Answer:
69
Explanation:
Perform addition first as;
(0.210 + 0.41) = 0.62 -------{2 significant figures}
Perform product
(1.10 x 102) = 112 -------------- {3 significant figures}
The combined product
0.62 x 112 =69.44
= 69 ----------2 significant figures
Rule
The least number of significant figures in the numbers performing the operation determines the number of significant figures in the answer.
A car is being driven down a level track with no net force on it. This motion is best described as
A)Linear motion
B)circular motion
C) periodic motion
D) projectile motion
A)Linear motion
If there is not net force on the car, then by the Newton Second Law, the acceleration is zero, and the only valid option for zero acceleration is A).
It takes 10 lb of force to stretch a spring 2 in. What is the stiffness (spring constant) of the spring
Answer:
875.59 N/m
Explanation:
From the question given above, the following data were obtained:
Force (F) = 10 lb
Extention (e) = 2 in
Spring constant (K) =?
Next, we shall convert 10 lb to Newton (N). This can be obtained as follow:
1 lb = 4.448 N
Therefore,
10 lb = 10 lb × 4.448 N / 1 lb
10 lb = 44.48 N
Thus, 10 lb is equivalent to 44.48 N
Next, we shall convert 2 in to metre (m). This can be obtained as follow:
1 in = 0.0254 m
Therefore,
2 in = 2 in × 0.0254 m / 1 in
2 in = 0.0508 m
Thus, 2 in is equivalent to 0.0508 m.
Finally, we shall determine the spring constant i.e the stiffness of the spring as follow:
Force (F) = 44.48 N
Extention (e) = 0.0508 m.
Spring constant (K) =?
F = Ke
44.48 = K × 0.0508
Divide both side by 0.0508
K = 44.48 / 0.0508
K = 875.59 N/m
Therefore the spring constant i.e the stiffness of the spring is 875.59 N/m
Suppose that you wanted to use either conservation of momentum or conservation of kinetic energy to predict the outcome when a large car collides with a smaller car in a demolition derby. Which of the two quantities would be more appropriate for your calculation? Explain please.
Answer:
Conservation of momentum
Explanation:
Out of the two options given, the more appropriate one is the law of conservation of momentum. This is particularly so because momentum deals with the velocity or speed of the car while considering its mass. So, if a large car is to collide with another car, small one, then the momentum is what is considered. Energy on the other hand is talking about the work that is done, while the momentum considers the mass of the car while it is moving and considering its speed also
Two masses sit at the top of two frictionless inclined planes that have different angles,__deleted9917f34947e1359b5705bbac0d9227f8ec5df862078213a2db19c72f00189109deleted__ 0N86-C1-52-40-A837-22820 50% Part (a) What can be said about the speeds of the two masses at the bottom of their respective paths
Answer:
v = [tex]\sqrt{2gh}[/tex]
the speed in the two planes will be the same since it does not depend on the angle of the same
Explanation:
In this exercise we are told that the two inclined planes have no friction force, so we can apply the conservation of energy for each one, we will assume that the initial height in the two planes is the same
starting point. Highest part of each plane
Em₀ = U = m g h
final point. Lowest part of each plane
[tex]Em_{f}[/tex] = K = ½ m v²
as there is no friction, the mechanical energy is preserved
Em₀ = Em_{f}
mg h = ½ m v²
v = [tex]\sqrt{2gh}[/tex]
As we can see, the speed in the two planes will be the same since it does not depend on the angle of the same
Hooke's law describes a certain light spring of unstretched length 32.1 cm. When one end is attached to the top of a doorframe and a 7.94 kg object is hung from the other end, the length of the spring is 42.4 cm. (a) Find its spring constant. .7554 Correct: Your answer is correct. kN/m (b) The load and the spring are taken down. Two people pull in opposite directions on the ends of the spring, each with a force of 178 N. Find the length of the spring in this situation. .7923 Incorrect: Your answer is incorrect.
Answer:
[tex]0.75623\ \text{kN/m}[/tex]
[tex]55.6\ \text{cm}[/tex]
Explanation:
m = Mass of object = 7.94 kg
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
x = Change in length of the spring = [tex](42.4-32.1)\ \text{cm}[/tex]
F = Force on the spring
Force of gravity on the object and the force on the spring will be equal
[tex]mg=kx\\\Rightarrow k=\dfrac{mg}{x}\\\Rightarrow k=\dfrac{7.94\times 9.81}{0.424-0.321}\\\Rightarrow k=756.23\ \text{N/m}[/tex]
The spring constant is [tex]756.23\ \text{N/m}=0.75623\ \text{kN/m}[/tex]
F = 178 N
Force on spring is given by
[tex]F=kx\\\Rightarrow x=\dfrac{F}{k}\\\Rightarrow x=\dfrac{178}{756.23}\\\Rightarrow x=0.235\ \text{m}=23.5\ \text{cm}[/tex]
The length of the spring will be [tex]32.1+23.5=55.6\ \text{cm}[/tex].