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3 years ago

How much work output can a 150hp snowmobile do in 15 seconds if it is 18 % efficiency?

Physics

1 answer:

Sav [38]3 years ago

3 0

The work output is $3.02\cdot 10^5 J$

Explanation:

The power of the snowmobile is

$P=150 hp$

Keeping in mind that

$1 hp = 746 W$

We can convert it into Watts:

$P=(150)(746)=111,900 W$

The energy used by the snowmobile can be found as follows:

$E=Pt$

where

P = 111,900 W is the power used

t = 15 s is the time interval

Substituting,

$E=(111,900)(15)=1.68\cdot 10^6 J$

However, the snowmobile is 18% efficient: this means that only 18% of this energy is converted into useful work. Therefore, the work output is

$W=0.18E=(0.18)(1.68\cdot 10^6)=3.02\cdot 10^5 J$

Learn more about power and work:

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B) Desirable friction exists between the tires and the road, helping the bike grip the road.

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3 years ago

An air-gap, parallel plate capacitor with area A and gap width d is connected to a battery that maintains the plates at potentia

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Answer:

The new potential energy decreases by the factor of 2 to the old potential energy.

Explanation:

Capacitance of a parallel plate capacitor is given by the relation :

C = (ε₀A)/d

Here ε₀ is vacuum permittivity, A is area of the capacitor plate and d is the distance between them.

Potential energy of the capacitor, U = $\frac{1}{2}CV^{2}$

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According to the problem, the distance between the plates get double but area remains same. So,

d₁ = 2d

Here d₁ is new distance between the plates.

Hence, new capacitance is :

C₁ = (ε₀A)/d₁ = (ε₀A)/2d = C/2

The capacitor have same potential difference that is V. Hence, the new potential energy is :

U₁ = $\frac{1}{2}C_{1} V^{2}$ = $\frac{1}{2}\frac{C}{2} V^{2}$

U₁ = U/2

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3 years ago

A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f

konstantin123 [22]

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

$A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m$

$E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s$

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3 years ago

How do you think electricity was discovered

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Read 2 more answers

A golf ball (mass 0.045 kg) is hit with a club from a tee. The plot shows the force on the ball as a function of time in millise

almond37 [142]

Answer:

16.3 m/s

Explanation:

We are given that

Mass of golf ball=m=0.045 kg

We have to find the speed right after the hit.

$b_1=(16-1)\times 10^{-3}=15\times 10^{-3} s$