what is the force that every mass experts on every other mass called?

Answers

Answer 1

Answer: The forces of gravity

Explanation:  The consequence of this phenomenon is that every mass exerts a so-called "force of mutual attraction" on every other mass. The attractive force that the celestial bodies exert on other masses by virtue of their total mass is called the force of gravity.

Hope this helps


Related Questions

Two streams merge to form a river. One stream has a width of 8.3 m, depth of 3.2 m, and current speed of 2.2 m/s. The other stream is 6.8 m wide and 3.2 m deep, and flows at 2.4 m/s. If the river has width 10.4 m and speed 2.8 m/s, what is its depth?

Answers

Answer:

The depth of the resulting stream is 3.8 meters.

Explanation:

Under the assumption that streams are formed by incompressible fluids, so that volume flow can observed conservation:

[tex]\dot V_{1} + \dot V_{2} = \dot V_{3}[/tex] (1)

All volume flows are measured in cubic meters per second.

Dimensionally speaking, we can determine the depth of the resulting stream ([tex]h_{3}[/tex]), in meters, by expanding (1) in this manner:

[tex]w_{1}\cdot h_{1}\cdot v_{1} + w_{2}\cdot h_{2}\cdot v_{2} = w_{3}\cdot h_{3}\cdot v_{3}[/tex]

[tex]h_{3} = \frac{w_{1}\cdot h_{1}\cdot v_{1}+w_{2}\cdot h_{2}\cdot v_{2}}{w_{3}\cdot v_{3}}[/tex] (2)

[tex]v_{1}, v_{2}[/tex] - Speed of the merging streams, in meters per second.

[tex]h_{1}, h_{2}[/tex] - Depth of the merging streams, in meters.

[tex]w_{1}, w_{2}[/tex] - Width of the merging streams, in meters.

[tex]w_{3}[/tex] - Width of the resulting stream, in meters.

[tex]v_{3}[/tex] - Speed of the resulting stream, in meters per second.

If we know that [tex]w_{1} = 8.3\,m[/tex], [tex]h_{1} = 3.2\,m[/tex], [tex]v_{1} = 2.2\,\frac{m}{s}[/tex], [tex]w_{2} = 6.8\,m[/tex], [tex]h_{2} = 3.2\,m[/tex], [tex]v_{2} = 2.4\,\frac{m}{s}[/tex], [tex]w_{3} = 10.4\,m[/tex] and [tex]v_{3} = 2.8\,\frac{m}{s}[/tex], then the depth of the resulting stream is:

[tex]h_{3} = \frac{(8.3\,m)\cdot (3.2\,m)\cdot \left(2.2\,\frac{m}{s} \right) + (6.8\,m)\cdot (3.2\,m)\cdot \left(2.4\,\frac{m}{s} \right)}{(10.4\,m)\cdot \left(2.8\,\frac{m}{s} \right)}[/tex]

[tex]h_{3} = 3.8\,m[/tex]

The depth of the resulting stream is 3.8 meters.

A bullet 2cm log is fired at 420m/s and passes straight a 10cm thick board exiting at 280m/s
a) what is the average acceleration of the bullet through the board?
b)what is the total time the bullet is in contact with the board?
c)what minimum thickness could the board have if it was supposed to bring the bullet to a stop?

Answers

Solving for the acceleration of the bullet

acceleration = (vf^2 – vi^2) / 2d

acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)

acceleration = (78400 - 176400) / 0.24 m

acceleration = -98000 / 0.24

acceleration = -408333 m/s^2

Solving for contact time with board

t^2 = 2d/a

t^2 = 2 * 0.12 m / 408333 m/s^2

t^2 = 0.24 m / 408333 m/s^2

t^2 = 5.8775558 x 10^-7

t = 0.0007666 s or 767 microseconds


(I was only able to do A and B)

Answer:

Explanation:

