Answer:
Explanation:
The speed of hare = .15 x 25 = 3.75 m /s . Let tortoise took t second to complete the race .
Distance traveled by it = .15 t
Distance traveled by hare = .15 t - .35 m
Time taken by hare to complete this distance
= t - 5 x 60 s
Speed of hare
= Distance / time
( .15t-.35 ) / t - 300 , so
![\frac{.15t-.35}{t-300} = 3.75](https://tex.z-dn.net/?f=%5Cfrac%7B.15t-.35%7D%7Bt-300%7D%20%3D%203.75)
t = 312.40
= 5 minutes 12.4 seconds
Distance of race
312.4 x speed of tortoise
= 312.4 x .15
= 46.85 m
Answer:energy which a body possesses by virtue of being in motion.
Explanation:
Answer:
Part a)
charge on each sphere is -1.95 micro coulomb
Part b)
For first sphere
excess charge
For second sphere
excess charge
For third sphere
absent charge
For third sphere
absent charge
Explanation:
Part a)
Since all the spheres are of identical size so the total charge of the sphere will divide equally on them
So we have
![q = \frac{Q_1 + Q_2 + Q_3 + Q_4}{4}](https://tex.z-dn.net/?f=q%20%3D%20%5Cfrac%7BQ_1%20%2B%20Q_2%20%2B%20Q_3%20%2B%20Q_4%7D%7B4%7D)
![q = \frac{2.2 \mu C + 4.2 \mu C - 7.4 \mu C - 6.8 \mu C}{4}](https://tex.z-dn.net/?f=q%20%3D%20%5Cfrac%7B2.2%20%5Cmu%20C%20%2B%204.2%20%5Cmu%20C%20-%207.4%20%5Cmu%20C%20-%206.8%20%5Cmu%20C%7D%7B4%7D)
![q = -1.95 \mu C](https://tex.z-dn.net/?f=q%20%3D%20-1.95%20%5Cmu%20C)
So charge on each sphere is -1.95 micro coulomb
Part b)
For first sphere
initial charge = 2.2 micro coulomb
final charge = -1.95 micro coulomb
excess charge = -1.95 - 2.2 = -4.15 micro coulomb
Q = Ne
![N = \frac{4.15\times 10^{-19}}{1.6 \times 10^{-19}}](https://tex.z-dn.net/?f=N%20%3D%20%5Cfrac%7B4.15%5Ctimes%2010%5E%7B-19%7D%7D%7B1.6%20%5Ctimes%2010%5E%7B-19%7D%7D)
![N = 2.6 \times 10^{13}](https://tex.z-dn.net/?f=N%20%3D%202.6%20%5Ctimes%2010%5E%7B13%7D)
For second sphere
initial charge = 4.2 micro coulomb
final charge = -1.95 micro coulomb
excess charge = -1.95 - 4.2 = -6.15 micro coulomb
Q = Ne
![N = \frac{6.15\times 10^{-19}}{1.6 \times 10^{-19}}](https://tex.z-dn.net/?f=N%20%3D%20%5Cfrac%7B6.15%5Ctimes%2010%5E%7B-19%7D%7D%7B1.6%20%5Ctimes%2010%5E%7B-19%7D%7D)
![N = 3.8 \times 10^{13}](https://tex.z-dn.net/?f=N%20%3D%203.8%20%5Ctimes%2010%5E%7B13%7D)
For third sphere
initial charge = -7.4 micro coulomb
final charge = -1.95 micro coulomb
absent charge = -1.95 + 7.4 = 5.45 micro coulomb
Q = Ne
![N = \frac{5.45\times 10^{-19}}{1.6 \times 10^{-19}}](https://tex.z-dn.net/?f=N%20%3D%20%5Cfrac%7B5.45%5Ctimes%2010%5E%7B-19%7D%7D%7B1.6%20%5Ctimes%2010%5E%7B-19%7D%7D)
![N = 3.4 \times 10^{13}](https://tex.z-dn.net/?f=N%20%3D%203.4%20%5Ctimes%2010%5E%7B13%7D)
For third sphere
initial charge = -6.8 micro coulomb
final charge = -1.95 micro coulomb
absent charge = -1.95 + 6.8 = 4.85 micro coulomb
Q = Ne
![N = \frac{4.85\times 10^{-19}}{1.6 \times 10^{-19}}](https://tex.z-dn.net/?f=N%20%3D%20%5Cfrac%7B4.85%5Ctimes%2010%5E%7B-19%7D%7D%7B1.6%20%5Ctimes%2010%5E%7B-19%7D%7D)
![N = 3.03 \times 10^{13}](https://tex.z-dn.net/?f=N%20%3D%203.03%20%5Ctimes%2010%5E%7B13%7D)
To determine how far the car goes, we have to know the distance covered by the 75 revolutions of the tires of the car. To do this, we have to calculate for the perimeter of the tire, then multiply it to the number of revolutions.
Distance =
![75 \pi d](https://tex.z-dn.net/?f=75%20%5Cpi%20d)
where pi*d is the circumference of the tire
Distance = 75 * pi * 0.90 m
Distance = 212.06 m
The car covers a distance of
212.06 m as it slows down.
Answer:
a) v2f = 1.2 m/s, b) h = 7.35 10⁻² m and c) ΔK = -4196.4 J
,
Explanation:
a) This problem must be solved with the conservation of the moment. Let's define the system as the one formed by the bullet plus the block, in this system all the forces are internal therefore the moment conserves, let's write the moment in two moments before and after the crash. In general these shocks are very fast, so it can be assumed that the box does not move during the crash.
The data they give us is the mass of the bullet (m = 0.010 kg), the initial and final velocities of the bullet (v1o = 1000 m / s and v1f = 400 m / s) and the block gives us the mass M = 5 kg and its initial velocity v2o = 0 m / s
Before the crash
po = m v₁₀
After the crash
pf = m
+ M ![v_{2f}](https://tex.z-dn.net/?f=v_%7B2f%7D)
p₀ = pf
m v₁₀ = m
+ M ![v_{2f}](https://tex.z-dn.net/?f=v_%7B2f%7D)
v2f = m (v₁₀ -
) / M
v2f = 0.010 (1000-400) / 5
v2f = 1.2 m / s
b) Having the speed of the block we can use the law of conservation of energy to find the height. Let's write the mechanical energy of the block just after the crash and at the point of maximum height
Initial. Just after the crash
v = ![v_{2f}](https://tex.z-dn.net/?f=v_%7B2f%7D)
Em1 = K = ½ M v²
Final. At maximum height
Em₂ = U = M g h
Em₁ = Em₂
½ M v² = M g h
h = ½ v² / g
h = ½ 1.2² / 9.8
h = 0.0735 m
h = 7.35 10⁻² m
c) Let's calculate the kinetic energy before and after the crash
Before
K₀ = ½ m v₁₀²
K₀ = ½ 0.01 1000²
K₀ = 5000 J
Final
= ½ m v1f² + ½ M v2f²
= ½ 0.010 400² + ½ 5 1.2²
= 800 + 3.6
= 803.6 J
We can give the amount of energy that is lost as the subtraction of the two energies or as the fraction of lost engoa
ΔK =
-K₀
ΔK = 803.6 - 5000
ΔK = -4196.4 J
/ K₀ = 803.6 / 5000
/ K₀ = 0.16