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3 years ago

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A 10-meter-long spear is thrown at a relativistic speed through a 10-meter-long pipe (both measured when at rest.) when the spea

r passes through the pipe,
a. the pipe shrinks so the spear extends at both ends.
b. both shrink equally so the pipe barely covers the spear.
c. the spear shrinks so the pipe completely covers it.
d. any of the above, depending on the motion of the observer
e. none of the above

1 answer:

Mekhanik [1.2K]3 years ago

8 0

The correct answer is a

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How long does it take an automobile traveling 72.5 km/h to become even with a car that is traveling in another lane at 54.5 km/h

Novosadov [1.4K]

Answer:

7.83

Explanation:

Just using the speed ,distance and time relation..Let time be T

Use this formula and find the distance covered by automobile .

Do same for car ..

and at last subtract the distance of automobile to car =141.

i.e (Distance of automobile -Dist.of car )=141

4 0

4 years ago

how much time do it take for a cheetah running at a speed of 31.67m/s to travel a distance of 1361.81m

ExtremeBDS [4]

V=D/T
V is speed
D is distance
T is time
V=D/T
31.67 m/s = 1361.81 m / T
31.67/31.67=1361.81/31.67
T=43 seconds
Hope this helps

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4 years ago

A merry-go-round rotates from rest with an angular acceleration of 1.04 rad/s2. How long does it take to rotate through (a) the

Andrei [34K]

Answer:

(a) 4.38 s.

(b) 1.817 s

Explanation:

(a)

Using

θ = ω₀t +1/2αt² ................ Equation 1

Where θ = number of revolution, t = time, α = angular acceleration, ω₀ = angular velocity.

Given: θ = 1.59 rev = 1.59×2π = 9.992 rad, ω₀ = 0 rad/s, α = 1.04 rad/s².

Substitute into equation 1

9.992 = 0(t) + 1/2(1.04)(t²)

t² = (2×9.992)/1.04

t² = 19.984/1.04

t = √(19.215)

t =4.38 s.

(b)

also using

θ = ω₀t +1/2αt²............... Equation 1

Given: θ =3.18 rev = 3.18×2π = 19.97 rad, ω₀ = 0 rad/s, α = 1.04 rad/s².

Substitute into equation 1

19.97 = 0(t) + 1/2(1.04)(t²)

t² = 19.97×2/1.04

t = √(38.40)

t = 6.197 s

The time require = 6.197-4.38 = 1.817 s

3 0

4 years ago

A spherical shell is rolling without slipping at constant speed on a level floor. What percentage of the shell's total kinetic e

IgorC [24]

Answer:

41.667 per cent of the total kinetic energy is translational kinetic energy.

Explanation:

As the spherical shell is rolling without slipping at constant speed, the system can be considered as conservative due to the absence of non-conservative forces (i.e. drag, friction) and energy equation can be expressed only by the Principle of Energy Conservation, whose total energy is equal to the sum of rotational and translational kinetic energies. That is to say:

$E = K_{t} + K_{r}$

Where:

$E$ - Total energy, measured in joules.

$K_{r}$ - Rotational kinetic energy, measured in joules.

$K_{t}$ - Translational kinetic energy, measured in joules.

The spherical shell can be considered as a rigid body, since there is no information of any deformation due to the motion. Then, rotational and translational components of kinetic energy are described by the following equations:

Rotational kinetic energy

$K_{r} = \frac{1}{2}\cdot I_{g}\cdot \omega^{2}$

Translational kinetic energy

$K_{t} = \frac{1}{2}\cdot m \cdot R^{2}\cdot \omega^{2}$

Where:

$I_{g}$ - Moment of inertia of the spherical shell with respect to its center of mass, measured in $kg\cdot m^{2}$ .

$\omega$ - Angular speed of the spherical shell, measured in radians per second.

$R$ - Radius of the spherical shell, measured in meters.

After replacing each component and simplifying algebraically, the total energy of the spherical shell is equal to:

$E = \frac{1}{2}\cdot (I_{g} + m\cdot R^{2})\cdot \omega^{2}$

In addition, the moment of inertia of a spherical shell is equal to:

$I_{g} = \frac{2}{3}\cdot m\cdot R^{2}$

Then, total energy is reduced to this expression:

$E = \frac{5}{6}\cdot m \cdot R^{2}\cdot \omega^{2}$

The fraction of the total kinetic energy that is translational in percentage is given by the following expression:

$\%K_{t} = \frac{K_{t}}{E}\times 100\,\%$

$\%K_{t} = \frac{\frac{1}{2}\cdot m \cdot R^{2}\cdot \omega^{2} }{\frac{5}{6}\cdot m \cdot R^{2}\cdot \omega^{2} } \times 100\,\%$

$\%K_{t} = \frac{5}{12}\times 100\,\%$

$\%K_{t} = 41.667\,\%$

41.667 per cent of the total kinetic energy is translational kinetic energy.

7 0

4 years ago

Un recipiente contiene 224 dm3 de Ozono de masa 4.561 Kg a 51.09 grados celsius. Calcula la presión del Ozono

kari74 [83]

Answer:

Por lo tanto, la presión del ozono es:

$P=0.011\: atm$

Explanation:

Podemos usar la ecuacion de los gases ideales;

$PV=nRT$ (1)

Tenemos:

El volumen V = 224 dm³ = 224 L

La temperatura T = 51.09 C = 324.09 K

La masa es m = 4.561 kg

Lo necesitamos ahora es calvular n que es el numero de moles;

recordemos que el peso molecular del ozono M = 48 g/mol.

$n=\frac{m}{M}=\frac{4.561}{48}=0.095\: mol$

Finalmente, usando la ecuacion 1 despejamos la presion P

$P=\frac{nRT}{V}$

$P=\frac{0.095*0.082*324.09}{224}$

Por lo tanto, la presion del ozono es:

$P=0.011\: atm$

Espero te haya ayudado!

5 0

3 years ago