The new frequency of oscillation when the car bounces on its springs is 0.447 Hz
<h3>Frequency of oscillation of spring</h3>
The frequency of oscillation of the spring is given by f = (1/2π)√(k/m) where
- k = spring constant and
- m = mass on spring
Now since k is constant, and f ∝ 1/√m.
So, we have f₂/f₁ = √(m₁/m₂) where
- f₁ = initial frequency of spring = 1.0 Hz,
- m₁ = mass of driver,
- f₂ = final frequency of spring and
- m₂ = mass on spring when driver is joined by 4 friends = 5m₁
So, making f₂ subject of the formula, we have
f₂ = [√(m₁/m₂)]f₁
Substituting the values of the variables into the equation, we have
f₂ = [√(m₁/m₂)]f₁
f₂ = [√(m₁/5m₁)]1.0 Hz
f₂ = [√(1/5)]1.0 Hz
f₂ = 1.0 Hz/√5
f₂ = 1.0 Hz/2.236
f₂ = 0.447 Hz
So, the new frequency of oscillation when the car bounces on its springs is 0.447 Hz
Learn more about frequency of oscillation of spring here:
brainly.com/question/15318845
The similarity is that they both are types of bonds in molecules.
Ionic bonds are between a metal and a nonmetal.
Covalent bonds are between two nonmetals.
(6.2 × 10⁴) × (3.3 × 10²)
(6.2 × 3.3) × (10⁴ × 10²)
20.46 × 10⁽⁴⁺²⁾
20.46 × 10⁶ ⇒ The value 20.46 is not in standard form, so we need to rewrite this in standard form
2.046 × 10 × 10⁶
2.046 × 10⁽⁶⁺¹⁾
2.046 × 10⁷
Answer:
a) y = 162.6 m
, b) R = 928.64 m
Explanation:
We will solve this exercise using projectile launch kinematics, as the initial velocities and angle are equal on the planet and the Earth, let's look for the gravedd coinage of the planet
Earth
R = v₀² sin 2θ / g
R g = v₀² sin 2θ
In the planet
4.8 R = v₀² sin 2θ /
4.8 R = v₀² sin 2θ
4.8 R = R g
= g / 4.8
= 9.8 / 4.8
= 2.04 m / s²
Now we can answer the questions
a) The maximum height
Vy² = ² - 2 g y
For ymax the vertical speed is zero ( = 0)
sin θ = /
= sinθ
= 44.9 sin 35
= 25.75 m/s
y = ²/2
y = 25.75² / (2 2.04)
y = 162.6 m
b) the scope
R = v₀² sin 2θ /
R = 44.9² sin 2 35 / 2.04
R = 928.64 m
Answer:
Positive
Explanation:
Work is defined as the product of displacement of an object produced by the applied force and the component of force along that direction. Let us consider a situation in which the angle between the applied force and the displacement produced by it is at an angle θ. Let the magnitude of displacement be s and that of the applied force be F. Now, we have to find the component of force along the direction of the displacement. For that we have to resolve the force in to two components- one along the direction of displacement and other perpendicular to it. Since the angle between force and displacement is θ, the component of force along the direction of displacement will be Fcosθ and that perpendicular to it will be Fsinθ. Thus, b the definition of work, work done = Fcosθ x s = Fscosθ.
Now, coming to our question, the force here is gravitational force of attraction which is along downward direction. It is given in the question itself that the stone is falling down. Since the displacement and the applied force is along the same direction, angle between them, θ = 0.
Thus, work done = Fscosθ = Fscos0 = Fs (since cos0 =1)
Fs > 0
Thus, the work done is positive