Earths atmosphere, Temperature, Plant life and Oxygen I guess
Initial velocity =0, a=3m/s2, final velocity =18m/s, 18=3t, t=6 sec
Answer:
a) T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)
b) T = 295.37 K
Explanation:
Given;
Initial temperature of tea T1 = 31 C
Initial temperature of ice T2 = 0 C
Mass of tea m1 = 0.89 kg
Mass of ice m2 = 0.075kg
The heat capacity of both water and tea c = 4186 J/(kg⋅K)
the latent heat of fusion for water is Lf = 33.5 × 10^4 J/kg
And T = the final temperature of the mixture
Heat loss by tea = heat gained by ice
m1c∆T1 = m2c∆T2 + m2Lf
m1c(T1-T) = m2c(T-T2) + m2Lf
m1cT1 - m1cT = m2cT - m2cT2 + m2Lf
m1cT + m2cT = m1cT1 + m2cT2 - m2Lf
T(m1c + m2c) = m1cT1 + m2cT2 - m2Lf
T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)
Substituting the values;
T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)
T = (0.89×4186×31 + 0.075×4186×0 - 0.075×33.5 × 10^4)/(0.89×4186 + 0.075×4186)
T = 22.37 °C
T = 273 + 22.37 K
T = 295.37 K
Answer:
The final velocity of the car is 2.02 m/s
Explanation:
Hi there!
The kinetic energy of the car as it runs along the first flat horizontal segment can be calculated using the following equation:
KE = 1/2 · m · v²
Where:
KE = kinetic energy
m = mass
v = velocity
Then, the initial kinetic energy will be:
KE = 1/2 · 0.100 kg · (2.77 m/s)²
KE = 0.384 J
When the car gains altitude, it gains potential energy. The amount of gained potential energy will be equal to the loss of kinetic energy. So let´s calculate the potential energy of the car as it reaches the top:
PE = m · g · h
Where:
PE = potential energy.
m = mass
g = acceleration due to gravity.
h = height.
PE = 0.100 kg · 9.8 m/s² · 0.184 m
PE = 0.180 J
Then, the final kinetic energy will be (0.384 J - 0.180 J) 0.204 J
Using the equation of kinetice energy, we can obtain the velocity of the car:
KE = 1/2 · m · v²
0.204 J = 1/2 · 0.100 kg · v²
2 · 0.204 J / 0.100 kg = v²
v = 2.02 m/s
The final velocity of the car is 2.02 m/s
Answer:
If the object is placed between Focus and centre of Curvature, the image forms beyond centre of curvature. If the object is placed at principal focus the image forms at infinity . If the object is placed between the pole and the principal focus, the image forms behind the mirror.