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4 years ago

What can you infer about the density of two objects if they object remains suspended in the liquid?

Physics

2 answers:

ZanzabumX [31]4 years ago

5 0

Answer:

Equal Densities

Explanation:

if the density of the object was greater than that of the liquid, it would sink to the bottom. if the density od the object was lesser than the liquid, it would float :)

4vir4ik [10]4 years ago

5 0

Answer:

The densities of the objects are equal

Explanation:

higher density objects sink to the bottom while lower density objects rise to the top

since the object remains suspended in the liquid, it has neither higher or lower density than the water because it is neither sinking nor floating

therefore, the object has equal densities with the liquid

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There was an airplane crash, every single person on board died, but yet two people survived. How is this possible?

serg [7]

If everyone died then the too people were not on the plane they were near by when the plane crashed and were not onbard.

8 0

3 years ago

Read 2 more answers

A car travels 70 km in two hours 104 km in the next 1.5 hours and then 79 km in two hours before reaching its destination what w

stiks02 [169]

Average speed = (total distance) / (total time)

Total distance = (70km + 104km + 79km) = 253 km

Total time = (2hr + 1.5hr + 2hr) = 5.5 hrs

Average speed = (253 km) / (5.5 hrs)

<em>Average speed = 46 km/hr</em>

4 0

4 years ago

In recent years, astronomers have found planets orbiting nearby stars that are quite different from planets in our solar system.

BartSMP [9]

Answer:

The value of g in Kepler-12b is $3.85 m/s^{2}$

Explanation:

To determine the value of g at Kepler-12b it is necessary to combine the equation of the weight and the equation for the Universal law of gravity:

$W = m.g$ (1)

Where m is the mass and g is the value of the gravity

$F = G \frac{M.m}{R^{2}}$ (2)

Equation (1) and equation (2) will be equal, since the weight is a force acting on the object as a consequence of gravity:

$m.g = G \frac{M.m}{R^{2}}$ (3)

Then g will be isolated from equation 3:

$g = G \frac{M.m}{m.R^{2}}$

$g = \frac{G.M}{R^{2}}$ (4)

The radius of Jupiter has a value of 69911000 meters, so its diameter can be determined by:

$d = 2R$ (5)

Where d and R are the diameter and radius of Jupiter

$d_{jupiter} = 2(69911000 m)$

$d_{jupiter} = 139822000 m$

Procedure for finding the radius of Kepler-12b:

For the case of Kepler-12b it has a diameter that is 1.7 times that of Jupiter

$d_{Kepler-12b} = 1.7(d_{jupiter})$

$d_{Kepler-12b} = 1.7(139822000 m)$

$d_{Kepler-12b} = 237697400 m$

By means of equation 5 the radius of Kepler-12b can be known:

$R_{Kepler-12b} = \frac{d}{2}$

$R_{Kepler-12b} = \frac{237697400 m}{2}$

$R_{Kepler-12b} = 118848700 m$

Procedure for finding the mass of Kepler-12b:

The mass of Jupiter has a value of $1.898x10^{27}$ kilograms

For the case of Kepler-12b it has a mass that is 0.43 times that of jupiter

$m_{Kepler-12b} = 0.43(m_{jupiter})$

$m_{Kepler-12b} = 0.43(1.898x10^{27} Kg)$

$m_{Kepler-12b} = 8.161x10^{26} Kg$

The value of g in Kepler-12b can be found by replacing its radius and mass in equation 4:

$g = \frac{(6.67x10^{-11} N.m^{2}/Kg^{2})(8.161x10^{26} Kg)}{(118848700 m)^{2}}$

$g = 3.85 m/s^{2}$

Hence, in Kepler-12b the gravity has a value of $3.85 m/s^{2}$

8 0

3 years ago

Assume that, when we walk, in addition to a fluctuating vertical force, we exert a periodic lateral force of amplitude 25 NN at

dexar [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told

The amplitude of the lateral force is $F = 25 \ N$

The frequency is $f = 1 \ Hz$

The mass of the bridge per unit length is $\mu = 2000 \ kg /m$

The length of the central span is $d = 144 m$

The oscillation amplitude of the section considered at the time considered is $A = 75 \ mm = 0.075 \ m$

The time taken for the undriven oscillation to decay to $\frac{1}{e}$ of its original value is t = 6T

Generally the mass of the section considered is mathematically represented as

$m = \mu * d$

=> $m = 2000 * 144$

=> $m = 288000 \ kg$

Generally the oscillation amplitude of the section after a time period t is mathematically represented as

$A(t) = A_o e^{-\frac{bt}{2m} }$

Here b is the damping constant and the $A_o$ is the amplitude of the section when it was undriven

So from the question

$\frac{A_o}{e} = A_o e^{-\frac{b6T}{2m} }$

=> $\frac{1}{e} =e^{-\frac{b6T}{2m} }$

=> $e^{-1} =e^{-\frac{b6T}{2m} }$

=> $-\frac{3T b}{m} = -1$

=> $b = \frac{m}{3T}$

Generally the amplitude of the section considered is mathematically represented as

$A = \frac{n * F }{ b * 2 \pi }$

=> $A = \frac{n * F }{ \frac{m}{3T} * 2 \pi }$

=> $n = A * \frac{m}{3} * \frac{2\pi}{25}$

=> $n = 0.075 * \frac{288000}{3} * \frac{2* 3.142 }{25}$

=> $n = 1810 \ people$

3 0

3 years ago

a radio controlled car rolls 10 m south then reverses direction and rolls 8 m north the car traveled a distance of 18 m and has

EastWind [94]

The displacement is 2 m south

Explanation:

Distance and displacement are two different quantities:

Distance is the total length of the path covered by an object during its motion, regardless of the direction. It is a scalar quantity
Displacement is a vector connecting the initial position to the final position of motion of an object. The magnitude of the displacement is the distance in a straight line between the two points

For the car in this problem, the motion is:

10 m south

8 m north

Taking north as positive direction, we can describe the two parts of the motion as

$d_1 = -10 m$