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2 years ago

Why do some objects fall faster than others?

Physics

1 answer:

monitta2 years ago

3 0

Answer:

Weight and Mass !!!!!!

Explanation:

Galileo discovered that objects that are more dense, or have more mass, fall at a faster rate than less dense objects, due to this air resistance. A feather and brick dropped together. Air resistance causes the feather to fall more slowly.

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A man can jump 1.5 m on earth. calculate the approximate

Olegator [25]

Answer:

18 m

Explanation:

G = Gravitational constant

m = Mass of planet = $\rho V$

$\rho$ = Density of planet

V = Volume of planet assuming it is a sphere = $\dfrac{4}{3}\pi r^3$

r = Radius of planet

Acceleration due to gravity on a planet is given by

$g=\dfrac{Gm}{r^2}\\\Rightarrow g=\dfrac{G\rho V}{r^2}\\\Rightarrow g=\dfrac{G\rho \dfrac{4}{3}\pi r^3}{r^2}\\\Rightarrow g=\dfrac{4G\rho\pi r}{3}$

So,

$g\propto \rho r$

Density of other planet = $\rho_p=\dfrac{1}{4}\rho_e$

Radius of other planet = $r_p=\dfrac{1}{3}r_e$

$\dfrac{g_e}{g_p}=\dfrac{\rho_e r_e}{\rho_p r_p}\\\Rightarrow \dfrac{g_e}{g_p}=\dfrac{\rho_e r_e}{\dfrac{1}{4}\rho_e\times \dfrac{1}{3}r_e}\\\Rightarrow \dfrac{g_e}{g_p}=12\\\Rightarrow g_p=\dfrac{g_e}{12}\\\Rightarrow g_p=\dfrac{9.8}{12}$

Since the person is jumping up the acceleration due to gravity will be negative.

From kinematic equations we have

$v^2-u^2=2g_es\\\Rightarrow u^2=v^2-2g_es\\\Rightarrow u^2=0-2\times -9.8\times 1.5\\\Rightarrow u^2=2\times 9.8\times 1.5$

On the other planet

$v^2-u^2=2g_ps\\\Rightarrow s=\dfrac{v^2-u^2}{2g_p}\\\Rightarrow s=\dfrac{0-(2\times 9.8\times 1.5)}{2\times -\dfrac{9.8}{12}}\\\Rightarrow s=18\ \text{m}$

The man can jump a height of 18 m on the other planet.

5 0

2 years ago

Question: A car (assumed to be a Ford Taurus) is traveling around a turn that is banked at 7 degrees. The turn has a radius of 2

AysviL [449]

Answer:

Question: A car (assumed to be a Ford Taurus) is traveling around a turn that is banked at 7 degrees. The turn has a radius of 29 m. The car has a mass of 1300 kg. The coefficient of static friction between the tires and the road is 0.68.

1. What is the "ideal speed?" That is, what speed would allow the car to make the turn without requiring friction?

2. What is the maximum speed the car can go around the turn without sliding?

6 0

2 years ago

A fence 8 ft high (w) runs parallel to a tall building and is 24 ft (d) from it. Find the length (L) of the shortest ladder t

AveGali [126]

Answer:

25.3 ft

Explanation:

The illustration of the problem is shown the attached image.

The length of the ladder can be calculated using the Pythagoras theorem:

$Hypotenuse^2 = opposite^2 + adjacent^2$

The hypotenuse is the length of the ladder.

$Hypotenuse = \sqrt{opposite^2 + adjacent^2}$

$L^2 = BC^2 + (24 + AE)^2$ ........1

Triangle ABC is similar to triangle AEF, hence:

$\frac{BC}{8} = \frac{AE + 24}{AE}$

BC = $\frac{8(AE + 24)}{AE}$ .................2

Substitute 2 into 1

$L^2$ = $(\frac{8(AE + 24)}{AE})^2$ + $(24 + AE)^2$

Let AE = x

$L^2$ = $(\frac{8x + 192}{x})^2 + (24 + x)^2$

= $(8 + \frac{192}{x})^2 + (24 + x)^2$

Minimize L with respect to x.

2 $\frac{dL}{dx}$ = $2(8 + \frac{192}{x})(-\frac{192}{x^2}) + 2(24 + x)$

5 0

2 years ago

A student conducts an experiment to test how the temperature affects the amount of sugar that can dissolve in water. In the expe

ZanzabumX [31]

B. The temperature of the water.

The independent variable is the variable that is changed to affect the dependent variable. In this instance, the temperature of the water is being changed to affect the amount of sugar that dissolves.

6 0

2 years ago

A point charge with a charge q1 = 2.30 μC is held stationary at the origin. A second point charge with a charge q2 = -5.00 μC mo

Alla [95]

Answer:

W = 2.74 J

Explanation:

The work done by the charge on the origin to the moving charge is equal to the difference in the potential energy of the charges.

This is the electrostatic equivalent of the work-energy theorem.

$W = \Delta U = U_2 - U_1$

where the potential energy is defined as follows

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$

Let's first calculate the distance 'r' for both positions.

$r_1 = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2} = \sqrt{(0.170 - 0)^2 + (0 - 0)^2} = 0.170~m\\r_2 = \sqrt{(x_2 - x_0)^2 + (y_2 - y_0)^2} = \sqrt{(0.250 - 0)^2 + (0.250 - 0)^2} = 0.353~m$

Now, we can calculate the potential energies for both positions.

$U_1 = \frac{kq_1q_2}{r_1^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.170)^2} = -3.57~J\\U_2 = \frac{kq_1q_2}{r_2^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.3530)^2} = -0.829~J$

Finally, the total work done on the moving particle can be calculated.

$W = U_2 - U_1 = (-0.829) - (-3.57) = 2.74~J$

4 0

2 years ago

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