Answer:
Explanation:
During a car collision momentum of vehicle ceases within a fraction of seconds so Force due to the impulse is huge.
Impulse is defined as the product of average force and time. If we can increase the period of collision for the same impulse then the average force imparted will be less.
If we can increase the time period then damage due to collision will be less.
Answer:
Vd = 1.597 ×10⁻⁴ m/s
Explanation:
Given: A = 3.90×10⁻⁶ m², I = 6.00 A, ρ = 2.70 g/cm³
To find:
Drift Velocity Vd=?
Solution:
the formula is Vd = I/nqA (n is the number of charge per unit volume)
n = No. of electron in a mole ( Avogadro's No.) / Volume
Volume = Molar mass / density ( molar mass of Al =27 g)
V = 27 g / 2.70 g/cm³ = 10 cm³ = 1 × 10 ⁻⁵ m³
n= (6.02 × 10 ²³) / (1 × 10 ⁻⁵ m³)
n= 6.02 × 10 ²⁸
Now
Vd = (6A) / ( 6.02 × 10 ²⁸ × 1.6 × 10⁻¹⁹ C × 3.9×10⁻⁶ m²)
Vd = 1.597 ×10⁻⁴ m/s
Answer:

Explanation:
Information we have:
velocities:
initial velocity:
(starts from rest)
final velocity: 
time:
Since we need the answer in
, we nees to convert the speed to meters per second:

We find the acceleration with the following formula:

substituting the known values:

the acceleration is 10.07
Answer:
979.6 kg/m³
Explanation:
We know pressure P = hρg where h = height of liquid = 10.5 m, ρ = density of liquid and g = acceleration due to gravity = 9.8 m/s²
So, density ρ = P/hg
Since P = 100.8 kPa = 100.8 × 10³ Pa
substituting the values of the variables into the equation for ρ, we have
ρ = P/hg
= 100.8 × 10³ Pa ÷ (10.5 m × 9.8 m/s²)
= 100.8 × 10³ Pa ÷ 102.9 m²/s²
= 0.9796 × 10³ kg/m³
= 979.6 kg/m³
So, the density of the liquid is 979.6 kg/m³