1250 J in 5 sec= 250 Joule(s) per second (1250/5 0
250 Joules per second = 250 Watts ( 1J/s = 1 Watt per definition)
250 Watts output = 250/0.65 efficiency = 384 Watts input
1 Horsepower = 732 Watts
Motors 1 Horsepower and under are made in certain step sizes like
3/4 , 1/2 , 1/3, 1/4, 1/16 1/20 of a Horsepower.
3/4 Horsepower is 549 Watts
1/2 Horsepower is 366 Watts
so you need to 3/4 horsepower motor to achieve 1250 J of work in 5 seconds.
Answer:
The new force is 1/4 of the previous force.
Explanation:
Given
---- 
--- 
Required
Determine the new force
Let the two particles be q1 and q2.
The initial force F1 is:
--- Coulomb's law
Substitute 2 for r1


The new force (F2) is

Substitute 4 for r2



Substitute 


The new force is 1/4 of the previous force.
From the given information in the question, the correct option is Option 1: 14 cm.
A non-stretched elastic spring has a conserved potential energy which gives it the ability to perform work. The elastic potential energy can be expressed as:
PE =
k 
Where PE is the energy, k is the spring constant and x is extension.
i. Given that: PE = 10 J and x = 10 cm, then;
PE =
k 
10 =
k 
20 = 100k
k = 0.2 J/cm
ii. To determine how far the spring is needed to be stretched, given that PE = 20 J.
PE =
k 
20 =
(0.2) 
40 = 0.2 
= 200
x = 
= 14.1421
x = 14.14 cm
So that;
x is approximately 14.00 cm.
Thus, the spring need to be stretched to 14.00 cm to give the spring 20 J of elastic potential energy.
For more information, check at: brainly.com/question/1352053.
Average velocity = (800+1600)/(4+10)
= 171.42m/s
Well I don't know. Let's actually LOOK at the picture and see if that helps.
A, B, C, and D all have the same TOTAL length, but A has the most waves crammed into that same total length.
By golly, that means the length of <u><em>each</em></u> wave in A must be shorter than each wave in B, C, or D.
The correct choice is <em> A </em>. Looking at the picture did the trick !