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vodomira [7]
3 years ago
7

Do all engines operate at less than 100 percent efficiency because they conduct heat

Physics
2 answers:
Alika [10]3 years ago
4 0

all engines operate at less than 100 percent efficiency because they EMIT heat. ;)

Julli [10]3 years ago
3 0

They EMIT heat is the correct answer



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A certain spring stretches 3 cm when a load of 15 n is suspended from it. how much will the spring stretch if 30 n is suspended
Alik [6]
Initially, the spring stretches by 3 cm under a force of 15 N. From these data, we can find the value of the spring constant, given by Hook's law:
k= \frac{F}{\Delta x}
where F is the force applied, and \Delta x is the stretch of the spring with respect to its equilibrium position. Using the data, we find
k= \frac{15 N}{3.0 cm}=5.0 N/cm

Now a force of 30 N is applied to the same spring, with constant k=5.0 N/cm. Using again Hook's law, we can find the new stretch of the spring:
\Delta x =  \frac{F}{k}= \frac{30 N}{5.0 N/cm}=6 cm
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3 years ago
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A physical science test book has a mass of 2.2kg. . What is the weight on the earth
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Weight = mg,          g ≈ 9.8 m/s²

Weight = 2.2 * 9.8 ≈ 21.56 N
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A 25.0-kg test rocket is fired vertically. When the engine stops firing, the rocket’s kinetic energy is 2017 J. After the fuel i
Dahasolnce [82]

Answer:

1. 2.12105 J. The final kinetic energy is. KE f mv2. (875.0 kg)(44.0 m/s)2 2.What is the velocity of the two hockey players after the collision? ... A 10.0-kg test rocket is fired vertically from.

Explanation:

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3 years ago
A charge of -2.65 nC is placed at the origin of an xy-coordinate system, and a charge of 2.00 nC is placed on the y axis at y =
stiks02 [169]

Answer:

A. Fnx = 5.71*10⁻⁵ N  ,  Fny= -3.67*10⁻⁵ N

B. Fn= 6.78 *10⁻⁵ N

C. α= 32.4° counterclockwise with the positive x+ axis

Explanation:

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Equivalences

1nC= 10⁻⁹C

1cm = 10⁻²m

Known data

k= 9*10⁹N*m²/C²

q₁= -2.65 nC =-2.65*10⁻⁹C

q₂= +2.00 nC = 2*10⁻⁹C

q₃= +5.00 nC= =+5*10⁻⁹C

d_{13} = \sqrt{(3.2)^{2} +(3.8)^{2} }

d_{13} =\sqrt{24.68} * 10⁻²m    = 4.9678* 10⁻²m

(d₁₃)² = 24.68*10⁻⁴m²

d₂₃ = 3.2 cm = 3.2*10⁻²m  

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.

The force (F₂₃) of q₂ on q₃ is repulsive because the charges have equal signs and the forces.

The force (F₁₃) of q₁ on q₃ is attractive because the charges have opposite signs.

Magnitudes of F₁₃ and F₂₃

F₁₃ = (k*q₁*q₃)/(d₁₃)²=( 9*10⁹*2.65*10⁻⁹*5*10⁻⁹) /(24.68*10⁻⁴)

F₁₃ = 4.8 *10⁻⁵ N

F₂₃ = (k*q₂*q₃)/(d₂₃)² =  ( 9*10⁹*2*10⁻⁹*5*10⁻⁹) /((3.2)²*10⁻⁴)

F₂₃ = 8.8 *10⁻⁵ N

x-y components of F₁₃ and F₂₃

F₁₃x= -4.8 *10⁻⁵ *cos β= - 4.8 *10⁻⁵(3.2/ (4.9678)= - 3.09*10⁻⁵ N

F₁₃y= -4.8 *10⁻⁵ *sin β= - 4.8 *10⁻⁵(3.8/(4.9678) =  - 3.67*10⁻⁵ N

F₂₃x  = F₂₃ =  +8.8 *10⁻⁵ N

F₂₃y = 0

x and y components of the total force exerted on q₃ by q₁ and q₂ (Fn)

Fnx= F₁₃x+F₂₃x =  - 3.09*10⁻⁵ N+8.8 *10⁻⁵ N= 5.71*10⁻⁵ N

Fny= F₁₃y+F₂₃y = - 3.67*10⁻⁵ N+0= - 3.67*10⁻⁵ N

Fn magnitude

F_{n} =\sqrt{(Fn_{x})^{2}+(Fn_{y})^{2}  }

F_{n} = \sqrt{(5.71)^{2}+(3.67)^{2}  } *10⁻⁵ N

Fn= 6.78 *10⁻⁵ N

Fn direction  (α)

\alpha =tan^{-1}( \frac{Fn_{y} }{Fn_{x} } )

\alpha =tan^{-1}( \frac{-3.67 }{5.71} )

α= -32.4°

α= 32.4° counterclockwise with the positive x+ axis

4 0
3 years ago
In what way are mercury and venus similar ?
Neporo4naja [7]

Answer:

they are both planets they are both made of rock and  metel

Explanation:

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3 years ago
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