(a)Solving for the acceleration of the bullet

acceleration = (vf^2 – vi^2) / 2d

acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)

acceleration = (78400 - 176400) / 0.24 m

acceleration = -98000 / 0.24

acceleration = -408333 m/s^2

(a)Solving for contact time with board

t^2 = 2d/a

t^2 = 2 * 0.12 m / 408333 m/s^2

t^2 = 0.24 m / 408333 m/s^2

t^2 = 5.8775558 x 10^-7

t = 0.0007666 s or 767 microseconds

An object of mass 80 kg is released from rest from a boat into the water and allowed to sink. While gravity is pulling the object down, a buoyancy force of 1/50 times the weight of the object is pushing the object up (weight=mg). If we assume that water resistance exerts a force on the abject that is proportional to the velocity of the object, with proportionality constant 10 N-sec/m, find the equation of motion of the object. After how many seconds will the velocity of the object be 40 m/s? Assume that the acceleration due to gravity is 9.81 m/sec^2.

Answers

Answer:

a) Fnet = mg - Fb - Fr

b) 8.67 secs

Explanation:

mass of object = 80 kg

Buoyancy force = 1/50 * weight ( 80 * 9.81 ) = 15.696

Proportionality constant = 10 N-sec/m

a) Calculate  equation of motion of the object

Force of resistance on object  due to water = Fr ∝ V

                                                                         = Fr = Kv = 10 V

Given that : Fb( due to buoyancy ) , Fr ( Force of resistance ) acts in the positive y-direction on the object  while mg ( weight ) acts in the negative y - direction on the object.

Fnet = mg - Fb - Fr

∴ Equation of motion of the object ( Ma = mg - Fb - Fr )

b) Calculate how long before velocity of the object hits 40 m/s

Ma = mg - Fb - Fr

a = 9.81 - 0.1962 - 0.125 V = 9.6138 - 0.125 V

V = u + at ---- ( 1 )

u = 0

V = 40 m/s

a = 9.6138 - 0.125 V

back to equation 1

40 = 0 + ( 9.6138 - 0.125 (40) ) t

40 = 4.6138 t

∴ t = 40 / 4.6138 = 8.67 secs

A 3.1-mole sample of an ideal gas is gently heated at constant temperature 320 K. It expands from initial volume 23 L to final volume V2. A total of 1.7 kJ of heat is added during the expansion process. What is V2? Let the ideal-gas constant R = 8.314 J/(mol • K).

Answers

From the ideal gas law,

PV = nRT   ==>   P = nRT/V

where P is the pressure exerted by the gas on the container. The work W done by this pressure as the volume of the gas changes from V₁ to V₂ is given by the integral,

[tex]W = \displaystyle \int_{V_1}^{V_2}P\,\mathrm dV \implies W = nRT \ln\left(\dfrac{V_2}{V_1}\right)[/tex]

and solving for V₂ gives

[tex]V_2 = V_1\exp\left(\dfrac{W}{nRT}\right)[/tex]

If you add 1.7 kJ of heat to the system, which does the aforementioned work, the gas will expand to a volume of

[tex]V_2 = (23\,\mathrm L)\exp\left(\dfrac{1.7\,\mathrm{kJ}}{(3.1\,\mathrm{mol})\left(8.314\frac{\rm J}{\mathrm{mol}\cdot\mathrm K}\right)(320\,\mathrm K)}\right) \approx \boxed{28 \,\mathrm L}[/tex]

Physics question plz help ASAP

Answers

Answer:

Option D.

Explanation:

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

Force applied (F) = 5 N

Extention (e) = 0.075 m

Spring constant (K) =?

The spring constant for the spring can be obtained as follow:

F = Ke

5 = K × 0.075

Divide both side by 0.075

K = 5 / 0.075

K = 67 N/m

Thus, the spring constant for the spring is 67 N/m

A 2.50-W beam of light of wavelength 124 nm falls on a metal surface. You observe that the maximum kinetic energy of the ejected electrons is 4.16 eV. Assume that each photon in the beam ejects a photoelectron. (a) What is the work function (in electron volts) of this metal

Answers

Answer:

φ = 13.43 x 10⁻¹⁹ J = 8.4 eV

Explanation:

Using the Einstein's Photoelectric equation:

Energy of Photon = Work Function + Kinetic Energy of Electron

[tex]\frac{hc}{\lambda} = \phi + K.E[/tex]

where,

h = Plank's Constant = 6.625 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength = 124 nm = 1.24 x 10⁻⁷ m

φ = work function = ?

