Answer:
7500 Newtons
Explanation:
Mass of the sportscar= 1500 kg
Acceleration of the sportscar= 5m/s^2
Hence, let the Force acting on it be F
Answer:
(a) 0.063 m/s
(b) 1.01 m/s
Explanation:
rate of volume flow, V = 4 x 10^-6 m^3/s
(a) radius, r = 4.5 x 10^-3 m
Let the speed of blood is v.
So, V = A x v
where A be the area of crossection of artery
4 x 10^-6 = 3.14 x 4.5 x 10^-3 x 4.5 x 10^-3 x v
v = 0.063 m/s
Thus, the speed of flow of blood is 0.063 m/s .
(b) Now r' = r / 4 = 4.5 /4 x 10^-3 m = 1.125 x 10^-3 m
Let the speed is v'.
So, V = A' x v'
4 x 10^-6 = 3.14 x 1.125 x 10^-3 x 1.125 x 10^-3 x v'
v' = 1.01 m/s
Thus, the speed of flow of blood is 1.01 m/s .
The behavior of an ideal gas at constant temperature obeys Boyle's Law of
p*V = constant
where
p = pressure
V = volume.
Given:
State 1:
p₁ = 10⁵ N/m² (Pa)
V₁ = 2 m³
State 2:
V₂ = 1 m³
Therefore the pressure at state 2 is given by
p₂V₂ = p₁V₁
or
p₂ = (V₁/V₂) p₁
= 2 x 10⁵ Pa
Answer: 2 x 10⁵ N/m² or 2 atm.
Answer:
1.69 T
Explanation:
Applying,
F = BvqsinФ.................. Equation 1
Where F = Force, B = magnetic field, v = velocity, q = charge on an electron, Ф = angle between the electron and the field.
make B the subject of the equation,
B = F/(vqsinФ)............. Equation 2
From the question,
Given: F = 2.0×10⁻¹³ N, v = 7.4×10⁵ m/s, Ф = 90°
Constant: q = 1.60×10⁻¹⁹ C
Substitute into equation 2
B = 2.0×10⁻¹³/(7.4×10⁵×1.60×10⁻¹⁹×sin90°)
B = 0.169×10
B = 1.69 T
