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Work is transfer of energy in an object when it travel some distance by external force,

Work= force × displacement

Answer: [tex]0.05\ mm[/tex]

Explanation:

Given

Cross-sectional area of wire [tex]A_1=4\ mm^2[/tex]

Extension of wire [tex]\delta l=0.1\ mm[/tex]

Extension in a wire is given by

[tex]\Rightarrow \delta l=\dfrac{FL}{AE}[/tex]

where, [tex]E=\text{Youngs modulus}[/tex]

[tex]\Rightarrow \delta_1=\dfrac{FL}{A_1E}\quad \ldots(i)[/tex]

for same force, length and material

[tex]\Rightarrow \delta_2=\dfrac{FL}{A_2E}\quad \ldots(ii)[/tex]

Divide (i) and (ii)

[tex]\Rightarrow \dfrac{0.1}{\delta_2}=\dfrac{A_2}{A_1}\\\\\Rightarrow \delta_2=0.1\times \dfrac{4}{8}\\\\\Rightarrow \delta_2=0.05\ mm[/tex]

Answer:

Explanation:

The frictionless surface implies that the speed of the spring is at a max. When the speed of the spring is at its max, the potential energy in the spring is 0. Use the equation for the Total Energy in a Spring/Mass System:

KE + PE = [tex]\frac{1}{2}kA^2[/tex] where KE is the Kinetic Energy available to the spring, PE is the potential energy available to the spring, and the sum of those is equal to one-half times the spring constant, k, times the amplitude of the spring's movement away from the equilibrium position. Sometimes this amplitude is the same as the displacement of the spring. This can be tricky. But since we are only given one value for the distance, we are going to use it as an amplitude. Keeping in mind that the PE is 0 when KE is at its max, then the equation becomes

KE + 0 = [tex]\frac{1}{2}kA^2[/tex] or to put it simpler terms:

KE = [tex]\frac{1}{2}kA^2[/tex] We need to find the value for KE before we can fully solve the problem we are being tasked with.

Filling in using the info given:

[tex]KE=\frac{1}{2}(2.0)(.06)^2[/tex] Notice I added another place of significance to the 2 because 1 simply isn't enough and the physics teacher in me can't handle that. Simplifying a bit:

[tex]KE=(.06)^2[/tex] because the k = 2 cancels out the 2 in the denominator of the 1/2. So

KE = 3.6 × [tex]10^{-3[/tex]

Now plug that in for KE and solve for v:

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

[tex]3.6*10^{-3}=\frac{1}{2}(5.0)v^2[/tex] and

[tex]v=\sqrt{\frac{2(3.6*10^{-3})}{5.0} }[/tex] gives us a velocity of

v= [tex]3.8*10^{-2[/tex]

Answer:

3.2075*10^16

Explanation:

Q=P/V just search up a converter and youll get 30V and so you do 15/30 which is a half and a single coulomb is 6.415*10^16 so you half it. I belive this is correct if you dont belive me wait for someone else smarter to answer and compare.

Answer:

Sedimentary rocks are porous.

Explanation:

So option A) Sedimentary is your answer. Have a great summer

First, calculate the components of its initial velocity v0:

v0x = v0cos10 = 24.6 m/s

v0y = v0sin10 = 4.34 m/s

The ball reaches its peak when vy = 0. Let's calculate the time it takes for vy to become zero:

vy = v0y - gt ---> t = v0y/g = 0.44 s

The horizontal distance it travels in this time is

x = v0xt = (24.6 m/s)(0.44 s)

= 10.8 m

Note that the net is 15 m away. After traveling a horizontal distance of 10.8 m, the height of the ball is

y = -(1/2)gt^2 + v0yt + 2.1

= -(4.9 m/s^2)(0.44 s)^2 + (4.34 m/s)(0.44 s) + 2.1 m

= -0.95 m + 1.9 m + 2.1 m

= 3.05 m

Note that this is the height of the ball at its peak. While the ball is well above the net at its peak, it is well short of its required horizontal distance to clear it. Instead, let's find the time it takes for the tennis ball to travel a horizontal distance of 15 m first:

x = v0xt ----> t = x/v0x = (15 m)/(24.6 m/s) = 0.61 s

Then calculate the height y when t = 0.61 s. If y > 0.9 m (height of the net), then the ball will clear the net.

y = -4.9t^2 + v0yt + 2.1

= -4.9(0.61 s)^2 + (4.34 m/s)(0.61 s) + 2.1 m

= -1.82 m + 2.65 m + 2.1 m

= 2.93 m

Yes, the ball will clear the net by 2.03 m.

