A lens of focal length 12cm forms an
three times the size of the
to the object. The distance between the object and the image is what
​

Answer:

Kinetic energy is the energy of motion

Answer:

Energy which a body possesses by virtue of being in motion

Explanation:

Answer:

  the vaulter has a height of 6.112 meters

Explanation:

Base runner: no question content.

__

Pole vaulter:

Initial kinetic energy* is ...

  KE = (1/2)mv^2 = (1/2)(45 kg)(11 m/s)^2 = 2722.5 J

Above the bar, her kinetic energy is ...

  KE = (1/2)mv^2 = (1/2)(45 kg)(1.1 m/s)^2 = 27.225 J

Then the amount of kinetic energy converted to potential energy is ...

  2722.5 J -27.225 J = 2695.275 J

This corresponds to a change in height of ...

  PE = mgh

  2695.275 J = (45 kg)(9.8 m/s^2)h

  h = 2695.275/(45·9.8) m = 6.112 m

Her altitude above the bar is 6.112 meters.

_____

* Here, we use the traditional equations for kinetic and potential energy, which use "m" to represent mass. When we fill in the numbers, we attach units to the numbers in which "m" represents "meters". We trust you can sort this without confusion.

Answer:

b

Explanation:

If a substance is boiled repeatedly and stirred, but the solute never mixes with the solvent then the chemical properties of the solute and solvent are different, therefore the correct answer is option B.

A chemical compound is a combination of two or more either similar or dissimilar chemical elements.

For example, H₂O is a chemical compound made up of two oxygen atoms and a single hydrogen atom.

As given in the problem a substance is boiled repeatedly and stirred, but the solute never mixes with the solvent,

The correct response is option B because the chemical characteristics of the solute and solvent are different if the material is repeatedly boiled and stirred without the solute and solvent mixing.

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Answer:

q3 = 21.9 nC

Explanation:

By the Gauss theorem you have that the electric flux in a Gaussian surface is given by:

[tex]\Phi_E=\frac{Q}{\epsilon_o}[/tex]      (1)

ФE: electric flux = -218Nm^2/C

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/(Nm^2)

You can consider the spherical shell as a Gaussian surface. Then, the net charge inside the surface is:

[tex]Q=-18.0nC+38.0nC+q_3[/tex]     (2)

where charge q3 is unknown charge of the third object:

You replace the equation (2) into the equation (1), and you solve for q3:

[tex]\epsilon_0 \Phi_E=-18.0*10^{-9}C+38.0*10^{-9}C+q_3\\\\\epsilon_0 \Phi_E=20*10^{-9}C+q_3\\\\q_3=(8.85*10^{-12}C^2/(Nm^2))(-218Nm^2/C)-20*10^{-9}C\\\\q_3=2.19*10^{-9}C=21.9nC[/tex]

hence, the charge of the third object is 21.9 nC

Answer:

im sure your already past this but it's E.

Explanation:

This is because in this case potential energy is linear to height, which means that the higher the more potential energy.

Answer:

Explanation:

capacitance of the capacitor C = 4πε₀R

= 1 x .103 / (9 x 10⁹ )  [ 1/ 4πε₀  = 9 x 10⁹ . ]

= .0114 x 10⁻⁹ F

potential  V = 4.5 x 10³ v

charge Q = CV

.0114 x 10⁻⁹  x 4.5 x 10³

= .0515 x 10⁻⁶ C

charge on one electron = 1.6 x 10⁻¹⁹

no of electrons to be removed

= .0515 x 10⁻⁶ / 1.6 x 10⁻¹⁹

= .032 x 10¹³

3.2 x 10¹¹ electrons.

Explanation:

The force of the roller-coaster track on the cart at the bottom is given by :

[tex]F=\dfrac{mv^2}{R}[/tex], m is mass of roller coaster

Case 1.

R = 60 m v = 16 m/s

[tex]F=\dfrac{(16)^2m}{60}=4.26m\ N[/tex]

Case 2.

R = 15 m v = 8 m/s

[tex]F=\dfrac{(8)^2m}{15}=4.26m\ N[/tex]

Case 3.

R = 30 m v = 4 m/s

[tex]F=\dfrac{(4)^2m}{30}=0.54m\ N[/tex]

Case 4.

