An object is placed 12.0 cm from a thin diverging lens with a focal length of 4 cm. Which one of the
following statements is true concerning the image?
A. The image is virtual and 3.0 cm from the lens.
B. The image is real and 6.0 cm from the lens.
C. The image is virtual and 12 cm from the lens.
D. The image is real and 12 cm from the lens.

Answer:

get egg and try to make in not crack when it falls by exerting the momentum of the fall into something other than the egg ex. make a box full of bubble wrap and put your egg in it

Explanation:

ALL of the quantities are unknown, because you haven't bothered to tell us anything that's known.

Answer:

the wooden animal will go 0.7068 m above its equilibrium

Explanation:

Given the data in the question;

mass of wooden animal m = 120 g = 0.12 kg

the animal oscillates up and down, T = 0.54 s

older sister pulls the animal 32 cm below its equilibrium position;

x = 32 cm = 0.32 m

g = 9.81 m/s²

We know that

k = mω²

where ω = 2π/T

So, k = m( 2π/T )²

we substitute

k = 0.12( 2π / 0.54 )²

k = 0.12 × (11.6355)²

k = 0.12 × 135.38486

k = 16.25 N/c

so Also,

kx²/2 = mgh

we solve for h

h = kx² / 2mg

we substitute

h = ( 16.25 × (0.32)²) / ( 2 × 0.12 × 9.81 )

h = 1.664 / 2.3544

h = 0.7068 m

Therefore, the wooden animal will go 0.7068 m above its equilibrium

Answer:

What is the chemical fórmula For the molecule modeled?

Explanation:

C6H12O2

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

I hope this helped!+*

Answer:

please give me brainlist and follow

Explanation:

4 degrees C turns out to be the temperature at which liquid water has the highest density. If you heat it or cool it, it will expand. ... Ice floats on top of lakes, preventing evaporation (and convection in the frozen layer), and lakes stay liquid underneath, allowing fish and other life to survive.

Answer:

Amplitude

Explanation:

The loudness of a sound depends on the amplitude of vibration producing the sound

Answer:

12294.31 m/s

Explanation:

Momentum = (mass)(velocity)

Momentum before = Momentum after

(momentum of bullet)+(momentum of block)=(momentum of bullet and block)

0.035v+50(0)=(0.035+50)(8.6)

0.035v=430.301

v=12294.3142857m/s

Answer:

1) m = 0.3137 kg

2a)T_f = -181.7°C

2b) T_f = -1176.97°C

Explanation:

1) We are given;

Length; l = 4.30 cm = 0.043 m

Width; w = 14.0 cm = 0.014 m

height; h = 27.0 cm = 0.027 m

density of gold; ρ = 19.3 × 10³ kg/m³

Formula for the density is known as;

ρ = mass/volume

Thus;

m =ρV

m = 19.3 × 10³ × (lwh)

m = 19.3 × 10³ × (0.043 × 0.014 × 0.027)

m = 0.3137 kg

2a) We are given;

Diameter of brass; L_br = 10 cm

Diameter of aluminum; L_al = 10.01 cm

Now, to some for change in temperature we will use the formula;

L_f = L_i + αL_i(Δt)

Where α is coefficient of expansion.

Now, for the ring to be removed from the rod, the final diameter of the brass has to be same as the aluminium.

Thus;

L_f(brass) = L_f(aluminium)

From table attached, α_brass ≈ 19 × 10^(-6) /°C

Also, α_aluminium ≈ 24 × 10^(-6) /°C

Thus;

L_f(brass) = 10 + (19 × 10^(-6) × 10 × (Δt))

Similarly,

L_f(aluminium) = 10.01 + (24 × 10^(-6) × 10.01 × (Δt))

Since L_f(brass) = L_f(aluminium), then;

10 + (19 × 10^(-6) × 10 × (Δt)) = 10.01 + (24 × 10^(-6) × 10.01 × (Δt))

Rearranging, we have;

