Explanation:
I will prepare my speech first and then I will give a speech on it.
A car starts from rest and for 15.0 s its wheels has a constant linear acceleration of 0.800 m/s^2 to the right. What is the angle through which each wheel rotated in 20.0 s interval if the radius of the tires is 0.330 m?
A wooden block moves at a constant speed on a rough horizontal surface. Draw a free-body diagram
clearly showing all the forces applied to the block; compare their magnitudes and directions.
The weight of the block and normal reaction are equal in magnitude but opposite in direction. Also, the frictional force and applied force are equal in magnitude but opposite in direction.
The forces acting on the wooden block moving at a constant speed include the following;
weight of block acting downwards, Wnormal reaction on the block acting upwards, Napplied force on the block acting towards positive x-direction, Ffrictional force acting towards negative x-direction, [tex]F_f[/tex]The free-body diagram is presented below;
N
↑
[tex]F_f[/tex] ← ⊕ → F
↓
W
The block is moving at a constant speed, we will have the following;
[tex]W = N \\\\F = F_f[/tex]
Thus, we can conclude that the weight of the block and normal reaction are equal in magnitude but opposite in direction. Also, the frictional force and applied force are equal in magnitude but opposite in direction.
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What do scientists call a substance that contains atoms from two or more different elements that are chemically bonded?
Atom
Compound
Element
Molecule
A 68 g plastic ball is moving to the left at 17 m/s . How much work must be done on the ball to cause it to move to the right at 17 m/s ?
Answer:
KE = 19.652 J to the right
Explanation:
KE = J or N*m
KE = (1/2)m*v²
KE = kinetic energy
m = mass(kg)
v = velocity (m/s)
KE = .5*0.068kg*(17m/s)²
KE = 9.826 J to the left
So double it to the right to get 17m/s.
KE = 19.652 J to the right
J = Joules or can be expressed as N*m which is Newton-meters.
A 1,000 kg truck is traveling at 3 m/s. Suddenly, the driver sees a herd of cows on the road ahead and applies the brakes. The truck's tires could fail after doing 5,000 J of work to slow the vehicle. Can the truck stop before the tires fail?
A. Yes, the total KE the tires need to transfer out of the system is less than 5,000 J.
B. Yes, the tires do not do any work, it is only the brakes that do work.
C. No, the truck had to stop suddenly and the quick change in KE will cause the tires to fail.
D. No, the total KE the tires need to transfer out of the system is more than 5,000 J.
This question involves the concepts of the law of conservation of energy and kinetic energy.
The correct option is "A. Yes, the total KE the tires need to transfer out of the system is less than 5,000 J".
According to the law of conservation of energy:
Loss in Kinetic Energy = Work done by the tires
[tex]\frac{1}{2}mv^2=W[/tex]
where,
W = work done by tires = ?
m = mass of the truck = 1000 kg
v = speed of the truck = 3 m/s
Therefore,
[tex]W=\frac{1}{2}(1000\ kg)(3\ m/s)^2[/tex]
W = 4500 J
Since the failure limit of work done by the tire is 5000 J, which is greater than the actual work done by the tire in this scenario. Hence, the tire will not fail in this case.
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The attached picture explains the law of conservation of energy.