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Softa [21]
1 year ago
15

a solid object is found to weigh 4.784.78n in air. when it is weighed while fully immersed in water, its apparent weight is 2.48

2.48n. what is the density of the object?
Physics
1 answer:
ololo11 [35]1 year ago
3 0

A solid object is found to weigh 4.784.78n in air. when it is weighed while fully immersed in water, its apparent weight is 2.482.48n. 983 is the density of the object.

The substance's density is defined as its mass per unit of volume (volumetric mass density or specific mass). Although the Latin letter D may also be used, the symbol for density that is most usually used is (the lower case Greek letter rho). where V is the volume, is the density, and m is the mass. Weight per unit volume is a common informal definition of density, however this is incorrect scientifically; the actual term is specific weight. The US oil and gas industry serves as one illustration of this. A pure substance's mass concentration in numbers is equal to its density. To make density comparisons between different systems of units easier, it is occasionally replaced by the dimensionless quantity "relative density" or "specific gravity," which is the ratio of the density of the material to that of a standard material, usually water. If a substance's relative density to water is less than one, it will float in it. Temperature and pressure have an impact on a substance's density. This variation is frequently not very noticeable for solids and liquids, but it is very noticeable for gases. As pressure is applied, an object's density rises, which reduces the object's volume. With a few rare exceptions, as temperature increases, a substance's density decreases as its volume grows.

To know more about density please refer: brainly.com/question/15164682

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Hippocampus

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Interactive LearningWare 4.1 reviews the approach taken in problems such as this one. A 1800-kg car is traveling with a speed of
Lubov Fominskaja [6]

Answer:

F= 4788 N

Explanation:

Because the car moves with uniformly accelerated movement we apply the following formula:

vf²=v₀²+2*a*d Formula (1)

Where:  

d:displacement in meters (m)  

v₀: initial speed in m/s    

vf: final speed in m/s  

a: acceleration in m/s²

Data

d=36.9 m

v₀=14.0 m/s m/s    

vf= 0  

Calculating of the acceleration of the car

We replace dta in the formula (1)

vf²=v₀²+2*a*d

(0)²=(14)²+2*a*(36.9)

-(14)²= (73.8) *a

a= - (196) /  (73.8)

a= - 2.66 m/s²

Newton's second law of the car in direction  horizontal (x):

∑Fx = m*ax Formula (2)

∑F : algebraic sum of the forces in direction x-axis (N)

m : mass (kg)

a : acceleration  (m/s²)

Data

m=1800 Fkg

a= - 2.66 m/s²

Magnitude of the horizontal net force (F) that is required to bring the car to a halt in a distance of 36.9 m :

We replace data in the formula (2)

-F= (1800 kg) * ( -2.66 m/s² )

F= 4788 N

6 0
3 years ago
Six artificial satellites complete one circular orbit around a space station in the same amount of time. Each satellite has mass
oee [108]

Answer:

The ranking of the net force acting on different satellite from largest to smallest is {F_E} > {F_F} > {F_A} = {F_B} = {F_D} > {F_C}

Explanation:

In order to get a good understanding of this solution we need to understand that the main concepts used to solve this problem are centripetal force and velocity of satellite.

Initially, use the expression of the velocity of satellite and find out its dependence on the radius of orbit. Use the dependency in the centripetal force expression.

Finally, we find out the velocity of the six satellites and use that expression to find out the force experienced by the satellite. Find out the force in terms of mass (m) and radius of orbit (L) and at last compare the values of force experienced by six satellites.

Fundamentals

The centripetal force is necessary for the satellite to remain in an orbit. The centripetal force is the force that is directed towards the center of the curvature of the curved path. When a body moves in a circular path then the centripetal force acts on the body.

The expression of the centripetal force experienced by the satellite is given as follows:

                    {F_{\rm{c}}} = \frac{{m{v^2}}}{L}

Here, m is the mass of satellite, v is the velocity, and L is the radius of orbit.

The velocity of the satellite with which the satellite is orbiting in circular path is given as follows:

                        v = \frac{{2\pi L}}{T}

Here, T is the time taken by the satellite.

