All categories

3 years ago

An engine creat a 2230N force that pushes a car forward 58.3m in 9.4s

Physics

1 answer:

Scilla [17]3 years ago

5 0

Answer:

13830.74 W

Explanation:

Applying

P = W/t.................... Equation 1

Where P = Power produce in the engine, W = Work done by the engine, t = time.

But,

W = F×d................... Equation 2

Where F = force, d = distance.

Substitute equation 2 into equation 1

P = (F×d)/t...................... Equation 3

From the question,

Given: F = 2230 N, d = 58.3 m, t = 9.4 s

Substitute these values into equation 3

P = (2230×58.3)/9.4

P = 130009/9.4

P = 13830.74 W

You might be interested in

2. A ladder 16cm long weighs 420N and it's centre of gravity is 7m from one end.It is carried horizontally by two boys,each hold

ipn [44]

Answer:

well the answers are 180 and 240 obviously

Explanation:

Download pdf

8 0

3 years ago

Read 2 more answers

Given that the CFL bulb is rated at 10% efficiency, if the LED bulb consumes half the amount of electrical power, for the same a

Butoxors [25]

Answer and explanation:

The efficiency of the LED bulb is 20% while The efficiency of the Inc bulb is 1%.

The lightbulb efficiency can be defined as the quotient between the amount of usable light and the amount of electrical power output. For the efficiency of the CFL Bulb:

$\epsilon_{CFL}=\frac{L_0}{P_0}=0.1$

Therefore:

$\epsilon_{led}=\frac{L_0}{0.5P_0}=2 \cdot0.1=0.2\\\epsilon_{INC}=\frac{L_0}{10P_0}=0.1 \cdot 0.1=0.01$

4 0

3 years ago

Two billiard balls of identical mass move toward each other. Assume that the collision between them is perfectly elastic. If the

anzhelika [568]

Answer:

-20.0 m/s and 30.0 m/s

Explanation:

Momentum is conserved:

m (30.0) + m (-20.0) = m v₁ + m v₂

30.0 − 20.0 = v₁ + v₂

10.0 = v₁ + v₂

Since the collision is perfectly elastic, energy is also conserved. Since there's no rotational energy or work done by friction, the initial kinetic energy equals the final kinetic energy.

½ m (30.0)² + ½ m (-20.0)² = ½ mv₁² + ½ mv₂²

(30.0)² + (-20.0)² = v₁² + v₂²

1300 = v₁² + v₂²

We now have two equations and two variables. Solve the system of equations using substitution:

1300 = v₁² + (10 − v₁)²

1300 = v₁² + 100 − 20v₁ + v₁²

0 = 2v₁² − 20v₁ − 1200

0 = v₁² − 10v₁ − 600

0 = (v₁ + 20) (v₁ − 30)

v₁ = -20, 30

If v₁ = -20, v₂ = 30.

If v₁ = 30, v₂ = -20.