Answer:
heat required in pan B is more than pan A
Explanation:
Heat required to raise the temperature of the substance is given by the formula
now we know that both pan contains same volume of water while the mass of pan is different
So here heat required to raise the temperature of water in Pan A is given as
Now similarly for other pan we have
So here by comparing the two equations we can say that heat required in pan B is more than pan A
Betelgeuse is one of the largest known stars and is probably at least the size of the orbits of Mars or Jupiter around the sun. That's a diameter about 700 times the size of the Sun or 600 million miles. For a star it has a rather low surface temperature (6000 F compared to the Sun's 10,000 F).
Answer:
m1/m2 = 0.51
Explanation:
First to all, let's gather the data. We know that both rods, have the same length. Now, the expression to use here is the following:
V = √F/u
This is the equation that describes the relation between speed of a pulse and a force exerted on it.
the value of "u" is:
u = m/L
Where m is the mass of the rod, and L the length.
Now, for the rod 1:
V1 = √F/u1 (1)
rod 2:
V2 = √F/u2 (2)
Now, let's express V1 in function of V2, because we know that V1 is 1.4 times the speed of rod 2, so, V1 = 1.4V2. Replacing in the equation (1) we have:
1.4V2 = √F/u1 (3)
Replacing (2) in (3):
1.4(√F/u2) = √F/u1 (4)
Now, let's solve the equation 4:
[1.4(√F/u2)]² = F/u1
1.96(F/u2) =F/u1
1.96F = F*u2/u1
1.96 = u2/u1 (5)
Now, replacing the expression of u into (5) we have the following:
1.96 = m2/L / m1/L
1.96 = m2/m1 (6)
But we need m1/m2 so:
1.96m1 = m2
m1/m2 = 1/1.96
m1/m2 = 0.51
An example is when u rub your pen on your hair hard that is friction
