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
Answer: 0.56 m/s
Explanation:
Hi, to answer this question we have to apply the formula of the conservation of momentum.
m1 v1 = m2 v2 (because the system is stationary at the beginning)
Where:
m1 = mass of the astronaut
v1= velocity of the astronaut
m2= mass of the satellite
v2= velocity of the satellite
Replacing with the values given and solving:
86 kg (2.35m/s) = 360 kg v2
202.1 kgm/s=360kg v2
202.1kgm/s /360kg =v2
v2 = 0.56 m/s
Feel free to ask for more if needed or if you did not understand something.
The correct option is C.
When the temperature of an object that is giving off light is increased, the particles in the object will move at a faster rate and there will be increased vibration of these molecules. This will makes the object to emit more light and to shine more brightly. Thus, the higher the temperature, the brighter the light that will be emitted.
Soory i think multiple of velocity and angle