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boyakko [2]
2 years ago
5

Compare the energy consumption of two commonly used items in the household. Calculate the energy used by a 1.40 kW toaster oven,

Wtoaster , which is used for 6.00 minutes, and then calculate the amount of energy that an 11.0 W compact fluorescent light (CFL) bulb, Wlight , uses when left on for 9.00 hours .
Physics
1 answer:
Nikolay [14]2 years ago
3 0

Explanation:

It is given that,

Power consumed by toaster oven, P=1.4\ kW

Time taken, t = 6 minutes = 0.1 hour

Energy used by toaster, W_{toaster}=P\times t

W_{toaster}=1.4\ kW\times 0.1\ h=0.14\ kWh

Similarly,

Power consumed by toaster oven, P=11\ W

Time taken, t = 9 minutes = 0.15 hour

Energy used by toaster, W_{toaster}=P\times t

W_{toaster}=11\ W\times 0.15\ h=1.65\ Wh=0.0016\ kWh

Hence, this is the required solution.

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An eagle carrying a trout flies above a lake along a horizontal path. The eagle drops the trout from a height of 6.1 m. The fish
Snezhnost [94]

Answer:

7.1 m/s

Explanation:

First, find the time it takes for the fish to reach the water.

Given in the y direction:

Δy = 6.1 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

6.1 m = (0 m/s) t + ½ (9.8 m/s²) t²

t = 1.12 s

Next, find the velocity needed to travel 7.9 m in that time.

Given in the x direction:

Δx = 7.9 m

a = 0 m/s²

t = 1.12 s

Find: v₀

Δx = v₀ t + ½ at²

7.9 m = v₀ (1.12 s) + ½ (0 m/s²) (1.12 s)²

v₀ = 7.1 m/s

4 0
3 years ago
Read 2 more answers
Suppose that the height of the incline is h = 14.7 m. Find the speed at the bottom for each of the following objects. 1.solid sp
tensa zangetsu [6.8K]

Answer:

1. 14.4 m/s  2. 13.2 m/s 3. 12.0 m/s 4. 13.9 m/s

Explanation:  

Assuming no friction present, the different objects roll without slipping, so there is a constant relationship between linear and angular velocity, as follows:

ω= v/r

If no friction exists, the change in total kinetic energy must be equal in magnitude to the change in the gravitational potential energy:

∆K = -∆U

 ½ *m*v² + ½* I* ω²  = m*g*h

Simplifying and replacing the value of the angular velocity:

½ * v² + ½ I *(v/r)² = g*h (1)

In order to answer the question, we just need to replace h by the value given, and I (moment of inertia) for the value for each different object, as follows:

  •  Solid Sphere I = 2/5* m *r²

                Replacing in (1):

                ½ * v² + ½ (2/5 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

                v = √(10/7*9.8 m/s2*14.7 m)  = 14. 4 m/s

  • Spherical shell I=2/3*m*r²

                Replacing in (1):

                ½ * v² + ½ (2/3 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

                v = √(6/5*9.8 m/s2*14.7 m)  = 13.2 m/s

  • Hoop   I= m*r²

                Replacing in (1):

                ½ * v² + ½ (m*r²) *(v/r)² = g*h

               Replacing by the value given for h, and solving for v:

               v = √(9.8 m/s2*14.7 m)  = 12.0 m/s

  • Cylinder I = 1/2 * m* r²

                 Replacing in (1):

                ½ * v² + ½ (1/2 *m*r²) *(v/r)²= g*h

                 Replacing by the value given for h, and solving for v:

                v = 2*√(1/3*9.8 m/s2*14.7 m)  = 13.9 m/s

5 0
3 years ago
Aristotle supported which of these views?
pishuonlain [190]
<span>The Sun and all the planets revolve around Earth.</span>
4 0
2 years ago
Read 2 more answers
Charge of uniform linear density (6.7 nC/m) is distributed along the entire x axis. Determine the magnitude of the electric fiel
ad-work [718]

Thw question is not complete. The complete question is;

Charge of uniform linear density (6.7 nCim) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y = 1.6 m. a. 32 N/C b. 150 NC c 75 N/C d. 49 N/C e. 63 NC

Answer:

Option C: E = 75 N/C

Explanation:

We are given;

Uniform linear density; λ = 6.7 nC/m = 6.7 × 10^(-9) C/m

Distance on the y-axis; d = 1.6 m

Now, the formula for electric field with uniform linear density is given as;

E = λ/(2•π•r•ε_o)

Where;

E is electric field

λ is uniform linear density = 6.7 × 10^(-9) C/m

r is distance = 1.6m

ε_o is a constant = 8.85 × 10^(-12) C²/N.m²

Thus;

E = (6.7 × 10^(-9))/(2π × 1.6 × 8.85 × 10^(-12))

E = 75.31 N/C ≈ 75 N/C

5 0
3 years ago
A converging-diverging nozzle has a throat area of 10 cm2 and an exit area of 28.96 cm2 . A normal shock stands in the exit when
Svetllana [295]

The tank pressure is 5.08 kPa and the mass flow rate is 2.6 kg/s.

The given parameters:

  • <em>Throat area of the nozzle, </em>A^*<em> = 10 cm² = 0.001 m²</em>
  • <em>The exit area of the nozzle, A = 28.96 cm² = 0.002896 m²</em>
  • <em>Air pressure at sea level = 101.325 kPa</em>

The ratio of the areas of the converging-diverging nozzle is calculated as follows;

= \frac{A}{A^*} \\\\= \frac{0.002896}{0.001} \\\\= 2.896

From supersonic isentropic table, at \frac{A}{A^*} = 2.896, we can determine the following;

M_e = 2.6 \ kg/s\\\\\frac{P_o}{P_e} = 19.954

The tank pressure is calculated as follows;

\frac{P_o}{P_e} = 19.954 \\\\P_e = \frac{P_o}{19.954} \\\\P_e = \frac{101.325 \ kPa}{19.954} \\\\P_e = 5.08 \ kPa

Thus, the tank pressure is 5.08 kPa and the mass flow rate is 2.6 kg/s.

Learn  more about converging-diverging nozzle design here: brainly.com/question/13889483

8 0
2 years ago
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