Explanation:
It is given that,
Mass of the truck, m = 2000 kg
Initial velocity of the truck, u = 34 km/h = 9.44 m/s
Final velocity of the truck, v = 58 km/h = 16.11 m/s
(a) Change in truck's kinetic energy,
(b) Change in momentum of the truck,
Hence, this is the required solution.
Answer:
114.19186 m
Explanation:
v = Velocity of girl at bottom = 7.1 m/s
= Coefficient of kinetic friction = 0.045
g = Acceleration due to gravity = 9.81 m/s²
d = Distance
N = Normal force = 783 N
The weight of the sled and girl is considered as the sum of the frictional force and weight
Hence we use the following equation where the kinetic energy and potential energy are conserved
The sled travels 114.19186 m on the level ground before coming to a rest
Answer:
Elastic collision
Explanation:
The interaction of two particles is called collision. There are two types of collision i.e. elastic collision and inelastic collision.
Two bouncy balls colliding with each other is an example of elastic collision. In this type of collision the momentum and kinetic energy of the bodies remains constant i.e.
and
Where
m₁ and m₂ are masses of two balls A and B
u₁ and u₂ are initial velocities of ball A and B
v₁ and v₂ are final velocities of ball A and B
Answer:
The observer could see to a depth of 4.38 m
Explanation:
Please check attachment for diagram.
Mathematically, from Snell law;
n1sin theta = n2 sin theta
1 * sin 90 = n2 * sin θR
where n2 = 1.33
1/1.33 = sin θR
Sin θR = 0.7519
θR = arc sin 0.7519
θR = 48.76
Now to get the height, we use the triangle
Using trigonometric ratio;
Tan( 90- θR) = H/5
H = 5 Tan( 90 - θR)
H = 5 Tan( 90-48.76)
H = 5 Tan41.24
H = 4.38 m
Given Information:
Magnetic field = B = 1×10⁻³ T
Frequency = f = 72.5 Hz
Diameter of cell = d = 7.60 µm = 7.60×10⁻⁶ m
Required Information:
Maximum Emf = ?
Answer:
Maximum Emf = 20.66×10⁻¹² volts
Explanation:
The maximum emf generated around the perimeter of a cell in a field is given by
Emf = BAωcos(ωt)
Where A is the area, B is the magnetic field and ω is frequency in rad/sec
For maximum emf cos(ωt) = 1
Emf = BAω
Area is given by
A = πr²
A = π(d/2)²
A = π(7.60×10⁻⁶/2)²
A = 45.36×10⁻¹² m²
We know that,
ω = 2πf
ω = 2π(72.5)
ω = 455.53 rad/sec
Finally, the emf is,
Emf = BAω
Emf = 1×10⁻³*45.36×10⁻¹²*455.53
Emf = 20.66×10⁻¹² volts
Therefore, the maximum emf generated around the perimeter of the cell is 20.66×10⁻¹² volts