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
Let loudness be L, distance be d, and k be the constant of variation such that the equation that would best represent the given above is,
L = k/(d^2)
For Case 1,
L1 = k/(d1^2)
For Case 2,
L2 = k/((d1/4)^2)
For k to be equal, L1 = 16L2.
Therefore, the loudness at your friend's position is 16 times that of yours.
Answer:
15,505 N
Explanation:
Using the principle of conservation of energy, the potential energy loss of the student equals the kinetic energy gain of the student
-ΔU = ΔK
-(U₂ - U₁) = K₂ - K₁ where U₁ = initial potential energy = mgh , U₂ = final potential energy = 0, K₁ = initial kinetic energy = 0 and K₂ = final kinetic energy = 1/2mv²
-(0 - mgh) = 1/2mv² - 0
mgh = 1/2mv² where m = mass of student = 70kg, h = height of platform = 1 m, g = acceleration due to gravity = 9.8 m/s² and v = final velocity of student as he hits the ground.
mgh = 1/2mv²
gh = 1/2v²
v² = 2gh
v = √(2gh)
v = √(2 × 9.8 m/s² × 1 m)
v = √(19.6 m²/s²)
v = 4.43 m/s
Upon impact on the ground and stopping, impulse I = Ft = m(v' - v) where F = force, t = time = 0.02 s, m =mass of student = 70 kg, v = initial velocity on impact = 4.43 m/s and v'= final velocity at stopping = 0 m/s
So Ft = m(v' - v)
F = m(v' - v)/t
substituting the values of the variables, we have
F = 70 kg(0 m/s - 4.43 m/s)/0.02 s
= 70 kg(- 4.43 m/s)/0.02 s
= -310.1 kgm/s ÷ 0.02 s
= -15,505 N
So, the force transmitted to her bones is 15,505 N
The work done is the product between the intensity of the force applied F, the amount of the displacement d of the book and the cosine of the angle
between the direction of the force and the direction of the displacement:
In our problem, the student is lifting the book, so he is applying a force directed upward, and the book is moving upward, so F and d are parallel and therefore the angle is zero, so
Therefore, the work done is
<u>Answer:</u>
Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.
<u>Explanation:</u>
Let the east point towards positive X-axis and north point towards positive Y-axis.
Speed of truck = 25 m/s north = 25 j m/s
Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s
= (1.43 i + 1.00 j) m/s
Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j
Magnitude of velocity = 26.04 m/s
Angle from positive horizontal axis = 86.85⁰
So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.
