1.
Answer:
a) It is less
Explanation:
By energy conservation we can say that initial potential energy of both child must be equal to the final kinetic energy of the two child.
Since initially they are at same height so we will say that initial potential energy will be given as
and MgH
so the child with greater mass has more energy and hence smaller child will reach with smaller kinetic energy
2.
Answer:
b. The two speeds are equal.
Explanation:
As we know by mechanical energy conservation law we have
since both child starts at same height so here they both will reach the bottom at same speed
3.
Answer:
c. The two accelerations are equal
Explanation:
Since we know that average acceleration of the motion is given as
since here initial and final speeds are same so they both must have same average acceleration here.
Answer:
I1 = 0.772 A
Explanation:
<u>Given</u>: R1 = 5.0 ohm, R2 = 9.0 ohm, R3 = 4.0 ohm, V = 6.0 Volts
<u>To find</u>: current I = ? A
<u>Solution: </u>
Ohm's law V= I R
⇒ I = V / R
In order to find R (total) we first find R (p) fro parallel combination. so
1 / R (p) = 1 / R1 + 1/ R2 ∴(P) stand for parallel
R (p) = R1R2 / ( R1 + R2)
R (p) = (5.0 × 9.0) / (5.0 + 9.0)
R (p) = 3.214 ohm
Now R (total) = R (p) + R3 (as R3 is connected in series)
R (total) = 3.214 ohm + 4.0 Ohm
R (total) = 7.214 ohm
now I (total) = 7.214 ohm / 6.0 Volts
I (total) = 1.202 A
This the total current supplied by 6 volts battery.
as voltage drop across R (p) = V = R (p) × I (total)
V (p) = 3.214 ohm × 1.202 A = 3.864 volts
Now current through 5 ohms resister is I1 = V (P) / R1
I1 = 3.864 volts / 5 ohm
I1 = 0.772 A
Answer:
18 times
Explanation:
According to the security purposes which is set under the rules and regulation OSHA, which describes all the rights to the worker.
In the boom hoist receiving system all the sheaves which are used should have a pitch diameter of rope not less than 18 times the diameter of the nominal rope which is used.
Answer:
The energy of this particle in the ground state is E₁=1.5 eV.
Explanation:
The energy 
of a particle of mass <em>m</em> in the <em>n</em>th energy state of an infinite square well potential with width <em>L </em>is:
In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:
So we can rewrite the energy in the ground state as:
Finally
