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E moko.habai meka lesi wadiy....
Mata nam wadiya physics ba. Eth uditha aiya or anu hadai.
A 250g bullet is moving at 330m/s hits and travels through a 1.2kg block of wood, 0.30m long.
If the bullet's speed upon leaving the block is 120m/s, find
1. the block's velocity after bullet exits
2. the impulse applied to the block
3. the impulse applied to the bullet
4. the time the bullet spent in the wood
5. the force applied to the bullet
6. the acceleration of the bullet
lesi wadida manda...
If the bullet's speed upon leaving the block is 120m/s, find
1. the block's velocity after bullet exits
2. the impulse applied to the block
3. the impulse applied to the bullet
4. the time the bullet spent in the wood
5. the force applied to the bullet
6. the acceleration of the bullet
lesi wadida manda...
A 250g bullet is moving at 330m/s hits and travels through a 1.2kg block of wood, 0.30m long.
If the bullet's speed upon leaving the block is 120m/s, find
1. the block's velocity after bullet exits
2. the impulse applied to the block
3. the impulse applied to the bullet
4. the time the bullet spent in the wood
5. the force applied to the bullet
6. the acceleration of the bullet
lesi wadida manda...
තෑන්ක්ස් මල්ලී ත්රෙඩ් එකට සප් කලාට
බලමු බලමු ගාන
1) 43.75 m/sA 250g bullet is moving at 330m/s hits and travels through a 1.2kg block of wood, 0.30m long.
If the bullet's speed upon leaving the block is 120m/s, find
1. the block's velocity after bullet exits
2. the impulse applied to the block
3. the impulse applied to the bullet
4. the time the bullet spent in the wood
5. the force applied to the bullet
6. the acceleration of the bullet
lesi wadida manda...
(v)2)52.5 kgm/s -> (I)
3) 52.5 kgm/s <-
4) 0.3*2/(330 + v) =(t)
5)I/t
6)-I/t*0.25
calculations waradi athi.. ekai symbols damme. F=ma dannath mathaka na dan
Last edited:
1) 43.75 m/s
2)52.5 kgm/s -> (I)
3) 52.5 kgm/s <-
4) 0.3*2/(330 + v) =(t)
5)I/t
6)-I/t*0.25
calculations waradi athi.. ekai symbols damme. F=ma dannath mathaka na dan
ela..ela....3rd eka habai -52.5kgm/s..
menna answers (mewa 2 significant digits walata thiyenne)..
1. 44 m/s
2. 52 kgm/s
3. -52 kgm/s
4. 1.3E-3 s
5. -4.0E4 N
6. 1.6E5 m/s^2
මේ ගන මම කලින් හදල තියෙනවා. කොහේදිද කියල මතක නෑ.අයියෝ උත්තරේ එන්නෙ නැතිඋනහමනෙ පුනරාවර්තන කියන එකේ තේරුම මතක්වුනේ...625
මටත් මතක විදියට උත්තරේ ඕක තමයි.
අහ් මල්ලි ඒක - කියන එක තමයි මම ඊතල වලින් දාල තියෙන්නේ.ඕහ් දැනුයි දැක්කේ මගේ 4 වෙනි උත්තරේ පොඩි වෙනසක් වෙන්න ඕන.. V එක වෙනස් වෙන්න ඕන.
ela..ela....3rd eka habai -52.5kgm/s..
menna answers (mewa 2 significant digits walata thiyenne)..
1. 44 m/s
2. 52 kgm/s
3. -52 kgm/s
4. 1.3E-3 s
5. -4.0E4 N
6. 1.6E5 m/s^2
V වෙනුවට කුට්ටියට සාපේක්ෂව උණ්ඩයේ ප්රවේගය.
Last edited:
me wage sankaya 3125k hadanna puluwanne,mata ehema haduwamanm awe ithuru 625
class sci{
public static void main(String[] args){
int a,b,c,d,e,fin=0;
String val=null;
for (a=1;a<=5;a++){
for (b=1;b<=5;b++){
for (c=1;c<=5;c++){
for (d=1;d<=5;d++){
for (e=1;e<=5;e++){
val = Integer.toString(a) + Integer.toString(b) + Integer.toString(c) + Integer.toString(d) + Integer.toString(e);
fin = Integer.parseInt(val) + fin;
}}}}}
fin = fin%1000;
System.out.println(fin);
}
}
Last edited:
අහ් මල්ලි ඒක - කියන එක තමයි මම ඊතල වලින් දාල තියෙන්නේ.ඕහ් දැනුයි දැක්කේ මගේ 4 වෙනි උත්තරේ පොඩි වෙනසක් වෙන්න ඕන.. V එක වෙනස් වෙන්න ඕන.
me wage sankaya 3125k hadanna puluwanne,mata ehema haduwamanm awe ithuru 625
class sci{
public static void main(String[] args){
int a,b,c,d,e,fin=0;
String val=null;
for (a=1;a<=5;a++){
for (b=1;b<=5;b++){
for (c=1;c<=5;c++){
for (d=1;d<=5;d++){
for (e=1;e<=5;e++){
val = Integer.toString(a) + Integer.toString(b) + Integer.toString(c) + Integer.toString(d) + Integer.toString(e);
fin = Integer.parseInt(val) + fin;
}}}}}
fin = fin%1000;
System.out.println(fin);
}
}
moko bn?man podi kale idan sankarana karanne ohoma thama
algorithm eka hithenwa ara monado nCr ehma mata baa
