Answer:
Explanation:
Given
length of wire 
change in length 
mass of wire 
Young's modulus for silver 
load on wire 

change in length is given by

Where A=area of cross-section




also wire is the shape of cylinder so cross-section is given by





Answer:
No concave lens cannot be used to make hand lens because in hand Lens we use convex lens so as to converge the rays of the light After refraction and helps to produce a magnified image of an object.
Hope it will help you :)❤
The spider will cross the driveway <u>25.71 in seconds</u>.
Why?
It's a conversion problem, so, in order to solve it, we need to convert the units of the given information.
Let's convert the given speed (in cm/s) to m/s.
We know that:

So, converting we have:

Then, calculating the time, we have:

We have that the spider will cross the driveway in 25.71 seconds.
Have a nice day!
A. Six electrons
B. six protons
C. they are composed of two up quarks and one down quark.