Answer:
The velocity of the object at the bottom is, v = 17.15 m/s
Explanation:
Given data,
The initial velocity of the object, u = 0
The height of the hill, h = 15 m
Let 'S' be the distance of the slope of the hill and 'Ф' be the slope of the hill formed with the ground.
The acceleration due to gravity component along the slope is given by,
a = g Sin Ф
The distance of the slope since height 'h' of the hill is given,
s = h / Sin Ф
Using the III equation of motion,
v² = 2 as (∵ u = 0)
v² = 2 x g Sin Ф x h / Sin Ф
= 2 gh
Therefore,
<em> v = √(2gh)</em>
Substituting the given values,
v = √(2x9.8x15)
= 17.15 m/s
Hence, the velocity of the object at the bottom is, v = 17.15 m/s
Answer" The average force on the wall be equal to 0.5 N.
Explanation:
To calculate the average force on the wall, we use the following equation:
where,
F = average force = ?N
m = mass of the ball = 0.05kg
= Initial velocity of the ball = -10m/s (negative sign because the ball rebounds)
= Final velocity of the ball = 10m/s
t = time taken by the ball = 2s
Putting values in above equation, we get:
![F=\frac{0.05kg[10-(-10)]m/s}{2s}\\\\F=\frac{0.05kg(20m/s)}{2s}\\\\F=0.5kgm/s^2=0.5N](https://tex.z-dn.net/?f=F%3D%5Cfrac%7B0.05kg%5B10-%28-10%29%5Dm%2Fs%7D%7B2s%7D%5C%5C%5C%5CF%3D%5Cfrac%7B0.05kg%2820m%2Fs%29%7D%7B2s%7D%5C%5C%5C%5CF%3D0.5kgm%2Fs%5E2%3D0.5N)
Hence, the average force on the wall will be equal to 0.5N.
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:
0.015m^3
Explanation:
1 m^3 = 1000 liters
x m^3 = 15 liters
Cross multiply
xm^3 x 1000 l = 15 l
Divide both sides by 1000
xm^3 x1000/1000 = 15/1000
xm^3 = 0.015m^3
Therefore 15 liter = 0.015m^3
