Ask question

Ask question

All categories

2 years ago

5

Aperson walk along actual path of radias 5m/s in 10 second what is the acceleration

1 answer:

AlekseyPX2 years ago

4 0

Answer:50m/s

Explanation:

You might be interested in

how much energy is required to move a 4.00-microcoulomb charge through a potential difference of 36.0 volts?

musickatia [10]

Answer:

All you gotta do is search it up. ask the same question you asked on here, type it into google, it'll teach you a lot more than your actual teachers.

Explanation:

4 0

2 years ago

How does energy from the sun affect the motion of molecules in a gas compared to molecules in a liquid?

AnnZ [28]

Molecules in a gas move faster than in a liquid.

hope it helps.

3 0

2 years ago

Read 2 more answers

A positron is an elementary particle identical to an electron except that its charge is . An electron and a positron can rotate

denis23 [38]

Explanation:

According to the energy conservation,

$F_{centripetal} = F_{electric}$

$\frac{mv^{2}}{r} = \frac{kq^{2}}{d^{2}}$

$v^{2} = \frac{kq^{2}r}{d^{2}m}$

= $\frac{9 \times 10^{9} N.m^{2}/C^{2} \times 1.6 \times 10^{-19} C \times 0.75 \times 10^{-9} m}{(1.50 \times 10^{-9}m)^{2} \times 9.11 \times 10^{-31} kg}$

= $8.430 \times 10^{10} m^{2}/s^{2}$

v = $\sqrt{8.430 \times 10^{10} m^{2}/s^{2}}$

= $2.903 \times 10^{5} m/s$

Formula for distance from the orbit is as follows.

S = $2 \pi r$

= $2 \times 3.14 \times 0.75 \times 10^{-9} m$

= $4.71 \times 10^{-9} m$

Now, relation between time and distance is as follows.

T = $\frac{S}{v}$

$\frac{1}{f} = \frac{S}{v}$

or, f = $\frac{v}{S}$

= $\frac{2.903 \times 10^{5} m/s}{4.71 \times 10^{-9} m}$

= $6.164 \times 10^{13} Hz$

Thus, we can conclude that the orbital frequency for an electron and a positron that is 1.50 apart is $6.164 \times 10^{13} Hz$ .

7 0

2 years ago

2. (Free-fall problem) A ball is thrown up in the air. 10 seconds later it reaches the top of its trajectory. What was the origi

Sever21 [200]

Explanation:

Using kinematics,

t = 10s

v = 0m/s

a = -9.8m/s^2

Therefore u = v - at = 0 - (-9.8)(10) = 98m/s.

8 0

2 years ago

A 77.0 kg rider sitting on a 7.3 kg bike is riding along at 9.3 m/s in the positive direction. The rider drags a foot on the gro

saw5 [17]

Answer with Explanation:

We are given that

Mass of rider=m=77 kg

Mass of bike =m'=7.3 kg

Initial velocity,u=9.3 m/s

Final velocity,v=6 m/s

A.Change in velocity=v-u=6-9.3=-3.3 m/s

Total mass,M=m+m'=77+7.3=84.3 kg

Change in momentum= $M(v-u)=84.3(-3.3)=-278.19 kgm/s$

B.Impulse=Ft=Change in momentum=-278.19kg m/s

C.Time,t=13.2 s

$v=u+at$

Using the formula

$6=9.3+13.2a$

$13.2a=6-9.3=-3.3$

$a=-\frac{3.3}{13.2}=-0.25 ms^{-2}$

F= $ma$

$F=84.3(-0.25)=-21.075 N$

D. $S=ut+\frac{1}{2}at^2$

$S=9.3(13.2)+\frac{1}{2}(-0.25)(13.2)^2$

$S=100.98 m$

5 0

2 years ago