Answer:
322.6 m/s
Explanation:
Given that there are two components of position;
x= vot
y component is;
y= 1.9 - gt^2/2
When the bullet hits the ground, y=0
1.9 -gt^2/2 =0
Where g = 10ms^-2
t= √2 × 1.9/10
t= 0.62 secs
Therefore,
x= vot
vo = x/t
Where, x= 200 m
vo= 200/0.62 =
vo= 322.6 m/s
Answer:
1) joule
2) ![kgm^{2}/s^{2}](https://tex.z-dn.net/?f=kgm%5E%7B2%7D%2Fs%5E%7B2%7D)
3) ![10\%](https://tex.z-dn.net/?f=10%5C%25)
Explanation:
1) Luminosity is the <u>amount of light emitted</u> (measured in Joule) by an object in a unit of<u> time</u> (measured in seconds). Hence in SI units luminosity is expressed as joules per second (
), which is equal to Watts (
).
This amount of light emitted is also called radiated electromagnetic power, and when this is measured in relation with time, the result is also called radiant power emitted by a light-emitting object.
Therefore, if we want to calculate luminosity the Joule as a unit will be used.
2) Work
is expressed as force
multiplied by the distane
:
Where force has units of
and distance units of
.
If we input the units we will have:
This is 1Joule (
) in the SI system, which is also equal to ![1 Nm](https://tex.z-dn.net/?f=1%20Nm)
3) The formula to calculate the percent error is:
![\% error=\frac{|V_{exp}-V_{acc}|}{V_{acc}} 100\%](https://tex.z-dn.net/?f=%5C%25%20error%3D%5Cfrac%7B%7CV_%7Bexp%7D-V_%7Bacc%7D%7C%7D%7BV_%7Bacc%7D%7D%20100%5C%25)
Where:
is the experimental value
is the accepted value
![\% error=\frac{|7.34 (10)^{-11} Nm^{2}/kg^{2}-6.67 (10)^{-11} Nm^{2}/kg^{2}|}{6.67 (10)^{-11} Nm^{2}/kg^{2}} 100\%](https://tex.z-dn.net/?f=%5C%25%20error%3D%5Cfrac%7B%7C7.34%20%2810%29%5E%7B-11%7D%20Nm%5E%7B2%7D%2Fkg%5E%7B2%7D-6.67%20%2810%29%5E%7B-11%7D%20Nm%5E%7B2%7D%2Fkg%5E%7B2%7D%7C%7D%7B6.67%20%2810%29%5E%7B-11%7D%20Nm%5E%7B2%7D%2Fkg%5E%7B2%7D%7D%20100%5C%25)
This is the percent error
Answer:
A force is a push or pull that acts on an object.
:)