If the boat's velocity is 18m/sec relative to the water in the river and not the shore, it would need to be added the river speed of 2.5m/sec to get a total of 20.5m/sec. The 20.5m/sec would then be the total velocity of the boat relative to the shore. From personal experience, I know that when one runs with the tide, one is adding the tide flow speed to one's boat speed (what it would be in neutral waters) to get a sometimes much faster speed.
Answer:
zero or 2π is maximum
Explanation:
Sine waves can be written
x₁ = A sin (kx -wt + φ₁)
x₂ = A sin (kx- wt + φ₂)
When the wave travels in the same direction
Xt = x₁ + x₂
Xt = A [sin (kx-wt + φ₁) + sin (kx-wt + φ₂)]
We are going to develop trigonometric functions, let's call
a = kx + wt
Xt = A [sin (a + φ₁) + sin (a + φ₂)
We develop breasts of double angles
sin (a + φ₁) = sin a cos φ₁ + sin φ₁ cos a
sin (a + φ₂) = sin a cos φ₂ + sin φ₂ cos a
Let's make the sum
sin (a + φ₁) + sin (a + φ₂) = sin a (cos φ₁ + cos φ₂) + cos a (sin φ₁ + sinφ₂)
to have a maximum of the sine function, the cosine of fi must be maximum
cos φ₁ + cos φ₂ = 1 +1 = 2
the possible values of each phase are
φ1 = 0, π, 2π
φ2 = 0, π, 2π,
so that the phase difference of being zero or 2π is maximum
Answer:
v₀ₓ = 15 m / s, 
= 5.2 m / s
v = 15.87 m / s
, θ = 19.1
Explanation:
This is a projectile launch problem. The horizontal speed that is constant throughout the entire path is worth 15 m / s, instead the vertical speed changes in value due to the acceleration of gravity, let's look for the initial vertical speed
Vy² =² - 2 g y
² = 
² + 2 g y
= √ (² + 2 gy
Let's calculate
= √ (1.25² + 2 9.8 1.3)
= √ (27.04)
= 5.2 m / s
The initial speed can be calculated by the initial speed
v = √ v₀ₓ² + ²
v = RA (15² + 5.2²)
v = 15.87 m / s
We look for the angle with trigonometry
tan θ = voy / vox
θ = tan⁻¹ I'm going / vox
θ = tan⁻¹ 5.2 / 15
θ = 19.1
The answer is
v₀ₓ = 15 m / s
= 5.2 m / s
Answer:
If they are metallic spheres they are connected to earth and a charged body approaches
non- metallic (insulating) spheres in this case are charged by rubbing
Explanation:
For fillers, there are two fundamental methods, depending on the type of material.
If they are metallic spheres, they are connected to earth and a charged body approaches, this induces a charge of opposite sign and of equal magnitude, then it removes the contact to earth and the sphere is charged.
If the non- metallic (insulating) spheres in this case are charged by rubbing with some material or touching with another charged material, in this case the sphere takes half the charge and when separated each sphere has half the charge and with equal sign.
<u>Answer:</u>
Cannonball will be in flight before it hits the ground for 2.02 seconds
<u>Explanation:</u>
Initial height from ground = 20 meter.
We have equation of motion , 
, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this the velocity of body in vertical direction = 0 m/s, acceleration = 9.8 
, we need to calculate time when s = 20 meter.
Substituting
So it will take 2.02 seconds to reach ground.
