Answer: False
Explanation: The circulatory system of the body consists of the heart, blood vessels and blood. The deoxygenated blood from the body is carried to the heart.
Here, the deoxygentaed blood is converted into oxygenated by removing carbon dioxide from them and making it oxygenated.
The impure blood from the body is collected by the right ventricle and transported to the lungs for purification and then transported to the body.
Answer:
19.1 deg
Explanation:
v = speed of the proton = 8 x 10⁶ m/s
B = magnitude of the magnetic field = 1.72 T
q = magnitude of charge on the proton = 1.6 x 10⁻¹⁹ C
F = magnitude of magnetic force on the proton = 7.20 x 10⁻¹³ N
θ = Angle between proton's velocity and magnetic field
magnitude of magnetic force on the proton is given as
F = q v B Sinθ
7.20 x 10⁻¹³ = (1.6 x 10⁻¹⁹) (8 x 10⁶) (1.72) Sinθ
Sinθ = 0.327
θ = 19.1 deg
If its just a number 2,300 is an answer
Answer:
(a) p = 3.4 kg-m/s (b) 37.78 N.
Explanation:
Mass of a basketball, m = 0.4 kg
Initial velocity of the ball, u = -5.7 m/s (as it comes down so it is negative)
It rebounds upward at a speed of 2.8 m/s (as it rebounds so positive)
(a) Change in momentum = final momentum - initial momentum
p = m(v-u)
p = 0.4 (2.8-(-5.7))
p = 3.4 kg-m/s
(b) Impulse = change in momentum
Ft = 3.4
We have, t = 0.09 s
Hence, this is the required solution.
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
