if a body is accelerating m change the velocity of 2 metre per second square when it was acted by five hundred Newton of force then calculate its mass.

Answers

Answer 1
Force=500NAcceleration=2m/s^2

using newtons law

[tex]\\ \rm\Rrightarrow Force=Mass(Acceleration)[/tex]

[tex]\\ \rm\Rrightarrow Mass=\dfrac{Force}{Acceleration}[/tex]

[tex]\\ \rm\Rrightarrow Mass=\dfrac{500}{2}[/tex]

[tex]\\ \rm\Rrightarrow Mass=250kg[/tex]


Related Questions

0
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ZOOLS
6) The mass of a motorcycle is 250 kg. What is?
A) Its weight on Earth in Newtons?
B) Its weight on the moon (in Newtons)?
ges
C) The mass of your motorcycle on the moon?

Answers

Answer:

Explanation:

Weight is actually a force. A force can change depending on its location. A mass remains constant no matter where it is.

A)

F = m * a

m = 250 kg

a = 9.81 m/s^2

F = 250 * 9.81 = 2452.5 N

B)

The acceleration due to gravity on the moon is roughly 1/6 what it is on earth. You can check its value in your notes.

a = 9.81 + (1/6) = 1.635

m = 250

F = 250 * 1.635

F = 408.75

C)

The mass is the same anywhere in the universe.

250 kg

A 207-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.750 rev/s in 2.00 s

Answers

Answer:

366 N

Explanation:

  τ = Iα

FR = ½mR²α

  F = ½mR(Δω/t)

  F = ½(207)(1.50)(0.75)(2π) /2.00

  F = 365.79919...

coin 1 is thrown upward from the top of 100m tower with a speed of 15m/s. coin 2 is dropped from the top of the tower 2.0second later. assume g is 10m/s. how far below the top of the tower des coin 1 pass coin 2

Answers

The height below the tower at which coin 1 pass coin 2 is 89.04 m.

The given parameters:

height of the tower, h = 100 m

initial velocity of coin 1, v = 15 m/s

time spent in air by coin 1 before coin 2 was dropped = 2s

To find:

the height below the tower at which coin 1 passes coin 2

Find the maximum height attained by coin 1 before falling to the ground:

[tex]v^2 = u^2 - 2gh\\\\where;\\\\v \ is \ the \ final \ velocity \ of \ coin \ 1 \ at \ maximum \ height, v \ = 0\\\\0 = (15^2) - 2(10)h\\\\20h = 225\\\\h = \frac{225}{20} \\\\h = 11.25 \ m[/tex]

Find the time taken for coin 1 to fall to the ground:

Total height of coin 1 above the ground, H = 11.25 m + 100 m = 111.25 m

[tex]t = \sqrt{\frac{2H}{g} } \\\\t = \sqrt{\frac{2\times 111.25}{10} } \\\\t = 4.72 \ s[/tex]

But the time taken for the coin 1 to reach 11.25 m above the tower:

[tex]t_1 = \sqrt{\frac{2h}{g} } \\\\t_1 = \sqrt{\frac{2\times 11.25}{10} } \\\\t_1 = 1.5 \ s[/tex]

Total time spent by coin 1 before reaching ground with respect to coin 2:

time = (1.5 s + 4.72 s) - 2 s

time = 4.22 s

Note: the 2 s was subtracted to keep both coins at a fair starting time below the tower.

Find the total time taken for coin 2 to fall to the ground:

Height of coin 2 above the ground = 100 m

Total time taken by coin 2 before falling to the ground is calculated as:

[tex]t_2 = \sqrt{\frac{2(100)}{10} } \\\\t_2 = 4.47s[/tex]

The time  at which coin 1 will pass coin 2 is 4.22 s.

Find the height below the tower when the time is 4.22 s.

[tex]h = \frac{1}{2} (10)(4.22)^2\\\\h = 89.04 \ m[/tex]

Thus, the height below the tower at which coin 1 pass coin 2 is 89.04 m.

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Một vật được ném lên trên theo phương thẳng đứng. Người quan sát
thấy vật đó đi qua vị trí có độ cao h hai lần và khoảng thời gian giữa hai lần đó là
t. Tìm vận tốc ban đầu và thời gian chuyển động của vật từ lúc ném đến khi vật
rơi về vị trí ban đầu.

