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sukhopar [10]
3 years ago
13

Plsssssss helppppppp

Physics
1 answer:
lara31 [8.8K]3 years ago
8 0

Answer:

B

Explanation:

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A hop is a springing from one foot but landing on the other foot? Question 4 options: True False
Rama09 [41]

Answer:

i thinc it's False

Explanation:

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6 0
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7 0
3 years ago
The weights in atwoods machine, starting at rest, attain a velocity of 2ft/sec in one sec. Find the ratio of the masses
Orlov [11]
Refer to the figure shown below.
Let m₁ and m₂ e the two masses.
Let a = the acceleration.
Let T =  tension over the frictionless pulley.

Write the equations of motion.
m₂g - T = m₂a            (1)
T - m₁g = m₁a            (2)

Add equations (1) and (2).
m₂g - T + T - m₁g = (m₁ + m₂)a
(m₂ - m₁)g = (m₁ + m₂)a

Divide through by m₁.
(m₂/m₁ - 1)g = (1 + m₂/m₁)a

Define r = m₂/m₁ as the ratio of the two masses. Then
(r - 1)g = (1 +r)a
r(g-a) = a + g
r = (g - a)/(g + a)

With  = 2 ft/s from rest, the acceleration is
a = 2/32.2 = 0.062 ft/s²
Therefore
r = (32.2 - 0.062)/(32.2 + 0.062) = 0.9962

Answer:
The ratio of masses is 0.9962 (heavier mass divided by the lighter mass).

8 0
3 years ago
At what wavelength would a star radiate the greatest amount of energy if the star has a surface temperature of 60,000 K?
kompoz [17]

Answer:

\lambda=4.81\times 10^{-8}\ m

Explanation:

We have,

The surface temperature of the star is 60,000 K

It is required to find the wavelength of a star that radiated greatest amount of energy. Wein's displacement law gives the relation between wavelength and temperature such that :

\lambda T=2.89\times 10^{-3}

Here,

\lambda = wavelength

\lambda=\dfrac{2.89\times 10^{-3}}{60000}\\\\\lambda=4.81\times 10^{-8}\ m

So, the wavelength of the star is 4.81\times 10^{-8}\ m.

7 0
3 years ago
Air enters a turbine operating at steady state at 8 bar, 1600 K and expands to 0.8 bar. The turbine is well insulated, and kinet
kobusy [5.1K]

Answer:

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

Explanation:

To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables

Mathematically this can be determined as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}

Where

Temperature at inlet of turbine

Temperature at exit of turbine

Pressure at exit of turbine

Pressure at exit of turbine

The steady flow Energy equation for an open system is given as follows:

m_i = m_0 = mm(h_i+\frac{V_i^2}{2}+gZ_i)+Q = m(h_0+\frac{V_0^2}{2}+gZ_0)+W

Where,

m = mass

m(i) = mass at inlet

m(o)= Mass at outlet

h(i)= Enthalpy at inlet

h(o)= Enthalpy at outlet

W = Work done

Q = Heat transferred

v(i) = Velocity at inlet

v(o)= Velocity at outlet

Z(i)= Height at inlet

Z(o)= Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

h_i = h_0 + WW = h_i -h_0

Using the relation T-P we can find the final temperature:

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}\\

\frac{T_2}{1600K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}\\ = 828.716K

From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK

W = h_i -h_0W = C_p (T_1-T_2)W = 1.005(1600 - 828.716)W = 775.140kJ/Kg

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

4 0
3 years ago
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