Eben331:
A man drives a car a distance of 200km at an average speed of 44km/h. What must be his average speed for the next 220km if he is to cover the total distance in 9 hours?

Eben331:
A man drives a car a distance of 200km at an average speed of 44km/h. What must be his average speed for the next 220km if he is to cover the total distance in 9 hours?

Gracy3535:

200=44
220=x
200x=9680
x=48.4km
which is correct

vicstiqs95:
A simple pendulum oscillate with an amplitude of 30° if the length of the string is 1.0m. calculate the velocity of the pendulum Bob at its lowest point.

Eben331:
A man drives a car a distance of 200km at an average speed of 44km/h. What must be his average speed for the next 220km if he is to cover the total distance in 9 hours?

Eben331:
1 A ball of mass 1kg is dropped from a height of 5m onto a floor and rebounds to a height of 1m. Find the loss of kinetic energy due to impact. (g=9.8m/s2) 2 An electron of mass 9.4*10^-31kg has a wavelength of 1.3*10^-10m. Calculate the velocity of the electron. (h=6.6*10^-34Js) 3 A compound microscope, with lenses of focal lengths of 3cm and 5cm for the objective and eyepiece respectively, has an object placed 2cm from the objective lens. If the first image formed 3cm from the eyepiece lens, calculate the distance of the final image from the eyepiece. Happy Weekend Bosses.

Hardebaryor:

For the first 200km, the time spent is
200÷44=4.54hrs

For the next 220km, the average speed is 220/9-4.54hrs
This is because the total time spent is 9hrs
220/4.45 =49.43km/hr. Am I right?

Hardebaryor:

For the first 200km, the time spent is
200÷44=4.54hrs

For the next 220km, the average speed is 220/9-4.54hrs
This is because the total time spent is 9hrs
220/4.45 =49.43km/hr. Am I right?

Eben331:
You are correct

Gracy3535:
1.V^2=2gh v^2=2*9.8*1 v^2={19.6} v=4.4m/s kE=1/2mv^2 kE=1/2*1*4.4^2 KE=9.68 2.V=Freq*Wavelength v=3.0*10^-8*1.3*10-10 v=3.0*1.3*10^-8-10 v=3.9*10-18 3. 1/f=1/u+1/v 1/5=1/2+1/v 1/v=1/5_1/2 1/v=2-5/10 1/v=-3/10 v=3.3cm

Gracy3535:

am i correct for the 3answers

Eben331:
Hint: First find vo(image of objective lens) nd ur new u will be vo, then find ve which is the final image.

Eben331:
1 A ball of mass 1kg is dropped from a height of 5m onto a floor and rebounds to a height of 1m. Find the loss of kinetic energy due to impact. (g=9.8m/s2) 2 An electron of mass 9.4*10^-31kg has a wavelength of 1.3*10^-10m. Calculate the velocity of the electron. (h=6.6*10^-34Js) 3 A compound microscope, with lenses of focal lengths of 3cm and 5cm for the objective and eyepiece respectively, has an object placed 2cm from the objective lens. If the first image formed 3cm from the eyepiece lens, calculate the distance of the final image from the eyepiece. Happy Weekend Bosses.

Eben331:
1 A ball of mass 1kg is dropped from a height of 5m onto a floor and rebounds to a height of 1m. Find the loss of kinetic energy due to impact. (g=9.8m/s2) 2 An electron of mass 9.4*10^-31kg has a wavelength of 1.3*10^-10m. Calculate the velocity of the electron. (h=6.6*10^-34Js) 3 A compound microscope, with lenses of focal lengths of 3cm and 5cm for the objective and eyepiece respectively, has an object placed 2cm from the objective lens. If the first image formed 3cm from the eyepiece lens, calculate the distance of the final image from the eyepiece. Happy Weekend Bosses.

Eben331:
come back bro

Damayogenius18:
...OK bro...only back to solve ur ques

Gracy3535:

i got 30m/s. Pls about dat oscillation wat z d formular u use

Damayogenius18:
...OK bro...only back to solve ur ques

Hardebaryor:
1. K.E= mv²/2. P.E=Mgh
But P.E=K.E. Hence K.E=mgh
Loss in K.E = mg(H-h) = 1×9.8(5-1) =9.8×4 =39.2J Not really sure of this answer
2. Landa(wavelength)= h/mv v=h/landa×velocity v=6.6×10^–34/9.4×10^–31× 1.3×10^–10 v=5.4×10^6m/s 3. Hmm.. I no know am

Eben331:
I first used h=l(1-costita) and later use v=_/2gh I dnt knw weda it is correct. Can u show how u got 30m/s?

Gracy3535:

when u r through wit them come and collect ur cane

Gracy3535:
dat ansa was 4 the hint u gave me

Eben331:
big ups to u bro.

dolapo74:
Please ,where did get this questions from?

Eben331:
but image distance is supposed to be in cm or m, you mistake ve for velocity. The ans is not correct ooo

Eben331:
is dat important?

Hardebaryor:
Help me with number 3 please

Eben331:
1 A ball of mass 1kg is dropped from a height of 5m onto a floor and rebounds to a height of 1m. Find the loss of kinetic energy due to impact. (g=9.8m/s2) 2 An electron of mass 9.4*10^-31kg has a wavelength of 1.3*10^-10m. Calculate the velocity of the electron. (h=6.6*10^-34Js) 3 A compound microscope, with lenses of focal lengths of 3cm and 5cm for the objective and eyepiece respectively, has an object placed 2cm from the objective lens. If the first image formed 3cm from the eyepiece lens, calculate the distance of the final image from the eyepiece. Happy Weekend Bosses.

Gracy3535:
sori jare i mix up d questions value i ll do it again nt nw

Damayogenius18:
...1..we can calculate it in many....easiest way...1/2mvi^2- 1/2mvf^2....vi=√2gh..√2 *10*5....vi= 10ms^-1.
Vf=√2 *1*10.. Vf=4.47.....1/2m(vi^2-vf^2)....
0.5 * 1(100-22)..=0.5 * 1 * 78...=39J...another method...mghinitiall=mghfinal +energy lost...energy lost = mghinitial - mghfinal...
Mg(hinitial- hfinal)...
1 * 10(5-1)...=10 *4= 40J...got 39j base or rounding up in first method...I use g= 10 for convenience...should be= 9.81