Answers

Answer:

Language -English plz I cant understand

PLEASE HELP ME WITH THIS PHYSICS QUESTION PLSSS!!!

Vf^2 = v0^2 + 2a (xf -x0)


Solve for v0

Answers

b. solve for Vo

[tex] vf ^{2} = vo^{2} + 2a(xf - xo) \\ vf ^{2} = vo ^{2} + 2axf - 2axo\\ vo ^{2} = vf ^{2} - 2axf + 2axo \\ vo = \sqrt{vf ^{2} - 2axf + 2axo } \\ vo = - \sqrt{vf^{2} - 2axf + 2axo } [/tex]

I hope I helped you ^_^

A parachutist relies on air resistance (mainly on her parachute) to decrease her downward velocity. She and her parachute have a mass of 58.0 kg and air resistance exerts a total upward force of 670 N on her and her parachute. What is the combined weight of the parachutist and parachute? Calculate the net force on the parachutist. What is the magnitude of the acceleration of the parachutist?

Answers

Answer:

1. F = 569 N

2. F_net = 101N

3. a = 1.74m/s²

Explanation:

Weight is a force measurement.

F = m*a

F = 58.0kg*9.81m/s²

F = 568.98 N

F_net = 670N+(-569N)

F_net = 101N

a = F/m

a = 101N/58.0kg

a = 1.74m/s²

The combined weight of the parachutist and parachute is equal to 568.98 N and the net force due to air resistance exerts a total upward force of 101.02N on her and her parachute, then the magnitude of the acceleration of the parachutist is 1.74 m/s².

What is Newton's second law?

Newton's second law states that the resultant force acting on a body is proportional to the rate of change of momentum of that body.

If a  parachutist relies on air resistance to decrease her downward velocity.

The mass of the parachutist and her parachute is 58 kg

The air resistance exerts a total upward force of 670 N

The combined weight of the parachutist and parachute, W = mg

W = 58 × 9.81

W = 568.98 N

The net force on the parachutist = 670 - 568.98 = 101.02 N

The acceleration of the Parachutist = Net force/mass

a = F/m

a = 101.02/58

a  = 1.74 m/s²

Thus, the magnitude of the acceleration of the parachutist would be 1.74m/s².

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g Suppose that the Red cart has a mass of 500 grams, and the blue cart has a mass of 250 grams, and they are connected with a spring of spring constant 100 N/m. If the Red cart is held, and the Blue cart is pulled away with a force 19.2 N, what will be the force on the Red cart (write your answer in newtons, magnitude only)

Answers

Answer:

19.2 N

Explanation:

After any transients dissipate, all forces must balance. The red cart never moves.

what additional load will be required to cause the extension of 2.0cm when an elastic wire extend by 1.0cm when a load of 20g range from it

Answers

Answer:

The additional load is 20g

Write down the principals of them lever in points .​

Answers

The major principal of leaver is

load × load distance = effort × effort distance

where,

effort dis= distance between effort and fulcrum

load distance = distance between load and fulcrum......

need help with question after number 40

Answers

Answer:

4 device 3 e e and 4 divide 3 + 5 barabar 11 33 size to 46 size 35 and size 49 browser


In the diagram, the amplitude of the wave is shown by:

A
B
C
D

Answers

Answer:

A.

Explanation:

Amplitude measures how much a wave rises or falls. This is illustrated by A.

In the diagram, the amplitude of the wave is shown by A.

What is Amplitude?The amplitude of a periodic variable is a measure of its change in a single period. The amplitude of a non-periodic signal is its magnitude compared with a reference value.

There are various definitions of amplitude, which are all functions of the magnitude of the differences between the variable's extreme values.

The amplitude of a variable is simply a measure of change relative to its central position. In contrast, magnitude is a measure of the distance or quantity of a variable irrespective of its direction.

Amplitude is a property that is unique to waves and oscillations.

Therefore, in the diagram, the amplitude of a wave is shown by A.

To learn more about amplitude, refer:

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which of the following can not happen when a light ray strikes a new medium

Answers

Answer:

amplification

Explanation:

reflection can happen

some amount of lighr get absorbed

something gets refracted

but amplification cant

3 of 3 : please help got an extra day for a test and i don’t get this (must show work) points and brainliest!

