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arsen [322]
3 years ago
9

Debby and Ben took different routes to travel from Point A to Point E. Debby took the route along A, B, C, D, and E. Ben took th

e route along A, G, F, and E. Debby and Ben both took 4 hours to complete their walk. Their routes are shown. (3 points)
Which statement is true about Debby and Ben's average speed?
Ben and Debby's average speed was the same.
Debby's average speed was twice Ben's average speed.
Ben's average speed was twice Debby's average speed.
Debby's average speed was three times Ben's average speed.

Physics
2 answers:
Nata [24]3 years ago
8 0
Bens was twice debbys speed.
svetlana [45]3 years ago
4 0

Answer:

Ben's average speed was twice Debby's average speed.

Explanation:

Ben covered a total distance of 16 miles (10+4+2) and Debby covered 8 miles (3+2+2+1) which is half of what Ben covered. As they both reached the place in the same amount of time it tells us Ben was faster.

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Name:
Brums [2.3K]

Answer:

1.a) 1 kJ

1.b) 4 kJ

     ratio 1:4

1.c) 4 times as before

2.a)  3.33 m/s2

Explanation:

1.a) bicycle's velocity =Displacement/time

                                   =100/20 m/s

                                   =5 m/s

bicycler's KE =1/2 *mass*(velocity)^2

                      =1/2*80*5^2

                       =1000 J = 1 kJ

1.b) bicycle's new velocity =200/20 m/s

                                   =10 m/s

bicycler's new KE =1/2*80*10^2

                             =4000 J = 4 kJ

Ratio= KE 1 :KE new

        = 1 :4

1.c)  when bicycler's speed was doubled it increased the KE by 4 times (2^2). because In KE we consider the square of the speed , so the factor we increase the speed , the KE will get increased with the square value of it

ex : speed is triple the prior value , then the KE is as 3^2 times as before. that is 9 times

2.a) car acceleration = (20-0)/6 m/s2

                                  = 3.33 m/s2

4 0
3 years ago
If a scale on Earth reads 650 N, what is your mass?
OlgaM077 [116]
If the scale reads 650N, then the mass of whoever it is standing on the scale is

         (weight) / (gravity)  =  (650N) / (9.8 m/s²)  =  66.3 kilograms  .

It's not MY mass, even if I'm the one standing on the scale. 
If I stand on a scale and it reads 650 N, the scale is broken.
4 0
3 years ago
How much electrical is use by a 350 W television that is operating for 25 minutes
Komok [63]

350Js^¹- *25*60s=525000J

7 0
3 years ago
Read 2 more answers
How many innings are in a regulation softball game?
Ilia_Sergeevich [38]

Answer:

A regulation game consists of 7 innings unless extended because of a tie score or unless shortened because the home team needs none or only a fraction of its 7th inning or unless 1 team is leading by 10 runs after 5 innings.

Explanation:

3 0
3 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0
3 years ago
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