The route followed by a hiker consists of three displacement vectors A with arrow, B with arrow, and C with arrow. Vector A with
1 answer:
Answer:
Explanation:
If we express all of the cordinates in their rectangular form we get:
A = (1404.77 , 655.06) m
Since we need C to be (0,0) we stablish that:
That way we make an equation system from both X and Y coordinates:
Replacing values:
With this system we can solve for both Db and Dc and get the answers to the question:
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Answer:
Periodic.
Explanation:
Electromagnetic waves is a propagating medium used in all communications device to transmit data (messages) from the device of the sender to the device of the receiver.
Generally, the most commonly used electromagnetic wave technology in telecommunications is radio waves.
Radio waves can be defined as an electromagnetic wave that has its frequency ranging from 30 GHz to 300 GHz and its wavelength between 1mm and 3000m. Therefore, radio waves are a series of repetitive valleys and peaks that are typically characterized of having the longest wavelength in the electromagnetic spectrum.
Basically, as a result of radio waves having long wavelengths, they are mainly used in long-distance communications such as the carriage and transmission of data.
Generally, a fixed speed is used for the propagation of traveling waves and this speed is usually denoted with the variable "v" or sometimes "c."
Furthermore, if the waveform of a traveling wave is repeated every time at specific intervals T, it is referred to as periodic wave.
Mathematically, the period of a traveling wave is given by the formula;
Where;
T is the time measured in seconds.
Answer:
Specific heat of brass is 0.40 J g⁻¹ °C⁻¹ .
Explanation:
Given :
Mass of brass, m₁ = 440 g
Temperature of brass, T₁ = 97° C
Mass of water, m₂ = 350 g
Temperature of water, T₂ = 23° C
Specific heat of water, C₂ = 4.18 J g⁻¹ °C⁻¹
Equilibrium temperature, T = 31° C
Let C₁ be the specific heat of brass.
Heat loss by brass = Heat gain by water
m₁ x C₁ x ( T₁ -T ) = m₂ x C₂ x ( T - T₁ )
Substitute the suitable values in above equation.
440 x C₁ x (97 - 31) = 350 x 4.18 x (31 - 23)
C₁ =
C₁ = 0.40 J g⁻¹ °C⁻¹