Answers

Explanation:

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

Multiply both sides by 2 and then divide by m to get

[tex]\dfrac{2qV}{m} = v^2[/tex]

Take the square root of both sides to get

[tex]v = \sqrt{\dfrac{2qV}{m}}[/tex]

A 0.3 kg mass attached to a 1.5 m long string is whirled in a horizontal circle at a speed of 6.0 m/s. What is the tension in the string? (neglect gfavity)

Answers

Answer:

Hi I hope this is correct!

Explanation:

You can use this formula to solve this question T = mv^2/R

m = 0.3 kg , v = 6.0 m/s , R = 1.5 m

T = (0.3 kg)(6.0 m/s)^2 / 1.5 m  

  = 7.2 Newtons

Hope this helps! Best of luck <3

A cowgirl ties one end of a 10.0 m long rope to a fence post (to the left of the cowgirl) and pulls on the other end so the rope is stretched horizontally with a tension of 140 N. The mass of the rope is

Answers

Answer:

Mass is zero if the whole rope is horizontal. Imaginary rope.

Explanation:

With any mass at all, only a small section of the rope will be truly horizontal. The rope curve will be a a catenary.

2. A parallel-plate capacitor has a capacitance of C. If the area of the plates is doubled and
the distance between the plates is doubled, what is the new capacitance?
A) C/4
B) C/2
C)C
D) 4C

Answers

(C)

Explanation:

The capacitance C of a parallel plate capacitor is given by

[tex]C = \epsilon_0 \dfrac{A}{d}[/tex]

Let C' be the new capacitance where the area and the plate separation distance are doubled. This gives us

[tex]C' = \epsilon_0\dfrac{A'}{d'} = \epsilon_0\left(\dfrac{2A}{2d}\right) = \epsilon_0 \dfrac{A}{d} = C[/tex]

A block of mass 2 kg is launched by compressing a spring of force constant 1200 N/m. The block slides on a frictionless surface, up a 1 meter tall ramp, then it enters a region of rough surface. It comes to a stop after traveling 3 meters over the rough surface. The coefficient of kinetic friction between the block and the rough surface is 0.40.

Required:
a. How many forces end up doing work on the block from release to stop?
b. What is the total non-conservative work done on the block?
c. What is the change in the spring potential energy of the block?

Answers

Answer:

zzyibgsdwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwe

Explanation:

Is the study of the moons places applied or pure science

Answers

Answer:

It is pure science

Explanation:

A basic knowledge for the discovery of unknown laws based on well controlled experiments and deductions from demonstrated facts or truths.

the rate of cooling determines ....... and ......​

Answers

Answer:

freezing point and melting point

The speed of a sound wave

A. Depends on wavelength.
B. Depends on the medium.
C. Depends on amplitude.
D. None of the above.

Answers

Answer:

B) the medium

Explanation:

B. Depends on the medium.

What is the volume of a metal block 3cm long by 2cm wide by 4cm high? What would be the volume of a block twice as long, wide, and high?

Answers

Answer:

Volume of a metal block = 24 cm^3

Volume of a block twice as long, wide and high = 192 cm^3

Explanation:

Volume of a block is measured in l*w*h and in the first block, the sides are 3, 2 and 4 and 3*2*4 = 24

Second block, just double each of the lengths to get 6*4*8 = 192

A roller coaster uses 800 000 J of energy to get to the top of the first hill. During this climb, it gains 500 000 J of potential energy and pauses (velocity = 0) for a fraction of a second at the very top before heading down the other side.

a) Draw a sankey diagram for a roller coaster's climb.

Answers

A roller coaster uses 800 000 J of energy to get to the top of the first hill. During this climb, it gains 500 000 J of potential energy and pauses for a fraction of a second at the very top before heading down the other side. At the top of the hill total, the kinetic energy of the roller coaster would be zero as the velocity is zero at the top of the hill, therefore the total mechanical energy is only because of potential energy.

What is mechanical energy?

Mechanical energy is the combination of all the energy in motion represented by total kinetic energy and the total stored energy in the system which is represented by total potential energy.

The expression for total mechanical energy is as follows

ME= KE+PE

As total mechanical energy is the sum of all the kinetic as well as potential energy stored in the system.As given in the problem a roller coaster uses 800000 J of energy to get to the top of the first hill. During this climb, it gains 500 000 J of potential energy which means 300000 J of energy is lost in the frictional energy while climbing the hill,

Thus at the top of the hill, the total energy of the roller coasters is only due to the potential energy.

