Answer:
16613 m/s
Explanation:
Given that
mass of the fly, m = 0.55 g = 0.55*10^-3 kg
Kinetic Energy of the fly, E = 7.6*10^4 J
Speed of the fly, v = ? m/s
We know that the Kinetic Energy is that energy that an object, in this case, the fly, possesses due to its motion.
The Kinetic Energy, KE of any object is represented by the formula
KE = 1/2 * m * v²
If we substitute the values in the relation, we have,
7.6*10^4 = 1/2 * 0.55*10^-3 * v²
v² = (15.2*10^4) / 0.55*10^-3
v² = 2.76*10^8
v = √2.76*10^8
v = 16613 m/s
Thus, the fly would need a speed of 16.6 km/s in order to have a Kinetic Energy of 7.6*10^4 J
Answer: Technician A is correct
Explanation:
The intake manifold is the compactment that all fuel and air supply to the cylinders. It's connected to the engine so it has to be disconnected while the exhaust manifold receives all the exhaust gases from the cylinders and releases the gas through a single or double exhaust gases outlet.
<span>A solution is oversaturated with solute. The thing that could be done to decrease the oversaturation is to add more solvent in order to decrease the concentration of the solute. You can also increase the temperature to increase solubility of the solute. Hope this answers the question.</span>
Answer:
the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
Explanation:
Given the data in the question;
Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J
Mass of proton = 1.673 × 10⁻²⁷ kg
Charge of proton = 1.602 × 10⁻¹⁹ C
distance d = 2 m
we know that
Kinetic Energy = Charge of proton × Potential difference ΔV
so
Potential difference ΔV = Kinetic Energy / Charge of proton
we substitute
Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )
Potential difference ΔV = 20287.14 V
Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;
E = Potential difference ΔV / distance d
we substitute
E = 20287.14 V / 2 m
E = 10143.57 V/m or 1.01 × 10⁴ V/m
Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m