction, and the positive y-axis to the S-N direction, we can write the following equation for the initial momentum along the x-axis:

       p_{ox} = p_{oAx} + p_{oBx}  (1)

  • We can do exactly the same for the initial momentum along the y-axis:

       p_{oy} = p_{oAy} + p_{oBy}  (2)

  • The final momentum along the x-axis, since the collision is inelastic and both objects stick together after the collision, can be written as follows:

       p_{fx} =  (m_{A} + m_{B} ) * v_{fx}  (3)

  • We can repeat the process for the y-axis, as follows:

       p_{fy} =  (m_{A} + m_{B} ) * v_{fy}  (4)

  • Since (1) is equal to (3), replacing for the givens, and since p₀Bₓ = 0, we can solve for vfₓ as follows:

       v_{fx} = \frac{p_{oAx}}{(m_{A}+ m_{B)}} = \frac{m_{A}*v_{oAx} }{(m_{A}+ m_{B)}} =\frac{17.0kg*8.00m/s}{46.0kg} =  2.96 m/s (5)

  • In the same way, we can find the component of the final momentum along the y-axis, as follows:

       v_{fy} = \frac{p_{oBy}}{(m_{A}+ m_{B)}} = \frac{m_{B}*v_{oBy} }{(m_{A}+ m_{B)}} =\frac{29.0kg*5.00m/s}{46.0kg} =  3.15 m/s (6)

  • With the values of vfx and vfy, we can find the magnitude of the final speed of the two-object system, applying the Pythagorean Theorem, as follows:

      v_{f} = \sqrt{v_{fx} ^{2} + v_{fy} ^{2}} = \sqrt{(2.96m/s)^{2} + (3.15m/s)^{2}} = 4.32 m/s (7)

  • The magnitude of the final total momentum is just the product of the combined mass of both objects times the magnitude of the final speed:

       p_{f} = (m_{A} + m_{B})* v_{f}  = 46 kg * 4.32 m/s = 198.8 kg*m/s (8)

  • Finally, the angle that the final momentum vector makes with the positive x-axis, is the same that the final velocity vector makes with it.
  • We can find this angle applying the definition of tangent of an angle, as follows:

       tg \theta = \frac{v_{fy}}{v_{fx}} = \frac{3.15 m/s}{2.96m/s} = 1.06 (9)

       ⇒ θ = tg⁻¹ (1.06) = 46.8º N of E

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3 years ago
What is the relationship between mass and weight in physics?
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We are aware that weight is the product of applied gravitational force and mass. W = MG thus, where W represents the weight, M the mass, and G the gravitational force. As a result, it might also mean that "an object's weight is directly proportionate to its mass."

<h3>What is mass?</h3>
  • Mass is a physical body's total amount of matter.
  • It also serves as a gauge for the body's inertia, or resistance to acceleration (change in velocity) in the presence of a net force.
  • The strength of an object's gravitational pull to other bodies is also influenced by its mass.
  • The kilogram is the primary mass unit in the SI (kg).
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<h3>What is weight?</h3>
  • The force exerted on an object by gravity is known as the weight of the object in science and engineering.
  • Weight is sometimes described as a vector quantity, or the gravitational force exerted on the object, in some common textbooks.
  • Others define weight as a scalar quantity, the gravitational force's strength.
  • Others define it as the strength of the force applied to a body as a result of systems designed to resist the effects of gravity; the weight is the amount that is determined, for instance, by a spring scale.

Learn more about mass here:

brainly.com/question/19694949

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7 0
2 years ago
A gas undergoes a process in a piston–cylinder assembly during which the pressure-specific volume relation is pv1.3 = constant.
Galina-37 [17]

Answer:

Change in specific internal Energy=250\ \rm Btu/lb

Explanation:

Given:

  • Mass of the gas, m=0.4 lb
  • Initial pressure and volume are p_1=160\ \rm lbf/in^2\ and\ v_1=1\ \rm ft^3\\
  • Final pressure and temperature are p_1=480\ \rm lbf/in^2
  • Heat transfer from the gas is 2.1 Btu

Since the process is isotropic we have

p_1v_1^{1.3}=p_2v_2^{1.3}\\160\times 1^{1.3}=480\times v_2^{1.3}\\v_2=0.43\ \rm ft^3\\

So the final volume of the gas is calculated.

Work in any isotropic is given by w

w=\dfrac{p_1v_1-P_2v_2}{n-1}\\\\w=\dfrac{160\times1-480\times0.43}{1.3-1}\\w=-154.67\ \rm Btu\\

According to the first law of thermodynamics we have

Q=\Delta U+w\\-2.1=\Delta U-154.67\\\Delta U=152.56\ \rm Btu\\

So the Specific Internal Change is given by

\Delta u=\dfrac{\Delta U}{m}\\\Delta u=\dfrac{152.56}{0.4}\\\Delta u=250\ \rm Btu/lb

So the specific Change in Internal energy is calculated.

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3 years ago
The expression below was formed by combining different gas laws.
kap26 [50]

Answer:

Avogadro's law.

Explanation:

Avogadro’s law states that, equal volumes of all gases at the same temperature and pressure contain the same number of molecules.

Mathematically,

V n

V = Kn where V = volume in cm3, dm3, ml or L; n = number of moles of gas;

K = mathematical constant.

The ideal gas equation is a combination of Boyle's law, Charles' law and Avogadro’s law.

V 1/P at constant temperature (Boyle’s law)

V T at constant pressure ( Charles’law)

V n at constant temperature and pressure ( Avogadro’s law )

Combining the equations yields,

V nT/P

Introducing a constant,

V = nRT/P

PV = nRT

Where P = pressure in atm, Pa, torr, mmHg or Nm-2; V = volume in cm3, dm3, ml or L; T = temperature in Kelvin; n = number of moles of gas in mol; R = molar gas constant = 0.082 dm3atmK-1mol-1

7 0
3 years ago
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