Answer:
The total energy, i.e. sum of kinetic and potential energy, is constant.
i.e. E = KE + PE
Initially, PE = 0 and KE = 1/2 mv^2
At maximum height, velocity=0, thus, KE = 0 and PE = mgh
Since, total energy is constant (KE converts to PE when the ball is rising),
therefore, KE = PE
or, 1/2 mv^2 = mgh
or, h = v^2 /2g = 13^2 / (2x9.8) = 8.622 m
Hope this helps.
Answer:
The pressure inside the container will be 3.3 atmospheres
Explanation:
The relationship between the temperature and pressure of a gas occupying a fixed volume is given by Gay-Lussac's law which states that the pressure of a given amount of gas is directly proportional to its temperature on the kelvin scale when the volume is kept constant.
Mathematically, it expressed as: P₁/T₁ = P₂/T₂
where P₁ is initial pressure, T₁ is initial temperature, P₂ is final pressure, T₂ is final temperature.
The above expression shows that the ratio of the pressure and temperature is always constant.
In the given question, the gas in the can attains the temperature of its environment.
P₁ = 3 atm,
T₁ = 25 °C = (273.15 + 25) K = 298.15 K,
P₂ = ?
T₂ = (55 °C = 273.15 + 55) K = 328.15 K
Substituting the values in the equation
3/298.15 = P₂/328.15
P₂ = 3 × 328.15/298.15
P₂ = 3.3 atm
Therefore, the pressure inside the container will be 3.3 atmospheres
The correct answer is option B, that is, the pitch changes from low to high.
The Doppler shift or the Doppler Effect refers to the variation in the wavelength or frequency of a wave in association with an observer who is traveling comparative to the source of the wave. A prime illustration of Doppler shift is the modification of the pitch heard when a vehicle sounding a horn comes towards and move away from an observer.
In comparison to the frequency emitted, the obtained frequency is greater at the time of approach, similar at the instant of passing by, and lower at the time of recession.
I think it’s 44.6 J, but I’m not to sure so hoped this helped /:).
Answer:
Density of the copper = 8.94g/cm^3
Student A results = 7.3gm/cm^3 ,9.4 gm/cm^3 , 8.3gm/cm^3
Student B results = 8.4 gm/cm^3 , 8.8 gm/cm^3 , 8gm/cm^3
From the observations we conclude that
Student A's result is accurate but not precise as the trials noted are not close to each other.
Student B's result is accurate and precise as the trials noted are close to each other.
Mean density of student A = 7.3 + 9.4 + 8.3 /3 = 8.33gm/cm^3
Mean density of student B = 8.4 + 8.8 + 8 /3 = 8.4 gm/cm^3
both the densities of A and B are 0.5 away from the actual density.
