Answer:
The speed of the baseball is approximately 19.855 m/s
Explanation:
From the question, we have;
The frequency of the microwave beam emitted by the speed gun, f = 2.41 × 10¹⁰ Hz
The change in the frequency of the returning wave, Δf = +3190 Hz higher
The Doppler shift for the microwave frequency emitted by the speed gun which is then reflected back to the gun by the moving baseball is given by 2 shifts as follows;
Where;
Δf = The change in frequency observed, known as the beat frequency = 3190 Hz
= The speed of the baseball
c = The speed of light = 3.0 × 10⁸ m/s
f = The frequency of the microwave beam = 2.41 × 10¹⁰ Hz
By plugging in the values, we have;
The speed of the baseball, ≈ 19.855 m/s
Explanation:
It is given that total distance is 3.65 m and times is 33.1 s. Therefore, calculate the average speed as follows.
Average speed =
=
= 0.11 m/s
Also, we know that
Average velocity =
Where, displacement means that how far final point is from the initial point, here, that is 0.0463 m.
Hence, calculate the average velocity as follows.
Average velocity =
= 0.0014 m/s
Thus, we can conclude that gnat's average speed is 0.11 m/s and magnitude of the gnat's average velocity is 0.0014 m/s.
Answer:
First option
Explanation:
If the ball is running in a circular motion then its velocity <em>v</em> will be tangency to the circular path.
In this problem the centripetal force that allows the circular movement is the tension <em>T</em> of the rope to which the ball is tied, if the child releases the rope then this tension becomes equal to zero and the circular movement is interrupted.
As the speed <em>v</em> of the ball is always tangential to the circumference at any point of the same, then at the instant in which the rope is released, the ball will follow the same trajectory that it had at that moment, that is, tangential to the circumference.
Observe the attached image.
Therefore the answe is: tangent to the circle
It is incorrect because the two types of ions in sodium phosphate cannot be seen.
Answer:
=30.22°C
Explanation:
The enthalpy change made the water to cool.
Enthalpy change = MC∅ where M is the mass of the water, C is the specific heat capacity of water and ∅ the temperature change.
ΔH=MC∅
∅=ΔH/MC
=(6.3×10⁴J)/(0.5kg×4186J/(kg°C))
=30.1
Final temperature = 35.00°C-30.1°C
=4.9°C