Answer:
v = 6t² + t + 2, s = 2t³ + ½ t² + 2t
59 m/s, 64.5 m
Explanation:
a = 12t + 1
v = ∫ a dt
v = 6t² + t + C
At t = 0, v = 2.
2 = 6(0)² + (0) + C
2 = C
Therefore, v = 6t² + t + 2.
s = ∫ v dt
s = 2t³ + ½ t² + 2t + C
At t = 0, s = 0.
0 = 2(0)³ + ½ (0)² + 2(0) + C
0 = C
Therefore, s = 2t³ + ½ t² + 2t.
At t = 3:
v = 6(3)² + (3) + 2 = 59
s = 2(3)³ + ½ (3)² + 2(3) = 64.5
Answer is 1 molecule of S
Answer:
45.3°C
Explanation:
Heat gained = mass × specific heat × increase in temperature
q = mC (T − T₀)
Given C = 0.128 J/g/°C, m = 94.0 g, q = 305 J, and T₀ = 20.0°C:
305 J = (94.0 g) (0.128 J/g/°C) (T − 20.0°C)
T = 45.3°C
Answer: n the laboratory, the life time of a particle moving with speed 2.8 x 10^10 cm\s is found to be 2.5 x 10^-7. Calculate the proper life of the ...
Explanation:
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
