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2 years ago

3. Planet 1 has mass M₁ and radius R₁. Planet 2 has mass M₂ and radius

Physics

1 answer:

Oksanka [162]2 years ago

8 0

(a) The initial speed must the object be launched so that it reaches

the surface of Planet 2 with zero speed is √[2G{ M₁/ R₁ + M₂/(R₂ + L) - M₁/ (R₁ +L) - M₂/ R₂}]

(b) An inequality between M₁ and M₂ that represents when (a)

can occur is { M₁/ R₁ + M₂/(R₂ + L) - M₁/ (R₁ +L) - M₂/ R₂} ≥ 0

<h3>What is gravity?</h3>

The force of attraction felt by a person which is directed at the center of a planet or Earth is called as the gravity.

The force of attraction is directly proportional to the product of masses of the object and inversely proportional to the square of distance between them.

F = GMm/R²

Given, Planet 1 has mass M₁ and radius R₁. Planet 2 has mass M₂ and radius R₂. The two planets are a distance of L apart, measured from surface to surface. An object is launched with some initial speed from the surface of Planet 1 directly towards Planet 2. For this problem, assume that Planets1 and 2 are stationary.

If v is the launch velocity, then initial total energy will be

T.E = 1/2 mv² + ( -GM₁m/ R₁ - G M₂m/(R₂ + L)

The final total energy will be

T.E =0 + ( -GM₁m/ (R₁ +L) - G M₂m/ R₂)

From energy conservation principle, we get

1/2 mv² + ( -GM₁m/ R₁ - G M₂m/(R₂ + L) = ( -GM₁m/ (R₁ +L) - G M₂m/ R₂)

v = √[2G{ M₁/ R₁ + M₂/(R₂ + L) - M₁/ (R₁ +L) - M₂/ R₂}]

(b) an inequality between M₁ and M₂ so that object reaches the surface of Planet 2 with zero speed is

[2G{ M₁/ R₁ + M₂/(R₂ + L) - M₁/ (R₁ +L) - M₂/ R₂}] =0

{ M₁/ R₁ + M₂/(R₂ + L) - M₁/ (R₁ +L) - M₂/ R₂} ≥ 0

Thus, this is an inequality between M₁ and M₂.

{ M₁/ R₁ + M₂/(R₂ + L) - M₁/ (R₁ +L) - M₂/ R₂} ≥ 0

M₁(R+L) + M₂R - M₁R -M₂(R+L) / R (R+L) ≥ 0

M₁(R+L) + M₂R - M₁R -M₂(R+L) ≥ 0

M₁L - M₂L ≥ 0

M₁ ≥ M₂

M₁ must be greater than M₂.

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Answer:

The beam used is a negatively charged electron beam with a velocity of

v = E / B

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3 years ago

A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w

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Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

directly proportional to its length i.e $R\propto l$
inversely proportional to its cross section area i.e $R\propto \frac{1}{A}$

Therefore

$R=\rho\frac{l}{A}$

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

$A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2$

$R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }$

$\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }$ ......(1)

Again for tungsten:

$R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }$

$\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }$ ........(2)

Given that $R_1=R_2$ and $l_1=l_2$

Dividing the equation (1) and (2)

$\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}$

$\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}$ [since $R_1=R_2$ and $l_1=l_2$ ]

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}$

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}$

$\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}$

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

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3 years ago

What is not changed when work is done by a machine?

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B) The amount of work done

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2 years ago

Read 2 more answers

For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par

Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

initial temperature of metals, = $T_m$
initial temperature of water, = $T_i$
specific heat capacity of copper, $C_p$ = 0.385 J/g.K
specific heat capacity of aluminum, $C_A$ = 0.9 J/g.K
both metals have equal mass = m

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

For copper metal, the quantity of heat transferred is calculated as;

$Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w ) = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t$

The change in heat energy for copper metal;

$\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t$

For aluminum metal, the quantity of heat transferred is calculated as;

$Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w) = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t$

The change in heat energy for aluminum metal ;

$\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t$

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

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