How is acceleration calculated?

A 1.94-m-diameter lead sphere has a mass of 5681 kg. A dust particle rests on the surface. What is the ratio of the gravitationa

scoundrel [369]

To develop this problem we will apply Newton's laws regarding gravitational forces, both in space and on earth. From finding this relationship, leaving the variable of the dust mass open, we will find the relationship of the forces between the two surfaces. Our values are,

$\text{Diameter of the lead sphere} = D=1.940m$

$\text{Mass of the lead sphere} = m_1 = 5681kg$

$\text{Mass of the dust particle} = m_2$

Distance between the center of lead sphere to dust particle

$r = \frac{D}{2}$

$r = \frac{1.940m}{2}$

$r = 0.97m$

Gravitational force of the sphere on the dust particle:

$F = \frac{Gm_1m_2}{r^2}$

$F = \frac{(6.67*10^{-11}N\cdot m^2/kg^2)(5681kg)(m_2)}{(0.97m)^2}$

$F = (4.027*10^{-7} N/kg)m_2$

Weight of the dust particle

$W = m_2 g$

$W = m_2 (9.8m/s^2)$

Ratio of F and W:

$\frac{F}{W} = \frac{(4.07*10^{-7}N/kg) m_2)}{m_2(9.8m/s^2)}$

$\frac{F}{W} = 4.153*10^{-8}$

Therefore the ratio is $4.153*10^{-8}$

3 0

3 years ago

A 0.25-kg ball attached to a string is rotating in a horizontal circle of radius 0.5 m. If the ball revolves twice every second,

Aloiza [94]

Answer:

19.74 N

Explanation:

mass of ball (m) = 0.25 kg

radius (r) = 0.5 m

time (t) = 2 revolutions per seconds = 1/2 = 0.5 second per revolution

find the tension in the string

tension (T) = $\frac{mv^{2} }{r}$

where velocity (v) = $\frac{2πr}{t}$

tension now becomes (T) = $\frac{m}{r} x (\frac{2πr}{t})^{2}$

tension (T) = $\frac{4π^{2}rm }{t^{2} }$

now substituting the values of mass (m), time (t) and radius (r) into the equation above we have

tension (T) = $\frac{4π^{2}x0.5x0.25}{0.5^{2} }$

tension (T) = $2π^{2}$ = $2x3.142^{2}$ = 19.74 N

4 0

3 years ago

State newton's universal law of gravitation

Lorico [155]

Every particle of mass is attracted to every other particle of mass. The magnitude of the force between two objects is proportional to the product of their masses, and inversely proportional to the square of the distance between them. The direction of the force is along the line between their centers.

(NOTE: Newton's 3rd law of motion tells us that gravitational forces always come in pairs. Between two objects, there are two forces ... one in each direction. Their strengths are equal ... Your weight on Earth is exactly equal to the Earth's weight on YOU.)

4 0

3 years ago

Since the density of air decreases with an increase in temperature, but the bulk modulus B is nearly independent of temperature,

Usimov [2.4K]

To develop this problem it is necessary to apply the concept related to the speed of sound waves in fluids.

By definition we know that the speed would be given by

$v=\sqrt{\frac{\beta}{\rho}}$

$\beta =$ Bulk modulus

$\rho =$ Density of air

From the expression shown above we can realize that the speed of sound is <em>inversely proportional</em> to the fluid in which it is found, in this case the air. When the density increases, the speed of sound decreases and vice versa.

According to the statement then, if the density of the air decreases due to an increase in temperature, we can conclude that the speed of sound increases when the temperature increases. <u>They are directly proportional.</u>

6 0

3 years ago

I have 20 assignments let on my physical assignment will pay after assignment is complete per assignment

Mama L [17]

In the name of love~

3 0

3 years ago