Duplain St. is 300 m long and runs from west to east between Baron and Burkey. If Keith is strolling east from Baron at an avera
1 answer:
Sue from Burkey and Keith from Baron will meet in 2 minutes
Answer: Option b
<u>Explanation:</u>
Time taken can be calculated when distance and the speeds are given. Here speeds of Keith and Sue are given. So, we have to find the relative speeds in order to calculate the time taken.
When two objects travel in same direction the relative speed will be the difference between speeds. Similarly when two objects travel in opposite direction, the relative speed will be the sum of given speeds.
Given:
Speed of Sue from Burkey is 6 km/hr and speed of Keith from Baron is 3 km/hr.
The distance between Burkey and Baron is 300 m.
From the formula,
where d is distance,s is speed and t is time
It can be derived that,
s = sum of given speeds = 3 km/hr + 6 km/hr = 9 km/hr
d = 300 m = 0.3 km
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Answer:
Explanation:
Radius of dee, r = 8 mm = 0.008 m
Electric field, e = 400 V/m
Magnetic field, B = 4.7 x 10^-4 T
mass of electron, m = 9.1 x 10^-31 kg
charge of electron, q = 1.6 x 10^-19 C
(a) Let v is the speed of electrons.
v = 661098.9 = 661099 m/s
(b)
e / m = 1.76 x 10^14 C / kg
(c) Let K be the kinetic energy
K = 0.5 x mv²
K = 0.5 x 9.1 x 10^-31 x 661099 x 661099
K = 1.99 x 10^-19 J
K = 1.24 eV
So, the potential difference is
V = 1.24 V
(d) if the acceleration voltage is doubled
V = 2 x 1.24 = 2.48 V
So, Kinetic energy
K = 2.48 eV
K = 2.48 x 1.6 x 10^-19 = 3.968 x 10^-19 J
Let v is the speed
K = 0.5 x mv²
3.968 x 10^-19 = 0.5 x 9.1 x 10^-31 x v²
v = 933856.5 m/s
Let the new radius is r.
r = 0.0113 m = 1.13 cm
Answer:
d = 100.8 ft
Explanation:
As we know that initial speed of the van is 40 miles then the stopping distance is given as 70 feet
here we know that
so here we have
now again if the speed is increased to 48 mph then let say the stopping distance is "d"
so we will have
now divide the above two equations
Answer:
Electric field at radius r inside the solid sphere is
Electric field at radius r between inner radius and outer radius of the shell is
E=0 N/C
Explanation:Given
The radius of the solid sphere is
The charge on the solid sphere is
The inner radius of the shell is
The outer radius of the shell is
The total charge on the shell is
PART(A)
The magnitude of electric field at radius r where \\The volumetric charge density of the solid sphere will be
The charge enclosed by the radius r inside the solid sphere is
rho=
According to gauss law
PART(B)
The electric field at radius r where
The shell is conducting so due to induction of charge
The charge induced on the inner surface of the shell is equal in magnitude of the total charge on the solid sphere but polarity will be changed because a conducting shell has no net electric field inside the shell
So,
The charge on the inner surface of the shell is
Due to conservation of the charge on the shell
The charge accumulated on the outer surface of the shell is
The charge enclosed by the radius r where
According to gauss law
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