K.E = Kinetic Energy of Electrons = (4.16 eV)([tex]\frac{1.6\ x\ 10^{-19}\ J}{1\ eV}[/tex]) = 2.6 x 10⁻¹⁹ J

Therefore,

[tex]\frac{(6.625\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{1.24\ x\ 10^{-7}\ m} = \phi + 2.6\ x\ 10^{-19}\\\\\phi=16.03\ x\ 10^{-19}\ J - 2.6\ x\ 10^{-19}\ J[/tex]

φ = 13.43 x 10⁻¹⁹ J = 8.4 eV

A long copper wire of radius 0.321 mm has a linear charge density of 0.100 μC/m. Find the electric field at a point 5.00 cm from the center of the wire. (in Nm2/C, keep 3 significant figures)

Answers

Answer:

[tex]E=35921.96N/C[/tex]

Explanation:

From the question we are told that:

Radius [tex]r=0.321mm[/tex]

Charge Density [tex]\mu=0.100[/tex]

Distance [tex]d= 5.00 cm[/tex]

Generally the equation for electric field is mathematically given by

[tex]E=\frac{mu}{2\pi E_0r}[/tex]

[tex]E=\frac{0.100*10^{-6}}{2*3.142*8.86*10^{-12}*5*10^{-2}}[/tex]

[tex]E=35921.96N/C[/tex]

Cho hệ thống thùng lắc có mô hình tại vị trí đang xét như hình vẽ

Answers

Answer:

I can't understand this language!!!

Answer:

vdhdbdnnsnsbdhhshzbhshsbbsbd is not ask you to be able and r in the exam qq and

Two loudspeakers placed 6.00 m apart are driven in phase by an audio oscillator having a frequency range from 1908 Hz to 2471 Hz. A point P is located 4.70 m from one loudspeaker and 3.60 m from the other speaker. At what frequency of the oscillator does the sound reaching point P interfere constructively

Answers

Answer:

2164 Hz

Explanation:

Since point P is 4.70 m away from one speaker and 3.60 m away from the other speaker, the path length difference ΔL = 4.70 m - 3.60 m = 1.1 m.

The path length difference ΔL = nλ for a constructive interference where n is an integer and λ = wavelength of sound from oscillator = v/f where v = speed of sound in air = 340 m/s and f = frequency of sound from oscillator.

So, ΔL = nλ = nv/f

So, the frequency from the oscillator is f = nv/ΔL

Substituting the values of v and ΔL into the equation, we have

f = nv/ΔL

f = n340 m/s/1.1 m

f = n309.09 /s

f = 309.09n Hz

We now insert values of n that will gives us a frequency in the range 1908 Hz to 2471 Hz.

The value of n that will give us a frequency in the range is n = 7

So, when n = 7,

f = 309.09n Hz

f = 309.091 × 7 Hz

f = 2163.64 Hz

f ≅ 2164 Hz

So, the frequency of the oscillator that will produce a constructive interference at P is 2164 Hz.

Data related to Meena’s and Malini’s journey is given below, plot a graph of their
respective journey on a graph paper. You have already plotted Meena’s Journey during the
summer vacation. On the same graph paper, now plot Malini’s Journey.
PDF task 2
Please do this for me urgent I can give you extra points if someone answers this in less than 1 hour.

Answers

Answer:

download the pdf

Explanation:

De que esta hecho el sol? plisss ayuda.no necesito un texto de 100 reglones, puede ser resumido en solo 2 renglones

Answers

El sol es una bola de gas así como las estrellas:)

26. A square loop whose sides are 6.0-cm long is made with copper wire of radius 1.0 mm. If a magnetic field perpendicular to the loop is changing at a rate of 5.0 mT/s, what is the current in the loop?

