Answer: The objects' temperatures are not changing
Explanation: the whole thing about a thermal equilibrium is that both objects temps are changing to be an equal temperature.
Answer:
A.) 411.6 Mega metres
B.) 411.6 × 10^18 Picometers
Explanation:
Given that One light-year is the distance light travels in one year. This distance is equal to 9461 1015 m.
After the sun, the star nearest to Earth is Alpha Centauri, which is about 4.35 light-years from Earth.
The distance travelled will be
Distance = 9461 1015 × 4.35
Distance = 411557915.3 m
To Express this distance in
a. megameters, divide the value by 1000000
411557915.3 / 1000000
411.5579153 Mega metre
411.6 Mega metres approximately
b. picometers, multiply by 10^12
That is, 411557915.3 × 10^12
411.6 × 10^18 Picometers
Answer: V = 15 m/s
Explanation:
As stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. The observed frequency the car will be experiencing will be addition of the two frequency. That is,
F = 2.1 × 10^10 + 1030 = 2.100000103×10^10Hz
Using doppler effect formula
F = C/ ( C - V) × f
Where
F = observed frequency
f = source frequency
C = speed of light = 3×10^8
V = speed of the car
Substitute all the parameters into the formula
2.100000103×10^10 = 3×10^8/(3×10^8 -V) × 2.1×10^10
2.100000103×10^10/2.1×10^10 = 3×108/(3×10^8 - V)
1.000000049 = 3×10^8/(3×10^8 - V)
Cross multiply
300000014.7 - 1.000000049V = 3×10^8
Collect the like terms
1.000000049V = 14.71429
Make V the subject of formula
V = 14.71429/1.000000049
V = 14.7 m/s
The speed of the car is 15 m/s approximately
Basically, we want to see the distance in terms of thickness. So, for this rectilinear motion at constant acceleration, the equation to be used is:
x = v₀t + 0.5at²
First, let's determine t using the equation:
a = |v - v₀|/t
300 m/s² = |0 - 92.8 km/h * 1,000 m/1 km * 1 h/3600 s|/t
t = 0.086 seconds
x = (92.8 km/h * 1,000 m/1 km * 1 h/3600 s)(0.086 s) + 0.5(300 m/s²)(0.086 s)²
x = 3.33 meters
Answer:
Here's what I get
Explanation:
A. Distance between A and B.
h = -½gt²
The stones go faster the farther they fall.
Stone A has already reached 5 m when B is released.
When B reaches 5 m, A has dropped further and is falling even faster.
The distance between the stones increases with time.
Figure 1 shows this effect in a graph of height vs. time.
B. Speed of Stone B
v² = 2gh =2 × ( -9.81 m·s⁻²) × (-5 m) = 98.1 m·s⁻²
v = 9.9 m/s
The stone is travelling at 9.9 m/s when it reaches 5 m.
C. Velocity vs time
v = -gt
Both stones accelerate at the same rate.
When Stone B has reached 10 m at time t, Stone A is falling much faster.
Fig. 2 shows this in a graph of velocity vs time.