Answer:
A-Light from the Sun reflects off the Moon as it orbits Earth
Answer:
a) 4.9 s
b) 167.8 m
Explanation:
Hello!
To solve this question we need to make use of the equations of motion of both the motorcycle xm(t) and the car xc(t) at t=5
Let us consider the position of the motorcycle at t=5 as the origin, that is:
xm(t+5) = vt + (1/2)at^2
xc(t+5) = vt + 60 m
where v = 22.0m/s and a=5m/s^2
We are looking for the time t' when the position of the car and the motorcycle are the same:
xm(t'+5)=xc(t'+5)
vt' + (1/2)at'^2 = vt' +60m
t' = √(120 m /a) = 4.89898... s
Since we are considering the origin of the cooordinate system at the position when the motorcycle starts to accelerate, the distance travelled by the motorcycle until it catches the car is given by:
xm(t'+5)= vt' + (1/2)at'^2
xm(9.89898s) = (22 * 9.89898 + 2.5 * 9.89898^2)m
xm(9.89898s)= 167.777... m
There is only one pressure this situation would be a "constant pressure" process
Answer:
Mix
Explanation:
A battery has two electrodes, at one end it has the anode and the other end has the cathode. Electrons travel through the circuit from the anode (negative) to the cathode (positive), and this is the driving force that provides electricity to flow through circuits.
The anode needs to have a low electron affinity because it needs to readily release electrons, and the cathode needs to have a high electron affinity because it needs to readily accept electrons.
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