Answers

Answer:

Explanation:

The formula for determining the Emf induced in a loop is:

[tex]\varepsilon = \dfrac{d \phi}{dt}[/tex]

[tex]\varepsilon = \dfrac{d (B*A)}{dt}[/tex]

[tex]\varepsilon = A \times \dfrac{dB}{dt}[/tex]

[tex]\varepsilon = (side (l))^2 \times \dfrac{dB}{dt}[/tex]

where;

square area A = ( l²)

l² = 6.0 cm = 6.0 × 10⁻²

[tex]\varepsilon = ( 6.0 \times 10^{-2})^2 \times 5.0 \times 10^{-3} \ T/S[/tex]

[tex]\varepsilon =18 \times 10^{6} \ V[/tex]

Recall that:

The resistivity of copper = [tex]1.68 \times 10^{-8}[/tex] ohm m

We can as well say that the length of the copper wire = perimeter of the square loop;

The perimeter of the square loop = 4L

Thus, the length of the copper wire  = 4 (6.0 × 10⁻² )m

= 24× 10⁻² m

Finally, the current in the loop is determined from the formula:

V = IR

where,

V = voltage

I = current and R = resistance of the wire

Making "I" the subject:

I = V/R

where;

[tex]R = \dfrac{\rho \times l}{A}[/tex]

[tex]R = \dfrac{\rho \times l}{\pi * r^2}[/tex]

[tex]R = \dfrac{1.68 *10^{-8} \times 24*10^{-2}}{\pi * (1*10^{-3})^2}[/tex]

[tex]R = 0.001283 \ ohms[/tex]

[tex]I = \dfrac{18*10^{-6}}{0.001283}[/tex]

I = 14.029 mA

: A fan is placed on a horizontal track and given a slight push toward an end stop 1.80 meters away. Immediately after the push, the fan of the cart engages and slows the cart with an acceleration of -0.45 m/s2. What is the maximum possible velocity (magnitude) the cart can have after the push so that the cart turns around just before it hits the end-stop

Answers

Answer:

The initial velocity is 1.27 m/s.

Explanation:

distance, s = 1.8 m

acceleration, a = - 0.45 m/s^2

final velocity, v = 0

let the initial velocity is u.

Use third equation of motion

[tex]v^2 = u^2 + 2 a s \\\\0 = u^2 - 2 \times 0.45\times 1.8\\\\u = 1.27 m/s[/tex]

We have that the Initial velocity  is mathematically given as

u=1.27m/s

Maximum possible velocity

Question Parameters:

a slight push toward an end stop 1.80 meters away

he fan of the cart engages and slows the cart with an acceleration of -0.45 m/s2

Generally the equation for the  third equation of motion    is mathematically given as

Vf^2 = Vi^2 + 2ad

Therefore

0=u^2+0.45*1.8

u=1.27m/s

For more information on velocity visit

https://brainly.com/question/17617636

Una persona de 76 kg está siendo retirada de un edificio en llamas mientras se muestra en la figura. Calcule la tensión
en las dos cuerdas si la persona está momentáneamente inmovil.
Ayuda por favor.

Answers

Answer:

T1 = 736.6 N, T2 = 193.5 N

Explanation:

W = 76 N

The tension is T1 and T2.

By use of Lami's theorem

[tex]\frac{T_1}{Sin100}=\frac{T_2}{Sin165}=\frac{W}{Sin 95}\\\\So, \\\\T_1 = \frac{76\times 9.8\times Sin 100}{Sin 95} = 736.6 N \\And\\T_2 = \frac{76\times 9.8\times Sin 165}{Sin 95} = 193.5 N \\[/tex]

An object moving with a constant
acceleration changes its velocity from
10ms' to 20 ms' in five seconds. What is the
distance travelled in five seconds ​

Answers

Answer:

Acceleration:

[tex]{ \tt{a = \frac{v - u}{t} }} \\ { \tt{a = \frac{20 - 10}{5} }} \\ { \tt{a = 2 \: m {s}^{ - 2} }}[/tex]

From third equation:

[tex]{ \bf{ {v}^{2} = {u}^{2} + 2as}} \\ { \tt{s = \frac{ {20}^{2} - {10}^{2} }{2 \times 2} }} \\ = { \tt{s = 75 \: m}}[/tex]

Answer:

Formula = m/s

Explanation:

The answer is 10 m / 5 seconds = 2 meters distance

The answer is 20 m / 5 seconds = 4 meters distance

A T-shirt is launched at an angle of 30° with an initial velocity of 25 m/s how long does it take to reach the peak? How long is it in the air for totally?

