Answer:
a. v₁ = 16.2 m/s
b. μ = 0.251
Explanation:
Given:
θ = 15 ° , r = 100 m , v₂ = 15.0 km / h
a.
To determine v₁ to take a 100 m radius curve banked at 15 °
tan θ = v₁² / r * g
v₁ = √ r * g * tan θ
v₁ = √ 100 m * 9.8 m/s² * tan 15° = 16.2 m/s
b.
To determine μ friction needed for a frightened
v₂ = 15.0 km / h * 1000 m / 1 km * 1h / 60 minute * 1 minute / 60 seg
v₂ = 4.2 m/s
fk = μ * m * g
a₁ = v₁² / r = 16.2 ² / 100 m = 2.63 m/s²
a₂ = v₂² / r = 4.2 ² / 100 m = 0.18 m/s²
F₁ = m * a₁ , F₂ = m * a₂
fk = F₁ - F₂ ⇒ μ * m * g = m * ( a₁ - a₂)
μ * g = a₁ - a₂ ⇒ μ = a₁ - a₂ / g
μ = [ 2.63 m/s² - 0.18 m/s² ] / (9.8 m/s²)
μ = 0.251
Answer: True
Explanation:
Because plants have an increase in vigar to grow
The force (F) of attraction or repulsion between two point charges (Q1 and Q2) is given by the following rule:
F = <span>(k * q1 * q2) / (r^2) where:
</span>q1 and q2 are the charges
k is coulomb's constant = 9 x 10^9<span> N. m</span>2/ C<span>2
</span>r is the distance between the two charges.
Applying the givens in the mentioned equation, we find that:
F = (9 x 10^9<span> x 0.07 x 10^6 x 2) / (0.0108)^2 = 1.08 x 10^19 n </span>
He was a British philosopher, and an important experimental and theoretical chemist. He is known for his discovery of hydrogen. He at that time called it "inflammable air".
The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.
<h3>What is Potential and Kinetic energy?</h3>
Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.
mass of rod, mab = 2.4kg
mass of rod, mbc = 4kg
conservation of energy
![T_{1} + V_{1} = T_{2} + V_{2}](https://tex.z-dn.net/?f=T_%7B1%7D%20%20%2B%20V_%7B1%7D%20%3D%20T_%7B2%7D%20%20%2B%20V_%7B2%7D)
![h_{ab} = h_{bc} = 0.18m](https://tex.z-dn.net/?f=h_%7Bab%7D%20%20%3D%20h_%7Bbc%7D%20%20%3D%200.18m)
potential energy at position 1,
![V1 = m_{ab} gh_{ab} + m_{bc} gh_{bc}](https://tex.z-dn.net/?f=V1%20%3D%20m_%7Bab%7D%20gh_%7Bab%7D%20%20%2B%20m_%7Bbc%7D%20gh_%7Bbc%7D)
V1 = 2.5 * 9.81 * 0.18 + 4 * 9.81 * 0.18
V1 = 11.30112
kinetic energy T1 at position 1 is zero
potential energy at position 2 is zero
K.E at position 2,
![T_{2} = \frac{1}{2} l_{ab} w^{2}_{ab} + \frac{1}{2} m_{bc} v^{2}_{G} + \frac{1}{2} lw^{2}_{bc}](https://tex.z-dn.net/?f=T_%7B2%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20l_%7Bab%7D%20w%5E%7B2%7D_%7Bab%7D%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20m_%7Bbc%7D%20v%5E%7B2%7D_%7BG%7D%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20lw%5E%7B2%7D_%7Bbc%7D)
![l_{ab} =\frac{m_{ab} l^{2}_{ab} }{3}](https://tex.z-dn.net/?f=l_%7Bab%7D%20%3D%5Cfrac%7Bm_%7Bab%7D%20l%5E%7B2%7D_%7Bab%7D%20%20%7D%7B3%7D)
= 1/3 *4 * (0.36)²
=0.10368kg m²
![l =\frac{m_{bc} l^{2}_{bc} }{12}](https://tex.z-dn.net/?f=l%20%3D%5Cfrac%7Bm_%7Bbc%7D%20l%5E%7B2%7D_%7Bbc%7D%20%20%7D%7B12%7D)
= 1/12 *4 * (0.6)²
=0.12kg m²
on putting the values in above equation we get,
T₂ = 1.0667vb²
0 + 11.30112 = 1.0667vb² + 0
vb = 3.2549 m/s
to learn more about Kinetic and potential energy go to - brainly.com/question/18963960
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