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Blizzard [7]
2 years ago
11

Joules are Nm, which is kg middot m^2/s^2 Use this relationship and 1 Btu = 1.0551 kJ to determine what Btu using primary Englis

h units (ft, lb_m, degree F, sec, etc.)
Physics
1 answer:
son4ous [18]2 years ago
3 0

Answer:

23851.35840\ lbft^2/s^2

Explanation:

1\ Btu=1.0051\ kJ\\\Rightarrow 1\ Btu=1.0051\times 1000 kgm^2/s^2

The English units that are to be used here are pounds (lb) and feet (ft)

1\ kg=2.20462\ lb

1\ m=3.28084\ ft

1005.1\times 2.20462\times 3.28084^2=23851.35840\ lbft^2/s^2

The answer is 1\ Btu=23851.35840\ lbft^2/s^2

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Define fundamental unit ? write its two example. ​
stealth61 [152]

the International System of Units, the fundamental units are: The meter (symbol: m), used to measure length. The kilogram (symbol: kg), used to measure mass. The second (symbol: s), used to measure time. The ampere (symbol: A), used to measure electric current.

8 0
3 years ago
The force between a pair of .005 charges is 750 N. What is the distance between them?
Ganezh [65]

Question: The force between a pair of 0.005 C is 750 N. What is the distance between them?

Answer:

17.32 m

Explanation:

From coulomb's Law,

F = kqq'/r²........................... Equation 1

Where F = Force between the force, q' and q = both charges respectively, k = coulomb's constant, r = distance between both charges.

make r the subject of the equation above

r = √(kqq'/F)..................... Equation 2

From the question,

Given: q = q' = 0.005 C, F = 750 N

Constant: k = 9.0×10⁹ Nm²/C².

Substitute these values into equation 2

r = √(9.0×10⁹×0.005×0.005/750)

r = √(300)

r = 17.32 m.

Hence the distance between the pair of charges = 17.32 m

6 0
3 years ago
Roots that grow horizontally are called
Romashka [77]
I believe they are called Rhizomes. 
8 0
3 years ago
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Thermal energy is the _____ of the kinetic and potential energies of all particles in an object.
Sphinxa [80]
Thermal energy is the force of the kinetic and potential energies of all particles in an object
3 0
2 years ago
A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall, and the other end is connec
juin [17]

Answer:

Explanation:

Given that:

angular frequency = 11.3 rad/s

Spring constant (k) = = \omega^2  \times m

k = (11.3)² m

k = 127.7 m

where;

x_1 = 0.065 m

x_2  = 0.048 m

According to the conservation of energies;

E_1=E_2

∴

\Big(\dfrac{1}{2} \Big) kx_1^2 =\Big(\dfrac{1}{2} \Big) mv_2^2 + \Big(\dfrac{1}{2} \Big) kx_2^2

kx_1^2 = mv_2^2 + kx_2^2

(127.7 \ m) \times 0.065^2 = v_2^2 + (127.7 \ m) \times 0.048^2

0.5395325 = v_2^2 +0.2942208 \\ \\ 0.5395325  - 0.2942208 = v_2^2 \\ \\  v_2^2 = 0.2453117 \\ \\  v_2 = \sqrt{0.2453117} \\ \\ \mathbf{ v_2 \simeq0.50 \ m/s}

4 0
2 years ago
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