R = 45 m v = 4 m/s

[tex]F=\dfrac{(4)^2m}{45}=0.36m\ N[/tex]

Case 5.

R = 30 m v = 16 m/s

[tex]F=\dfrac{(16)^2m}{30}=8.54m\ N[/tex]

Case 6.

R = 15 m v =12 m/s

[tex]F=\dfrac{(12)^2m}{15}=9.6m\ N[/tex]

Ranking from largest to smallest is given by :

F>E>A=B>C>D

Answer:

f' = 358442.3 Hz

Explanation:

This is a problem about the Doppler's effect. To find the perceived frequency of the observers you use the following formula:

[tex]f'=f(\frac{v}{v\pm v_s})[/tex]

v: speed of sound = 342 m/s

vs : speed of the source (jet) = 1220 km/h (1h/3600s)*(1000 m/ 1km) = 338.88 m/s  (it is convenient to convert the units of vs to m/s)

f: frequency emitted by the source = 3270 Hz

f': perceived frequency

Due to the jet is getting closer to the observers, the sing of the denominator in equation (1) is minus (-). Then, you replace the values of f, vs and v:

[tex]f'=(3270s^{-1})(\frac{342m/s}{342m/s-338.88m/s})=358442.3 Hz[/tex]

Hence, the perceived frequency by the observers is 358442.3 Hz

Answer:

[tex]f_o=359466.42Hz[/tex]

Explanation:

You can solve this problem using doppler effect formula. The Doppler effect is the phenomenon by which the frequency of the waves perceived by an observer varies when the emitting focus or the observer itself moves relative to each other

Doppler effect general case is given by:

[tex]f_o=f\frac{v\pm v_o}{v \mp v_s} \\\\Where:\\\\f=Actual\hspace{3}frequency\\f_o=Observed\hspace{3}frequency\\v=Speed\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves\\v_o=Velocity\hspace{3}of\hspace{3}the\hspace{3}observer\\v_s=Velocity\hspace{3}of\hspace{3}the\hspace{3}source[/tex]

Now, you can use the following facts:

[tex]+v_o[/tex] Is used when the observer moves towards the source

[tex]-v_o[/tex] Is used when the observer moves away from the source

[tex]-v_s[/tex] Is used when the source moves towards the observer

[tex]+v_s[/tex] Is used when the source moves away from the observer

Since the source is alone in motion towards the observer, the formula is given by:

[tex]f_o=f\frac{v}{v-v_s} \\\\Where:\\\\v_s=1220km/h\approx 338.8888889m/s\\v=342m/s\\f=3270Hz[/tex]

Therefore:

[tex]f_o=(3270) \frac{342}{342-338.8888889} = 359466.42Hz[/tex]

Answer:

It may turn into sedimentary rock when exposed to pressure.

Explanation:

When successive layer of sand is deposited on top of sand, the bottom layers go deeper and deeper. They are exposed to high pressures and temperatures and turn into hard rocks.

Answer:

The velocity of the ions relative to the earth observer is  [tex]v = 0.63 c[/tex]

Explanation:

From the question  we are told that

     The speed of the alien ship is  [tex]v_a = 0.3c[/tex]

     The speed of the iron gun is  [tex]v_g = 0.4c[/tex]

The motion of the ions  relative to the earths observer can be mathematically represented as

       [tex]v = \frac{v_a + v_g}{[\frac{1+ v_av_g}{c^2} ]}[/tex]

substituting values

     [tex]v = \frac{ 0.3c + 0.4c}{[\frac{1+ 0.3c * 0.4c}{c^2} ]}[/tex]

 =>   [tex]v = 0.63 c[/tex]

Answer:

If one places ones's hands into a sink full of hot water, heat is transferred from the hot water to the hands. Heat only flows when there is a temperature gradient, i.e when there is a difference in temperature between a hot and a cold body. Heat is always transferred from a hot body to a cold body and never the reverse in a normal system.

At the boundary of the hand and water, heat transfer is by conduction between the hot water molecules and the molecules of the hands. This heat is then further transferred to the internal body organs and body fluid through either convection or conduction. This heat transfer is maintained until both the water and the hands are at the same temperature, assuming the owner is able to withstand the heat.

Answer:

Reflection is when light bounces off an object, while refraction is when light bends while passing through an object.

Explanation:

I just learned about this 2 weeks ago actually.

