please help
What variables have the most effect on the strength of the gravitational attraction between two
massive objects?

Answer:

The centripetal acceleration of the girl is 2.468 m/s²

Explanation:

Given;

number of turns, = ¹/₄ Revolution

distance traveled by the girl, d = 25 m

time of motion, t = 5.0 s

The linear speed of the of the girl is calculated as;

[tex]v = \omega \ r\\\\v =(\frac{1}{4}rev \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1}{5 \ s} ) (25 \ m)\\\\v = (0.3142 \ \frac{rad}{s} )(25 \ m)\\\\v = 7.855 \ m/s[/tex]

The centripetal acceleration of the girl is calculated as;

[tex]a_c = \frac{v^2}{r} \\\\a_c = \frac{(7.855)^2}{25} \\\\a_c = 2.468 \ m/s^2[/tex]

Therefore, the centripetal acceleration of the girl is 2.468 m/s²

Answer:

What am I suppose to solve

Explanation:

Answer:

58.469 kg.m/s are moving in the xy plane

Answer:

7.22 × 10²⁹ kg

Explanation:

For the material to be in place, the gravitational force on the material must equal the centripetal force on the material.

So, F = gravitational force = GMm/R² where M = mass of neutron star, m = mass of object and R = radius of neutron star = 17 km

The centripetal force F' = mRω² where R = radius of neutron star and ω  = angular speed of neutron star

So, since F = F'

GMm/R² = mRω²

GM = R³ω²

M = R³ω²/G

Since ω = 500 rev/s = 500 × 2π rad/s = 1000π rad/s = 3141.6 rad/s = 3.142 × 10³ rad/s and r = 17 km = 17 × 10³ m and G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²

Substituting the values of the variables into M, we have

M = R³ω²/G

M = (17 × 10³ m)³(3.142 × 10³ rad/s)²/6.67 × 10⁻¹¹ Nm²/kg²

M = 4913 × 10⁹ m³ × 9.872 × 10⁶ rad²/s²/6.67 × 10⁻¹¹ Nm²/kg²

M = 48,501.942 × 10¹⁵ m³rad²/s² ÷ 6.67 × 10⁻¹¹ Nm²/kg²

M = 7217.66 × 10²⁶ kg

M = 7.21766 × 10²⁹ kg

M ≅ 7.22 × 10²⁹ kg

OK please your picture not perfect please try again

Answer:

W = 181.26 J

Explanation:

Given that,

The force acting on the cart, F = 20 N

It is at an angle of 25 degrees above the horizontal to  move a crate a distance of 10m across the floor.

We need to find work done by the student. The work done by the student is given by :

[tex]W=Fd\cos\theta\\\\W=20\times 10\times \cos25\\W=181.26\ J[/tex]

So, the required work done is 181.26 J.

Answer:

Energy stored in an object due to its position is Potential Energy. · Energy that a moving object has due to its motion is Kinetic Energy.

Explanation:

Answer:

Ff = F₀ *(576/196)

Explanation:

       [tex]F_{o} = \frac{k*q^{2} }{d^{2}} (1)[/tex]

       [tex]F_{f} = \frac{k*(q/14)^{2} }{(d/24)^{2}} = \frac{k*q^{2}}{d^{2}} * \frac{(24)^{2}}{(14)^{2}} (2)[/tex]

       ⇒[tex]F_{f} =\frac{k*q^{2}}{d^{2}} * \frac{(24)^{2}}{(14)^{2}} = F_{o} * \frac{576}{196} (3)[/tex]

Answer:

no kinetic energy

hope this helps! :-D

Explanation:

the monk is not moving

Answer: Terraces on moderate to steep irregular slopes pro- ... sure of infertile or toxic soils. ... Following are terms used to define distances mea- ... the soil in the entire field will be disturbed to con-.

