Let
a = acceleration.
v = Speed
r = position
t = time.
To find the answer, you must write the equations of speed, and position known and clear the variables that are needed to find the acceleration.
The answer is a = 7.98m / s ^ 2.
I attach the solution.
Answer:
<h2>9.3kN</h2>
Explanation:
Step one:
given data
mass of bullet= 0.02kg
speed v=700m/s
time taken =1.5ms= 0.0015 seconds
Step two:
we know that from the first law
F=ma-----1 first law of motion
also, we know that
a=v/t----2
put a=v/t in equation 1 we have
F=mv/t
Step three:
substitute our given data to find force
F=0.02*700/0.0015
F=14/0.0015
F=9333.33N
F=9.3kN
<u>The average force exerted is 9.3kN</u>
M=meter, km=kilometer, mm=millimeter, mg=micrometer, cm=centimeter
Answer:
i will like to know the answer but idk
Explanation:
Answer:
Explanation:
a ) If the image of this object is viewed with the eyepiece adjusted for minimum eyestrain (image at the far point of the eye) , the image from object lens must have been formed at the focus of eye lens . So the objective image must have been formed at 19.5 - 2.75 = 16.75 cm from the object lens.
b ) Let the object distance be u
For object lens
v = 16.75 cm , f = .35 cm
1/v - 1/u = 1/f
1/16.75 - 1/u = 1/ .35
.0597 - 1/u = 2.857
1/u = - 2.7973
u = .3575 cm
c ) Angular magnification
= 
v₀ and u₀ are image and object distance for object lens , D = 25 cm and f_e is focal length of eye lens
= (16.75 / .3575) x( 25 / 2.75)
= 46.85 x 9.09
= 426
