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tangare [24]
3 years ago
9

A pendulum consists of a stone with mass m swinging on a string of length L and negligible mass. The stone has a speed of v0 whe

n it passes its lowest point. (a) Write an expression for the speed of the stone when the string is at some angle theta with the vertical. (b) What is the greatest angle with the vertical that the string will reach during the stone's motion? (c) If the potential energy of the pendulum-Earth system is taken to be zero at the stone's lowest point, what is the total mechanical energy of the system? State all your answers in terms of the given variables and g
Physics
1 answer:
Tcecarenko [31]3 years ago
6 0

Answer:

m

Explanation:

You might be interested in
A 30-cm-diameter, 4-m-high cylindrical column of a house made of concrete ( k = 0.79 W/m⋅K, α = 5.94 × 10 −7 m2/s, rho = 1600kg
PilotLPTM [1.2K]

Answer:

a) Time it will taken for the column surface temperature to rise to 27°C is  

17.1 hours

b) Amount of heat transfer is 5320 kJ  

c) Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ

Explanation:

Given that;

Diameter D = 30 cm

Height H = 4m

heat transfer coeff h = 14 W/m².°C

thermal conductivity k = 0.79 W/m.°C

thermal diffusivity α  = 5.94 × 10⁻⁷ m²/s

Density p = 1600 kh/m³

specific heat Cp = 0.84 Kj/kg.°C

a)

the Biot number is

Bi = hr₀ / k

we substitute

Bi = (14 W/m².°C × 0.15m) / 0.79 W/m.°C

Bi = 2.658

From the coefficient for one term approximate of transient one dimensional heat conduction The constants λ₁ and A₁ corresponding to this Biot number are,  

λ₁ = 1.7240

A₁ = 1.3915

Once the constant J₀ = 0.3841 is determined from corresponding to the constant λ₁

the Fourier number is determined to be  

[ T(r₀, t) -T∞ ] / [ Ti - T∞]  = A₁e^(-λ₁²t') J₀ (λ₁r₀ / r₀)

(27 - 28) / (14 - 28)   = (1.3915)e^-(17240)²t (0.3841)  

t' = 0.6771

Which is above the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can be used. Then the time it will take for the column surface temperature to rise to 27°C becomes  

t =  t'r₀² / ₐ

= (0.6771 × 0.15 m)² /  (5.94 x 10⁻⁷ m²/s)

= 23,650 s

= 7.1 hours

Time it will taken for the column surface temperature to rise to 27°C is  

17.1 hours

b)

The heat transfer to the column will stop when the center temperature of column reaches to the ambient temperature, which is 28°C.  

Maximum heat transfer between the ambient air and the column is

m = pV

= pπr₀²L

= (1600 kg/m³ × π × (0.15 m)² × (4 m)

= 452.389 kg

Qin = mCp [T∞ - Ti ]

= (452.389 kg) (0.84 kJ/kg.°C) (28 - 14)°C

= 5320 kJ  

Amount of heat transfer is 5320 kJ  

(c)

the amount of heat transfer until the surface temperature reaches to 27°C is

(T(0,t) - T∞) / Ti - T∞  = A₁e^(-λ₁²t')

= (1.3915)e^-(1.7240)² (0.6771)

= 0.1860

Once the constant J₁ = 0.5787 is determined from Table corresponding to the constant λ₁, the amount of heat transfer becomes  

(Q/Qmax)cyl = 1 - 2((T₀ - T∞) / ( Ti - T∞)) ((J₁(λ₁)) / λ₁)

= 1 - 2 × 0.1860 × (0.5787  / 1.7240)  

= 0.875

Q = 0.875Qmax

Q = 0.875(5320 kJ)  

Q = 4660 kJ

Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ

6 0
2 years ago
Which is more work, pushing with 115 N over 15m or lifting 20N over 10 m?
Nikitich [7]

Work = (force) x (distance)

You could look at the two cases, and see right away that
the first one has more force acting through more distance,
so it must be more work.  But since I just gave you the formula
for Work, let's calculate the amount of it for both cases:

First case:  Work = (115 N) x (15 m) =   1,725 joules

Second case:  Work = (20 N) x (10 m)  =  200 joules

The first case involves 8.625 times as much work as the second case.

3 0
3 years ago
Read 2 more answers
A projectile is launched horizontally from a height of 8.0 m. The projectile travels 6.5 m before hitting the ground.
spayn [35]

Answer:

5.09 m/s

Explanation:

Use the height to find the time it takes to land:

y = y₀ + v₀ᵧ t + ½ gt²

0 = 8.0 m + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 1.28 s

Now use the horizontal distance to find the initial velocity.

x = x₀ + v₀ₓ t + ½ at²

6.5 m = 0 m + v₀ (1.28 s) + ½ (0 m/s²) (1.28 s)²

v₀ = 5.09 m/s

7 0
3 years ago
An alpha particle (charge +2e) travels in a circular path of radius .5m in a magnetic field of 1.0 T. Find the (a) period, (b) s
navik [9.2K]

Given Information:

Radius = r = 0.5 m

Magnetic field = 1.0 T

Required Information:

Period = T = ?

Speed = v = ?

Kinetic energy = KE = ?

Answer:

Period = 0.13x10⁻⁶ seconds

speed = 24.16x10⁶ m/s

Kinetic energy = 12.11 MeV

Explanation:

(a) period

The time period of alpha particle is related to its orbital speed as

T = 2πr/v  eq. 1

According to newton's law

F = ma

Force due to magnetic field is given by

F = qvB

qvB = ma

qvB = m(v²/r)

qB = mv/r

v = qBr/m  eq. 2

substitute the eq. 2 in eq. 1

T = 2πr/qBr/m

r cancels out

T = 2π/qB/m

T = 2πm/qB

T = 2π*6.65x10⁻²⁷/2*1.602x10⁻¹⁹*1

T = 0.13x10⁻⁶ seconds

(b) speed

From equation 1

T = 2πr/v

v = 2πr/T

v = 2π*0.5/0.13x10⁻⁶

v = 24.16x10⁶ m/s

(c) kinetic energy (in electron volts)

Kinetic energy is given by

KE = 0.5mv²

KE = 0.5*6.65x10⁻²⁷*(24.16x10⁶)²

KE = 1.94x10⁻¹² J

since 1 electron volt has 1.602x10⁻¹⁹ J

KE = 1.94x10⁻¹²/1.602x10⁻¹⁹

KE = 12.11 MeV

5 0
3 years ago
What can parents do if they find a mistake in the student records?      A. They can’t do anything B. Make a correction themselve
Lelechka [254]
D. Ask the school to make the correction
5 0
3 years ago
Read 2 more answers
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