This is called refraction. Some common examples of refraction are a straw appearing bent or cut once it enters water in a glass.
The answers for this question are:
a. You push a box until it moves. = unbalanced
b. You push a box but it doesn't move. = balanced
c. <span>You stop pushing a box and it slows down. = unbalanced
As a general explanation for all the items, forces are considered balanced when they cancel each other out. This means that no net force is produced. A and C are unbalanced because one force was able to overcome the force exerted by the object.</span>
Answer:
E = 75 J
Explanation:
First, we will calculate the total power consumed by the five lamps:
Now, the energy supply can be calculated as follows:
where,
E = Energy = ?
t = time = 5 s
Therefore,
E = (15 W)(5 s)
<u>E = 75 J</u>
Answer:
Explanation:
Using Boyles law
Boyle's law states that, the volume of a given gas is inversely proportional to it's pressure, provided that temperature is constant
V ∝ 1 / P
V = k / P
VP = k
Then,
V_1 • P_1 = V_2 • P_2
So, if we want an increase in pressure that will decrease volume of mercury by 0.001%
Then, let initial volume be V_1 = V
New volume is V_2 = 0.001% of V
V_2 = 0.00001•V
Let initial pressure be P_1 = P
So,
Using the equation above
V_1•P_1 = V_2•P_2
V × P = 0.00001•V × P_2
Make P_2 subject of formula by dividing be 0.00001•V
P_2 = V × P / 0.00001 × V
Then,
P_2 = 100000 P
So, the new pressure has to be 10^5 times of the old pressure
Now, using bulk modulus
Bulk modulus of mercury=2.8x10¹⁰N/m²
bulk modulus = P/(-∆V/V)
-∆V = 0.001% of V
-∆V = 0.00001•V
-∆V = 10^-5•V
-∆V/V = 10^-5
Them,
Bulk modulus = P / (-∆V/V)
2.8 × 10^10 = P / 10^-5
P = 2.8 × 10^10 × 10^-5
P = 2.8 × 10^5 N/m²
In a closed system, the loss of momentum of one object is same as________ the gain in momentum of another object
according to law of conservation of momentum, total momentum before and after collision in a closed system in absence of any net external force, remains conserved . that is
total momentum before collision = total momentum after collision
P₁ + P₂ = P'₁ + P'₂
where P₁ and P₂ are momentum before collision for object 1 and object 2 respectively.
P'₁ - P₁ = - (P'₂ - P₂)
so clearly gain in momentum of one object is same as the loss of momentum of other object
