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olasank [31]
3 years ago
6

An object is dropped from an altitude of one Earth radiusabove Earth's surface. If M is the mass of Earth and R is itsradius the

speed of the object just before it hits Earth isgiven by: A. √GM/R B. √GM/2R C. √2GM/R D. √GM/R2 E. √GM/2R2
Physics
1 answer:
Montano1993 [528]3 years ago
4 0

Answer:

The speed of the object just before it hits Earth is  \sqrt{\dfrac{GM}{R}}

(A) is correct option.

Explanation:

Given that,

M = mass of earth

R = radius of earth

The potential energy at height above the surface of the earth

P.E=-\dfrac{GmM}{R+h}

The kinetic energy at height above the surface of the earth

K.E = 0

The total energy at height above the surface of the earth

E = K.E+P.E

E = -\dfrac{GmM}{R+h}....(I)

The total energy at the surface of the earth

E'=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}....(II)

We need to calculate the speed of the object  just before it hits Earth

From equation (I) and (II)

-\dfrac{GmM}{R+h}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}

Here, h = R

-\dfrac{GmM}{2R}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}

v= \sqrt{\dfrac{GM}{R}}

Hence, The speed of the object just before it hits Earth is  \sqrt{\dfrac{GM}{R}}.

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7 0
3 years ago
Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin
tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of m_2 would be four times that of m_1.

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2.

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is T, then the angular velocity of the SHM would be

\displaystyle \omega = \frac{2\pi}{T}.

Assume that the mass starts with a zero displacement and a positive velocity. If A represent the amplitude of the SHM, then the displacement of the mass at time t would be:

\mathbf{x}(t) = A\sin(\omega\cdot t).

The velocity of the mass at time t would be:

\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t).

The acceleration of the mass at time t would be:

\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t).

Let m represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time t would be:

\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t),

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant k will be equal to:

\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}.

Since \displaystyle \omega = \frac{2\pi}{T}, it can be concluded that:

\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}.

For the first mass m_1, if the time period is T_1, then the spring constant would be:

\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2.

Similarly, for the second mass m_2, if the time period is T_2, then the spring constant would be:

\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2.

Since the two springs are the same, the two spring constants should be equal to each other. That is:

\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2.

Simplify to obtain:

\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2.

6 0
3 years ago
Please help me!
sergeinik [125]

C. The Densities are equal.

<h3>What is density?</h3>

Density is mass per unit volume or mass of a unit volume of a material substance.

If m1, V1 and D1 = mass, volume  and density respectively of ball C

m2, V2 and D2 = mass, volume and density respectively of ball D

According to the Question ,

V_{1} = 3V_{2}  , m_{2}  = \frac{1}{3} (m_{1} ) \\ \\= m_{1} = 3m_{2}

Therefore,

\frac{D_{1} }{D_{2} }  = (\frac{m_{1} }{V_{1} } )* (\frac{m_{2} }{V_{2} } )\\ \\= (\frac{3m_{2} }{3V_{2} })*(\frac{V_{2} }{m_{2} }) \\\\= 1

Hence, D1 = D2

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6 0
2 years ago
A beaker of mass 1.3 kg containing 2.8 kg of water rests on a scale. A 3.7 kg block of a metallic alloy of density 4600 kg/m3 is
Oksana_A [137]

1) Force read on the upper scale: 33.4 N

2) Force read on the lower scale: 43.0 N

Explanation:

1)

The reading on the upper scale is equal to the net force acting on the block of metallic alloy. The net force is given by:

F=W-B (1)

where

W is the weight of the block (downward)

B is the buoyant force (upward)

The weight of the block is given by

W=mg

where m = 3.7 kg is the mass of the block and g=9.8 m/s^2 is the acceleration of gravity.

The buoyant force is given by

B=\rho_w V g

where

\rho_w=1000 kg/m^3 is the water density

V is the volume of the block

The volume of the block can be written as

V=\frac{m}{\rho_b}

where \rho_b=4600 kg/m^3 is the density of the block.

Substituting everything into eq.(1), we find:

F=mg-\rho_w \frac{m}{\rho_b}g=(3.7)(9.8)-(1000)\frac{1.3}{4600}(9.8)=33.4 N

2)

Here we want to find the force on the lower scale.

The force on the lower scale is equal to the difference between the total weight of the system (given by the weight of the beaker + the weight of the water + the weight of the block) and the upper net force exerted on the upper scale, therefore:

F' = m_B g + m_wg+m_b g - F

where:

m_B=1.3 kg is the mass of the beaker

m_w=2.8 kg is the mass of the water

m_b = 3.7 kg is the mass of the block

F=33.4 N is the upper net force

Substituting and solving, we find:

F'=(1.3+2.8+3.7)(9.8)-33.4=43.0 N

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6 0
3 years ago
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