Answer:
The crate was being lifted by a height of 1.48 meters.
Explanation:
In an attempt o move a crate;
Force applied = 2470 N
Work done by the force = 3650 J
We know that the work done is defined as the force used to move an object to a distance.
Given the Force used and the work done by that Force, we need to find out the distance the crate was lifted to.
Work done is defined as:
Work = Force*distance covered in the direction of the force
3650 = 2470*distance
distance = 3650/2470
distance = 1.48 meters
Answer:
The value of developed electric force is 
Solution:
As per the question:
Mass of the droplet = 1.8 mg = 
Charge on droplet, Q = 
Distance between the 2 droplets, D = 0.40 cm = 0.004 m
Now, the Electrostatic force given by Coulomb:
The magnitude of force is too low to be noticed.
Answer: Got It!
<em>Explanation:</em> Guide A Starts From Rest With Pin P At The Lowest Point In The Circular Slot, And Accelerates Upward At A Constant Rate Until It Reaches A Speed Of 175 Mm/s At The ... In the design of a timing mechanism, the motion of pin P in the fixed circular slot is controlled by the guide A, which is being elevated by its lead screw.
Answer:
Total Work done =0.65 joule
Explanation:
Work done is given Mathematically as
W=F *d
Where w=work done in joules
F=applied force
d= distance moved
The work done to move the toy accros the first meter is
W1=0.5*1
W1=0.5joule
The work done to move the toy across the next 2m at an angle of 30° is
.W2=0.5*2cos30
W2=0.5*2*0.154
W2=0.154joule
Hence total work done is
W1+W2=0.5+0.154
Total Work done =0.65 joule
It would be 17 m/s
If we use
V2 = V1 + a*t
Sub in 5 for v1
2m/s*2 for a
And
6 for t
That should give you the answer.