Answer:

its false

Explanation:

because if an leaf floats down from a tree it is not considered an object for a free-fall

Answer:

[tex]a)-3346.8\;N \\ b)-4937.04\;N-m[/tex]

Explanation:

a) - In Free-body diagram :

At point D, the free body diagram of a man :

[tex]D = Fn\\Fn=T+mg\\put\;values\;in\;it\\ .\;\;\;\;=300+(80)(9.81)=1084.8\;N[/tex]

[tex]Mg=200\times9.81=1962\;N[/tex]

[tex]\sum Fx=0\; where\; Ax=0[/tex]

[tex]\sum Fy=0\; where\; Ay-Mg-Fn-T=0[/tex]

Then, put the value in the equation.

[tex]Ay=3346.8\;N[/tex]

b)-

[tex]Ma=Mg(AE)+Fn(AD)+T=4937.04\;N-m[/tex]

Answer:

a)y = 485 m ,  v = 220 m / s , b)  y = 2954.39 m , c)   t_total = 51 s ,

d) v = 240.59 m / s

Explanation:

a) We can use vertical launch ratios for this exercise

the speed of the rocket the run out the fuel is

        v = v₀ + a t

the rocket departs with initial velocity v₀ = 0

        v = a t

        v = 55 4

        v = 220 m / s

the height at this point is

        y = y₀ + v₀t + ½ a t²

        y = y₀ + 1/2 a t²

        y = 45 + ½ 55 4²

        y = 485 m

b) the maximum height of the rocket is when its speed is zero

for this part we will use as the initial speed the speed at the end of the fuel (v₀´ = 220 m / s) and the height of y₀´ = 485 m

        v² = v₀´² + 2 g (y-y₀´ )

         0 = v₀´² + 2 g (y-y₀´ )

         y = y₀´ + v₀´² / 2g

         y = 485 + 220 2/2 9.8

         y = 2954.39 m

c) the time that the rocket is in the air is the acceleration time t₁ = 4 s, plus the rise time (t₂) plus the time to reach the ground (t₃)

let's calculate the rise time

           v = v₀´- g t

           v = 0

            t₂ = v₀´ / g

            t₂ = 220 / 9.8

            t₂ = 22.45 s

Now let's calculate the time it takes to get from this point (y₀´´ = 2954.39 m) to the floor

           y = y₀´´ + v₀´´ t - ½ g t²

           0 = y₀´´ - ½ g t²

          t = √ (2 y₀´´ / g)

          t = √ (2 2954.39 / 9.8)

          t = 24.55 s

the total flight time is

       t_total = t₁ + t₂ + t₃

       t_total = 4 + 22.45 + 24.55

        t_total = 51 s

d) the veloicda right now

       v = vo + g t

       v = 9.8 24.55

       v = 240.59 m / s

Answer:

WT =  3.32*10^34 J

Explanation:

The work done by the gravitational attraction between the Sun and the Earth in one complete orbit of the Earth can be calculated by using the following formula:

[tex]W_T=\int F_g dr[/tex]  (1)

Fg: gravitational force between Sun and Earth

The gravitational force is given by:

[tex]F_g=G\frac{m_sm_e}{r^2}[/tex]  (2)

G: Cavendish's constant = 6.674*10^-11 m^3 kg^-1 s^-2

ms: mass of the sun = 1.989*10^30 kg

me: mass of the Earth = 5.972 × 10^24 kg

r: distance between Earth and Sun, this value is a constant r = R = 149,597,870 km

You replace the formula (2) in (1):

[tex]W_T=\int G\frac{m_sm_e}{R^2}dr=G\frac{m_sm_e}{R^2}\int dr\\\\W_T=G\frac{m_sm_e}{R^2}(2\pi R)=2\pi G\frac{m_sm_e}{R}[/tex]

Next, you replace the values of all variables and solve obtain WT:

[tex]W_T=2\pi (6.674*10^{-11}m^3kg^{-1}s^{-2})\frac{(1.989*10^{30}kg)(5.972*10^{24}kg)}{(149597870*10^3 m)}\\\\W_T=3.32*10^{34}J[/tex]

hence, the work done on the Earth, in one orbit, is 3.32*10^34 J

H= P × t

1kW= 1000w

10 min = 600s

H= 1000×600=600,000J

=> 143.40kcal

Answer:

Potential Energy is when the energy is kept in place the object isnt moving so it isnt creating Kinetic energy.

