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krek1111 [17]
3 years ago
5

The sun is important to is because of the heat it provides. The sun is just one of ________ of stars in the universe.

Physics
1 answer:
marishachu [46]3 years ago
3 0
The sun is just one of the uncountable stars in the universe, as there are more than billions of galaxies containing billions of stars. 
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Most motorcycles could win a drag race against most cars. Use<br>Newton's 2nd law to explain this.​
shutvik [7]

Answer:

is has be deducted that acceleration of an object depends only on the the mass of the object. car has a greater mass that a motorcycle.

Explanation:

the newton second law states that when a net force act on an object,the object accelerates in the direction of the net force. the acceleration is directly proportional to the net force and inversely to the mass.

mathematically, F = ma    .......................... (eqn 1)

where, F = force, m=mass and  a = acceleration

F = m(ΔV/t)    .......................... (eqn 2)

acceleration =change in velocity divide time

t = time

v= velocity

Ft = mΔV

also a∝1/m

this same second can also mean the rate of change in momentum of an object is directly proportional to the resultant force.

Drag race is a function of acceleration, therefore, a car weighs more than a motorcycle.

the reason while motorcycle moves more faster than a car is due to the weight of the motorcycle as compare to the car. if the same acceleration is applied, and there weigth are difference, this will affect the force (causing an increased resultant force for the car), whereas time used will be affected.

for example:

a motorcycle has a weight 0f 181kg

a car has a weight 0f 1590kg

from  a∝1/m

they both will moves at a different acceleration

the force will be different

Ft = mΔV

5 0
3 years ago
A force of 20 N acts upon a 5 kg block. What is the acceleration of the object.
horsena [70]

Answer:

4m/s/s

Explanation:

a=f/m

m=5kg

f=20N

20/5=4

(N=kg-m/s/s)

7 0
2 years ago
Read 2 more answers
You need to design a 60.0-Hz ac generator that has a maximum emf of 5500 V. The generator is to contain a 150-turn coil that has
olchik [2.2K]

Answer:

So magnetic field will be equal to 0.1144 T  

Explanation:

We have given frequency f = 60 Hz

Maximum emf e = 5500 volt

Number of turns N = 150

Area A=0.85m^2

Emf generated in ac generator is given e=NBA\omega sin(\omega t)

For maximum emf sin(\omega t)=1

So maximum emf will be equal to e=NBA\omega

B=\frac{e}{NA\omega }=\frac{5500}{150\times 2\times 3.14\times 60\times 0.85}=0.1144T

So magnetic field will be equal to 0.1144 T

3 0
3 years ago
Which of the following statements are true?
yawa3891 [41]
A. False, because the SI unit for frequency is Hertz, 1 hertz equals 1 wave passing a fixed point in 1 second,

B. True, f and Nu (v)

C. To find the frequency of a wave, divide velocity of wave by wavelength

D. False, the period of a wave is measured in frequency (Hertz)
8 0
3 years ago
The parasailing system shown uses a winch to pull the rider in towards the boat, which is traveling with a constant velocity. Du
leonid [27]

Answer:

The magnitude of the force is  F_{net}= 1.837 *10^4N

the direction is 57.98° from the horizontal plane in a counter clockwise direction

Explanation:

From the question we are told that

      At t = 0 , \theta = 20^o

      The rate at which the angle increases is w = 2 \ ^o/s

Converting this to revolution per second  \theta ' = 2 \ ^o/s * \frac{\pi}{180} =0.0349\ rps

     The length  of the rope is defined by

                      r = 125- \frac{1}{3}t^{\frac{3}{2} }    

    At \theta  =30^o , The tension on the rope T = 18 kN

      Mass of the para-sailor is M_p = 75kg

Looking at the question we see that we can also denote the equation by which the length is defied as an an equation that define the linear displacement

  Now the derivative of displacement is velocity

   So

           r' = -\frac{1}{3} [\frac{3}{2} ] t^{\frac{1}{2} }

represents the velocity, again the derivative of velocity gives us acceleration

So

         r'' = -\frac{1}{4} t^{-\frac{1}{2} }

Now to the time when the rope made angle of 30° with the water

      generally angular velocity is mathematically represented as

                      w = \frac{\Delta \theta}{\Delta t}

Where \theta is the angular displacement

      Now considering the interval between 20^o \ to \ 30^o we have

                 2 = \frac{30 -20 }{t -0}

making t the subject

             t = \frac{10}{2}

               = 5s

Now at this time the displacement is

             r = 125- \frac{1}{3}(5)^{\frac{3}{2} }  

                = 121.273 m

The linear velocity is

             r' = -\frac{1}{3} [\frac{3}{2} ] (5)^{\frac{1}{2} }

                = -1.118 m/s

The linear acceleration is

          r'' = -\frac{1}{4} (5)^{-\frac{1}{2} }

              = -0.112m/s^2

Generally radial acceleration is mathematically represented by

         \alpha _R = r'' -r \theta'^2

              = -0.112 - (121.273)[0.0349]^2

              = 0.271 m/s^2

Generally angular acceleration  is mathematically represented by

                 \alpha_t = r \theta'' + 2 r' \theta '

Now \theta '' = \frac{d (0.0349)}{dt}  = 0

So

             \alpha _t = 121.273 * 0  + 2 * (-1.118)(0.0349)

                   = -0.07805 m/s^2

The net resultant  acceleration is mathematically represented as

                a = \sqrt{\alpha_R^2 + \alpha_t^2  }

                  = \sqrt{(-0.07805)^2  +(-0.027)^2}

                  = 0.272 m/s^2

Now the direction of the is acceleration is mathematically represented as

                  tan \theta_a = \frac{\alpha_R }{\alpha_t }

                       \theta_a = tan^{-1} \frac{-0.271}{-0.07805}

                           = 73.26^o

               

The force on the para-sailor along y-axis is mathematically represented as

               F_y = mg + Tsin 30^o + ma sin(90- \theta )

                    = (75 * 9.8) + (18 *10^3) sin 30 + (75 * 0.272)sin(90-73.26)

                    = 9.74*10^3 N

The force on the para-sailor along x-axis is mathematically represented as

              F_x = mg + Tcos 30^o + ma cos(90- \theta )    

             = (75 * 9.8) + (18 *10^3) cos 30 + (75 * 0.272)cos(90-73.26)

             = 1.557 *10^4 N

The net resultant force is mathematically evaluated as

                      F_{net} = \sqrt{F_x^2 + F_y^2}

                             =\sqrt{(1.557 *10^4)^2  + (9.74*10^3)^2}

                            F_{net}= 1.837 *10^4N

The direction of the force is

              tan \theta_f = \frac{F_y}{F_x}

                   \theta_f = tan^{-1} [\frac{1.557*10^4}{9.74*10^3} ]

                       = tan^{-1} (1.599)

                       = 57.98^o

     

                     

7 0
3 years ago
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