Explanation:

Start with what you know and list your knowns and unknowns

F = ma

F= 3N

m = 6kg

a =?

3N = 6kg x a

solve for a

3N / 6kg = a

Answer:

E = 1.1 10⁶ N / C

Explanation:

In this case they indicate that we can approximate the membrane as a parallel plate capacitor, we can use

            E = [tex]\frac{\sigma}{\epsilon_o }[/tex]

note that in this case the electric field created by each plate goes in the same direction, they are added

let's calculate

            E =  [tex]\frac{10^{-5}}{8.85 \ 10^{-12}}[/tex]

            E = 1.1 10⁶ N / C

Answer:

1.84 m/s

Explanation:

Applying,

The kinetic energy of the pinball = Elastic energy of the spring

mv²/2 = ke²/2

mv² = ke².................. Equation 1

Where m = mass of the pin ball, v = velocity of the pin ball, k = force constant of the spring, e = extension of the spring.

make v the subject of the equation

v = √(ke²/m)................ Equation 2

From the question,

Given: e = 0.15 m, k = 30 N/m, m = 0.20 kg

Substitute these values into equation 2

v = √[(30×0.15²)/0.2]

v = 1.84 m/s

Answer:

15m

Explanation:

W=f×s

[tex]s = \frac{w}{f} \\ s = \frac{900j}{60n} \\ s = 15m [/tex]

Given:

Force,

Work done,

We know,

→ [tex]W = F\times s[/tex]

or,

→ [tex]s = \frac{W}{F}[/tex]

By putting the values, we get

      [tex]= \frac{900}{60}[/tex]

      [tex]= 15 \ m[/tex]

Thus the above response i.e., "Option c" is right.

Learn more about work done here:

https://brainly.com/question/8625856

Answer:

10 is the correct answer

Answer:

Total Distance: 420 meters

Diameter: 42 cm

Notice the units meters vs cm

420÷ 42 = 10 total revolutions

Answer:    

Explanation:

Answer:

Explanation:

The formula to find Potential Energy is PE = mgh, where m is the mass of the object, g is gravity, and h is the height from which the object can potentially fall. Because this is linear, then PE will increase as either the mass or the height increase (gravity is constant at 9.8 m/s/s). If the mass of the ball being dropped doesn't change, then the only thing that determines this ball's max PE is the height from which it is dropped; max PE ALWAYS occurs at the highest point from which an object can potentially fall. So your answer is "At the top".

Answer:

It’s like medium uk

Explanation:

The size of electron charge is estimated to be 1.6 × 10-¹⁹

Answer:

84.05

Explanation:

F=f+F

F=49.05+35

=84.05

Answer:

9.5 m/s

Explanation:

Distance, S = 150m

Acceleration, a = 0.3 m/s^2

Initial velocity, u = 0 m/s

Final velocity, v

Use kinematics equation

v^2 - u^2 = 2aS

v^2 - 0 = 2*0.3*150 = 90

v = sqrt(90) = 9.49 m/s

Answer:

 a = 10 ρ_air g [tex]\frac{\Delta P}{m P_o}[/tex]

Explanation:

Let's solve this problem in parts, with the initial data the camera rises slowly, so we can assume that at constant speed, we apply the equilibrium condition

         B - W = 0

The thrust is given by Archimedes' law

         B = rho_aire g V_body

The volume of the body can be found from the ideal gas ratio

         P V = n R T

let's use the subscript "o" for the initial concisions

         V₀ = (nR) T₀/P₀               1

we substitute

          10 ρ_air g (nR) T₀ /P₀ = W

when the storm approaches the pressure decreases, P

If we use the ideal gas equation

           V = (nR) T₀ / P

we combine this equation with equation 1

            V = V₀P₀ / P

if we write the pressure

             P = P₀ -ΔP

we substitute

            V = [tex]\frac{V_oP_o}{P_o - \Delta P} =V_o \ ( 1 - \frac{\Delta P}{P_o} )^{-1}[/tex]

we expand serie  and eliminate higher order terms

            V = V₀ ([tex]1+ \frac{\Delta P}{Po}[/tex])

with this expression we can write the thrust

             B = B₀ + ΔB

Newton's second law for the new conditions is

             B - W = m a

             (B₀ + ΔB) - W = ma

              ΔB = m a

              a = ΔB / m

              a = 10 ρ_air g [tex]\frac{\Delta P}{m P_o}[/tex]

This is the initial acceleration of the camera

Answer:

Breaker

Explanation: A wave is a motion or disturbance that transfers energy from one location to another without any permanent displacement of the particles involved in the wave motion.

