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lisov135 [29]
3 years ago
6

You walk forward at 1.5 m/s for 8 s. Your friend decides to walk faster and starts out at 2.0 m/s for the first 4 s. Then she sl

ows down and walks forward at 1.0 m/s for the next 4 s. Make a distance-time graph of your motion and your friend's motion. Who walked farther?
Physics
2 answers:
natima [27]3 years ago
8 0
The guy above is wrong. s=d/t true but your speed 1.5m/s =m/8s, D= 12m and her speed 2.0m/s=m/4s m=8m, 1.0m/s=m/4s= 4m. . . 8m+4m=12m you both walked the same distance, the only difference was she sped up .5m/s more in the first 4 seconds and then down 1.0m/s the last 4sseconds.
amid [387]3 years ago
3 0
The first person went further. s = d/t

Hope this helps you out APHDenmark! :D
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O ônibus espacial,preso sobre o avião está em movimento ou repouso?
marissa [1.9K]
Movimento es correcto
6 0
3 years ago
What year did Badminton become a full-medal Olympic sport?
Basile [38]

Answer:

1992

Explanation:

Badminton made its debut as a demonstration sport at the 1972 Olympic Games in Munich. It was not until the 1992 Games in Barcelona that it was officially included on the Olympic programme, with men's and women's singles and doubles events.

6 0
3 years ago
Not in book
umka2103 [35]

Answer:

x=2.4365\ m

and

x=-1.4365\ m

Explanation:

Given:

  • first charge, q_1=5\times 10^{-3}\ C
  • second charge, q_2=3\times 10^{-3}\ C
  • position of first charge, x_1=-2\ m
  • position of second charge, x_2=-1\ m

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

E_1=E_2

  • since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}

\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}

3(r^2+1+2r)=5r^2

2r^2-6r-3=0

r=3.4365 \&\ r=-0.4365

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

x=-1+3.4365=2.4365\ m

and

x=-1-0.4365=-1.4365\ m

6 0
2 years ago
Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
dalvyx [7]

Answer:

Explanation:

a )  V = 3 cos(0.5t)

differentiating with respect to t

dv /dt = -3 x .5 sin0.5t

= -1.5 sin0.5t.

acceleration = - 1.5 sin 0.5t

when t = 3 s

acceleration = - 1.5 sin 1.5

= - 1.496 ms⁻²

v = 3 cos.5t

b )  dx/dt = 3 cos 0.5 t

dx = 3 cos 0.5 t dt

integrating on both sides

x = 3 sin .5t / .5

x = 6 sin0.5t

At t = 2 s

x = 6 sin 1

x = 5.05 m

4 0
3 years ago
3) A dock worker pushes a 72 kg crate up a 2.0 m high,
Vlad [161]

Work done on the crate is 1411.2 J

Explanation:

Work done is defined as the product of force and the distance moved by the object. The unit of work done is in joules and denoted by the symbol J.

                                     Work done = F * d

where F represents the force and d represents the distance moved by the object.

mass = 72 kg , distance moved by the object is given by 2.0 m

Force F = mass * gravity = 72 * 9.8

             = 705.6 N =706 N.

Work done = 706 * 2.0 = 1412 J.

                   

7 0
3 years ago
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