A campfire being lighted and plants converting carbon-dioxide and water into glucose and oxygen are both forms of chemical change.
Therefore, the answer is:
B. Both are examples of chemical change.
To know if their research is correct or matches their hypothesis.
Explanation:
It is given that,
Resistance, R = 20 ohms
Voltage of the battery, V = 10 V
We can find current flowing through the circuit using Ohm's law as follows :
V = IR
It can be seen from the Ohm's law, that the current is directly proportional to the voltage. It means that if the battery is replaced by a battery that provides a larger voltage, the current through the circuit will be more than 0.5 A i.e. it increases.
Answer:
t=2.14s
Explanation:
Since it asks us for the time it takes from the moment it starts accelerating, the first 5 seconds of movement mentioned is irrelevant here.
Next it'd be simple just to use the kinematic equation involving distance, time, and velocity:
∆d=V₀t+at²/2
From here we can move the change in distance over and use the quadratic formula to solve for time.
My work is in the attachment. Comment with any questions.
Complete question is;
a. Two equal sized and shaped spheres are dropped from a tall building. Sphere 1 is hollow and has a mass of 1.0 kg. Sphere 2 is filled with lead and has a mass of 9.0 kg. If the terminal speed of Sphere 1 is 6.0 m/s, the terminal speed of Sphere 2 will be?
b. The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1. The masses remain 1.0 kg and 9.0 kg, The terminal speed (in m/s) of Sphere 2 will now be
Answer:
A) V_t = 18 m/s
B) V_t = 10.39 m/s
Explanation:
Formula for terminal speed is given by;
V_t = √(2mg/(DρA))
Where;
m is mass
g is acceleration due to gravity
D is drag coefficient
ρ is density
A is Area of object
A) Now, for sphere 1,we have;
m = 1 kg
V_t = 6 m/s
g = 9.81 m/s²
Now, making D the subject, we have;
D = 2mg/((V_t)²ρA))
D = (2 × 1 × 9.81)/(6² × ρA)
D = 0.545/(ρA)
For sphere 2, we have mass = 9 kg
Thus;
V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρA))]
V_t = 18 m/s
B) We are told that The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1.
Thus;
Area of sphere 2 = 3A
Thus;
V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρ × 3A))]
V_t = 10.39 m/s