Lead-202 has a half-life of 53,000 years. How long will it take for 15/16 of a sample of lead-202 to decay?

Find the mass of a 52.2N bucket.

Goshia [24]

Answer:

m = 5.22 kg

Explanation:

The force acting on the bucket is 52.2 N.

We need to find the mass of the bucket.

The force acting on the bucket is given by :

F = mg

g is acceleration due to gravity

m is mass

$m=\dfrac{F}{g}\\\\m=\dfrac{52.2}{10}\\\\=5.22\ kg$

So, the mass of the bucket is 5.22 kg.

3 0

3 years ago

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

4 years ago

Is work done in this example: A student lifting a heavy backpack off of the floor.

vagabundo [1.1K]

I don't know what u mean

4 0

3 years ago

Read 2 more answers

Where are overcurrent protective devices normally installed in a branch circuit?

love history [14]

Overcurrent protective devices are normally installed in a branch circuit from where the conductors receive their supply.

<h3>What is resistance?</h3>

Resistance is the obstruction of electrons in an electrically conducting material. The mathematical relation for resistance can be understood with the help of the empirical relation provided by Ohm's law.

V=IR

V is the voltage,I is the current and R is the resistance

The vercurrent protective devices are normally installed in a branch circuit from where the conductors receive their supply.

Thus, In a branch circuit, overcurrent safety devices are often located from where the conductors get their supply.

Learn more about resistance from here, refer to the link;

brainly.com/question/14547003

#SPJ4

5 0

1 year ago

A highway patrolman traveling at the speed limit is passed by a car going 20 mph faster than the speed limit. After one minute,

Oksi-84 [34.3K]

Answer:

It will take 40 seconds to catch the speeding car

Explanation:

Initial speed of the patrolman = 55 mph

after one minute speed of patrol man = 115 mph

now acceleration of patrolman is given by

$a = \frac{v_f - v_i}{t}$

$a = \frac{115 - 55}{1/60} = 3600 m/h^2$

now at this acceleration the distance covered by patrolman in "t" time is given as

$d = v_i t + \frac{1}{2}at^2$

$d = 55t + 1800t^2$

now we know the speed of the speeding car is given as

$v' = (55+20) mph$

now in the same time distance covered by it

$d = 75 t$

now since the distance covered is same

$75 t = 55t + 1800 t^2$

$t = \frac{20}{1800} = \frac{1}{90} h$

$t = 40 seconds$

7 0

4 years ago

Read 2 more answers