Answer:
[tex]E=101955.8volt/m[/tex]
The electric field direction is toward the right
Explanation:
From the question we are told that
Initial X-co-ordinate of proton [tex]X_1=20.0cm => \frac{20}{100}m[/tex]
Initial speed of Proton [tex]V_1= 3.5*10^6 m/s[/tex]
Final X-co-ordinate of proton [tex]X_2=80.0cm => \frac{80.0}{100}m[/tex]
Final speed of Proton [tex]V_2=0[/tex]
Generally the mass of Proton is given by
[tex]m_P=1.67*10^-27[/tex]
Generally the kinetic energy of the proton is mathematically given by
[tex]K.E_p=1/2mv^2[/tex]
[tex]K.E_p=1/2*1.6*10^-^2^7*(3.5*10^6)^2[/tex]
[tex]K.E_p=9.8*10^-^1^5[/tex]
Generally the change in electric potential [tex]\triangle V[/tex] is mathematically given by
[tex]\triangle V =\frac{K.E_p}{q}[/tex]
Charge on a proton [tex]q=1.602*10^-^1^9[/tex]
[tex]\triangle V =\frac{9.8*10^-^1^5}{1.602*10^-^1^9}[/tex]
[tex]\triangle V =61173.5volts[/tex]
Generally the equation for magnitude of an electric field is mathematically given by
[tex]E=\frac{\triangle V}{\triangle d}[/tex]
Where
[tex]d=0.8m-0.2m\\d=0.6m[/tex]
Therefore
[tex]E=\frac{61173.5}{0.8-0.2}[/tex]
[tex]E=\frac{61173.5}{0.6}[/tex]
[tex]E=101955.8volt/m[/tex]
The direction of the charge is towards the right
An electron traverses a vacuum tube with a length of 2 m in 2 X 10- 4
sec. What is the average speed of the
electron during this time?
Answer:
Average speed = 10,000 m/s
Explanation:
Given the following data;
Distance = 2m
Time = 0.0002secs
To find the average speed;
Average speed = distance/time
Average speed = 2/0.0002
Average speed = 10,000 m/s
Therefore, the average speed of the
electron is 10,000 meters per seconds.
A 22.0 kg child is riding a playground merry-go- round that is rotating at 40.0 rev/min. What centripetal force must
Answer:
F = 482.51 N
Explanation:
Given that,
Mass of a child, m = 22 kg
Angular velocity of the merry-go-round, [tex]\omega=40\ rev/min[/tex]
Let the radius of the path, r = 1.25 m
We need to find the centripetal force acting on the child. The formula for the centripetal force is given by :
[tex]F=m\omega^2r\\\\=22\times (4.18879)^2\times 1.25\\\\=482.51\ N[/tex]
So, the required centripetal force is 482.51 N.
The human nervous system can propagate nerve impulses at about 102 m>s. Estimate the time it takes for a nerve impulse to travel 2 m from your toes to your brain.
Answer:
t = 0.196 s
Explanation:
The speed of a pulse is determined by the characteristics of the medium, its density and its resistance to stress, as long as these remain the speed will be constant for which we can use the kinetic expressions of the uniform movement
v = x / t
t = x / v
calculate
t = 2/102
t = 0.196 s
2. A 2500 kg car is slowed down uniformly from an initial velocity of 20.0 m/s to
the north by a 6250 N braking force acting opposite the car's motion. Use the
impulse-momentum theorem to answer the following questions:
a. What is the car's velocity after 2.50 s?
b. How far does the car move during 2.50 s?
c. How long does it take the car to come to a complete stop?
Answer:
13.75m/s; 42.2m; 8s
Explanation:
(a) the car's velocity after 2.50 s is 13.75 m/s
(b) The distance traveled by the car is 42.18 m
(c) the time taken for the car to come to complete stop is 8 s.
The given parameters;
mass of the car, m = 2500 kg
initial velocity of the car, u = 20 m/s
breaking applied on the car, f = 6250 N
The acceleration of the car is calculated as follows;
[tex]F = ma \\\\a = \frac{F}{m} = \frac{6250}{2500} = 2.5 \ m/s^2[/tex]
(a) Using impulse-momentum theorem, the car's velocity after 2.5 s is calculated as follows;
[tex]F = \frac{m(u-v)}{t} \\\\m(u-v) = Ft\\\\u-v = \frac{Ft}{m} \\\\v = u - \frac{Ft}{m} \\\\v = 20 - \frac{6250 \times 2.5}{2500} \\\\v = 13.75 \ m/s[/tex]
(b) The distance traveled by the car during the 2.5 s;
[tex]v^2 = u^2 - 2as\\\\2as = u^2 - v^2\\\\s = \frac{u^2 - v^2}{2a} \\\\s = \frac{20^2 - 13.75^2}{2\times 2.5} \\\\s = 42.18 \ m[/tex]
(c) The time taken for the car to come to a complete stop;
when the car stop's the final velocity, v = 0
v = u - at
0 = 20 - 2.5t
2.5t = 20
[tex]t = \frac{20}{2.5} \\\\t = 8 \ s[/tex]
Thus, the time taken for the car to come to complete stop is 8 s.
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