Answer:
Explanation:
Volume of block A = 10 x 6 x 1 = 60 cm³
Mass of block A = 630 g
density of mass A = mass / density
= 630 / 60 = 10.5g / cm³
Volume of block B = 5 x 5 x 3 = 75 cm³
Mass of block A = 604 g
density of mass A = mass / density
= 604 / 75 = 8.05 g / cm³
Since density of both A and B are less than that of mercury , both will float in mercury.
Answer:
number of electrons = 2.18*10^18 e
Explanation:
In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.
Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:
(1)
i1 = 0.40 A
i2 = 0.75 A
you solve the equation i3 from the equation (1):

Next, you take into account that 1A = 1C/s = 6.24*10^18
Then, you have:

The number of electrons that trough the wire 3 is 2.18*10^18 e/s
A) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction<span>
m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N</span><span> ... y direction
Net force ^2 = F1^2 + F2^2 = (20N)^2 + (15n)^2 = 625N^2 =>
Net force = √625 = 25N
F = m*a => a = F/m = 25.0 N /5.00 kg = 5 m/s^2
Answer: 5.00 m/s^2
b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal.
</span>
<span>m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N</span><span> ... 60 degress above x direction
Components of F2
F2,x = F2*cos(60) = 15N / 2 = 7.5N
F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N
Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N
Total force in y = F2,y = 13.0 N
Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =
= 30.42N
a = F /m = 30.42 N / 5.00 kg = 6.08 m/s^2
Answer: 6.08 m/s^2
</span>
The correct answer is: 13900589.
Answer:
Explanation:
Given that,
A lady falling has a final velocity of 4m/s
v = 4m/s
Mass of the lady is 60kg.
m = 60kg
Using conservation of energy, the potential energy of the body from the point where the lady is dropping is converted to the final kinetic energy of the lady.
Therefore,
P.E = K.E(final) = ½mv²
P.E = ½ × 60 × 4²
P.E = 480 J.