elastic wire extend by 1.ocm when a load on 20g range from It, what additional load will it be required Cause the futher extension of 2.0cm​

Answer:

c

Explanation:

Liquid plastic is frozen in a physical change that increases its volume,it can be known about the plastic that Its mass will remain the same, therefore the correct answer is option C.

Anything which has mass and occupies space is known as matter ,mainly there are four states of matter solid liquid gases, and plasma.

These different states of matter have different characteristics according to which they vary their volume and shape.

It is known about plastic that its mass will remain the same when liquid plastic is frozen, by increasing its volume.

Liquid plastic is frozen in a physical change that increases its volume,it can be known about the plastic that Its mass will remain the same,  therefore the correct answer is C.

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Answer:

chk picture for eqn

Explanation:

Answer:

4.73 × 10^5

Explanation:

The comparison of the speeds and kinetic energy of the identical balls are as follows

The speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal

The reason for the above comparison results areas follows;

Known parameters;

First ball is thrown straight up

Second ball is thrown 30° above the horizontal

Third ball it thrown 30° below the horizontal

The fourth ball is thrown straight down

Unknown:

Comparison of the speed and kinetic energy of the four balls

Method:

The kinetic energy, K.E. = (1/2) × m × v²

The velocity of the ball, v = u × sin(θ)

Where;

u = The initial velocity of the ball

θ = The reference angle

θ = 90°

∴ [tex]v_y[/tex] = u

The final velocity of the ball as it strikes the ground is v₂ = u² + 2gh

Where;

h = The height of the cliff

∴ Kinetic energy of first ball, K.E.₁ = (1/2) × m × (u₁² + 2gh)²

K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²

Kinetic energy  K.E.₂ = (1/2) × m × ((0.5·u₂)² + 2·g·h)²

K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²

Kinetic energy K.E.₃ = (1/2) × m × ((0.5·u₃)² + 2·g·h)²

Kinetic energy K.E.₄ = (1/2) × m × (u₄² + 2gh)²

Therefore, as the ball strike the ground, the speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal

u₁ = u₄, K.E₁ = K.E.₄, u₂ = u₃, K.E₂ = K.E.₃

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Answer:

-4/7

Explanation:

Given the following

A=4i-10j and B= 7i+5j

A+ bB = 4i-10j + (7i+5j)b

A+ bB =  4i-10j + 7ib+5jb

A+ bB =

The vector along the x-axis is expressed as i + 0j

If the vector A+ bB is pointing in the direction of the x-axis then;

[tex]A+ bB * \frac{i+0j}{|i+0j|} = 0 \\ (4+7b)i-(10-5b)j* \frac{i+0j}{\sqrt{1^2+0^2} } = 0\\(4+7b)i-(10-5b)j *(i+0j) = 0\\4+7b-0 =0\\7b=-4\\b = -4/7[/tex]

Hence the value of b is -4/7

The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.

According to the statement, we have following system of vectorial equations:

[tex]\vec A = 4\,\hat {i} - 10\,\hat{j}[/tex] (1)

[tex]\vec {B} = 7\,\hat{i} + 5\,\hat{j}[/tex] (2)

[tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] (3)

By applying (1) and (2) in (3):

[tex](4\,\hat{i}-10\,\hat{j}) + \beta\cdot (7\,\hat{i}+5\,\hat{j}) = c\,\hat{i}[/tex]

[tex](4+7\cdot \beta)\,\hat{i} +(-10+5\cdot \beta)\,\hat{j} = c\,\hat{i}[/tex]

And we get two scalar equations after analyzing each component:

[tex]4+7\cdot \beta = c[/tex] (4)

[tex]-10+5\cdot \beta = 0[/tex] (5)

We solve for [tex]\beta[/tex] in (5):

[tex]\beta = 2[/tex]

And for [tex]c[/tex] in (4):

[tex]c = 4+7\cdot (2)[/tex]

[tex]c = 18[/tex]

The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.

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Answer

E - 1/2 K x^2      potential energy of compressed spring

E = 1/2 * 4 N / m * (.5 m)^2 = 2 * .5^2 N-m = .5 N-m

momentum p= m x v = 15 x -4 = -60 N.s

Answer:

d = 6.32 m

Explanation:

Given that,

The mass of a puck, m = 2 kg

It is pushed straight north with a constant force of 5N for 1.50 s and then let go.

We need to find the distance covered by the puck when move from rest in 2.25 s.

We know that,

F = ma

[tex]a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2[/tex]

Let d is the distance moved in 2.25 s. Using second equation of motion,

[tex]d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m[/tex]

So, it will move 6.32 m from rest in 2.25 seconds.

Explanation:

Hey there!