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Differences between LED and CFL bulb..​

Answers

Explanation:

CFL bulbs were made to take the place of incandescent bulbs, which generate light as a result of heat. ... LED (light-emitting diode) is a type of bulb that produces light using a narrow band of wavelengths. LED lighting is more energy efficient than CFL bulb.

krichoffs law of current questions​

Answers

Answer:

Explanation:

       Kirchhoff's Current Law, often shortened to KCL, states that “The algebraic sum of all currents entering and exiting a node must equal zero.

           #I AM ILLITERATE

A woman pulls on a 6.00-kg crate, which in turn is connected to a 4.00-kg
crate by a light rope. The light rope remains taut. Compared to the 6.00-kg crate,
the lighter 4.00-kg crate

Please explain why any of these multiple choices is correct!

Answers

Answer:

B. is subject to a smaller net force but same acceleration.

Explanation:

F = m*a

So because our force applied is constant from the women pulling on the rope which means the acceleration is the same on both the 4kg create and the 6kg create. The only thing that changes here is the mass of the creates, so there is more tension force between the women and the 6kg create then there is between the 4kg create and the 6kg. It takes less force to move the 4kg create therefore the tension force is less between the two creates.

The net force on both crates is the same and the acceleration of both crates is the same.

The given parameters;

mass of the crate, m = 6 kgmass of the second crate, = 4 kg

The force on the 4kg crate is calculated as follows;

[tex]F_{4kg } = T + F[/tex]

The force on the 6kg crate is calculated as follows;

[tex]F_{6 kg} = -T + F[/tex]

The net force on both crates is calculated as follows;

[tex]\Sigma F= -T + F - (T + F)\\\\\Sigma F= -2T[/tex]

Thus, we can conclude that the net force on both crates is the same and the acceleration of both crates is the same.

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Suppose an electron is trapped within a small region and the uncertainty in its position is 24.0 x 10-15 m. What is the minimum uncertainty in the electron's momentum

Answers

Answer:

Uncertainty in position (∆x) = 24 × 10⁻¹⁵ mUncertainty in momentum (∆P) = ?Planck's constant (h) = 6.26 × 10⁻³⁴ Js

[tex]\longrightarrow \: \: \sf\Delta x .\Delta p = \dfrac{h}{4\pi} [/tex]

[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} {4 \times \frac{22}{7} } [/tex]

[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} { \frac{88}{7} } [/tex]

[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34} \times 7} { 8 } [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 8 \times 24 \times {10}^{ - 15} } [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 192 \times {10}^{ - 15} } [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} \times {10}^{15} } { 192} [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ -19} } { 192} [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 2} \times {10}^{ -19} } { 192} [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 21} } { 192} [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = 22.822\times {10}^{ - 21} [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = 2.2822 \times {10}^{1} \times {10}^{ - 21} [/tex]

[tex]\longrightarrow \: \: \underline{ \boxed{ \red{ \bf\Delta p = 2.2822 \times {10}^{ - 20} \: kg/ms}}}[/tex]

Present the ways to observe interference patterns on thin films. Why the thickness of thin films should be in scale of wavelength?

Answers

the president of The proposal ahr stay Sheba SBR Abba and and I send svrvs you svrvs

if C is The vector sum of A and B C = A + B What must be true about The directions and magnitudes of A and B if C=A+B? What must be tre about the directions and magnitudes of A and B if C=0? ​

Answers

Check attached photo

Check attached photo

Answer:

Explanation:

1. If C = A + B then the lines A and B may have the same magnitude or they may not. The direction of A for example may be northwest ↖️ and the direction of B must be south ⬇️ because the arrow of A and the point of B must connect. Then C’s direction is west ⬅️ because it shouldn’t be as equilibrium.

2. If C = 0 t means the force is at equilibrium. That means all forces add up to zero. A’s direction for example may be northeast ↗️ and the direction of B may be south ⬇️ and the direction of C must be west if it has to be at equilibrium.

The magnitude of A and B must be equal

Help! xoxo thank you

Answers

Like charges repel and opposite charges attract.

Which items are discounted?

—-
Ignore selected.

Answers

Answer:

nail file

facial tissues

body wash

deodorant

shampoo

Explanation:

if I'm going by the discount meaning correctly

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