Answers

Answer:

The launched angle θ = 30 degrees, the initial velocity Vo = 20 m/s, the initial horizontal velocity Vox= ?, the initial vertical velocity Voy = ?, the time of flight t = ? the maximum height h = ?

Vox = Vo * (cos of 30 degrees)

Voy = Vo * (sin of 30 degrees)

t = 2 * (Voy / g)

h = Voy * 0.5 t - 1/2 g * (0.5t)^2

I have given the equations for you to use, just plug – in the values and then solve in a step by step manner.

Answer:

approximately 15.68 meters.

Explanation:

Here is how;

First, let's calculate the time of flight for the t-shirt. We can use the vertical motion equation:

y = y0 + v0y * t - 0.5 * g * t^2

where:

y is the vertical displacement (27.7 m)

y0 is the initial vertical position (0 m)

v0y is the vertical component of the initial velocity (v0 * sin(theta))

g is the acceleration due to gravity (9.8 m/s^2)

t is the time of flight

Plugging in the values:

27.7 = 0 + (25.8 * sin(63.6°)) * t - 0.5 * 9.8 * t^2

Simplifying the equation, we get a quadratic equation:

4.9t^2 - (25.8 * sin(63.6°))t + 27.7 = 0

Solving this quadratic equation will give us the time of flight, t. Using the quadratic formula, we find that:

t ≈ 1.23 s

Now, let's find the horizontal displacement of the t-shirt using the horizontal motion equation:

x = x0 + v0x * t

where:

x is the horizontal displacement

x0 is the initial horizontal position (0 m)

v0x is the horizontal component of the initial velocity (v0 * cos(theta))

t is the time of flight

Plugging in the values:

x = 0 + (25.8 * cos(63.6°)) * 1.23

Calculating this:

x ≈ 14.92 m

The t-shirt falls short of reaching the person by the horizontal distance of:

Shortfall = 30.6 m - 14.92 m

Calculating this:

Shortfall ≈ 15.68 m

Therefore, the t-shirt will be approximately 15.68 meters short of reaching the person.

The standard unit of brightness is called the candela.
True
False

Answers

Answer:

TRUE

Explanation:

A roller coaster has a vertical loop with radius 25.7 m. With what minimum speed should the roller-coaster car be moving at the top of
the loop so that the passengers do not lose contact with the seats?
m/s

Answers

Answer:

15.88m/s

Explanation:

At the top of the roller coaster you will have three forces acting on the roller-coaster. See the image below. Fc is the centripetal force (for an object in circular motion), Fg is the gravitational force, and Fn is the normal force. To achieve the minimum speed we assume the roller-coaster is barely touching the vertical loop and so the normal force is zero. This leaves two acting forces.

[tex]F_g = F_c\\mg = \frac{m\times v^2}{r}\\v = \sqrt{gr} = \sqrt{9.81 \times 25.7} = 15.88 m/s[/tex]

Why is oiling done time and again in a sewing machine?

Answers

Answer:

to prevent friction on the surfaces

Answer:

Explanation:

Oiling reduces friction between parts with relative motion between them.

Repeated oiling is needed as the film of oil reducing the friction becomes thinner with time as some of the oil gets pushed off of the areas of motion where it can no longer be useful.

Oil also becomes oxidized which reduces the oil's ability to decrease friction.

Oil can also be fouled by dirt, lint or other material. This added material becomes coated in oil and typically gets sequestered away from the moving parts reducing the oil available for lubrication purposes.

What Are the type's of Tidal turbines?

Answers

Answer:

Types of tidal turbines

Axial turbines.

Crossflow turbines.

Flow augmented turbines.

Oscillating devices.

Venturi effect.

Tidal kite turbines.

Turbine power.

Resource assessment.

Answer:

Axial turbines

Crossflow turbines

flow augmented turbines

Describing Uses ñ Olivia wants to find out whether a substance will fluoresce. She says she should put it in a microwave oven. Do you agree with her? Why or why not?

Answers

I don’t agree what if it blows up?

The US currently produces about 27 GW of electrical power from solar installations. Natural gas, coal, and oil powered installations produce about 740 GW of electrical power. The average intensity of electromagnetic radiation from the sun on the surface of the earth is 1000 W/m2 . If solar panels are 30% efficient at converting this incident radiation into electrical power, what is the total surface area of solar panels responsible for the 27 GW of power currently produced

Answers

Answer:

The total surface area is "90 km²".

Explanation:

Given:

Power from solar installations,

= 27 GW

Other natural installations,

= 740 GW

Intensity,

[tex]\frac{F}{At}=\frac{P}{A}=1000 \ W/m^2[/tex]

%n,

= 30%

Now,

⇒ %n = [tex]\frac{out.}{Inp.}\times 100[/tex]

then,

⇒ [tex]Inp.=\frac{27}{30}\times 100[/tex]

           [tex]=90 \ GW[/tex]

As we know,

⇒ [tex]I=\frac{P}{A}[/tex]

by substituting the values, we get

[tex]1000=\frac{90\times 10^9}{A}[/tex]

    [tex]A = \frac{90\times 10^9}{10^3}[/tex]

        [tex]=90\times 10^6[/tex]

        [tex]=90 \ km^2[/tex]

A boy walks from point C to point D which is 50 m apart. Then, he walks back to point C. what is his displacement of his whole journey ?
A.25 m
B.75 m
C.50 m
D.0 m

Answers

Answer: D. 0 m

Explanation:

Concept:

Here, we need to know the concept of displacement.

Displacement is defined to be the change in position of an object.

The difference between displacement and distance is the total movement of an object without any regard to direction, while displacement is the pure change of position.

If you are still confused, please refer to the attachment below for a graphical explanation.

Solve:

STEP ONE: the boy walks from point C to point D (a distance of 50 m)

C ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ D

                                              50 m

STEP TWO: the boy walks from point D to point C (a distance of 50 m)

D ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ C

                                               50 m

STEP THREE: find the displacement

The boy started with point C

The boy ended with point C

He did not change his position throughout the journey.

Therefore, his displacement is 0 m.

Hope this helps!! :)

Please let me know if you have any questions

A ball of mass 0.50 kg is rolling across a table top with a speed of 5.0 m/s. When the ball reaches the edge of the table, it rolls down an incline onto the floor 1.0 meter below (without bouncing). What is the speed of the ball when it reaches the floor?


PLEASE EXPLAIN HOW YOU GOT THE ANSWER THANK YOU SO MUCH

Answers

Answer:

0

Explanation:

The speed of the ball when it reaches the floor is 0 because when an object is at rest or in uniform motion, it has no speed/velocity

The final speed of the ball when it reaches the floor is 7.10 m/s.

What is the conservation of energy?

The conservation of energy is a fundamental principle in physics that states that energy cannot be created or destroyed, but only converted from one form to another or transferred from one system to another. In other words, the total amount of energy in a closed system remains constant over time, even though it may be converted from one form to another.

This principle is based on the first law of thermodynamics, which states that the total energy of a closed system is always conserved, and can only be changed by the transfer of heat, work, or matter into or out of the system. The conservation of energy has important applications in various fields of physics, including mechanics, thermodynamics, and electromagnetism, and is a fundamental principle in the understanding of the natural world.

Here in the Question,

We can use the conservation of energy to solve this problem. Initially, the ball has kinetic energy due to its motion on the tabletop, but no potential energy since it is at a constant height. When the ball rolls off the edge of the table, it loses some kinetic energy due to friction but gains potential energy as it moves upward. When it reaches the floor, it has gained potential energy but lost kinetic energy due to friction. We can assume that the energy lost due to friction is converted to thermal energy, so the total energy of the system is conserved.

Let's start by calculating the potential energy gained by the ball as it moves from the edge of the table to the floor:

ΔPE = mgh

where ΔPE is the change in potential energy, m is the mass of the ball, g is the acceleration due to gravity, and h is the vertical distance traveled by the ball.