According to the question;

A person can lift mass of 60 kg on earth.

mass(m1) = 60kg

acceleration due to gravity on earth (a) = 9.8m/s²

Now;

force (f) = m.a

= 60*9.8

= 588 N

Since, there is application of same magnitude of force on moon,

mass(m) =?

acceleration due to gravity on moon (a) = 1.67m/s²

Now;

force (f) = m.a

588 = m*1.67

m = 352.09 kg

Therefore, the person who can lift the mass of 60 kg on earth can lift mass of 352 kg on moon.

Hope it helps!

Answer:

The speed of proton is 2.1 x 10^5 m/s .

Explanation:

potential difference, V = 234 V

let the initial speed of the proton is v.

The kinetic energy of proton is

KE = q V

[tex]0.5 mv^2 = e V \\\\0.5\times 1.67\times 10^{-27} v^2 = 1.6\times 10^{-19} \times 234\\\\v=2.1\times 10^5 m/s[/tex]

Explanation:

Work cannot be increased by using a machine of some kind.

Work cannot be increased by using a machine of some kind.

A machine is any device in which the effort applied at one end overcomes a load at the other end.

Machines are generally used to perform different tasks faster.

However, a simple machine can not be used to increase the amount of work done at any time.

Force, speed and torque can all be increased using machines.

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Answer:

Answer:30 cm

Answer:30 cmExplanation:

Answer:30 cmExplanation:Given=ROC= 60cm

Answer:30 cmExplanation:Given=ROC= 60cmObject be placed so that the rays that came from the object to them mirror are reflected from the mirror, and, then travel parallel to each other= 30cm at focus.

The index of refraction of the plastic is approximately 1.461

The known values in the question are;

The thickness of the piece of plastic placed on the dot = 1.30 cm

The height to which the microscope objective is raised to bring the dot back to focus = 0.410 cm

The unknown values in the question are;

The index of refraction

Strategy;

Calculate the refractive index by making use of the apparent height and real height method for the black dot under the thick piece of plastic

[tex]\mathbf{ Refractive \ index, n = \dfrac{Real \ depth}{Apparent \ depth}}[/tex]

The real depth of the dot below the piece of plastic, d₁ = 1.30 cm

The apparent depth of the dot, d₂ = The actual depth - The height to which the microscope is raised

Therefore;

The apparent depth of the dot, d₂ = 1.30 cm - 0.410 cm = 0.89 cm

[tex]The \ refractive \ index, \ n = \dfrac{d_1}{d_2}[/tex]

Therefore, n = 1.30/0.89 ≈ 1.461

The refractive index of the plastic block, n ≈ 1.461

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Answer:

Gap left = Change in length on heating

Gap=Initial length×Coefficient of linear expansion×change in temperature

Gap=10×0.000012×15m

⟹Gap=0.0018 m

this is an example u have to put your equation in it

Answer:

The distance between mirror and you is 2 ft.

Explanation:

diameter, d = 8 ft

radius of curvature, R = 4 ft

magnification, m = 0.5

focal length, f = R/2 = 4/2 = 2 ft

let the distance of object is u and the distance of image is v.

[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\\\frac{1}{2}=\frac{1}{v}+\frac{1}{u}\\\\v = \frac {2 u}{u - 2}[/tex]

Use the formula of magnification

[tex]m = \frac{v}{u}\\\\0.5 =\frac { u}{u - 2}\\ \\u - 2 = 2 u \\\\u = -2 ft[/tex]

Answer:

The rise in temperature is 0.06 K.

Explanation:

mass of bullet, m = 15 g

initial speed, u = 865 m/s

final speed, v = 534 m/s

mass of water, M = 13.5 kg

specific heat of water, c = 4200 J/kg K

The change in kinetic energy

[tex]K = 0.5 m(u^2 - v^2)\\\\K = 0.5\times 0.015\times (865^2-534^2)\\\\K = 3473 J[/tex]

According to the conservation of energy, the change in kinetic energy is used to heat the water.

K = m c T

where, T is the rise in temperature.

3473 = 13.5 x 4200 x T

T = 0.06 K

The corresponding frequency of this sinewave, in kHz, expressed to 3 significant figures is: 155 kHz.

Given the following data:

Note: μs represents microseconds.

Conversion:

1 μs = [tex]1[/tex] × [tex]10^-6[/tex] seconds

645 μs = [tex]645[/tex] × [tex]10^-6[/tex] seconds

To find corresponding frequency of this sinewave, in kHz;

Mathematically, the frequency of a waveform is calculated by using the formula;

[tex]Frequency = \frac{1}{Period}[/tex]

Substituting the value into the formula, we have;

[tex]Frequency = \frac{1}{645 * 10^-6}[/tex]

Frequency = 1550.39 Hz

Next, we would convert the value of frequency in hertz (Hz) to Kilohertz (kHz);

Conversion:

1 hertz = 0.001 kilohertz

1550.39 hertz = X kilohertz

Cross-multiplying, we have;

X = [tex]0.001[/tex] × [tex]1550.39[/tex]

X = 155039 kHz

To 3 significant figures;

Frequency = 155 kHz

Therefore, the corresponding frequency of this sinewave, in kHz is 155.