The velocity of the satellite with which the satellite is orbiting in circular path is given as follows;

                    v = \frac{{2\pi L}}{T}

Since, all the satellites complete the circular orbit in the same amount of time. The factor of   \frac{{2\pi }}{T}   is not affected the velocity value for the six satellites. Therefore, we can write the expression of v given as follows:

Substitute  v = \frac{{2\pi L}}{T} in the force expression {F_{\rm{c}}} = \frac{{m{v^2}}}{L}   as follows:

                              \begin{array}{c}\\{F_c} = \frac{{m{{\left( {\frac{{2\pi L}}{T}} \right)}^2}}}{L}\\\\ = \frac{{4{\pi ^2}}}{{{T^2}}}mL\\\end{array}

Since, all the satellites complete the circular orbit in the same amount of time. The factor of \frac{{4{\pi ^2}}}{{{T^2}}}  not affect the force value for six satellites.Therefore, we can write the expression of {F_c}  given as follows:

        {F_c} = kmL

Here, k refers to constant value and equal to  \frac{{4{\pi ^2}}}{{{T^2}}}

    {F_A} = k{m_A}{L_A}

Substitute 200 kg for {m_A}   and 5000 m for LA in the expression                                  {F_A} = k{m_A}{L_A}

\begin{array}{c}\\{F_A} = k\left( {200{\rm{ kg}}} \right)\left( {5000{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}

The force acting on satellite B from their rocket is given as follows:{F_B} = k{m_B}{L_B}

Substitute 400 kg for {m_B} and 2500 m for in the expression {F_B} = k{m_B}{L_B}

\begin{array}{c}\\{F_B} = k\left( {400{\rm{ kg}}} \right)\left( {2500{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}

The force acting on satellite C from their rocket is given as follows:{F_C} = k{m_C}{L_C}

Substitute 100 kg for {m_C}and 2500 m for in the above expression  {F_C} = k{m_C}{L_C}

\begin{array}{c}\\{F_C} = k\left( {100{\rm{ kg}}} \right)\left( {2500{\rm{ m}}} \right)\\\\ = 0.25 \times {10^6}k{\rm{ N}}\\\end{array}

The force acting on satellite D from their rocket is given as follows:{F_D} = k{m_D}{L_D}

Substitute 100 kg for {m_D} and 10000 m for {L_D} in the expression{F_D} = k{m_D}{L_D}

\begin{array}{c}\\{F_D} = k\left( {100{\rm{ kg}}} \right)\left( {10000{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}

The force acting on satellite E from their rocket is given as follows:{F_E} = k{m_E}{L_E}

Substitute 800 kg for {m_E}  and 5000 m for  {L_E} in the expression {F_E} = k{m_E}{L_E}

\begin{array}{c}\\{F_E} = k\left( {800{\rm{ kg}}} \right)\left( {5000{\rm{ m}}} \right)\\\\ = 4.0 \times {10^6}k{\rm{ N}}\\\end{array}

The force acting on satellite F from their rocket is given as follows:{F_F} = k{m_F}{L_F}

Substitute 300 kg for {m_F} and 7500 m for {L_F} in the expression {F_F} = k{m_F}{L_F}

\begin{array}{c}\\{F_F} = k\left( {300{\rm{ kg}}} \right)\left( {7500{\rm{ m}}} \right)\\\\ = 2.25 \times {10^6}k{\rm{ N}}\\\end{array}

The value of forces obtained for the six-different satellite are as follows.

\begin{array}{l}\\{F_A} = {10^6}k{\rm{ N}}\\\\{F_B} = {10^6}k{\rm{ N}}\\\\{F_C} = 0.25 \times {10^6}k{\rm{ N}}\\\\{F_D} = {10^6}k{\rm{ N}}\\\\{F_E} = 4.0 \times {10^6}k{\rm{ N}}\\\\{F_F} = 2.25 \times {10^6}k{\rm{ N}}\\\end{array}

     The ranking of the net force acting on different satellite from largest to smallest is {F_E} > {F_F} > {F_A} = {F_B} = {F_D} > {F_C}

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