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2 years ago

Power of an electric motor is 1 h.p. what does it mean ?

Physics

1 answer:

const2013 [10]2 years ago

8 0

Answer: HP = Horse Power.

Explanation: it is the unit given to tell the motor's particular power and 1hp = 746 watts.

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Human-Powered Flight Human-powered aircraft require a pilot to pedal, as in a bicycle, and produce a sustained power output of a

alukav5142 [94]

Answer:

# of Snickers bars 2

Explanation:

Power output= 0.30 HP

=0.3*746

= 0.30 HP (746 W=1.00 HP)

= 224 W

time required 2 h 49 m = 10140 seconds

Since power is work divided by time, then work is:

Work done by the jet = P*t

= 224 *(10140)

= 2.3 MJ (2.3 x $10^{6}$ J)

Converting MJ to Cal

2.3 MJ=549 Cal

# of Snickers bars = 549 Cal / 280 Cal

= 2.0 bars (rounded from 1.96)

8 0

3 years ago

Under what condition will the results of an experiment based on a hypothesis most likely lead to new experimentation?

MakcuM [25]

Failed experiments, uncontrolled variables, invalid data, and generalized human error

7 0

3 years ago

The vertebral region is _________ to the scapular region.

Lisa [10]

Answer:

The answer is medial!

Explanation:

The vertebral region is medial to the scapula.

Hope This Helps!

-Justin:)

7 0

2 years ago

Calculate the electric field associated to an electric dipole for two charges separated 10-8 m with a dipole moment of 10-33 C m

Alex Ar [27]

Answer:

18 N/C

Explanation:

Given that:

Electric field constant, k = 9*10^9 N/c

Distance, r = 10^-8 m

Dipole moment, p = 10^-33

Using the relation for electric field due to dipole :

E = [2KP / r³]

E = (2 * (9*10^9) * 10^-33) ÷ (10^-8)^3

E = (18 * 10^9 * 10^-33) ÷ 10^-24

E = [18 * 10^(9-33)] ÷ 10^-24

E = (18 * 10^-24) / 10^-24

E = 18 * 10^-24+24

E = 18 * 10^0

E = 18 N/C

5 0

3 years ago

A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat does Wnc = 75.0 J of work on the

Eva8 [605]

Answer:

The speed of the ball is 42.5 m/s

Explanation:

The initial kinetic energy of the ball is:

$K_1=\frac{1}{2} m v_0^2=\frac{1}{2}*0.140 kg*(35.0 m/s)^2$ = 85.75 J

The speed of the ball after leaving the bat is:

$K_2=K_1+W_{nc}\\ \frac{1}{2}mV^2= 85.75 J + 75 J\\ (\frac{1}{2}mV^2)2=( 160.75 J)2\\ mV^2= 321.5 J\\ V^2= \frac{321.5 J}{0.140kg} \\ V=\sqrt{\frac{321.5 J}{0.140kg}}$

V=47.92 m/s

Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:

$V_f^2-V^2=-2gh$

$V_f^2-(47.92 m/s)^2=-2*9.81m/s^2*25m$

$V_f^2=-2*9.81m/s^2*25m+(47.92 m/s)^2$

$V_f=\sqrt{-2*9.81m/s^2*25m+(47.92 m/s)^2}$

$V_f=42.5m/s$

3 0

3 years ago

Power of an electric motor is 1 h.p. what does it mean ?​

Power of an electric motor is 1 h.p. what does it mean ?