Wave motion can occur in various media such as water, air etc a wave is described as a breaker when it steepens and before it finally stops or losses its energy/collapses.

Answer:

pp i hate khan academy pls help me

Answer: A: 70/2=35

B: 35/2=17

C: 17.5/2=8.75

D: 8.75 of C-14 will be left

E: 5,730 years

F: 5,631

Explanation:

that’s all I got, hope I helped kinda

Answer:

A. Increase the distance between the objects.

D. Decrease the mass of one of the objects.

Explanation:

The smaller the mass of an object the less gravity it has and the farther away two objects are the less gravitational force

The direction of acceleration due to gravity is always towards earth, going downwards.

mark me brainliesttt :))

Answer:

The correct answer is - At the maximum height, the velocity and acceleration vectors are perpendicular to each other.

Explanation:

When the football reaches maximum height, then the vertical component of velocity will be zero and therefore the only component of velocity left will be the horizontal component, and acceleration of the object will be downward, due to gravity.

So at the maximum height, there is horizontal velocity only which means velocity is horizontal and acceleration is vertical thus, the velocity and acceleration are perpendicular to each other.

Answer:

im guessing red giant-

Answer:

[tex]W=17085KJ[/tex]

Explanation:

From the question we are told that:

Height [tex]H=16m[/tex]

Radius [tex]R=3[/tex]

Height of water [tex]H_w=9m[/tex]

Gravity [tex]g=9.8m/s[/tex]

Density of water [tex]\rho=1000kg/m^3[/tex]

Generally the equation for Volume of water is mathematically given by

 [tex]dv=\pi*r^2dy[/tex]

 [tex]dv=\frac{\piR^2}{H^2}(H-y)^2dy[/tex]

Where

   y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

 [tex]dw=(pdv)g (H-y)[/tex]

Substituting dv

 [tex]dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)[/tex]

 [tex]dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy[/tex]

Therefore

 [tex]W=\int dw[/tex]

 [tex]W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy[/tex]

 [tex]W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)[/tex]

 [tex]W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0[/tex]

 [tex]W=3420.84*0.25[2401-65536][/tex]

 [tex]W=17084965.5J[/tex]

 [tex]W=17085KJ[/tex]

'

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Answer:

391.5 J

Explanation:

The amount of work done can be calculated using the formula:

Looking at the formula, we can see that the mass of the object does not affect the work done on it.

Substitute the force applied and the displacement of the object into the equation.

The amount of work done on the object is 391.5 J in order to move it 4.5 meters with an applied force of 87 Newtons.

Mass doesn't matter on amount of work done .We can calculate amount of work done through Force and Displacement

[tex]\boxed{\sf W=Fd}[/tex]

[tex]\\ \sf\longmapsto Work\:Done=87(4.5)[/tex]

[tex]\\ \sf\longmapsto Work\:done=391.5J[/tex]

Answer:

a. Acceleration, a = 0.28 m/s²

b. Distance, S = 156 meters

Explanation:

Given the following data;

Initial velocity = 30 km/h

Final velocity = 45 km/h

Time = 15 seconds

a. To find the acceleration;

Conversion:

30 km/h to m/s = 30*1000/3600 = 8.33 m/s

45 km/h to m/s = 45*1000/3600 = 12.5 m/s

Mathematically, acceleration is given by the equation;

[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]

Substituting into the equation;

[tex]a = \frac{12.5 - 8.3}{15}[/tex]

[tex]a = \frac{4.2}{15}[/tex]

Acceleration, a = 0.28 m/s²

b. To find the distance travelled, we would use the second equation of motion given by the formula;

[tex] S = ut + \frac {1}{2}at^{2}[/tex]

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

[tex] S = 8.3*15 + \frac {1}{2}*(0.28)*15^{2}[/tex]

[tex] S = 124.5 + 0.14*225[/tex]

[tex] S = 124.5 + 31.5 [/tex]

S = 156 meters

Answer:

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

Explanation:

The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:

[tex]I = m\cdot (\vec{v}_{2} - \vec{v_{1}})[/tex] (1)

Where:

[tex]I[/tex] - Impulse, in kilogram-meters per second.

[tex]m[/tex] - Mass, in kilograms.

[tex]\vec{v_{1}}[/tex] - Initial velocity of the hockey park, in meters per second.

[tex]\vec{v_{2}}[/tex] - Final velocity of the hockey park, in meters per second.

If we know that [tex]m = 0.2\,kg[/tex], [tex]\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right][/tex] and [tex]\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right][/tex], then the impulse applied by the stick to the park is approximately:

[tex]I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right][/tex]

[tex]I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right][/tex]

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.