This Question is not complete
Complete Question:
a. A hawk flies in a horizontal arc of radius 11.3 m at a constant speed of 5.7 m/s. Find its centripetal acceleration.
Answer in units of m/s2
b. It continues to fly along the same horizontal arc but increases its speed at the rate of 1.34 m/s2. Find the magnitude of acceleration under
these new conditions.
Answer in units of m/s2
Answer:
a. 2.875m/s²
b. 3.172m/s²
Explanation:
a. The formula for centripetal acceleration = (speed²) ÷ radius
Centripetal acceleration = (5.7m/s)²÷ 11.3m
Centripetal acceleration = 2.875m/s²
b. Magnitude of acceleration can be calculated by finding the sum of the vectors for the both the centripetal acceleration and the increase in the speed rate.
Centripetal acceleration ( acceleration x) = 2.875m/s²
Increase in the speed rate ( acceleration n) = 1.34m/s²
Magnitude of acceleration = √a²ₓ + a²ₙ
=√( 2.875m/s²)²+ (1.34m/s²)²
= √ 10.06m/s²
= 3.172m/s²
Answer:
3.69 m/s
Explanation:
Forces :
mgsin Θ - mumgcosΘ = ma
g x sinΘ - mu x g x cosΘ = a
9.8 x sin 21 - 0.53 x 9.8 x cos 21 = a
a = -1.337 m/s²
so you have final velocity = 0 m/s
initial velocity = ? m/s
Given d = 5.1 m
By kinematics
vf² = vo² + 2ad
0 = vo² + 2 x -1.337*5.1
vo = 3.69 m/s
Under general relativity, there is no 'before the Big Bang'. The problem is that time is itself a part of the universe and is affected by matter and energy. Because of the huge densities just after the Big Bang, time itself is warped in such a way that it cannot go back before that event. It is somewhat like asking what is north of the north pole.
The conservation of matter and energy states that the total amount of mass and energy at one time is the same at any other time. Notice how time is a crucial part of this statement. To even talk about conservation laws, you have to have time.
The upshot is that the Big Bang did not break the conservation laws because time itself is part of the universe and started at the Big Bang and because the conservation laws need to have time in their statements.
Answer:
570 N
Explanation:
Draw a free body diagram on the rider. There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.
The rider is moving at constant speed, so acceleration is 0.
Sum of the forces in the x direction:
∑F = ma
F cos 30° - T cos 15° = 0
F = T cos 15° / cos 30°
Sum of the forces in the y direction:
∑F = ma
F sin 30° - W - T sin 15° = 0
W = F sin 30° - T sin 15°
Substituting:
W = (T cos 15° / cos 30°) sin 30° - T sin 15°
W = T cos 15° tan 30° - T sin 15°
W = T (cos 15° tan 30° - sin 15°)
Given T = 1900 N:
W = 1900 (cos 15° tan 30° - sin 15°)
W = 570 N
The rider weighs 570 N (which is about the same as 130 lb).
