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jekas [21]
3 years ago
10

Phil is riding a scooter and pushes off the ground with his foot. this causes him to accelerate at 12 m /s. Phil weighs 600 N. h

ow strong was his push off the ground?​
Physics
1 answer:
Dvinal [7]3 years ago
7 0

Answer:

734.16 kg m/s^{2}

Explanation:

The problem is asking for the Force of pushing off the ground.

  • The formula of Force is: F = mass x acceleration

Given = <em>Mass</em>: 600 newtons (N)

             <em>Acceleration</em>: 12 m/s^{2}

We have to convert the mass into kg first. Remember that <u>1 kg is equal to 9.80665 newtons.</u>

Let x be the<em> mass in newtons</em>.

Let's convert: \frac{1 kg}{9.80665 N} x \frac{x}{600 N} = \frac{600}{9.80665} = 61.18 kg

Phil's weight is 61.18 kg

Let's go back to finding the force.

F = m x a

F = 61.18 kg x 12 m/s^{2}

F = 734.16 kg m/s^{2}

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Two sound waves, from two different sources with the same frequency, 540 Hz, travel in the same direction at 330 m s . The sourc
oee [108]

Answer:

The value is \Delta  \phi   =   4.12 \ rad

Explanation:

From the question we are told that

    The frequency of each sound is  f_1 = f_2 = f =  540 \  Hz

      The speed of the sounds is  v = 330 \  m/s

       The  distance of the first source from the point considered is  a = 4.40 \  m

        The distance of the second source from the point considered is  b  = 4.00  \  m

Generally the phase angle made by the first sound wave at the considered point is mathematically represented as

           \phi_a =  2 \pi [\frac{a}{\lambda}  + ft]

Generally the phase angle made by the first sound wave at the considered point is mathematically represented as

           \phi_b =  2 \pi [\frac{b}{\lambda}  + ft]          

Here b is the distance o f the first wave from the considered point  

Gnerally the phase diffencence is mathematically represented as  

           \Delta \phi= \phi_a - \phi_b  =  2 \pi [\frac{ a}{\lambda}  + ft ] - 2 \pi [\frac{b}{\lambda}  + ft ]      

=>      \Delta  \phi   =   \frac{2\pi [ a - b]}{ \lambda }

Gnerally the wavelength is mathematically represented as

        \lambda  =  \frac{v}{f}

=>     \lambda  =  \frac{330}{540}

=>     \lambda  =  0.611 \ m

=>    \Delta  \phi   =   \frac{2* 3.142 [ 4.40 - 4.0 ]}{  0.611  }

=>    \Delta  \phi   =   4.12 \ rad

     

5 0
3 years ago
During the deceleration of an ascending elevator, the normal force on the feet of a passenger is _____ her weight. During the de
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Answer: Smaller than ; larger than

Explanation:

When the elevator is moving in the upward direction, then the force acting on it is negative in nature because of

N= mg +ma, (g is gravity and a is acceleration)

here ma is negative so the N= mg-ma

Hence, it feels smaller than its original weight.

When the elevator is moving downward , then the force acting will be positive in nature

N= mg+ma,

here ma will be positive so it feels larger the original weight of passenger.

7 0
3 years ago
Water flows through a water hose at a rate of Q1 = 860 cm3/s, the diameter of the hose is d1 = 1.85 cm. A nozzle is attached to
Snowcat [4.5K]

Answer:

a) A1 =  \frac{\pi (d1)^{2} }{4}

b) A1 = 2.688 cm^{2}

c) Q1 = A1 x v1

d) v1 = 3.1994 m/s

e) A2 = \frac{A1 X v1}{v2}

f)  A2 = 0.7963cm^{2}

Explanation:

a) Area = \pi r^{2}

r = \frac{d}{2}

thus,

area = \pi (\frac{d}{2})^{2}

A1 =  \frac{\pi (d1)^{2} }{4}[/tex]b) d1 = 1.85 cmsubstituting in the above equation,A1 =  [tex]\frac{\pi (d1)^{2} }{4}

A1 =  \frac{\pi (1.85)^{2} }{4}

A1 = 2.688 cm^{2}

c) Flow rate = Area x velocity ( refer brainly.com/question/13997998)

Q1 = A1 x v1

d) From the above equation,

v1 = \frac{Q1}{A1} = \frac{860}{2.688} = 319.94 cm/s = 3.1994 m/s

e) Since the flow rate Q1 is constant throughout the hose, Av is a constant.

i.e. A1 x v1 = A2 x v2

thus,

A2 = \frac{A1 X v1}{v2}

f) v2 = 10.8 m/s.

substituting the values in the above equation,

A2 = \frac{2.688 X 3.1994}{10.8}  = 0.7963cm^{2}

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saveliy_v [14]

" 50 W " means " 50 Watts "  or  " 50 Joules per second ".

(500,002 J) / (50 J/sec)  =  <em>10,000.04 seconds</em>

(That's 2 hours 46 minutes 40.04 seconds.)

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It's 8 times as much as before.
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