10.01 - 10 = (19 × 10^(-6) × 10 × (Δt)) - (24 × 10^(-6) × 10.01 × (Δt))

0.01 = Δt(-50.24 × 10^(-6))

Δt = 0.01/(-50.24 × 10^(-6))

Δt ≈ -199°C

Thus, temperature at which the combination must be cooled to separate the two metals is;

T_f = T_i + Δt

T_f = 17.3 + (-199)

T_f = -181.7°C

2b) Diameter of aluminum is now;

L_al = 10.06 cm

Thus;

10.06 - 10 = (19 × 10^(-6) × 10 × (Δt)) - (24 × 10^(-6) × 10.01 × (Δt))

0.06 = Δt(-50.24 × 10^(-6))

Δt = 0.06/(-50.24 × 10^(-6))

Δt = -1194.27°C

T_f = 17.3 + (-1194.27)

T_f = -1176.97°C

Answer:

By talking about something that interesto you’

sorry if wrong

Explanation:

Answer:

B : using a spring scale and measuring the weight of the car

Explanation:

Answer:

Gravity

Explanation:

Gravity pulls things down to earth and it is a force

Answer:

[tex]0.0018833\ \text{m/s}[/tex]

Explanation:

[tex]d[/tex] = Distance of Andromeda Galaxy from Earth = [tex]2.54\times 10^7\ \text{ly}[/tex]

[tex]t[/tex] = Time taken = [tex]90\ \text{years}[/tex]

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

We have the relation

[tex]t=t_o\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow 90=2.54\times 10^7\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow \dfrac{90^2}{(2.54\times 10^7)^2}=1-\dfrac{v^2}{c^2}\\\Rightarrow 1-\dfrac{90^2}{(2.54\times 10^7)^2}=\dfrac{v^2}{c^2}\\\Rightarrow v=c\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}}[/tex]

[tex]c-v=c(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=3\times 10^8(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=0.0018833\ \text{m/s}[/tex]

The required answer is [tex]0.0018833\ \text{m/s}[/tex].

Answer:

False

Explanation:

No, it is not true that energy of the electrons ejected should have been proportional to the intensity of the light. Perhaps the kinetic energy of the ejected electrons is independent of the intensity of the incident radiation. This is the very fact that classical theory of electromagnetism fails to explain in photoelectric effect. The kinetic energy of the electrons remains constant even if the amplitude of the incident light is increased.

Answer:

It will go up to 93.75 m before it is moving at 20 m/s

Explanation:

As we know that

[tex]v^2 - u^2 = 2aS[/tex]

here v is the final speed i.e 20 m/s

u is the initial speed i.e 5 m/s

a is the acceleration due to gravity i.e 2 m/s^2

Substituting the given values in above equation, we get -

[tex]20^2 - 5^2 = 2*2*S\\S = 93.75[/tex]meters

Answer:

the answer is false.

Explanation:

i took the test and it is false trust me!!!!!!!!!

Answer; they are larger and made of rocky material

Answer:

C

Explanation:

C

im confused hold on imma send you a link to the answerExplanation:

Answer: 84 %

Explanation:

Answer:

v = -17.94 cm

Explanation:

Given that,

The candle is placed at a distance of 50 cm, u = -50 cm

The focal length of the diverging lens, f = -28 cm (negative in case of a concave lens)

We need to find the image distance. We know that the lens formula is as follows:

[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(-28)}+\dfrac{1}{(-50)}\\\\v=-17.94\ cm[/tex]

So, the image distance is equal to 17.94 cm.

Answer:

e. 425.9 Hz

Explanation:

The computation of the frequency is being played by the out-of-tune key is shown below;

Given that

Δf1  = x - 349.2 = 76.7.........(1)

Δf2 = 440 - x  = 14.1......(2)

Now solve (1) and (2)

(440 - x) - x + 349.2 = 14.1 - 76.7

789.2 + (-2x) = -62.6

x = 425.9 Hz

Hence, the frequency is being played by the out-of-tune key is 425.9 Hz

Therefore the option e is correct

Answer:

Waves carry energy from one place to another.

