50 g of liquid X at 10 Celcius and 200 g of liquid Y
mx*cx*(t-tx)+my*cy*(t-ty)=0
cx/cy = - my*(t-ty) : mx*(t-tx) = (my/mx) * (ty - t) / (t-tx)
cx/cy = 200/50*(40-15)/(15-10) = 20
cx/cy = 20
Answer:
<h2>You can do 6516-3775=2741</h2><h2 /><h2>The difference is 27141 kilometers</h2><h2 /><h2>Did this help?</h2><h2>⇒ Yes or No?</h2>
Answer:
3.95 m
Explanation:
m = 1 kg, h = 100 m, k = 125 N/m
Let the spring is compressed by y.
Use the conservation of energy
potential energy of the mass is equal to the energy stored in the spring
m x g x h = 1/2 x ky^2
1 x 9.8 x 100 = 0.5 x 125 x y^2
y^2 = 15.68
y = 3.95 m
Answer:
The dog catches up with the man 6.1714m later.
Explanation:
The first thing to take into account is the speed formula. It is
, where v is speed, d is distance and t is time. From this formula, we can get the distance formula by finding d, it is 
Now, the distance equation for the man would be:

The distance equation for the dog would be obtained by the same way with just a little detail. The dog takes off running 1.8s after the man did. So, in the equation we must subtract 1.8 from t.

For a better understanding, at t=1.8 the dog must be in d=0. Let's verify:

Now, for finding how far they have each traveled when the dog catches up with the man we must match the equations of each one.






The result obtained previously means that the dog catches up with the man 3.8571s after the man started running.
That value is used in the man's distance equation.


Finally, the dog catches up with the man 6.1714m later.
Answer:
Explanation:
Given
charge of first body 
charge of second body 
Particle 1 is at origin and particle 2 is at 
third Particle which charge +q must be placed left of
because it will repel the q charge while
will attract it
suppose it is placed at a distance of x m







