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wariber [46]
3 years ago
8

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:

brainly.com/question/7956557

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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Answer:

B. Oblique

Explanation:

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3 0
3 years ago
Arrange the phases of the Moon in order of increasing rising time, from the phase with the earliest rising time at 12:00 a.m. to
ZanzabumX [31]

Answer:

Lets assume the Sun rise time to be 06:00 AM. The rise time of different phases of Moon will be as follows:

12:00 AM : Waning Half

01:00 AM - 05:00 AM : Waning Crescent

06:00 AM : New Moon

07:00 AM - 11:00 AM : Waxing Crescent

12:00 PM :  Waxing Half

01:00 PM - 06:00 PM : Waxing Gibbous

06:00 PM : Full Moon

07:00 PM - 09:00 PM : Waning Gibbous

Explanation:

The Moon is the only celestial object which shows visible changes in its shape and rise and set time over a very short period of time i.e. just one day. One can observe it by observing the Moon daily. One will notice the change easily. This happens because of the geometry of the Sun, Earth and Moon. The Moon doesn't have its own light and shines because of the light of Sun.

At any given time half of the Moon would be illuminated by the Sun but how much of this illuminated portion is facing the Earth decides the phase of the Moon visible from the Earth. Due to this the Moon shows us various phases namely: New, Waxing Crescent, Waxing Half, Waxing Gibbous, Full, Waning Gibbous, Waning Half, Waning Crescent.

Also, the Moon revolves around the Earth completing the orbit in 29.5 Days. Everyday the Moon will change its position in the orbit. Due to this the rising time of Moon shifts by approximately 52 minutes daily. So, the New Moon rises with the Sun and Full Moon rises just after the sunset.

Lets assume the Sun rise time to be 06:00 AM. The rise time of different phases of Moon will be as follows:

12:00 AM : Waning Half

01:00 AM - 05:00 AM : Waning Crescent

06:00 AM : New Moon

07:00 AM - 11:00 AM : Waxing Crescent

12:00 PM :  Waxing Half

01:00 PM - 06:00 PM : Waxing Gibbous

06:00 PM : Full Moon

07:00 PM - 09:00 PM : Waning Gibbous

5 0
3 years ago
How do determine which is Y2, Y1, X2, and X1 on a graph. <br><br> And how do find rise over run.
fredd [130]
One point will be X1,Y1 and the other will be X2,Y2. It does not matter which is which except that X1 and Y1 have to be the same point and X2 and Y2 have to be the same point. For example, let's say you were given (2,3) and (6,8). No matter which point is X1,Y1 and the other is X2,Y2, the slope will still be 5/4. 

The rise is the change in y from one point to the other. The run would be the change in x from one point to the other.
7 0
4 years ago
What contributions did J.J. Thomson make to atomic history?
luda_lava [24]
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4 0
3 years ago
A 56 kg sprinter, starting from rest, runs 49 m in 7.0 s at constant acceleration.what is the sprinter's power output at 2.0 s,
alexgriva [62]
The sprinter is in uniform accelerated motion, and its initial velocity is zero, so the relationship betwen space (S) and time (t) is
S= \frac{1}{2} a t^2
where a is the acceleration. Using the data of the problem, we can find a:
a= \frac{2S}{t^2} = \frac{2 \cdot 49 m}{(7.0 s)^2} =2.0 m/s^2
So now we can solve the 3 parts of the problem.

a) power output at t=2.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(2.0 s)=4.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(4.0 m/s)^2=448 J

and so the power output is
P= \frac{E}{t} = \frac{448 J}{2.0 s} =224 W

b) power output at t=4.0s 
The velocity at t=4.0 s is
v(t)=at=(2.0 m/s^2)(4.0 s)=8.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(8.0 m/s)^2=1792 J

and so the power output is
P= \frac{E}{t} = \frac{1792 J}{4.0 s} =448 W

c) Power output at t=6.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(6.0 s)=12.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(6.0 m/s)^2=4032 J

and so the power output is
P= \frac{E}{t} = \frac{4032 J}{6.0 s} =672 W
8 0
3 years ago
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