The acceleration of the ball is 5 m/s^2. This can be calculated using a formula that relates the change in velocity, acceleration, and time. This formula is:
Vf = Vi + at
where:
Vf = final velocity
Vi = initial velocity
a = acceleration
t = time
Substituting the values gives:
30 = 20 + a(2)
<span>a = 5 m/s^2 --> Final Answer</span>
Answer:
from O toWl Q partical is accelerating with constant magnitude. and from Q to P is decelerating with constant magnitude.
Explanation:
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Answer:
The longest wavelength for closed at one end and open at the other is
y / 4 where y is the wavelength - that is node - antinode
The next possible wavelength is 3 y / 4 - node - antinode - node -antinode
y / 4 = 3 m y = 12 meters the longest wavelength
3 y / 4 = 3 m y = 4 meters 1 / 3 times as long
Answer:
F = 789 Newton
Explanation:
Given that,
Speed of the car, v = 10 m/s
Radius of circular path, r = 30 m
Mass of the passenger, m = 60 kg
To find :
The normal force exerted by the seat of the car when the it is at the bottom of the depression.
Solution,
Normal force acting on the car at the bottom of the depression is the sum of centripetal force and its weight.
N = 788.6 Newton
N = 789 Newton
So, the normal force exerted by the seat of the car is 789 Newton.
Answer:
a) 0.049 m
b) Yes, increase
Explanation:
Draw a free body diagram.
In the y direction, there are three forces acting on the feeder. Two vertical components of the tension forces in each rope pulling up, and weight force pulling down.
Apply Newton's second law to the feeder in the y direction.
∑F = ma
2Ty − mg = 0
Ty = mg/2
Let's say the distance the rope sags is d. The trees are 4m apart, so the feeder is 2m horizontally from either tree. Using Pythagorean theorem, we can find the length of the rope on either side:
L² = 2² + d²
L = √(4 + d²)
Using similar triangles, we can write a proportion using the forces and distances.
Ty / T = d / L
Substitute:
(mg/2) / T = d / √(4 + d²)
Solve for d:
Td = mg/2 √(4 + d²)
T² d² = (mg/2)² (4 + d²)
T² d² = (mg)² + (mg/2)² d²
(T² − (mg/2)²) d² = (mg)²
d² = (mg)² / (T² − (mg/2)²)
d = mg / √(T² − (mg/2)²)
Given m = 2.4 kg and T = 480 N:
d = (2.4) (9.8) / √(480² − (2.4×9.8/2)²)
d = 0.049 m
b) If a bird lands on a feeder, this will increase the tension in the rope to support the bird's weight.