Answer:

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Explanation:

Potential energy is the energy stored in a body due to the body’s positioning or state.

Answer:

1

Explanation:

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Answer:

62.4m

Explanation:

a) Horizontal distance traveled = x = [tex]v_x[/tex] * t

where,

[tex]v_x[/tex]  = horizontal velocity

and t = time in the air)

Time in the air t can be solved using the equation for y:

[tex]i_y=y_o+v_o_yt - 0.5gt^2[/tex]

where, [tex]y_o[/tex] = initial height i.e 0.85 m

[tex]v_o_y[/tex] = initial vertical velocity

and g = 9.8 m/s^2

When y = 0, the dog has hit the water.

So set y = 0 and solve for t.  

[tex]v_o_y[/tex] = 8.62 m/s [tex]\times[/tex] (sin 28) = 4.04 m/s

0 = 0.85 m + (4.04 m/s)t - 0.5gt²  

0.5t²-4.04t-0.85  

Solve this quadratic formula for t: the solutions are t = 8.2 s and t = -0.2. Reject the negative solution, so t = 8.2 s.

How far does the dog travel horizontally in 9.2 s?

x = [tex]v_x[/tex] * t  = (8.62 m/s) [tex]\times[/tex] (cos 28) [tex]\times[/tex]  8.2 s = 62.4 m  

She travels the horizontal distance before hitting the surface of the water is 62.4m

When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion. The object which follows the projectile motion.

a) Horizontal distance traveled = x = V(x) * t

where, V(x)  is the horizontal velocity and t = time in the air

Time in the air t can be solved using the equation for y:

hy =yo + Voy x t - 1/2gt²

where,  yo= initial height i.e 0.85 m and Voy = initial vertical velocity

and g = 9.8 m/s^2

When y = 0, the dock has hit the water.

So set y = 0 and solve for t.  

Voy = 8.62 m/s x (sin 28) = 4.04 m/s

Substitute the values, we get time as

0 = 0.85 m + (4.04 m/s)t - 0.5gt²  

0.5t²-4.04t-0.85 =0

Solving the quadratic formula for t, we have

time t = 8.2 s and t = -0.2.

As time can't be negative, so, t = 8.2 s.

The horizontal distance is

x = Vo * t  = (8.62 m/s)  (cos 28)   8.2 s = 62.4 m  

Thus, she travels 62.4 m.

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Answer:

Nuclear energy comes from tiny mass changes in nuclei as radioactive processes occur. In fission, large nuclei break apart and release energy; in fusion, small nuclei merge together and release energy.

I hoep that was helpful! Let me know:)

Explanation:

Answer:

distance covered is 4.5km or 1.25m

Explanation:

GIVEN DATA:

SPEED=v=3km/h

TIME=t=1.5h

TO FIND:

DISTANCE COVERED=d=?

SOLUTION:

As we know that

speed=distance covered /time taken

here we have to find distance

speed×time taken=distance covered

3km/h×1.5h=distance covered

4.5km=distance covered or in international system of unit the answer is 1.25m

Expand each vector into their component forms:

[tex]\vec A=(4.5\,\mathrm N)(\cos\theta_A\,\vec\imath+\sin\theta_A\,\vec\jmath)=(2.58\,\vec\imath+3.69\,\vec\jmath)\,\mathrm N[/tex]

Similarly,

[tex]\vec B=(-1.23\,\vec\imath+0.860\,\vec\jmath)\,\mathrm N[/tex]

[tex]\vec C=(-3.44\,\vec\imath-4.91\,\vec\jmath)\,\mathrm N[/tex]

Then assuming the resultant vector [tex]\vec R[/tex] is the sum of these three vectors, we have

[tex]\vec R=\vec A+\vec B+\vec C[/tex]

[tex]\vec R=(-2.09\,\vec\imath-0.368\,\vec\jmath)\,\mathrm N[/tex]

and so [tex]\vec R[/tex] has magnitude

[tex]\|\vec R\|=\sqrt{(-2.09)^2+(-0.368)^2}\,\mathrm N\approx2.12\,\mathrm N[/tex]

and direction [tex]\theta_R[/tex] such that

[tex]\tan\theta_R=\dfrac{-0.368}{-2.09}\implies\theta_R=-170^\circ=190^\circ[/tex]

Answer:

Explanation:

Velocity relative to earth = .99537 c

= .99537 x 3 x 10⁸ m /s

= 2.98611 x 10⁸ m /s

distance relative to earth = 40 km = 40 x 10³ m .

