Answer:
3 outcomes
Explanation:
Since the sum must be less than 7, then one of them cannot be 7 because 7 + something > 7. Also the minimum number is 2 so one of them cannot be 5 either because 5 + 2 = 7 is not less than 7.
So the only options we have here is 2 and 4. The outcome of this is
(2,2) so 2 + 2 = 4 < 7
(2,4) so 2 + 4 = 6 < 7
(4,2) so 4 + 2 = 6 < 7
Note that (4,4) is not possible because 4 + 4 = 8 > 7
So 3 outcomes have a sum less than 7 and contain at least one even number
Answer:
The resistance for the second resistor is R2 = 240 Ohms and the equivalent resistance is Requivalent = 280 Ohms.
Explanation:
The resistance of a ohmic resistor is influeced by the type of it's material and by the it's construction. The longer the wire the greater the resistance and the greater the cross-sectional the lower the resistance. This can be expressed by the following equation:
R = (p*L)/A
Where p is a constant for the material of the resistor, L is the length of the wire and A is the area of the cross-sectional. In our case we have a resistor R1 that has a resistance of 40 Ohms, while a second resistor R2 made with the same material but with double length and half cross sectional. If we say that R1 is:
R1 = (p*L)/A
Then R2 must be:
R2 = (p*3*L)/(A/2)
Because the only things that changed were the length and area of the cross-sectional. We can now relate both resistors to find the second resistance, using the equation for R2. So we have:
R2 = [3*(p*L)/A]*2 = 6*(p*L)/A = 6*R1
We know that R1 is 40 Ohms so R2 = 6*40 = 240 Ohms.
The equivalent resistance of a series connection is the sum of the individual resistances, so we have:
Requivalent = R1 + R2 = 40 + 240 = 280 Ohms.
Answer:
4.24m/s,45 degrees in forward direction
Explanation:
We can begin with the following formula for the resistance of a wire:
We assume that resistivity
of both wires is the same. So they differ in length
and cross-sectional area
.
We assume that both wires are perfect cylinders, which means that their cross section is given by the following formula:
.
Where r is the radius of a wire. Diameter is just 2*r.
With all of this information, we can write equations for the resistance of both wires and then just divide them to find the ratio.
For wire 2 we have:
For cross section A, we write
, because if the diameter of wire two is twice as big that means it's radius is twice as big. Keep in mind we are only interested in ratios here.
For wire 1 we write:
.
When we divide those two equations we get: