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3 years ago

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At time t=0 a 2150 kg rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. This forc

e obeys the equation Fx=At2, where t is time, and has a magnitude of 893.68 N when t=1.25 s. What impulse does the engine exert on the rocket during the 3.1 s interval starting 2.00 s after the engine is fired?

1 answer:

vredina [299]3 years ago

8 0

Answer:

two people who are not going to be able to make it to the office and I will be there at the house and I will be there in a few minutes and I'll be there in a few

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R S ( M ) = 2 G M c 2 , where G is the gravitational constant and c is the speed of light. It is okay if you do not follow the d

padilas [110]

The provided question's answer is "Schwarzschild radius".

The conversion factor between mass and energy is the speed of light squared.

GM/r stands for gravitational potential energy, also known as energy per unit mass.

GM/rc² then has "mass per unit mass" units. In other words, as mass/mass splits out in a dimensional analysis, "dimensionless per unit."

The derivation yields a formula for time or space coordinate ratios requiring sqrt(1 - 2GM/rc²). This number becomes 0 when r=2GM/c2, or the formula becomes infinite if in the denominator. However, there is no justification for using c² as a conversion factor there. Consider the initial expression sqrt(1 - 2GM/rc²).

Assume that m is used as the test particle's mass instead of 1. Then you have sqrt(m - 2GMm/rc² and mass units. This expression denotes that the rest energy of the test mass m you introduced into the gravitational field is "gone" at that radius.

The 2 would be absent if the gravitational field were Newtonian. However, at the event horizon, Einstein gravity is slightly stronger than Newton gravity, resulting in the factor 2 in qualitative terms.

So, the given equation is of Schwarzschild radius.

Learn more about Schwarzschild radius here:

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3 years ago

Which technique is the most useful for measuring the distance to a galaxy located 10 million light-years away?

devlian [24]

The technique which is the most useful for measuring the distance to a galaxy located 10 million light-years away is Cepheids

A Cepheid variable is a particular kind of star that pulses radially, varying in diameter and temperature while causing changes in brightness with a clearly discernible steady time and amplitude.

A light-year is defined as the total amount of distance covered by light in a year. It is usually a unit to measure very large distances which usually concerns the Outerspace

A galaxy is a system made up of stars, stellar debris, interstellar gas, dust, and other matter that is gravitationally bound. The name "galaxy" is derived from the Greek word "galaxies," which is an allusion to the Milky Way and literally means "milky."

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2 years ago

A block of mass 3.20 kg is placed against a horizontal spring of constant k = 865 N/m and pushed so the spring compresses by 0.0

Aliun [14]

Answer:

a) The initial elastic potential energy of the block-spring system is 28.113 joules.

b) The final speed of the block is approximately 4.192 meters per second.

Explanation:

a) By applying Hooke's law and definition of work, we define the elastic potential energy ( $U_{g}$ ), measured in joules, by the following formula:

$U_{g} = \frac{1}{2}\cdot k\cdot x^{2}$ (1)

Where:

$k$ - Spring constant, measured in newtons per meter.

$x$ - Deformation of the spring, measured in meters.

If we know that $k = 865\,\frac{N}{m}$ and $x = 0.065\,m$ , then the elastic potential energy is:

$U_{g} = \frac{1}{2}\cdot \left(865\,\frac{N}{m} \right) \cdot (0.065\,m)$

$U_{g} = 28.113\,J$

The initial elastic potential energy of the block-spring system is 28.113 joules.

b) According to the Principle of Energy Conservation, the initial elastic potential energy of the block-spring system becomes into translational kinetic energy, that is:

$U_{g} = \frac{1}{2}\cdot m\cdot v^{2}$ (2)

Where:

$m$ - Mass, measured in kilograms.

$v$ - Final speed, measured in meters per second.

Then, the final speed is cleared:

$v = \sqrt{\frac{2\cdot U_{g}}{m} }$

If we know that $U_{g} = 28.113\,J$ and $m = 3.20\,kg$ , then the final speed of the block is:

$v = \sqrt{\frac{2\cdot (28.113\,J)}{3.20\,kg} }$

$v \approx 4.192\,\frac{m}{s}$

The final speed of the block is approximately 4.192 meters per second.

3 0

3 years ago

WHAT IS THE NET FORCE REQUIRED TO GIVE AN AUTOMBILE OF MASS 1600KG AN ACCELERATION OF 4.5M/S2?

Fed [463]

Answer:

The required net force has a magnitude of 7200 N

Explanation:

Use Newton's 2nd Law to obtain the answer:

$F_{net}= m\,*\,a\\F_{net}=1600 \,*\,4.5 \,= 7200\,\,N$

4 0

4 years ago

Determine the moment of inertia Ixx of the mallet about the x-axis. The density of the wooden handle is 860 kg/m3 and that of th

Yuki888 [10]

Complete Question

Diagram for this shown on the first uploaded image

Answer:

The moment of inertia Ixx of the mallet about the x-axis is $I{xx}= 0.119 kg \cdot m^2$

Explanation:

From the question we are told that

The density `of wooden handle is $\rho_w = 860 kg/m^3$

The density `of soft-metal head is $\rho_s =8000kg/m^3$

Generally the mass of the wooden can be mathematically obtained with this formula

$m_w = \rho_w A_w l_w$

Where $A_w$ is mass of wooden handle which is mathematically obtain with the formula

$A_w = \frac{\pi}{4} d^2_w$

Where $d_w$ is the diameter of the wooden handle which from the diagram is

$27mm = \frac{27}{1000} = 0.027m$

So $A_w = \frac{\pi}{4} * 0.027^2$

$l_w$ is the length of the the wooden handle which is given in the diagram as $l_w = 315mm = \frac{315}{1000} = 0.315m$

Substituting these value into the formula for mass

$m_w = 860 * (\frac{\pi}{4} * 0.027^2 ) *0.315$

$= 0.155kg$

Generally the mass of the soft-metal head can be mathematically obtained with this formula

$m_s = \rho_s A_s l_s$

Where $A_s$ is mass of soft-metal head which is mathematically obtain with the formula

$A_s = \frac{\pi}{4} d^2_s$

Where $d_s$ is the diameter of the soft-metal head which from the diagram is

$36mm = \frac{36}{1000} = 0.036m$

So $A_s = \frac{\pi}{4} * 0.036^2$

$l_s$ is the length of the the soft-metal head which is given in the diagram

as $l_s = 90mm = \frac{90}{1000} = 0.090m$

Substituting these value into the formula for mass

$m_s = 8000 * (\frac{\pi}{4} * 0.036^2 ) *0.090$

$=0.733kg$

Generally the mass moment of inertia about x-axis for the wooden handle is

$(I_{xx})_w = [\frac{1}{3}m_w + l_w^2 ]$

Substituting values

$(I_{xx})_w = [\frac{1}{3}*0.155 + 0.315^2 ]$

$=5.12*10^{-3}kg \cdot m^2$

Generally the mass moment of inertia about x-axis for the soft-metal head is

$(I_{xx})_s = [\frac{1}{12}m_s l_s ^2 + b^2]$

Where b is the distance from the centroid to the axis of the head which is mathematically given as

$b=l_w +\frac{d_s}{2}$

Substituting values

$b = 0.315 + \frac{0.036}{2}$

$= 0.336m$

Now substituting values into the formula for mass moment of inertia about x-axis for soft-metal head

$(I_{xx})_s = [\frac{1}{12} *0.733* 0.090^2 + 0.336^2]$

$=0.113 kg \cdot m^2$

Generally the mass moment of inertia about x-axis is mathematically represented as

$I_{xx} = (I_{xx})_w + (I_{xx})_s$

$= [\frac{1}{3}m_w + l_w^2 ] + [\frac{1}{12}m_s l_s ^2 + b^2]$

Substituting values

$I_{xx} = 5.12*10^{-3} +0.113$

$I{xx}= 0.119 kg \cdot m^2$

8 0

3 years ago