Time measured by scientist

= 40 x 10³ /  2.98611 x 10⁸

= 13.395 x 10⁻⁵ s

= 134 μs.

Answer:

C

Explanation:

A feather falls from one end of a tube to the other inside a vacuum.

A. All living things and objects

Answer:

A: All living things and objects. I Hope This Helped!

Explanation:

Describes Matter.

Answer:

Q1: B.2 Q2: B.Waxing crescent Q3: A.Waxing Gibbous

Explanation:

Answer:

The lost in kinetic energy is   [tex]KE_l = 125.5 \ J[/tex]

Explanation:

From the question we are told that

   The mass of the first skater is  [tex]m_1 = 30 \ kg[/tex]

    The speed of the first skater is  [tex]v_1 = 3 \ m/s[/tex]

     The mass of the second skater is [tex]m_2 = 35 \ kg[/tex]  

     The  speed of the second skater is  [tex]v_2 = 1 \ m/ s[/tex]

     The final speed of both skater are [tex]v_f = 1.9 m/s[/tex]

The initial kinetic energy of both skaters is mathematically represented as

     [tex]KE_i = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2[/tex]

substituting values  

    [tex]KE_i = \frac{1}{2} * 30 * 3^2 + \frac{1}{2} * 35 * 1^2[/tex]

     [tex]KE_i = 242.5 \ J[/tex]

The final kinetic energy of both skaters is mathematically represented as

       [tex]KE_f = \frac{1}{2} * (m_1 + m_2 ) v_f^2[/tex]

substituting values

      [tex]KE_f = \frac{1}{2} * (30 + 35 ) * 1.9^2[/tex]

      [tex]KE_f = 117 \ J[/tex]

The lost in kinetic energy is

      [tex]KE_l = 242.5 -117[/tex]

       [tex]KE_l = 125.5 \ J[/tex]

Answer:

It will take the wheel 278.9 s to come to a stop

Explanation:

Mass of the potter's wheel, M = 30 kg

Diameter of the potter's wheel, d₁ = 60 cm = 0.6 m

Radius, r₁ = d/2 = 0.6/2

r₁ = 0.3 m

The moment of inertia of the wheel, [tex]I = 0.5Mr_1^{2}[/tex]

[tex]I = 0.5*30*0.3^{2}\\I = 1.35 kg.m^2[/tex]

d₂ = 14 cm = 0.14 m

r₂ = 0.14/2 = 0.07 m

Angular velocity, [tex]\omega = 180 rpm[/tex]

[tex]\omega = \frac{180*2\pi }{60} \\\omega = 18.85 rad/s[/tex]

Frictional Force, F = 1.3 N

The torque generated:

[tex]\tau = F*r_{2}\\\tau = 1.3*0.07\tau = 0.091 Nm[/tex]

Torque can also be calculated as:

[tex]\tau = I \alpha\\\tau = I \frac{\omega }{t} \\0.091 = 1.35*\frac{18.8 }{t} \\t = (18.8*1.35)/0.091\\t = 278.9 s[/tex]

The time taken for the potter's wheel to come to a stop is 280 s.

The given parameters;

The momentum of inertia of the potter's wheel is calculated ;

[tex]I = \frac{1}{2} Mr^2\\\\I = (0.5)(30)(0.3)^2\\\\I = 1.35 \ kgm^2[/tex]

The angular speed of the potter's wheel is calculated as follows;

[tex]\omega = 180 \ \frac{rev}{\min} \ \times \ \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s} \\\\\omega = 18.85 \ rad/s[/tex]

The time taken for the wheel to come to a stop is calculated as;

[tex]Fr = I \alpha \\\\Fr= I \times \frac{\omega}{t} \\\\t = \frac{I \omega }{Fr} \\\\[/tex]

d = 14 cm, r = 7 cm = 0.07 m

[tex]t = \frac{1.35 \times 18.85}{1.3 \times 0.07} \\\\t = 279.6 \ s\\\\t\approx 280 \ s[/tex]

Thus, the time taken for the potter's wheel to come to a stop is 280 s.

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