Explanation:

Because waves carry energy, some waves are used for communication, eg radio and television waves and mobile telephone signals.

Answer:

Time taken by Mr. smith = 4.80 hour (Approx.)

Explanation:

Given:

Distance travel by Mr. smith = 523 kilometer

Average speed of Mr. smith = 109 km/hr

Find;

Time taken by Mr. smith

Computation:

Time taken = Distance cover / Speed

Time taken by Mr. smith = Distance travel by Mr. smith / Average speed of Mr.

smith

Time taken by Mr. smith = 523 / 109

Time taken by Mr. smith = 4.798 hr

Time taken by Mr. smith = 4.80 hour (Approx.)

Answer:

[tex]u=14.48m/s[/tex]

Explanation:

From the question we are told that:

Height of window [tex]h=2m[/tex]

Height of window off the ground [tex]h_g=7.5m[/tex]

Time to fall and drop [tex]t=1.3s[/tex]

Generally the Newton's equation motion  is mathematically given by

 [tex]s=ut+\frac{1}{2}at^2[/tex]

Where

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

[tex]2=u1.3-\frac{1}{2}*9.8*1.3^2[/tex]

[tex]2=u1.3-8.281[/tex]

[tex]u=7.91m/s^2[/tex]  

Generally the Newton's equation motion  is mathematically given by

[tex]2as=v^2-u^2[/tex]

Where

[tex]-2gh_g=v^2-u^2[/tex]

[tex]-2*9.8*7.5=(7.91)^2-u^2[/tex]

[tex]-147=62.5681-u^2[/tex]

[tex]u=\sqrt{209.5681}[/tex]

[tex]u=14.48m/s[/tex]

Therefore the  ball’s initial speed

[tex]u=14.48m/s[/tex]

Answer:

C

Explanation:

The men bow to the women and the women curtsy to the men.

(took on test and got it right)

Answer:

[tex]0.0035\ \text{m}[/tex]

Explanation:

y = Distance from the center point

d = Separation between slits = 0.3 mm

D = Distance between slit and screen = 1.5 m

[tex]\lambda[/tex] = Wavelength = 700 nm

m = Order = 1

We have the relation

[tex]d\dfrac{y}{D}=m\lambda\\\Rightarrow y=\dfrac{Dm\lambda}{d}\\\Rightarrow y=\dfrac{1.5\times 1\times 700\times 10^{-9}}{0.3\times 10^{-3}}\\\Rightarrow y=0.0035\ \text{m}[/tex]

The distance from the screen at which the first bright fringe beyond the center fringe appear is [tex]0.0035\ \text{m}[/tex].

Answer:

d= 23.25 m

Explanation:

       [tex]E_{o} = K_{transo} + K_{roto} (1)[/tex]

       [tex]E_{f} = m*g*h (2)[/tex]

       [tex]h = d* sin 37 (3)[/tex]

       [tex]E_{f} = m*g* d * sin 37 (4)[/tex]

       [tex]v = \omega * R (5)[/tex]

       [tex]K_{rot} = \frac{1}{2}* I * \omega^{2} (6)[/tex]

       [tex]E_{o} = K_{transo} + K_{roto} = \frac{1}{2}* m* v^{2} +(\frac{1}{2}* \frac{1}{2}) *m*r^{2}*(\frac{v}{r}) ^{2} = \\ \frac{3}{4} * m * v^{2} (7)[/tex]

       [tex]d =\frac{3}{4} * \frac{v^{2}}{g*sin37} = \frac{3}{4}*\frac{(\omega*r)^{2}}{g*sin 37} (8)[/tex]

       [tex]d =\frac{3}{4}*\frac{(\omega*r)^{2}}{g*sin 37} = \frac{3}{4} *\frac{(12.0536rad/sec*1.12m)^{2}}{9.8 m/s2*0.601} = 23. 25 m.[/tex]