The equation that relates distance, velocities, acceleration, and time is,
d = V₀t + 0.5gt²
where d is distance,
V₀ is the initial velocity,
t is time, and
g is the acceleration due to gravity (equal to 9.8 m/s²)
(1) Dropped rock,
(3 x 10² m ) = 0(t) + 0.5(9.8 m/s²)(t²)
The value of t from this equation is 24.73 s
(2) Thrown rock with V₀ = 26 m/s
(3 x 10² m) = (26)(t) + 0.5(9.8 m/s²)(t²)
The value of t from the equation is 5.61 s
The difference between the tim,
difference = 24.73 s - 5.61 s
difference = 19.12 s
<em>ANSWER: 19.12 s</em>
The mean may be calculated by summing the values of the refractive index and dividing the sum by the number of experiments. This is:
Mean = (1.45 + 1.56 + 1.54 + 1.44 + 1.54 + 1.53)/6
Mean = 1.51
The mean absolute error is the sum of the absolute values of errors divided by the number of trials:
MAE = (|1.45-1.51|+|1.56-1.51|+|1.54-1.51|+|1.44-1.51|+|1.54-1.51|+|1.53-1.51|)/6
MAE = 0.043
The fractional error is the MAE divided by the actual value:
Fractional error = 0.043 / 1.51
Fractional error = 43/1510
The percentage error is the fractional error multiplied by 100:
Percentage error = 2.85%
Answer: (a) The magnitude of its temperature change in degrees Celsius is 
.
(b) The magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is 
.
Explanation:
(a) Expression for change in temperature is as follows.
= 15.1 K
= 
= 
=
Therefore, the magnitude of its temperature change in degrees Celsius is .
(b) Change in temperature from Celsius to Fahrenheit is as follows.
F = 1.8C + 32
C = 
Since, K = C + 273
or, 
= 1.8 (15.1)
= 
or, =
Thus, we can conclude that the magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is .
Based on the description, it will be most likely that he believes in Oscillating model (also known as Cyclic model) of the universe
Based on this model, the gravitational force will prevent further expansion by pulling back the matters and undergo a bounce
hope this helps
Answer:
9.36*10^11 m
Explanation
Orbital velocity v=√{(G*M)/R},
G = gravitational constant =6.67*10^-11 m³ kg⁻¹ s⁻²,
M = mass of the star
R =distance from the planet to the star.
v=ωR, with ω as the angular velocity and R the radius
ωR=√{(G*M)/R},
ω=2π/T,
T = orbital period of the planet
To get R we write the formula by making R the subject of the equation
(2π/T)*R=√{(G*M)/R}
{(2π/T)*R}²=[√{(G*M)/R}]²,
(4π²/T²)*R²=(G*M)/R,
(4π²/T²)*R³=G*M,
R³=(G*M*T²)/4π²,
R=∛{(G*M*T²)/4π²},
Substitute values
R=9.36*10^11 m
