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Debora [2.8K]
2 years ago
15

Is horsepower the only thing to be considered when purchasing a piece of equipment?Explain.​

Physics
2 answers:
Novosadov [1.4K]2 years ago
8 0

Answer:

Mostly yes

Explanation:

But you should consider,brand value, material used and lasting of the equipment also

  • 1HP=720W

HP stands for horsepower

It's mostly used in pumps

Juli2301 [7.4K]2 years ago
3 0
Not necessarily. It should definitely be a factor, but the age of the piece should be in question as well as the brand, materials used and how fuel-efficient it is.
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A certain tuning fork vibrates at a frequency of 215 Hz while each tip of its two prongs has an amplitude of 0.832 mm. (a) What
Marysya12 [62]

Explanation:

It is given that,

Frequency of vibration, f = 215 Hz

Amplitude, A = 0.832 mm

(a) Let T is the period of this motion. It is given by the following relation as :

T=\dfrac{1}{f}

T=\dfrac{1}{215}

T=4.65\times 10^{-3}\ s

(b) Speed of sound in air, v = 343 m/s

It can be given by :

v=f\times \lambda

\lambda=\dfrac{v}{f}

\lambda=\dfrac{343\ m/s}{215\ Hz}

\lambda=1.59\ m

Hence, this is the required solution.

5 0
4 years ago
If someone is driving 100 miles in 60 minutes then drives 150 miles in 100 minutes west, what is his acceleration rate.
Sliva [168]

Answer:

his acceleration rate is -0.00186 m/s²

Explanation:

Given;

initial position of the car, x₀ = 100 miles = 160, 900 m ( 1 mile = 1609 m)

time of motion, t₀ = 60 minutes = 60 mins x 60 s = 3,600 s

final position of the car, x₁ = 150 miles = 241,350 m

time of motion, t₁ = 100 minutes = 100 mins x 60 s = 6,000 s

The initial velocity is calculated as;

u = 160, 900 m / 3,600 s

u = 44.694 m/s

The final velocity is calculated as;

v = 241,350 m / 6,000 s

v = 40.225 m/s

The acceleration is calculated as;

a = \frac{\Delta V}{\Delta t} = \frac{v- u}{t_1 - t_ 0} = \frac{40.225  - 44.694}{6000-3600} = -0.00186 \ m/s^2\\\\

Therefore, his acceleration rate is -0.00186 m/s²

6 0
3 years ago
Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the qu
Luba_88 [7]

Answer:

V_1=8 V_2

Explanation:

Given that:

  • Area of the plate of capacitor 1= Area of the plate of capacitor 2=A
  • separation distance of capacitor 2, d_2=d
  • separation distance of capacitor 1, d_1=2d
  • quantity of charge on capacitor 2, Q_2=Q
  • quantity of charge on capacitor 1, Q_1=4Q

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

C=\frac{k.\epsilon_0.A}{d}.....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}

From eq. (1)

For capacitor 2:

C_2=\frac{k.\epsilon_0.A}{d}

For capacitor 1:

C_1=\frac{k.\epsilon_0.A}{2d}

C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]

We know, potential differences across a capacitor is given by:

V=\frac{Q}{C}..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}

V_2=\frac{Q.d}{k.\epsilon_0.A}

& for capacitor 1:

V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}

V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}

V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]

V_1=8 V_2

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True or False? <br> Metrics are based on the British system of weights and measures.
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The answer I'm going with is false
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