Answer:
18 ohms
Explanation:
V = I(R1 + R2)
5V = (0.167A)(12 ohms + R2)
Solving for R2
R2 = 18 ohms
Answer:
speed cannot be used to calculate the temperature
Answer:
The speed of the block is 8.2 m/s
Explanation:
Given;
mass of block, m = 2.1 kg
height above the top of the spring, h = 5.5 m
First, we determine the spring constant based on the principle of conservation of potential energy
¹/₂Kx² = mg(h +x)
¹/₂K(0.25)² = 2.1 x 9.8(5.5 +0.25)
0.03125K = 118.335
K = 118.335 / 0.03125
K = 3786.72 N/m
Total energy stored in the block at rest is only potential energy given as:
E = U = mgh
U = 2.1 x 9.8 x 5.5 = 113.19 J
Work done in compressing the spring to 15.0 cm:
W = ¹/₂Kx² = ¹/₂ (3786.72)(0.15)² = 42.6 J
This is equal to elastic potential energy stored in the spring,
Then, kinetic energy of the spring is given as:
K.E = E - W
K.E = 113.19 J - 42.6 J
K.E = 70.59 J
To determine the speed of the block due to this energy:
KE = ¹/₂mv²
70.59 = ¹/₂ x 2.1 x v²
70.59 = 1.05v²
v² = 70.59 / 1.05
v² = 67.229
v = √67.229
v = 8.2 m/s
In order to create a charged object you need to transfer electrons either away or to the object by induction, conduction, or friction
Induction is without contact(like bringing a charged object to a electroscope charges the leaves at the bottom)
Conduction is with contact(like the previous answer, wore touching transfers the charge from a source to the object)
Friction(rubbing a balloon on wool)
To solve this problem we will apply the linear motion kinematic equations. With the data provided we will calculate the time of the first object to fall. Later we will get the time difference between the two. This difference will allow us to find the free fall distance. Through the distance we will find the initial velocity, that is,



The second object is thrown downward at one second later and it meets the first object at the water is


The distance of the object will travel due to free fall acceleration is



The distance of the object will travel due to its initial velocity is




Therefore the initial speed of the second object is 21.06m/s