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:
7.9
Explanation:
When we put the metal piece in the liquid (which is in the graduated cylinder), how much it goes up is equal to the volume of the piece we inserted.
So now we know that the volume of that piece of unknown metal is 7mL (which is the same as 7
).
Density is
.
So the density of that piece of metal is 
Which leaves us with a final density of 7.9
Answer:
Option D is the correct answer.
Explanation:
Since value of angular acceleration is constant, the body has only centripetal acceleration.
Centripetal acceleration

We have radius = 7.112 cm = 0.07112 m
Frequency, f = 1975 rpm = 32.92 rps
Angular frequency, ω = 2πf = 2 x π x 32.92 = 206.82 rad/s
Substituting in centripetal acceleration equation,

Option D is the correct answer.
Answer:
3.59 m/s
Explanation:
We are given that
Mass of lineman,m=85 kg
Mass of receiver,m'=90 kg
Speed of receiver,v'=5.8 m/s
Speed of lineman,v=4.1 m/s

We have to find the their velocity immediately after the tackle.
Initial momentum,
According to law of conservation of momentum
Initial momentum=Final momentum=


Answer:
Explanation:
Given
Weight of person
At highest point Magnitude of the normal force
net force at highest point
where
centripetal force
Normal Force
Negative sign shows force is in upward direction
At bottom point centripetal force is towards the bottom