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ella [17]
3 years ago
8

Two balls of equal size are dropped from the same height from the roof of a building. the mass of ball a is twice that of ball b

, m subscript a equals 2 m subscript
b. when the two balls reach the ground, how do their kinetic energies compare?
Physics
2 answers:
Sindrei [870]3 years ago
8 0
The mass of ball a is twice the mass of ball b:
m_a = 2 m_b
This means that the initial potential energy of ball a (U_a = m_a gh=2 m_b gh) is twice the potential energy of ball b (U_b = m_b  g h):
U_a = 2 U_b
When the two balls reach the ground, the potential energy of each ball has converted into kinetic energy (since now their altitude is h=0), because the total mechanical energy of each ball must be  conserved. Therefore:
K_a = U_a
K_b = U_b
and so the kinetic energy of ball a must be twice the kinetic energy of ball b:
K_a = 2 K_b
statuscvo [17]3 years ago
6 0

Answer:

5.5 m/s

Explanation:

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A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum
ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

\rho (r) \  \alpha  \  \delta (r -R)

\rho (r) = k \  \delta (r -R) \ \  at \ \  (r = R)

\rho (r) = 0\ \ since \ r< R  \ \ or  \ \ r>R---- (1)

To find the constant k, we  examine the total charge Q which is:

Q = \int \rho (r) \ dV = \int \sigma \times dA

Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2

∴

\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta  \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2

\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2

(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2

Thus;

k * 4 \pi  \int ^{R}_{0}  \delta (r -R) * r^2dr = \sigma \times  R^2

k * \int ^{R}_{0}  \delta (r -R)  r^2dr = \sigma \times  R^2

k * R^2= \sigma \times  R^2

k  =   R^2 --- (2)

Hence, from equation (1), if k = \sigma

\mathbf{\rho (r) = \delta* \delta (r -R)  \ \  at   \ \  (r=R)}

\mathbf{\rho (r) =0 \ \  at   \ \  rR}

To verify the units:

\mathbf{\rho (r) =\sigma \ *  \ \delta (r-R)}

↓         ↓            ↓

c/m³    c/m³  ×   1/m            

Thus, the units are verified.

The integrated charge Q

Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \  sin \theta  \ dr \ d\theta \  d \phi  \\ \\  Q = \int ^{2 \pi}_{0} \  d \phi  \int ^{\pi}_{0} \ sin \theta  \int ^R_{0} \rho (r) r^2 \ dr

Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  \int ^R_0  * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  *R^2    since  ( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )

\mathbf{Q = 4 \pi R^2  \sigma  }

6 0
3 years ago
For years, the tallest tower in the United States was the Phoenix Shot Tower in Baltimore, Maryland. The shot tower was used fro
serious [3.7K]

Answer:

The answer to your question is:

a) t = 3.81 s

b) vf =  37.4 m/s

Explanation:

Data

height = 71.3 m = 234 feet

t = 0 m/s

vf = ?

vo = 0 m/s

Formula

h = vot + 1/2gt²

vf = vo + gt

Process

a)

               h = vot + 1/2gt²

             71.3 = 0t + 1/2(9.81)t²

             2(71.3) = 9,81t²

              t² = 2(71.3)/9.81

              t² = 14.53

              t = 3.81 s

b)

      vf = 0 + (9.81)(3.81)

      vf = 37.4 m/s

3 0
3 years ago
Which is greater the attraction of the earth for 1 kg of aluminum or aluminum or attraction of 1kg of aluminum for the earth?
OleMash [197]
Those forces are exactly equal.

Gravity always works as a pair of <em>EQUAL</em> forces ... one in each direction
between two masses.  Your weight on the Earth is exactly the same as
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The center of the Hubble space telescope is 6940 km from Earth’s center. If the gravitational force between Earth and the telesc
Law Incorporation [45]
The gravitational force between two objects is given by:
F=G \frac{m_1 m_2}{r^2}
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is the separation between the two objects

The distance of the telescope from the Earth's center is r=6940 km=6.94 \cdot 10^6 m, the gravitational force is F=9.21 \cdot 10^4 N and the mass of the Earth is m_1=5.98 \cdot 10^{24} kg, therefore we can rearrange the previous equation to find m2, the mass of the telescope:
m_2 =  \frac{Fr^2}{Gm_1}= \frac{(9.21 \cdot 10^4 N)(6.94\cdot 10^6)^2}{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})} =11121 kg
6 0
3 years ago
Read 2 more answers
What is the correct order of the layers' density from lowest density to highest?
Orlov [11]

Answer:

C. crust, mantle, core

Explanation:

density increases as you travel from the crust to the inner core

the crust is on top

next is the mantle

and then the core

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3 years ago
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