Answer:
move at constant velocity.
Explanation:
Newton's first law (also known as law of inertia) states that:
"when the net force acting on an object is zero, the object will keep its state of rest or if it is moving, it will continue moving at constant velocity".
In the case of the probe, friction in deep space is negligible, therefore when the engine is shut down, there are no more forces acting on the probe: the net force therefore will be zero, so the probe will move at constant velocity.
Answer:
v = 6i + 12j + 4k
Explanation:
Find the magnitude of the direction vector.
√(3² + 6² + 2²) = 7
Normalize the direction vector.
3/7 i + 6/7 j + 2/7 k
Multiply by the magnitude of v.
v = 14 (3/7 i + 6/7 j + 2/7 k)
v = 6i + 12j + 4k
You do this one just like the other one that I just solved for you.
For this one ...
The density of the object is 2.5 gm/cm³.
We know that every cm³ of it we have contains 2.5 gm of mass.
We have to find out how many cm³ we have.
The question tells us: We have 2.0 cm³.
Each cm³ of space that the object occupies contains 2.5 gm of mass.
So the 2.0 cm³ that we have contains (2 x 2.5 gm) = 5 gms.
That's the mass of our object.
Muscles supply the force. They exert force by getting shorter (contracting).
A muscle can't get longer by itself. There needs to be another muscle
pulling in the opposite direction.
Answer:
Explanation:
Given that,
First Capacitor is 10 µF
C_1 = 10 µF
Potential difference is
V_1 = 10 V.
The charge on the plate is
q_1 = C_1 × V_1 = 10 × 10^-6 × 10 = 100µC
q_1 = 100 µC
A second capacitor is 5 µF
C_2 = 5 µF
Potential difference is
V_2 = 5V.
Then, the charge on the capacitor 2 is.
q_2 = C_2 × V_2
q_2 = 5µF × 5 = 25 µC
Then, the average capacitance is
q = (q_1 + q_2) / 2
q = (25 + 100) / 2
q = 62.5µC
B. The two capacitor are connected together, then the equivalent capacitance is
Ceq = C_1 + C_2.
Ceq = 10 µF + 5 µF.
Ceq = 15 µF.
The average voltage is
V = (V_1 + V_2) / 2
V = (10 + 5)/2
V = 15 / 2 = 7.5V
Energy dissipated is
U = ½Ceq•V²
U = ½ × 15 × 10^-6 × 7.5²
U = 4.22 × 10^-4 J
U = 422 × 10^-6
U = 422 µJ