PHYSICS COLLEGE

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg), who is sitting at the very bottom end, and whom he holds onto when he arrives. Laughing, John & William leave the slide horizontally and land in the muddy ground near the foot of the slide. (A) If John starts out 1.8 m above William, and the slide is essentially frictionless, how fast are they going when they leave the slide? (B) Thanks to the mud he acquired, John will now experience an average frictional force of 105 N as he slides down. How much slower is he going when he reaches the bottom than when friction was absent?

Answers

Answer 1

Answer:

Answer:

$v=3.564\ m.s^(-1)$

$\Delta v =2.16\ m.s^(-1)$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

Using the equation of motion:

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2* (9.8* (1.8)/(3))* 3$

$v_J=5.94\ m.s^(-1)$

Now using the law of conservation of momentum at the bottom of the slide:

Sum of initial momentum of kids before & after collision must be equal.

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30* 5.94+0=(30+20)v$

$v=3.564\ m.s^(-1)$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

Now we find the force along the slide due to the body weight:

$F=m_J.g.sin\theta$

$F=30* 9.8* (1.8)/(3)$

$F=176.4\ N$

Hence the net force along the slide:

$F_R=71.4\ N$

Now the acceleration of John:

$a_j=(F_R)/(m_J)$

$a_j=(71.4)/(30)$

$a_j=2.38\ m.s^(-2)$

Now the new velocity:

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2* 2.38* 3$

$v_J_n=3.78\ m.s^(-1)$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^(-1)$

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Answers

What is pseudo force?

A pseudo force, also called a fictitious force or an inertial force, is an apparent force that acts on all bodies whose motion is described using a non-inertial frame of reference, such as a rotating reference frame.

Seven seconds after a brilliant flash of lightning, thunder shakes the house. How far was the lightning strike from the house? Seven seconds after a brilliant flash of lightning, thunder shakes the house. How far was the lightning strike from the house? Much farther away than two kilometers Much closer than one kilometer About two kilometers away About one kilometer away It is impossible to predict.

Answers

Answer:

About two kilometers away

$\rm distance=2.401\ km$

Explanation:

Given:

The time gap between the light and sound to travel to the house, $t=7\ s$

Since the clouds are formed in the troposphere region of the atmosphere which extends from 8 kilometers to 12 kilometers above the earth-surface and the velocity of light is 300000 kilometers per second so it is visible almost instantly, hence we neglect the time taken by the light to travel to the house from the clouds.

∴Distance between the lightning-strike and the house:

$\rm distance=v* t$

we have the speed of sound as: $v=343\ m.s^(-1)$

So,

$\rm distance=343* 7$

$\rm distance=2401\ m$

$\rm distance=2.401\ km$

What is science? Plz help I cant find a answer thats not really complicated and i have to fit it within 3 sentences.

Answers

Science is the study of our universe, and our own planet. It is the study of the biological, chemical and physical world we live in. It is a process of discovering by experimenting and looking for patterns.

Circuit A in a house has a voltage of 218 V and is limited by a 45-A circuit breaker. Circuit B is at 120.0 V and has a 25-A circuit breaker.What is the ratio of the maximum power delivered by circuit A to that delivered by circuit B?

Answers

Answer:

3.27

Explanation:

Electric Power: This can be defined as the rate at which electric energy is consumed. The unit of power is Watt (W).

Mathematically, electric power is represented as

P = VI ..................................... Equation 1.

Where P = power, V = voltage, I = Current.

For Circuit A,

P₁ = V₁I₁ ................................... Equation 2

Where P₁ = maximum power delivered by circuit A, V₁ = Voltage of circuit A, I₁ = circuit breaker rating of circuit A.

Given: V₁ = 218 V, I₁ = 45 A.

Substituting into equation 2

P₁ = 218×45

P₁ = 9810 W.

For Circuit B,

P₂ = V₂I₂............................. Equation 3

Where P₂ = maximum power delivered by the circuit B, V₂ = voltage of circuit B, I₂ = circuit breaker rating of circuit B

Given: V₂ = 120 V, I₂ = 25 A.

Substitute into equation 3

P₂ = 120(25)

P₂ = 3000 W.

Ratio of maximum power delivered by circuit A to that delivered by circuit B = 9810/3000

= 3.27.

Thus the ratio of maximum power delivered by circuit A to circuit B = 3.27

The universe is filled with photons left over from the Big Bang that today have an average energy of about 2 × 10−4 eV (corresponding to a temperature of 2.7 K). As derived in lecture, the number of available energy states per unit volume for photons is ????(????)????????

Answers

Answer:

The number of available energy states per unit volume is $4.01*10^(48)$

Explanation:

Given that,

Average energy $E=2*10^(-4)\ eV$

Photon = $4*10^(-5)\ eV$

We need to calculate the number of available energy states per unit volume

Using formula of energy

$g(\epsilon)d\epsilon=(8\pi E^2dE)/((hc)^3)$

Where, E = energy

h = Planck constant

c = speed of light

Put the value into the formula

$g(\epsilon)d\epsilon=(8*\pi*2*10^(-4)*4*10^(-5)*1.6*10^(-19))/((6.67*10^(-34)*3*10^(8))^3)$

$g(\epsilon)d\epsilon=4.01*10^(48)$

Hence, The number of available energy states per unit volume is $4.01*10^(48)$

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Answers

Explanation:

Assuming the substances are fluids that do not mix, the lighter substance (ρ = 1.3 g/cm³) will float on top of the heavier substance (ρ = 2.0 g/cm³). This is due to Archimedes' Principle, which explains buoyancy.

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