Answer:
The magnitude of the force that the 6.3 kg block exerts on the 4.3 kg block is approximately 41.9 N
Explanation:
Forces on block 4.3 kg are:
63N to the right and R21 (contact force from the 6.3 kg block) to the left
Net force on 4.3 kg block is: 63 N - R21
Forces on the 6.3 kg block are:
R12 to the right (contact force from the 4.3 kg block) and 11 N to the left.
So net force on the 6.3 kg block is: R12 - 11 N
According to the action-reaction principle the contact forces R21 and R12 must be equal in magnitude (let's call them simply "R").
Then, since the blocks are moving with the SAME acceleration, we equal their accelerations:
a1 = (63 N - R)/4.3 = (R - 11 N)/6.3 = a2
solve for R by cross multiplication
6.3 (63 - R) = 4.3 (R - 11)
396.9 - 6.3 R = 4.3 R - 47.3
369.9 + 47.3 = 10.6 R
444.2 = 10.6 R
R = 444.2 / 10.6
R = 41.90 N
Answer:
20 m/s
Explanation:
The vertical component of the velocity of the ball is equal to the length of the perpendicular of a triangle such that it makes an angle of 30 degrees from the base and the length of the hypotenuse is 40 m/s. Sin(∅)*hypotenuse is the length of the perpendicular.
Vertical component of velocity is given by 40Sin(30°)=20
Using Ohm's Law, we can derived from this the value of resistance. If I=V/R, therefore, R = V/I
Substituting the values to the given,
P = Power = ?
R = Resistance = ?
V = Voltage = 2.5 V
I = Current = 750 mA
R = V/I = 2.5/ (750 x 10^-3)
R = 3.33 ohms
Calculating the power, we have P = IV
P = (750 x 10^-3)(2.5)
P = 1.875 W
The power consumption is the power consumed multiply by the number of hours. In here, we have;
1.875W x 4 hours = 7.5 watt-hours
Shsbbsnsjsbsnsnsnsnsnsndndndnd
Answer:
The eccentricity of an elliptical orbit is defined by the ratio e = c/a, where c is the distance from the center of the ellipse to either focus. The range for eccentricity is 0 ≤ e < 1 for an ellipse; the circle is a special case with e = 0.
Explanation:
