What is the acceleration when a force of 2 N is applied to a ball that has a mass of 0.60 kg?

Answers

Answer 1

Answer:

3.333

Explanation:

Acceleration is force divided by mass. So divide the force, 2, by the mass, 0.60, and you will get 3.333. I hope this helps :)

Answer 2

The acceleration of an object is the force divided by time. The acceleration of the ball of 0.60 kg when a force of 2N is applied, is 3.33 m/s².

What is acceleration ?

Acceleration of an object is the measure of rate of change in its velocity. Like velocity acceleration is a vector quantity having both magnitude and direction.

Acceleration is defined as the ratio of change in velocity to the change in time. However, according to Newton's second law of motion fore applied on an object is the product of its mass and acceleration.

Hence, F = m a .

Given that, force applied on the ball = 2 N

mass of the ball = 0.60 Kg.

Then acceleration a = force/mass

a = 2 N/ 0.60 Kg = 3.33 m/s²

Therefore, the acceleration of the ball is 3.33 m/s².

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Related Questions

A girl standing on a bridge throws a stone vertically downward with an initial velocity of 15.0 m/s into the river below. If the stone hits the water 2.00 seconds later, what is the height of the bridge above the water

Answers

Answer:

the height of the bridge above the water is 49.6 m.

Explanation:

Given;

initial velocity of the stone, u = 15 m/s

time of motion of the stone, t = 2 s

The height of the bridge above the water is calculated from the following kinematic equation as follows;

h = ut + ¹/₂gt²

h = (15 x 2) + ¹/₂(9.8)(2²)

h = 30 + 19.6

h = 49.6 m

Therefore, the height of the bridge above the water is 49.6 m.

I need help please will mark brainliest

Answers

Answer:

200

Explanation:

20m/s*10sec=200

When a wave hits an object,energy from the wave is both absorbed and reflected off the object
A. True
B. False

P.S pls help

Answers

I believe the statement is true. Have a good day.

Solids have a definite shape and volume this is because

Answers

i’m so confused it has a definite shape because It’s a solid

list 5 types of food that should be consumed daily in a healthy diet.Give an example of each type.


Answers

Answer:

vegetables and legumes or beans

fruit

lean meats and poultry, fish, eggs, tofu, nuts and seeds, legumes or beans

grain (cereal) foods, mostly wholegrain or high cereal fibre varieties

milk, yoghurt, cheese or alternatives, mostly reduced fat.

Explanation:

Foods are grouped together because they provide similar amounts of key nutrients. For example, key nutrients of the milk, yoghurt, cheese and alternatives group include calcium and protein, while the fruit group is a good source of vitamins, especially vitamin C.

The five food groups are:
vegetables and legumes or beans.
fruit.
lean meats and poultry, fish, eggs, tofu, nuts and seeds, legumes or beans.
grain (cereal) foods, mostly wholegrain or high cereal fibre varieties.
milk, yoghurt, cheese or alternatives, mostly reduced fat.

How do we know that an object has accelerated?

Answers

A change in velocity or a change in direction, or both! Hope this helps

The resolution of a lens can be estimated by treating the lens as a circular aperture. The resolution is the smallest distance between two point sources that produce distinct images. This is similar to the resolution of a single slit, related to the distance from the middle of the central bright band to the firstorder dark band; however, the aperture is circular instead of a rectangular slit which introduces a scale factor. Suppose the Hubble Space Telescope, 2.4 m in diameter, is in orbit 90.4 km above Earth and is turned to look at Earth. If you ignore the effect of the atmosphere, what is the resolution of this telescope for light of wavelength 557 nm?

Answers

Answer:

y = 2.56  10⁻² m

Explanation:

The resolution of this telescope is given by the Rayleigh criterion, for the phenomenal diffraction the first minimum for a linear slit is in

              a sin θ = λ

in general the angles are very small, so we approximate

              sin θ = θ

we substitute

              θ = λ / a

in the case of circular slits we must use polar coordinates, which introduces a numerical factor, leaving the equation

             θ = 1.22 [tex]\frac{\lambda }{D}[/tex]

where D is the diameter of the circular opening

In this case they indicate the lens diameter D = 2.4 m, the observation distance r = 90.4 km = 90.4 10³ m

how angles are measured in radians

            θ = y / r

we substitute

              y / r = 1.22\frac{\lambda }{D}

              y = 1.22 \frac{\lambda r }{D}

let's calculate

             y = [tex]1.22 \frac{ 557 \ 10^{-9} \ \ 90.4 \ 10^{3} }{2.4}[/tex]

             y = 2.56  10⁻² m

this is the minimum distance that can differentiate two objects on Earth

If the magnetic force is 3.5 × 10–2 N, how fast is the charge moving?

Answers

Answer:

D

Explanation:

Took it on edg

The speed of the charge at the given magnetic force and field is determined as 1.1 x 10⁴ m/s.

Speed of the charge

The speed of the charge is calculated as follows;

F = qvBsinθ

v = F/qBsinθ

where;

F is the magnetic forceB is magnetic fieldv is speed of the charge

v = (3.5 x 10⁻²)/(8.4 x 10⁻⁴ x 6.7x 10⁻³ x sin35)

v = 10,842.33 m/s

v ≅ 1.1 x 10⁴ m/s

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High-voltage power lines are a familiar sight throughout the country. The aluminum wire used for some of these lines has a cross-sectional area of 4.8 x 10-4 m2. What is the resistance of 14 kilometers of this wire

Answers

Answer:

Explanation:

For resistance of a wire , the formula is as follows

R = ρ L / S

where ρ is specific resistance , L is length and S is cross sectional area

Given L = 14 000 m ,

S = 4.8 x 10⁻⁴ m²

specific resistance of aluminum = 2.8 x 10⁻⁸ ohm-meter

Putting the values in the formula

R = 2.8 x 10⁻⁸ x 14 x 10³ /  (4.8 x 10⁻⁴ )

R = 0.8167 ohm .

= .82 ohm .

The current supplied by a battery in a portable device is typically about 0.122 A. Find the number of electrons passing through the device in two hours.

Answers

Ight hunch nvm cuh ong

Giving 50 points exactly 50 point and making brainles and if answered wrong I will report and get you kicked out
Draw a picture showing how Heat is added, released, and transferred from one object to another. Also, draw a picture explaining how sublimation or deposition works. please you can draw in your book and take a picture and post it

Answers

Answer:

i hope this helps i did'nt quiet understand the secound one. I hope youcan see my picture well .

Explanation:

sincerily, MEMC3891

Acceleration figures for cars usually are given as the number of seconds needed to go from 0.0 to 97 km/h. Convert 97 km/h into m/s.

Answers

Answer:

26.9444m/s

pls brainliest

If the kinetic energy of the 40kg box is 784 J, what is the velocity before it strikes the ground?

Answers

Answer:

Explanation:

[tex]KE=\frac{1}{2}mv^2[/tex]

[tex]784=\frac{1}{2}(40)v^2[/tex]

[tex]784=20v^2[/tex]

[tex]39.2=v^2[/tex]

[tex]v=6.26m/s[/tex]

Ropes 3 m and 5 m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 3 kg. The ropes, fastened at different heights, make angles of 52° and 40° with the horizontal. Find the tension in each wire and the magnitude of each tension. (Use g = 9.8 m/s2 for the acceleration due to gravity. Round your answers to two decimal places.)

Answers

Answer:

Explanation:

Let the tension in ropes be T₁ and T₂ . Ropes are making angle of 52 and 40 degree with the horizontal . The vertical component of tension will add up together to balance the weight of the decoration

T₁ sin52 + T₂ sin40 = 3 x 9.8

.788 T₁ + .643 T₂ = 29.4

The horizontal component of tension will add up to zero because the decoration piece is at rest .

T₁ cos52 + T₂ cos40 = 0

.615 T₁ -  .766 T₂ = 0

T₁ =  1.24 T₂

Substituting this value of T₁ in earlier equation , we have

.788 x 1.24 T₂ + .643 T₂ = 29.4

1.62 T₂ = 29.4

T₂ = 18.15 N

T₁ = 1.24 x 18.15 = 22.51 N .

Suppose there is a 3Mbps uplink and 10Mbps downlink between a geostationary satellite and the base station on earth. If the propagation speed is the speed of light (3 * 10^8 m/sec), the packet size is 20Mb, and the distance from the satellite to earth is 36,000 km, what is the uplink propagation delay of the link in milliseconds

Answers

Answer:

[tex]120\ \text{ms}[/tex]

Explanation:

Distance between the satellite and Earth = [tex]36000\ \text{km}[/tex]

Speed of light = [tex]3\times 10^8\ \text{m/s}[/tex]

Propagation delay is given by distance by the speed of light

[tex]\dfrac{36000\times 10^3}{3\times 10^8}[/tex]

[tex]=0.12\ \text{s}\times 10^3[/tex]

[tex]=120\ \text{ms}[/tex]

The uplink propagation delay of the link is [tex]120\ \text{ms}[/tex].

How can you describe the orbital period (planetary year) of each planet?

Answers

Answer:

How can you describe the orbital period (planetary year) of each planet?

A year is defined as the time it takes a planet to complete one revolution of the Sun, for Earth this is just over 365 days. This is also known as the orbital period. Unsurprisingly the the length of each planet's year correlates with its distance from the Sun.

Explanation:

Answer: The orbital period is how long it takes a plane to fulfill its  "revolution"(365 days)

Explanation:

Question 2 of 10 How many oxygen (O) atoms are in a molecule of C3 H4 03? A. 3 OB. 4 C. 10 ОО D. 1​

Answers

Answer:

a. 3

Explanation:

Answer:

A. 3

Explanation:

AP3X

Question: How did NASA use Newton's Laws to land the Perseverance lander,
safely
on Mars?


Please help asap.

Answers

Answer:

If the thrust is increased, the aircraft accelerates and the velocity increases. This is the second part sited in Newton's first law; a net external force changes the velocity of the object. The drag of the aircraft depends on the square of the velocity. So the drag increases with increased velocity.

Write any two uses of plane mirrors?​

Answers

Answer:

Uses of plane mirrors

They are used in periscopes, for signalling, in kaleidoscopes, to see round dangerous bends, in meters, as mirror tiles, in a sextant, in an overhead projector, an SLR camera, car wing mirrors, in microscopes and as reflecting number plates to mention only some!

Explanation:

Hope it is helpful....

Answer:

two uses are:

they are using for looking glassthey are used to make periscope

A skier pushes off the top of a hill with an initial speed of 3.30 m/s. How fast will she be moving after dropping 5.00 meters in elevation if friction is negligible?

Answers

Answer:

eeeeeeeeeeeeeeeeeeeee

The most common stars in the universe are red dwarf stars. Red dwarf stars use their hydrogen fuel very slowly. Our universe also contains stars classified as blue supergiants. The table below lists the color and approximate surface temperatures of four stars. Based on the information from the table above, which of the following stars has the greatest absolute brightness?


The Sun


Vega


Bellatrix


Betelgeuse

Answers

Answer:

sun

Explanation:

A skateboarder travels on a horizontal surface with an initial velocity of 4.2 m/s toward the south and a constant acceleration of 2.6 m/s^2 toward the east. Let the x direction be eastward and the y direction be northward, and let the skateboarder be at the origin at t=0.

Required:
a. What is her x position at t=0.60s?
b. What is her y position at t=0.60s?
c. What is her x velocity component at t=0.60s?
d. What is her y velocity component at t=0.60s?

Answers

Answer:

Explanation:

Let the skateboarder's movement in x -direction be taken into consideration .

a )

initial velocity in x direction u = 0

acceleration a = 2.6 m /s²

time t = .6 s

displacement in x direction

s = ut + 1/2 a t²

= 0 + .5 x 2.6 x .6²

= .468 m

= 46.80 cm

c )

velocity after .6 s

v = u + at

= 0 + 2.6 x .6

= 1.56 m /s

Let the skateboarder's movement in y -direction be taken into consideration .

b ) initial velocity in y direction u = - 4.2 m /s ( velocity is towards south or - y direction )

acceleration a = 0

time t = .6 s

displacement in y direction

s = ut + 1/2 a t²

= - 4.2 x .6  + 0

= - 2.52  m

2.52 m towards south .

d )

velocity after .6 s

v = u + at

= - 4.2  + 0

= - 4.2  m /s

or 4.2 m /s towards south .

what is the acceleration of a satellite moving in a circular orbit around the earth of radius 2r​

Answers

Explanation:

You do the radius times the circumference of the earth

A stone of mass 1kg is thrown at 10m/s upwards making an angle of 37°with the horizontal from a building that is 20m high. Using the law of conservation of energy calculate the speed wjen the stone hits the ground.

Answers

Answer:

31.68 m/s

Explanation:

The law of conservation of energy states that energy is not lost or gained it is just converted, in this example, since it is not given any resistance from the wind, you'd have two variables Speed on the Y-axis and on the X-axis, since both of them would result in the same decrease and increase with against gravity, it doesn't matter the value of both.

As the stone continues to go upwards it will continue to lose speed due to de-acceleration from the gravity acting on it, similarly, it will continue to gain Potential energy, instead of kinetic energy, when it reaches its highest point the speed on Y will be "0" and the free fall will start, since the up and down movement will be equal in time the and acceleration would be equal -9.81 m/s and 9.81 m/s because the only acceleration you have is gravity, you only need to calculate how much speed will gain a rock accelerating at 9.81 m/s falling 20 m:

[tex]H=\frac{1}{2}g*t^{2} \\\frac{20}{1/2*9.81} =t^{2} \\\\t^{2} =4.08\\t=\sqrt{4.08} \\t=2.21[/tex]

Now we just add the time accelerating:

[tex]Vf=Vo+at\\Vf=10 m/s+ 2.21*9.81\\Vf=10 m/s+21.68\\Vf=31.68 m/s[/tex]

Which statement about the sun's energy is correct?

It is entirely re-radiated back into space.
A part of it is destroyed by greenhouse gases.
A part of it is absorbed by atmospheric gases.
It makes Earth too hot for plants and animals to survive.

Answers

I believe it’s A. I know for sure it isn’t D.

Answer:

The answer is (A)

Explanation:

We know this because The suns energy is entirely re-radiated back into space.

We know also that the answer is not (D) It makes Earth too hot for plants and animals to survive.

Atmospheric gases are gases located in the Earth's atmosphere

The green house effect is a natural process that warms the Earth's surface

I also know the answer because I took the test... FLVS exam 3.09 right?

I took it and got this question right... So i know you will! GOOD LUCK!!

Mark me brainlyest please:)

pls help everything is in the pic​

Answers

Answer:

c

Explanation:

A constant electric field of 5.00 N/C points along the positive x-direction. An electron, initially at rest, moves a distance of 2.00 m in this space. How fast is the electron moving after its 2.00 m journey

Answers

Answer:

1.875 x 10⁶ m /s .

Explanation:

Force on electron = E e where E is electric field and e is charge on electron

acceleration generated = Ee / m where m is mass of the electron .

Putting the values

acceleration generated = 5 x 1.6 x 10⁻¹⁹ / 9.1 x 10⁻³¹

= .879 x 10¹² m /s²

v² = u² + 2 as , initial velocity u = 0 , displacement s = 2 m

v² = 0 + 2 x .879 x 10¹² x 2

v = 1.875 x 10⁶ m /s .

Astronomers discover a planet orbiting around a star similar to our sun that is 35 light years away. How fast must a rocket ship go if the round trip is to take no longer than 70 years in time for the astronauts aboard

Answers

Answer:

[tex]v = 0.7071c[/tex]

Explanation:

Given

Distance to the planet = 35 light years. So, the entire distance is: 2 * 35 = 70.

[tex]\triangle{x'} = 70[/tex]

[tex]T_0 = 70\ years[/tex] i.e time of travel of the ship.

For the observer on earth, the time is:

[tex]T' = \gamma T_0[/tex]

The required speed so that it does not take more than 70 years is then calculated using:

[tex]\triangle x' = vT'[/tex]

Substitute [tex]T' = \gamma T_0[/tex]

[tex]\triangle x' = v\gamma T_0[/tex]

[tex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/tex]

So, we have:

[tex]\triangle x' = \frac{vT_0}{\sqrt{1 - v^2/c^2}}[/tex]

Make v the subject of formula.

Square both sides

[tex]\triangle x'^2 = \frac{v^2T^2_0}{1 - v^2/c^2}[/tex]

Cross Multiply

[tex](1 - \frac{v^2}{c^2}) *\triangle x'^2 = v^2T^2_0[/tex]

Divide both sides by [tex]\triangle x'^2[/tex]

[tex](1 - \frac{v^2}{c^2}) = \frac{v^2T^2_0}{\triangle x'^2}[/tex]

Divide through by [tex]v^2[/tex]

[tex](\frac{1}{v^2} - \frac{v^2}{v^2*c^2}) = \frac{v^2T^2_0}{v^2\triangle x'^2}[/tex]

[tex]\frac{1}{v^2} - \frac{1}{c^2} = \frac{T^2_0}{\triangle x'^2}[/tex]

Make [tex]\frac{1}{v^2}[/tex] the subject

[tex]\frac{1}{v^2} = \frac{T^2_0}{\triangle x'^2} + \frac{1}{c^2}[/tex]

Inverse both sides

[tex]v^2 = \frac{1}{\frac{T^2_0}{\triangle x'^2} + \frac{1}{c^2}}[/tex]

Take square root of both sides

[tex]v = \sqrt{\frac{1}{\frac{T^2_0}{\triangle x'^2} + \frac{1}{c^2}}}[/tex]

Substitute values for [tex]T_0[/tex] and [tex]\triangle x[/tex]

[tex]v = \sqrt{\frac{1}{\frac{70^2}{(70c)^2} + \frac{1}{c^2}}}[/tex]

[tex]v = \sqrt{\frac{1}{\frac{70^2}{70^2*c^2} + \frac{1}{c^2}}}[/tex]

[tex]v = \sqrt{\frac{1}{\frac{1}{c^2} + \frac{1}{c^2}}}[/tex]

[tex]v = \sqrt{\frac{1}{\frac{2}{c^2}}}[/tex]

[tex]v = \sqrt{\frac{c^2}{2}}[/tex]

[tex]v = c\sqrt{\frac{1}{2}}[/tex]

[tex]v = c * 0.7071[/tex]

[tex]v = 0.7071c[/tex]

An air-filled capacitor consists of two parallel plates, each with an area of A , separated by a distance d . A V potential difference is applied to these plates. What is the magnitude of the electric field between the plates

Answers

Answer:

  E = V / d

Explanation:

In a charged capacitor an electric field is established that goes from the positive to the negative plate, this field is constant,

the potential difference is

           D = E d

in this case they do not give the difference in potential V and the distance between the plates d

           E = V / d

If the gravitational constant is extremely weak, how is the force of gravity on earth so strong?

Answers

If its gravity is too strong our blood will be pulled down into our legs, our bones might break, and we could even be pinned helplessly to the ground. Finding the gravitational limit of the human body is something that's better done before we land on a massive new planet.
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