1. Dump the container out.
2. Sift the large items out until the sugar and gravel is by itself.
3. Continue separating the other items.
Answer:
8.8 cm
31.422 cm/s
Explanation:
m = Mass of block = 0.6 kg
k = Spring constant = 15 N/m
x = Compression of spring
v = Velocity of block
A = Amplitude
As the energy of the system is conserved we have

Amplitude of the oscillations is 8.8 cm
At x = 0.7 A
Again, as the energy of the system is conserved we have

The block's speed is 31.422 cm/s
Answer:
Explanation:
(A) True: It is true.
In junction law, the current entering at a junction is equal to teh current leaving at the junction.
(B) False: It is false.
The kirchhoff's junction law is based on the conservation of charge.
(C) True: It is true.
Energy is used in the circuit.
(D) True: It is true.
It is based on the conservation of charge.
<u>Answer:</u> The specific heat of ice is 2.11 J/g°C
<u>Explanation:</u>
When ice is mixed with water, the amount of heat released by water will be equal to the amount of heat absorbed by ice.

The equation used to calculate heat released or absorbed follows:

......(1)
where,
q = heat absorbed or released
= mass of ice = 12.5 g
= mass of water = 85.0 g
= final temperature = 22.24°C
= initial temperature of ice = -15.00°C
= initial temperature of water = 25.00°C
= specific heat of ice = ?
= specific heat of water = 4.186 J/g°C
Putting values in equation 1, we get:
![12.5\times c_1\times (22.24-(-15))=-[85.0\times 4.186\times (22.24-25)]](https://tex.z-dn.net/?f=12.5%5Ctimes%20c_1%5Ctimes%20%2822.24-%28-15%29%29%3D-%5B85.0%5Ctimes%204.186%5Ctimes%20%2822.24-25%29%5D)

Hence, the specific heat of ice is 2.11 J/g°C
The electric force on the proton is:
F = Eq
F = electric force, E = electric field strength, q = proton charge
The gravitational force on the proton is:
F = mg
F = gravitational force, m = proton mass, g = gravitational acceleration
Since the electric force and gravitational force balance each other out, set their magnitudes equal to each other:
Eq = mg
Given values:
q = 1.60×10⁻¹⁹C, m = 1.67×10⁻²⁷kg, g = 9.81m/s²
Plug in and solve for E:
E(1.60×10⁻¹⁹) = 1.67×10⁻²⁷(9.81)
E = 1.02×10⁻⁷N/C