EXPLAIN the measures that can be taken to ensure safety in a storm.


Answer 1

Hello There!

Great Question You Asked.

I Will Provide A Number Of Steps To Below To Show You Measures of Being Safe

Number 1.

Find shelter immediately. If you find yourself caught in a lightning storm, the key to minimizing danger is to get inside a protective structure. While most people seek shelter if lightning appears to be near, people commonly wait too long to seek shelter. If you can detect lightning, it may be close enough to strike you so always make sure you are safe.

Number 2.

Try to stay away from windows because windows provide a direct path for lightning.

Number 3.

Don’t touch anything metal or electrical.

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The correct frequency of exercise in order for it to be effective is


there are three specific frequencies that we need to care about most. The first one is the overall exercise frequency.In the overall exercise frequency,you have to decide on how many times you will work out per week in total. This include weight training workouts, cardio workouts. The second one is the weight training frequency. In the weight training frequency, this is base on the fact that majority of the most highly proven and intelligently design workout program in existence are all built doing three to four weight training workouts per week.

What can parents do if they find a mistake in the student records?      A. They can’t do anything B. Make a correction themselves C. Transfer the child to another school D. Ask the school to make the correction


Answer: Option (D) is the correct answer.


If parents find a mistake in the student records then the parents should talk to the respective teacher of their child and bring this thing into the teacher's notice.

By doing so, the teacher will also in future keep in mind to verify the records correctly once again before forwarding the record to parents or other higher authorities.

Also, the parents can talk to the school so that correction can be made into the students's records.    

D. Ask the school to make the correction

Air at 30°C and 2MPa flows at steady state in a horizontal pipeline with a velocity of 25 m/s. It passes through a throttle valve where the pressure is reduced to 0.3 MPa. The pipe is the same diameter upstream and downstream of the valve. What is the outlet temperature and velocity of the gas? Assume air is an ideal gas with a temperature-independent CP  7R/2, and the average molecular weight of air is 28.8.




At initial condition

P=2 MPa


V=25 m/s

At final condition

P=0.3 MPa

Now from first law for open system

We know that for air

h= 1.010 x T  KJ/kg


Now from mass balance

We also know that


Now from equation 1 and 2

So we can say that

This is the outlet velocity.

Now by putting the values in equation 2


This is the outlet temperature.


What is the momentum of a 0.01 kg toy car moving at 5.0 m/s?





Plug the given info into the formula: 

Final answer: P=0.05kg/m/s

SCIENCE HELP If you are applying a force to an object and another person is applying another force to the object what is going to happen to the object? Construct your hypothesis using an “If…, then…” statement.


if the vector sum of the force applied by you and the force applied by another person is zero, then the object will not accelerate since the net force on the object is zero in this case. hence the object will not move at all or if it was already moving , will move at constant velocity.

if the vector sum of the force applied by you and the force applied by another person is non-zero, then the object will not accelerate since there is net force on the object in this case. hence the object will accelerate.


Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charges. (a) if the two charges have the same sign, find the locationof all points (if there are any) at which (i) the potential (relative to infinity) is zero (is the electric field zero at these points


let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as-

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero  between the two charges,but we can find the points situated on the axial  line but  outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining  two charges  where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.


Defention of reversibily


Capable of being reversed or of reversing: as a : capable of going through a series of actions (as changes) either backward or forward 
Example : water ----> ice 
melts into water again

At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.5 m/s2. At the same instant a truck, traveling with a constant speed of 8.7 m/s, overtakes and passes the automobile?



6.96 s



  • u = initial speed of the automobile = 0 m/s
  • a = constant acceleration of the automobile =
  • v = constant speed of the truck = 8.7 m/s


  • t = time instant at which the automobile overtakes the truck.

At the moment the automobile and the truck both meat each other the distance travel by both vehicles must be the same.

Since t = 0 s is the initial condition. So, they both meet again at t = 6.96 s such that the automobile overtakes the truck.


A very long uniform line of charge has charge per unit length 4.80 μC/m and lies along the x-axis. A second long uniform line of charge has charge per unit length -2.40 μC/m and is parallel to the x-axis at y = 0.400 m. What is the net electric field (magnitude and direction) at the following points on the y-axis: (a) y = 0.200 m and (b) y = 0.600 m?



a) 647 kN/C

b) -71.9 kN/C


Using Gauss law relating electric flux to the charge contained inside a surface we arrive at this equation for the electric field around a infinite charged line:

E = ρ / (2 * π * r * e0)

The electric field on the XY plane will be the sum of the electric fields caused by both lines:

E(y) = ρ1 / (2 * π * (y - Y1) * e0) + ρ2 / (2 * π * (y - Y2) * e0)


Y1 = 0 (the position in Y where the first line is)

Y2 = 0.4

e0 = 8.85*10^-12

ρ1 and ρ2 are the charges per unit of length of the lines

At point y = 0.2

E(0.2) = 4.8*10^-6 / (2 * π * (0.2) * 8.85*10^-12) -2.4*10^-6 / (2 * π * (0.2 - 0.4) * e8.85*10^-12) = 647 kN/C

At point y = 0.6

E(0.6) = 4.8*10^-6 / (2 * π * (0.6) * 8.85*10^-12) -2.4*10^-6 / (2 * π * (0.2 - 0.6) * e8.85*10^-12) = -71.9 kN/C

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