Explanation:

Answer:

h1/h2 = [tex]\frac{2R^2}{3R^2 + h^2}[/tex]

Explanation:

Using two rolls of tissue paper : One roll dropped normally while the other drops as some holds onto a sheet of the toilet paper ( I.e. the tissue paper drops rotating about its axis )

Determine the ratio of heights  h1/h2

mass of tissues = same

radius of tissues = same

h1 = height of tissue 1

h2 = height of tissue 2

For the first tissue ( Tissue that dropped manually )

potential energy = kinetic energy

mgh = 1/2 mv^2  

therefore the final velocity ( v^2 ) = 2gH  ----- ( 1 )

second tissue ( Tissue that dropped while rotating )

gh = [tex]\frac{v^2}{u}[/tex] ( 3 + [tex]\frac{u^2}{R^2}[/tex] ) ------ ( 2 )

To determine the ratio of heights we will equate equations 1 and 2

hence :

gh = [tex]\frac{2gH}{u}[/tex] ( 3 + [tex]\frac{u^2}{R^2}[/tex] )

∴ h1/h2 = [tex]\frac{2R^2}{3R^2 + h^2}[/tex]

Answer:

a) ΔV = 25.59 V, b)  ΔV = 25.59 V,  c)  v = 7 10⁴ m / s,  v/c= 2.33 10⁻⁴ ,

v/c% = 2.33 10⁻²

Explanation:

a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy

starting point. Where the electrons come out

          Em₀ = U = e DV

final point. Where they hit the target

          Em_f = K = ½ m v2

energy is conserved

          Em₀ = Em_f

         e ΔV = ½ m v²

         ΔV = [tex]\frac{1}{2}[/tex] mv²/e     (1)

If the speed of light is c and this is 100% then 1% is

         v = 1% c = c / 100

         v = 3 10⁸/100 = 3 10⁶6 m/ s

let's calculate

         ΔV = [tex]\frac{1}{2} \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }[/tex]

         ΔV = 25.59 V

b) Ask for the potential difference for protons with the same kinetic energy as electrons

             [tex]K_e = K_p[/tex]

              K_p = ½ m v_e²

              K_p = [tex]\frac{1}{2}[/tex]  9.1 10⁻³¹ (3 10⁶)²

              K_p = 40.95 10⁻¹⁹ J

we substitute in equation 1

              ΔV = Kp / M

              ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹

              ΔV = 25.59 V

notice that these protons go much slower than electrons because their mass is greater

c) The speed of the protons is

             e ΔV = ½ M v²

             v² = 2 e ΔV / M

             v² = [tex]\frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }[/tex]

              v² = 49,035 10⁸

               v = 7 10⁴ m / s

Relation

        v/c = [tex]\frac{7 \ 10^4 }{ 3 \ 10^8}[/tex]

        v/c= 2.33 10⁻⁴

The coaster starts at rest, so the kinetic energy (KE) at point A is 0. It is situated 33 m above ground, so its potential energy (PE) at A is

mgh = (3000 kg) (9.80 m/s²) (33 m) = 970,200 J

The total energy is the same, 970,200 J.

Assuming no energy is lost to friction or sound etc, energy is conserved throughout the coaster's motion, so the total energy should be the same at each point.

At point B, the coaster has dropped to a height of 10 m, so it has PE

mgh = (3000 kg) (9.80 m/s²) (10 m) = 294,000 J

which means it must have KE

970,200 J = KE + 294,000 J   →   KE = 676,200 J

which gives the coast a speed v at point B of

1/2 mv ² = 1/2 (3000 kg) v ² = 676,200 J   →   v ≈ 21.2 m/s

At point C, the coaster has a speed of 16.0 m/s, so it has KE

1/2 mv ² = 1/2 (3000 kg) (16.0 m/s)² = 384,000 J

and hence PE

970,200 J = 384,000 J + PE   →   PE = 586,200 J

This lets us determine the height h at C:

mgh = (3000 kg) (9.80 m/s²) h = 586,200 J   →   h ≈ 19.939 m

which means the loop has diameter h - 10 m ≈ 9.94 m.

