Explanation:
Given:
Acceleration of car = 10 m/s
Distance travelled in 5 sec = 150 m
To find:
Distance travelled in the next 5 seconds
Concept:
There are 2 ways to app this kind of questions .
Either , we can find the total distance travelled in 10 seconds and then subtract 150 m from it.
Whereas , you can find out the final velocity at the end of the 5th seconds. Using equation of motion, then get the distance travelled in next 5 seconds.
Calculation:
v² = u² + 2as
=> ( u + at)² = u² + 2as
=> u² + 2uat + a²t² = u² + 2as
=> 2uat + (at)² = 2as
=> 2u (10)(5) + (10 × 5)² = 2 (10)(150)
=> 100u = 3000 - 2500
=> 100u = 500
=> u = 5 m/s
Distance travelled in 10 seconds :
s = ut + ½at²
=> s = (5 × 10) + ½(10)(10)²
=> s = 50 + 500
=> s = 550 m
Distance travelled in the 2nd half will be :
d = 550 - 150
=> d = 400 m
So final answer is :
Answer: Total work done on the block is 3670.5 Joules.
Step by step:
Work done:
With F the force, d the displacement, and theta the angle of action (which is 0 since the block is pushed along the direction of displacement, and cos 0 = 1)
Given:
F = 75 N
m = 31.8 kg
Final velocity 
In order to calculate the Work we need to determine the displacement, or distance the block travels. We can use the information about F and m to first figure out the acceleration:
Now we can determine the displacement from the following formula:
Here, the initial displacement is 0 and initial velocity is also 0 (at rest):
Now we still have "t" as unknown. But we are given one more bit of information from which this can be determined:
(using vf as final velocity, and tf as final time)
So it takes about 6.44 seconds for the block to move. This allows us to finally calculate the displacement:
and the corresponding work:
Answer:
0.15A
Explanation:
The parameters given are;
R=20.0 Ω
C= 2.50 μF
V= 3.00 V
f= 2.48×10^-3 Hz
Xc= 1/2πFc
Xc= 1/2×3.142 × 2.48×10^-3 × 2.5 ×10^-6
Xc= 25666824.1
Z= 1/√(1/R)^2 +(1/Xc)^2
Z= 1/√[(1/20)^2 +(1/25666824.1)^2]
Z= 1/√(2.5×10^-3) + (1.5×10^-15)
Z= 20 Ω
But
V=IZ
Where;
V= voltage
I= current
Z= impedance
I= V/Z
I= 3.00/20
I= 0.15A
Well, first off, Newtons second law of motion <span>deals with the motion of accelerating and decelerating objects.
W</span>e already know that from everyday life examples such as simply pushing a car that if 2 people push a car on a flat road it will accelerate faster than if one person was pushing it... Therefore, there is a relationship between the size of the force and the acceleration.
Now onto the third law of motion. First of all, what is the third law of motion? Well, a force is a push or a pull that acts upon an object as a results of its interaction with another object. Forces result from interactions! According to Newtons third law, whenever one object, and another object interact with each other, they exert forces upon each other. "For every action, there is an equal and opposite reaction." The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. So, how is this important to everyday life you may ask?
<span>Well, the action-reaction force pairs are found everywhere in your body.
For example, right now as I am typing, my tendons are exerting forces on bones, and those bones exert reaction forces on the tendons, as muscles contract, pulling my fingers on the keys. I press on those keys, and they press back on my fingers. See? Since i'm pressing on the keys, the press back on me. Its opposite from each other, as stated in the quite above. "</span><span>For every action, there is an equal and opposite reaction." </span>
Energy and Work Energy is the capacity to do work or to produce heat. Internal energy is the sum of kinetic energy and the potential energy. ... The KE would increase because heating something causes an increase in temperature.
