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strojnjashka [21]
2 years ago
15

Calculate the following expression: 2.36 + 3.38 + 0.355 + 1.06 =

Physics
2 answers:
mario62 [17]2 years ago
5 0

the answer wil be 2.36+3.38+0.355+1.06=7.155

Salsk061 [2.6K]2 years ago
3 0
After adding them all up you get 7.155
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Which is NOT something that all green plants have in common?
Anvisha [2.4K]

Answer:

Number 4

Explanation:

7 0
2 years ago
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The current in a lamp is 0.5 ampere when it is plugged into a standard wall outlet. What is the resistance of the lamp when it i
lidiya [134]

The resistance of the lamp plugged in to a standard wall outlet with a current of 0.5 amps is 240 Ω (ohms)

Explanation:

In the United States Of America the standard voltage is 120 v and their frequency is 60 Hz

Standard wall outlet voltage is 120 V

The current in the lamp is 0.5 ampere

Resistance (R) = V/ I

      = 120/0.5

      = 240Ω (ohms)

Thus the resistance of the lamp plugged in to a standard wall outlet with a current of 0.5 amps is 240 Ω (ohms).

8 0
3 years ago
Air flows through an adiabatic turbine that is in steady operation. The air enters at 150 psia, 900oF, and 350 ft/s and leaves a
Nonamiya [84]

Answer:

1486.5\frac{Btu}{s}

Explanation:

The inlet specific volume of air is given by:

v_1=\frac{RT_1}{P_1}\\\\v_1=\frac{(0.3704\frac{psia.ft^3}{lbm.R})(1360R)}{150psia}\\\\v_1=3.358\frac{ft^3}{lbm} \ \ \ \  \ \  \ \ \...i

The mass flow rates is expressed as:

\dot m=\frac{1}{v_1}A_1V_1\\\\\dot m=\frac{1}{3.358ft^3/psia}(0.1ft^2)(350ft/s)\\\\\dot m=10.42\frac{lbm}{s}

The energy balance for the system can the be expresses in the rate form as:

E_{in}-E_{out}=\bigtriangleup \dot E=0\\\\E_{in}=E_{out}\\\\\dot m(h_1+0.5V_1^2)=\dot W_{out}+\dot m(h_2+0.5V_2^2)+Q_{out}\\\\\dot W_{out}=\dot m(h_2-h_1+0.5(V_2^2-V_1^2))=-m({cp(T_2-t_1)+0.5(V_2^2-V_1^2)})\\\\\\\dot W_{out}=-(10.42lbm/s)[(0.25\frac{Btu}{lbm.\textdegree F})(300-900)\textdegree F+0.5((700ft/s)^2-(350ft/s)^2)(\frac{1\frac{Btu}{lbm}}{25037ft^2/s^2})]\\\\\\\\=1486.5\frac{Btu}{s}

Hence, the mass flow rate of the air is 1486.5Btu/s

5 0
2 years ago
A calorimeter is used to determine the specific heat capacity of a test metal. If the specific heat capacity of water is known,
denis23 [38]

Answer:

initial and final temperatures of both the water and metal, mass of the metal, and mass of the water

Explanation:

Heat lost by the metal, Q = mc(t_{2} - t_{1})

Heat gained by the water in the calorimeter, Q_{w} = m_{w}c_{w}(t_{2w} - t_{1w})

For energy to be conserved in the system, the heat lost by the metal will equal the heat gain by the water in the calorimeter.

        mc(t_{2} - t_{1}) = m_{w}c_{w}(t_{2w} - t_{1w})

Where,

m is the mass of the metal

c is specific heat capacity of the metal

t₂ is the final temperature of the metal

t₁ is the initial temperature of the metal

m_{w} is the mass of the water

c_{w} is specific heat capacity of water

t_{2w} is the final temperature of water

t_{1w} is the initial temperature of water

From the question given, specific heat capacity of the water is known, the quantities to be measured are;

Initial and final temperatures of both the water and metal,

Mass of the metal, and mass of the water

8 0
2 years ago
Playing tetherball, a boy hits the 0.45 kg ball, moving toward him initially at 4.6 m/s, in the opposite direction at 2.5 m/s. H
saul85 [17]

Answer:

HI

Explanation:

8 0
2 years ago
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