find not true when the water freezes...

Answer:

Gravity controls the movement of the planets around the sun, holds together stars grouped in galaxies, and galaxies grouped in clusters.  The Universal Law of Gravitation depends on two things.  First it depends on mass of each object  and the second factor is the distance between two objects.  If the mass of one object is Larger, the gravitational pull towards it will be larger  and the smaller distance, the larger the gravitational pull will be between the objects. Therefore the Larger planets have more moon and the inner planets have less.

Explanation:

Answer:

12 m

Explanation:

From the question given above, the following data were obtained:

Mass (m) of bag = 25 kg

Potential energy (PE) = 2940 J

Height (h) =?

Objects carried to a particular height will always experience an acceleration due to gravity of 9.8 m/s².

With the above in mind, we can obtain the height to which the load is lifted to as shown below:

Mass (m) of bag = 25 kg

Potential energy (PE) = 2940 J

Acceleration due to gravity (g) = 9.8 m/s².

Height (h) =?

PE = mgh

2940 = 25 × 9.8 × h

2940 = 245 × h

Divide both side by 245

h = 2940 / 245

h = 12 m

Therefore, the worker lifts the load to a height of 12 m.

Answer:

The new force is 2/3 of the original force

Explanation:

Coulomb's Law

The electrical force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

[tex]\displaystyle F=k\frac{q_1q_2}{d^2}[/tex]

Where:

[tex]k=9\cdot 10^9\ N.m^2/c^2[/tex]

q1, q2 = the objects' charge

d= The distance between the objects

Suppose the first charge is doubled (2q1) and the second charge is one-third of the original charge (q2/3). Now the force is:

[tex]\displaystyle F'=k\frac{2q_1*q_2/3}{d^2}[/tex]

Factoring out 2/3:

[tex]\displaystyle F'=\frac{2}{3}k\frac{q_1*q_2}{d^2}[/tex]

Substituting the original force:

[tex]F'=\frac{2}{3}F[/tex]

The new force is 2/3 of the original force

Answer: 0.81N

Explanation: Fwx = .165x 9.81 Silicon 30 = 0.81N

Answer:

3rd law beacuse there a flies

Answer:

elements

not really an explanation

Answer:

2s

Explanation:

Given parameters:

Distance  = 30m

Speed  = 15m/s

Unknown:

Time before Laurence caught it = ?

Solution:

To solve this problem;

    Speed  = [tex]\frac{disance }{time}[/tex]  

 Time taken  = [tex]\frac{distance }{speed }[/tex]   = [tex]\frac{30}{15}[/tex]  = 2s

The time it takes is 2s

Given:

We will solve:

Speed  = [tex]\frac{distance}{time}[/tex]

Time taken  = [tex]\frac{distance}{time}[/tex] = [tex]\frac{30m}{15m/s}[/tex] = 2s

The time it takes is 2s

Answer:

The torque necessary to bring the wheel to rest in 40 seconds is 10.4 N·m

Explanation:

The question is with regards to rotational motion

The rotary motion parameters are;

The mass of the wheel = 50 kg

The radius of the wheel = 0.4 m

The rate of rotation of the wheel = 480 rpm

The time in which the wheel is to be brought to rest = 40 s

The rotational rate of the wheel in rotation per second is given as follows;

480 r.p.m = 480 r.p.m × 1 minute/(60 seconds) = 8 revolution/second

1 revolution = 2·π radians

Therefore, we have the angular velocity, ω, given as follows;

ω = 2·π × 8 revolutions/second ≈ 50.3 rad/s

The angular acceleration, α, is given as follows;

[tex]\alpha = \dfrac{\Delta \omega}{\Delta t} = \dfrac{\omega _2 - \omega_1}{t_2 - t_1}[/tex]

Whereby the wheel is brought to rest from its initially constant rotational motion in 40 seconds, we have;

ω₁ ≈ 50.3 rad/s, ω₂ = 0 rad/s, and t₂ - t₁ = 40 seconds

Plugging in the values for the variables of the equation for the angular acceleration, "α", we get;

[tex]\alpha = \dfrac{0 - 50.3 \ rad/s}{40 \ s} \approx 1.3 \ rad/s^2[/tex]

The torque on the wheel, τ, is given as follows;

τ = m·r²·α

Where;

m = The mass of the object = 50 kg

r = The radius of the wheel = 0.4 m

α = The acceleration of the wheel ≈ 1.3 rad/s²

Therefore;

τ = 50 kg × (0.4 m)² × 1.3 rad/s² ≈ 10.4 N·m

The torque necessary to bring the wheel to rest in 40 seconds = τ ≈ 10.4 N·m.

Answer:

-10.048 N m

Explanation:

Answer: B. Mechanical Waves

Explanation:

Answer:

option 2

Explanation:

you are very welcom