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Crank
3 years ago
11

A homeowner has a new oil furnace which has an efficiency of 60%. For every 100 barrels of oil used to heat his house, how much

(in barrels of oil) goes up the chimney as waste heat?
Physics
1 answer:
nasty-shy [4]3 years ago
4 0

Answer:

below

Explanation:

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If an object undergoes a change in momentum of 10 kg m/s in 3 s ,then the force acting on it is
Paha777 [63]

Answer:

Force = 3.333 Newton

Explanation:

Given the following data;

Change in momentum = 10 Kgm/s

Time = 3 seconds

To find the force acting on it;

In Physics, the change in momentum of a physical object is equal to the impulse experienced by the physical object.

Mathematically, it is given by the formula;

Force * time = mass * change in velocity

Impulse = force * time

Substituting into the formula, we have;

10 = force * 3

Force = 10/3

Force = 3.333 Newton

8 0
3 years ago
Birdman is flying horizontally at a
zavuch27 [327]

Answer:

X=92.49 m

Explanation:

Given that

u= 21 m/s

h= 97 m

Time taken to cover vertical distance h

h= 1/2 g t²

By putting the values

97 = 1/2 x 10 x  t²          ( g = 10 m/s²)

t= 4.4 s

The horizontal distance

X= u .t

X= 21 x 4.4

X=92.49 m

3 0
3 years ago
An astronaut on a strange new planet having no atmosphere finds that she can jump up to a maximum height of 27 m when her initia
vekshin1
The magnitude is 4.5
7 0
3 years ago
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8 0
3 years ago
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The parallel plates in a capacitor, with a plate area of 8.00 cm2 and an air-filled separation of 2.70 mm, are charged by a 8.70
MrRa [10]

Answer:

a)  ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J,  d)  W = 150 10⁻¹² J

Explanation:

Let's find the capacitance of the capacitor

         C = \epsilon_o \frac{A}{d}

         C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³

         C = 2.62 10⁻¹² F

for the initial data let's look for the accumulated charge on the plates

          C = \frac{Q}{\Delta V}

          Q₀ = C ΔV

           Q₀ = 2.62 10⁻¹² 8.70

           Q₀ = 22.8 10⁻¹² C

a) we look for the capacity for the new distance

          C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³

          C₁ = 1.04 10⁻¹² F

       

          C₁ = Q₀ / ΔV₁

          ΔV₁ = Q₀ / C₁

          ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²

          ΔV₁ = 21.9 V

b) initial stored energy

          U₀ = \frac{Q_o}{ 2C}

          U₀ = (22.8 10⁻¹²)²/(2  2.62 10⁻¹²)

          U₀ = 99.2 10⁻¹² J

c) final stored energy

          U_f = (22.8 10⁻¹²) ² /(2  1.04 10⁻⁻¹²)

          U_f = 249.9 10⁻¹² J

d) the work of separating the plates

as energy is conserved work must be equal to energy change

          W = U_f - U₀

          W = (249.2 - 99.2) 10⁻¹²

          W = 150 10⁻¹² J

note that as the energy increases the work must be supplied to the system

6 0
3 years ago
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