Hurry

g The “size” of the atom in Rutherford’s model is about 8 × 10−11 m. Determine the attractive electrostatics force between a ele

Tju [1.3M]

Answer:

$3.6\cdot 10^{-8} N$

Explanation:

The electrostatic force between the proton and the electron is given by:

$F=k\frac{q_p q_e}{r^2}$

where

$k=9.00\cdot 10^9 Nm^2 C^{-2}$ is the Coulomb constant

$q_p = 1.6\cdot 10^{-19} C$ is the magnitude of the charge of the proton

$q_e = 1.6\cdot 10^{-19}C$ is the magnitude of the charge of the electron

$r=8\cdot 10^{-11}m$ is the distance between the proton and the electrons

Substituting the values into the formula, we find

$F=(9\cdot 10^9 ) \frac{(1.6\cdot 10^{-19})^2}{(8\cdot 10^{-11})^2}=3.6\cdot 10^{-8} N$

8 0

3 years ago

If a water wave completes one cycle in 2 seconds, what is the period of the wave?

Anika [276]

The definition of the period of a wave is the time that it takes for the wave to complete one cycle. As stated in the problem, the certain water wave completes one cycle in 2 seconds. From this information alone, we can conclude that the period of the wave is indeed, 2 seconds.

8 0

3 years ago

Read 2 more answers

A student is running at her top speed of 5.4 m/s to catch a bus, which is stopped at the bus stop. When the student is still a d

olya-2409 [2.1K]

Answer:

a) t=8.19s; x=44.2m

b) v=1.401 m/s

c) see attachment

d) The second solution is a later time at which the bus catches the student. v=9.40 m/s

e) No, she won't.

f) v=3.63m/s;t=21.2; x=77m

Explanation:

a) The motion of both the bus and the student can be explained by the equation $x= v_{0} +\frac{1}{2}at^{2}$ . Since the student is not accelerating, but rather maintaining a constant speed; the particular equation that describes the motion of the student is: $x_{student} = 5.4 \frac{m}{s} * t$ . Meanwhile, since the bus starts its motion at an initial velocity of zero, the equation that describes its motion is: $x_{bus} =\frac{1}{2} * 0.171 \frac{m}{s^{2} } * t^{2}$ . The motion of the student relative to that of the bus can be described by the equation: $x_{s} =x_{bus} +38.5m$ . By replacing terms in the last equation we end up with the following quadratic equation: $(0.0855 \frac{m}{s^{2} } * t^{2} )-(5.4\frac{m}{s} *t)+38.5m =0$ . Solving the quadratic equation will yield two solutions; t1=8.19s and t2=55.0s. By plugging in t1 onto the equation that describes the motion of the student we will find the distance runned by her, x=44.2m.

b) The velocity of the bus can be modeled by the equation $v^{2} = v_{0} ^{2} +2ax$ . Since the initial velocity of the bus is zero, the first term of the equation cancels. Next, we solve for v and plug in the acceleration of 0.171 m/s2 and the distance 5.74m (traveled by the bus, note: this is equal to the 44.2m travelled by the student minus the 38.5m that separated the student from the bus at the beginning of the problem).

c) The equations that make up the x-t graph are: $x = 5.4 \frac{m}{s} * t$ and $x =\frac{1}{2} * 0.171 \frac{m}{s^{2} } * t^{2} + 38.5$ ; as described in part a.

d) The first solution states that the student would have to run 44.2 m in 8.19 s in order to catch the bus. But, at that point, the student has a greater speed than that of the bus. So if both were to keep he same specified motion, the student would run past the bus until it reaches a velocity greater than that of the student. At which point, the bus will start to narrow the distance with the student until it finally catches up with the student 55 seconds after both started their respective motion.

e) No, because the quadratic equation $(0.0855 \frac{m}{s^{2} } * t^{2} )-(3.5\frac{m}{s} *t)+38.5m =0$ has no solution. This means that the two curves that describe the distance vs time graph for both student and bus do not intersect.

f) This answer is reached by finding b of the quadratic equation. The minimum that b can be in order to find a real answer to the quadratic equation is found by solving $b^{2} -4ac=0$ . If we take a = 0.0855 and c = 38.5, then we find that the minimum speed that the student has to run at is 3.63 m/s. If we then solve the quadratic equation $(0.0855 \frac{m}{s^{2} } * t^{2} )-(3.63\frac{m}{s} *t)+38.5m =0$ ,we will find that the time the student will run is 21.2 seconds. By pluging in that time in the equation that describes her motion: $x_{student} = 3.63 \frac{m}{s} * t$ we find that she has to run 77 meters in order to catch the bus.