This question is incomplete, the complete question is;
Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine: the angular velocity of Bxy rotating frame (ω).
Answer:
the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s
Explanation:
Given the data in the question and image below and as illustrated in the second image;
distance S = 40 m
V
= 54 km/hr
V
= 72 km/hr
α = 100 m
now, angular velocity of Bxy will be;
ω = V / α
so, we substitute
ω = ( 54 × 1000/3600) / 100
ω = 15 / 100
ω = 0.15 rad/s
Therefore, the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s
Answer:
Explanation:
We shall apply the formula for velocity in case of elastic collision which is given below
v₁ = (m₁ - m₂)u₁ / (m₁ + m₂) + 2m₂u₂ / (m₁ + m₂)
m₁ and u₁ is mass and velocity of first object , m₂ and u₂ is mass and velocity of second object before collision and v₁ is velocity of first velocity after collision.
Here u₁ = 22 cm /s , u₂ = - 14 cm /s . m₁ = 7.7 gm , m₂ = 18 gm
v₁ = ( 7.7 - 18 ) x 22 / ( 7.7 + 18 ) + 2 x 18 x - 14 / ( 7.7 + 18 )
= - 8.817 - 19.6
= - 28.4 cm / s
By tightening a string you are actually putting more stress on the string you are giving it a new frequency that isn't natural.
Hope this helps
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Answer:
f = 692 N
Explanation:
given data:
f =800N
a =1.2 m s^{2}
m= 90 kg
from newton's second law
net force 
therefore we have from above equation
ma =F - f
putting all value to get force of friction
1.2*90 = 800 - f
f = 692 N
