Mechanical energy is the energy that is possessed by an object due to its motion or due to its position. It can either be kinetics or potential. In this problem you know it starting position so you can calculate it's potential energy (PE):
<span>PE=mass∗gravity∗height=0.3kg∗9.8m/s2∗1.8m=?
</span>The answer will typically be given in joules:
1J=kg∗m2s2 Could be wrong... But I believe it is 5.3...? as a final product.
- m1=1500kg
- m_2=3000kg
- v_1=5m/s
- v_2=7m/s
Using law of conservation of momentum
Answer:
The quantity of electrons that flows past a given point is 3.0 C.
Explanation:
An electric current (I) is the ratio of the quantity of charges (Q) that flows through a point to the time taken (t).
i.e I = 
It is measured in Ampere's by the use of an ammeter in the laboratory. The quantity of charge that flow through a given point is measured in Coulombs, while time is measured in seconds.
Given that; I = 1.5A and t = 2s, find Q.
Q = It
= 1.5 × 2
= 3.0 C
The quantity of electrons that flows past a given point is 3.0 C.
First let us assign variables,
d = distance travelled
t = time it took
v = velocity of the commercial airline
In linear physics, the equation for velocity is given as:
v = d / t
Rewriting for d:
d = v t
We know that the distance to and from south America are equal
therefore:
d1 (going) = d2 (return)
Let us say that velocity of air is v3. Since going to South
America, the wind is against the direction of the plane and the return trip is
the opposite, therefore:
(v1 - v3) t1 = (v1 + v3) t2
(v1 – v3) 4 = (v1 + v3) 3.53
4 v1 – 4 v3 = 3.53 v1 + 3.53 v3
0.47 v1 = 7.53 v3
v1 = 16.02 v3
Since we also know that:
(v1 - v3) t1 = 784
(16.02 v3 – v3) * 4 = 784
60.085 v3 = 784
v3 = 13.05 mph
Therefore the speed of the plane in still air, v1 is:
v1 = 16.02 * 13.05
<span>v1 = 209.03 mph (ANSWER)</span>
<span> </span>
Answer:
d
Explanation:
because is a reducing agent, O 2 is an oxidizing agent.