ΔPE = (0.50 kg)(9.81 m/s^2)(1.0 m) = 4.905 J

Now we can use the conservation of energy to find the final kinetic energy of the ball, which will allow us to calculate its final speed:

KEi + ΔPEi = KEf + ΔPEf

where KEi and ΔPEi are the initial kinetic and potential energies of the ball, respectively, and KEf and ΔPEf are the final kinetic and potential energies of the ball, respectively.

Since the ball is not bouncing, we can assume that its initial and final potential energies are zero. Therefore:

KEi = KEf + ΔKE

where ΔKE is the change in kinetic energy due to friction.

We can assume that the coefficient of kinetic friction between the ball and the incline is constant, and use the work-energy principle to find ΔKE:

Wfric = ΔKE

where Wfric is the work done by friction.

The work done by friction can be expressed as:

Wfric = ffricd

where ffric is the force of friction and d is the distance traveled by the ball on the incline.

The force of friction can be expressed as:

ffric = μmg

where μ is the coefficient of kinetic friction, and m and g have their usual meanings.

Putting it all together, we get:

KEi = KEf + ffricd

KEi = KEf + μmgd

(1/2)mv^2 = (1/2)mu^2 + μmgd

v^2 = u^2 + 2gd

where u is the initial speed of the ball on the tabletop, and v is the final speed of the ball on the floor.

Plugging in the given values, we get:

v^2 = (5.0 m/s)^2 + 2(9.81 m/s^2)(1.0 m)

v^2 = 50.405

v = 7.10 m/s

Therefore, the final speed of the ball when it reaches the floor is 7.10 m/s.

To learn more about  the Law of Conservation of Momentum click:

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#SPJ2

A cylindrical wire has a length of 2.80 m and a radius of 1.03 mm. It carries a current of current of 1.35 A, when a a voltage of 0.0320 V is applied across the ends of the wire. From this information calculate the resistance of the wire.

Answers

Answer:

0.023 Ohms

Explanation:

Given data

Length= 2.8m

radius= 1.03mm

current I= 1.35 A

voltage V= 0.032V

We know that from Ohm's law

V= IR

Now  R= V/I

Substitute

R= 0.032/1.35

R= 0.023 Ohms

Hence the resistance is 0.023 Ohms

A green object will absorb ____________________ light and reflect ____________________ light. (ref: p.447-455)

Answers

Answer:

A green object will absorb all light except for green light and reflect blue and yellow light.

A 15kg mass suspended from a ceiling is pulled aside with a horizontal force, F. Calculate the value of the tension.​

Answers

Answer:

147 Newtons

Explanation:

To find tension, you can use the formula Tension = (mass)(gravity)

*Gravity's acceleration = 9.8 m/s^2 because of Newton's law of universal gravitation*

T = (15kg)(9.8m/s^2)

  = 147 Newtons

Hope this helps! Best of luck <3

If a car drives 10 mph South, this is an example of a:
A. Displacement
B. Velocity
C. Speed
D. Distance

Answers

Answer:

杰杰伊杜杜杜伊格富尔杰迪耶赫分离福音

Explanation:

莱德利 · 赫耶尔伊 3uritievrirjrirhruebwkwieheoo2hfjcbvi3hd

Answer:

B velocity

Explanation:

A ball of mass m is dropped from a height h above the ground. neglecting air resistance then determine the speed of the ball when it is at a height y above the ground and determine the speed of the ball at y if at the instant of release it already has an initial upward speed vi at the initial altitude h.​

Answers

Answer:

Explanation:

kinematic equation (g will have a negative value if we assume UP is positive)

v² = u² + 2as

a) v = √(0² + 2(g)(y - h))

b) v = √(vi² + 2(g)(y - h))

how much energy will be created if a man of mass of 50 kg is destroyed completely. ​

Answers

Answer:

the energy released or created is  4.5 x 10¹⁸ J

Explanation:

Given;

mass of the given matter (man), m = 50 kg

Apply Einstein's mass defect equation.

When a mass of any given matter is completely destroyed, the energy released is calculated as follows;

E = mc²

where;

E is the released or created

c is the speed of light = 3 x 10⁸ m/s

E = (50) x (3 x 10⁸)²

E = 4.5 x 10¹⁸ J

Therefore, the energy released or created is  4.5 x 10¹⁸ J

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