Answer:
2666 kg
0.11567 m/s²
Explanation:
m = Mass of boat
a = Acceleration of boat
From Newton's second law
Force

Force on the first boat is 333.25 N

Hence, mass of the second boat is 2666 kg
Combined mass = 2666+215 = 2881 kg

The acceleration on the combined mass is 0.11567 m/s²
Answer:
E = 1.25×10¹³ N/m²
Explanation:
Young's modulus is defined as:
E = stress / strain
E = (F / A) / (dL / L)
E = (F L) / (A dL)
Given:
F = 10 kg × 9.8 m/s² = 98 N
L = 1 m
dL = 10⁻⁵ m
A = π/4 (0.001 m)² = 7.85×10⁻⁷ m²
Solve:
E = (98 N × 1 m) / (7.85×10⁻⁷ m² × 10⁻⁵ m)
E = 1.25×10¹³ N/m²
Round as needed.
Answer:
497.00977 N
3742514.97005
Explanation:
= Density of water = 1000 kg/m³
C = Drag coefficient = 0.09
v = Velocity of dolphin = 7.5 m/s
r = Radius of bottlenose dolphin = 0.5/2 = 0.25 m
A = Area
Drag force

The drag force on the dolphin's nose is 497.00977 N
at 20°C
= Dynamic viscosity = 
Reynold's Number

The Reynolds number is 3742514.97005
This question is incomplete, the complete question is;
A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp.
How much of its original total energy (in J) survives as KE when it reaches the ground? m = 9.9 kg h = 4.9 m d = 5 m μ = 0.3 θ = 36.87°
Answer:
the amount of its original total energy (in J) that survives as KE when it reaches the ground will is 358.975 J
Explanation:
Given that;
m = 9.9 kg
h = 4.9 m
d = 5 m
μ = 0.3
θ = 36.87°
Now from conservation of energy, the energy is;
Et = mgh
we substitute
Et = 9.9 × 9.8 × 4.9
= 475.398 J
Also the loss of energy i
E_loss = (umg cosθ) d
we substitute
E_loss = 0.3 × 9.9 × 9.8 × cos36.87° × 5
= 116.423 J
so the amount of its original total energy (in J) that survives as KE when it reaches the ground will be
E = Et - E_loss
E = 475.398 J - 116.423 J
E = 358.975 J
If an object is projected with vertical speed given as

now the time of flight of the object that time in which it comes back on ground

now here we will have


now the range of projectile is given as


now here we know that


now the range is given as


now in order to have maximum range we can say


so we will have

so now we can say

so both speed must be same to have maximum horizontal range