Answer:
1.matter made of mass and volume
2.matter is any substance that has mass and takes up a space by having a volume
3.table, coffee, and perfume
4.solid has a strong bonds, liquid has weak bond, and gas no bond
5
Answer:
85 K
Explanation:
T₀ = initial temperature of the gas = 77 K
T = final temperature of the gas = ?
n = number of moles of monoatomic gas = 1.00 mol
R = universal gas constant = 8.314
c = heat capacity at constant volume = (1.5) R = (1.5) (8.314) = 12.5
Q = Amount of heat absorbed = 100 J
Amount of heat absorbed is given as
Q = n c (T - T₀)
100 = (1) (12.5) (T - 77)
T = 85 K
Answer:
Barret True, the speed decreases with increasing time
Explanation:
The equation they give us to describe the movement is of the form
y = yo + v₀ t + ½ a t²
The given equation is
y = -4 - 9 t + 2 t²
We can match term to term and find the constants
y₀ = -4 m
v₀ = -9 m / s
a = 2 m / s²
With this data we can build the equation of speed and time
v = v₀ + a t
v = - 9 + 2 t
With this expression we see that as time increases the speed decreases since the speed and acceleration have direction is opposite
Now we can analyze the students' observations
Amadeo False, we see that the behavior is the opposite
Barret True, the speed decreases with increasing time
Chinue False With the equation we have all the data to build the speed equation as a function of time
The brightness of the light in the lamp would increase
Answer:
The one falling from the greatest height will have the greatest speed.
h = 1/2 g t^2 time for ball to fall distance h
h2 / h1 = t2^2 / t1^2 dividing equations
h2 / t2^2 = h1 / t1^2
Let v be the average speed (v2 = h2 / t2)
1 / t2 * v2 = 1 / t1 * v1
v2 / v1 = t2 / t1 the one taking the longest to fall has the greater av. speed
Check:
h4 / h1 = t4^2 / t1^2 or
t4 / t1 = (h4 / h1)^1/2
In this case t4 / t1 = (4 / 1)^1/2 = 2 or twice the average speed
t1 = (2 h / g)^1/2 = .2^1/2 = .447 using g = 10
t4 = (2 h / g)^1/2 = .8^1/2 = .894
v1 = 1 / .447 = 2.24 m/s average speed
v4 = 4 / .894 = 4.47 or twice the average speed