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Answer:

T = 1/f = 1/50(s)

ω = 2πf = 100π (rad/s)

(vote 5 sao nhó :3 )

Answer:

It will travel Vx * t = 30 m/s * 2 s = 60 m

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.

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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i'm pretty sure the answer is A wider

Electric lines of force where intensity of electric field is maximum when its wider.

The physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them, is known as an electric field (also known as an E-field. It can also refer to a system of charged particles' physical field.

Electric charges and time-varying electric currents are the building blocks of electric fields. The electromagnetic field, one of the four fundamental interactions (also known as forces) of nature, manifests itself in both electric and magnetic fields.

Electrical technology makes use of electric fields, which are significant in many branches of physics. For instance, in atomic physics and chemistry, the electric field acts as an attracting force to hold atoms' atomic nuclei and electrons together.

Therefore, Electric lines of force where intensity of electric field is maximum when its wider.

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Answer:

approaches infinity

Explanation:

There are two momentums, the classical momentum which is equal to the product of mass and velocity, and the relativistic momentum, the one we should look at when we work with high speeds, and this happens because massive objects have a speed limit, in this case, we are approaching the speed of light, so we need to work with the relativistic momentum instead of the classical momentum.

The relativistic momentum can be written as:

[tex]p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]

where

u = speed of the object relative to the observer, in this case we have that u tends to c, the speed of light.

m = mass of the object

c = speed of light.

So, as u tends to c, we will have:

[tex]\lim_{u \to c} p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]

Notice that when u tends to c, the denominator on the first term tends to zero, thus, the relativistic momentum of the object will tend to infinity.

Then the correct option is infinity, as the particle speed approaches the speed of light, the relativistic momentum of the particle tends to infinity.

Answer:The radioactive waste

Explanation:Fission is the splitting of a heavy unstable nucleus into two Lighter nuclei

Answer:

The smallest and largest areas could be 6400 m and 7500 m, respectively.

Explanation:

The area of a rectangle is given by:

[tex] A = l*w [/tex]

Where:

l: is the length = 100 m

w: is the width

We can calculate the smallest area with the lower value of the width.

[tex] A_{s} = 100 m*64 m = 6400 m^{2} [/tex]                            

And the largest area is:

[tex] A_{l} = 100 m*75 m = 7500 m^{2} [/tex]  

Therefore, the smallest and largest areas could be 6400 m and 7500 m, respectively.            

I hope it helps you!                        

Answer:

the largest areas that the field could be is [tex]A_l[/tex]=7587.75 m

the smallest areas that the field could be is [tex]A_s[/tex]=6318.25 m

Explanation:

to the find the largest and the smallest area of the field measurement error is to be considered.

we have to find the greatest possible error, since the measurement was made nearest whole mile, the greatest possible error is half of 1 mile and that is 0.5m.

therefore to find the largest possible area we add the error in the mix of the formular for finding the perimeter with the largest width as shown below:

[tex]A_l[/tex]= (L+0.5)(W+0.5)

(100+0.5)(75+0.5) = (100.5)(75.5) = 7587.75 m

To find the smallest length we will have to subtract instead of adding the error factor value of 0.5 as shown below:

[tex]A_s[/tex]= (L-0.5)(W-0.5)

(100-0.5)(64-0.5) = (99.5)(63.5) = 6318.25 m

Answer:

233.33°F

Explanation:

(385K - 273.15) * 9/5 + 32 = 233.33°F

Answer:

the initial velocity of the car is 12.04 m/s

Explanation:

Given;

force applied by the break, f = 1,398 N

distance moved by the car before stopping, d = 25 m

weight of the car, W = 4,729 N

The mass of the car is calculated as;

W = mg

m = W/g

m = (4,729) / (9.81)

m = 482.06 kg

The deceleration of the car when the force was applied;

-F = ma

a = -F/m

a = -1,398 / 482.06

a = -2.9 m/s²

The initial velocity of the car is calculated as;

v² = u² + 2ad

where;

v is the final velocity of the car at the point it stops = 0

u is the initial velocity of the car before the break was applied

0 = u² + 2(-a)d

0 = u² - 2ad

u² = 2ad

u = √2ad

u = √(2 x 2.9 x 25)

u =√(145)

u = 12.04 m/s

Therefore, the initial velocity of the car is 12.04 m/s

Answer:

About 37.70 radians.

Explanation:

1 revolution = 2[tex]\pi[/tex] radians

∴ 6 revolutions = (6)(2[tex]\pi[/tex] radians)

6 revolutions = 37.6991 or ≈ 37.70 radians