At point D, the coaster is 15 m above the ground so its PE at D is

mgh = (3000 kg) (9.80 m/s²) (15 m) = 441,000 J

and so its KE is

970,200 J = KE + 441,000 J   →   KE = 529,200 J

and hence has speed v at D

1/2 mv ² = 1/2 (3000 kg) v ² = 529,200 J   →   v ≈ 18.9 m/s

Answer:

The mass of an object is a measure of the object's inertial property, or the amount of matter it contains. The weight of an object is a measure of the force exerted on the object by gravity, or the force needed to support it. The pull of gravity on the earth gives an object a downward acceleration of about 9.8 m/s2.

Answer:

Explanation:

The mass is essentially "how much stuff" is in an object. ... Weight: There is a gravitational interaction between objects that have mass. If you consider an object interacting with the Earth, this force is called the weight. The unit for weight is the Newton (same as for any other force).

Answer:

zero

Explanation:

For a solid conducting sphere, charges are present on the surface of the sphere due to a phenomenon known as electrostatic sheilding. This affects the charge present in the body and makes it zero. However, the electrostatic potential appears to be equal to the whole present point that shows on the surface. The surface of a spherical conducting solid sphere is known as an equipotential surface. Thus, the potential difference between the two opposite points on the surface of the sphere will also be zero.

Answer:

:To find the Voltage, ( V ) [ V = I x R ] V (volts) = I (amps) x R (Ω)

To find the Current, ( I ) [ I = V ÷ R ] I (amps) = V (volts) ÷ R (Ω)

To find the Resistance, ( R ) [ R = V ÷ I ] R (Ω) = V (volts) ÷ I (amps)

To find the Power (P) [ P = V x I ] P (watts) = V (volts) x I (amps)

Answer:

v = 10.46 m/s

x =  30.134 m from house

Explanation:

given data

speed = 3.05 m/s

time = 7 s

height = 40.8 mm

solution

we get here first time required to fall that is t

t = [tex]\sqrt{\frac{2\times 40.8}{9.8}}[/tex]       ..................1

t = 2.88 s

now we take here initial speed that is v

so v × t = 3.05 × ( t+ 7)

v = [tex]\frac{3.05\times (2.88 + 7)}{2.88}[/tex]

v = 10.46 m/s

and

when she catch the bagel henrietta was at

x = 3.05 × ( 2.88 + 7)

x =  30.134 m from house

Answer: IT B TRUST ME MANN OK

Explanation: OK BYE TRUST

Answer:

A. estrogen

Explanation:

This is released in the female reproductive organ.

Answer: ITS 1 TRUST ME MAN BYE K

Explanation: OK BYE TRUST YEAH

Answer:

No, it is very unlikely to perform a controlled experiment, because you need to observe the amount or anything from something. Consider someone on the busy street of a New York neighborhood asking random people that pass by how many pets they have, then taking this data and using it to decide if there should be more pet food stores in that area.

Answer:

using W=1/2kW2

k=22N/m w=2.9

w=1/2×22×2.9×2.9

w=92.51Joules

Approximately 93J answer is C

Answer:

wat

Explanation:

Answer:

Um its the vbuck card on the 3 thrid level

Explanation:

Bc its a vbuck card you know sihdg;aig

Answer:

a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

Explanation:

Given the data in the question;

as the equation of standing wave on a string is fixed at both ends

y = 2AsinKx cosωt

but k = 2π/λ and ω = 2πf

λ = 4 × 0.150 = 0.6 m

and f =  v/λ = 260 / 0.6 = 433.33 Hz

ω = 2πf = 2π × 433.33 = 2722.69

given that A = 2.20 mm = 2.2×10⁻³

so [tex]V_{max1}[/tex] = A × ω

[tex]V_{max1}[/tex] = 2.2×10⁻³ × 2722.69 m/s

[tex]V_{max1}[/tex] =  5.9899 m/s

therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b)

A' = 2AsinKx

= 2.20sin( 2π/0.6 ( 0.075) rad )

= 2.20 sin(  0.7853 rad ) mm

= 2.20 × 0.706825 mm

A' = 1.555 mm = 1.555×10⁻³

so

[tex]V_{max2}[/tex] = A' × ω

[tex]V_{max2}[/tex] = 1.555×10⁻³ × 2722.69

[tex]V_{max2}[/tex] = 4.2338 m/s

Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s