A system can only achieve a lower energy state by:_______

Answer:

A. bent

Explanation:

Water molecule (H2O) is said to contain four valence electron pairs (2 bond pairs and 2 lone pairs). The presence of lone pair of electrons in the water molecule influences its molecular geometry or shape.

Since water has two lone pairs of electrons, which repel each other according to the VSEPR theory, water molecule is said to have a BENT molecular geometry.

Answer:

the answer is c which is a+2 charge

Explanation:

Beryllium is in group 2A. It's nearest noble gas is Helium, which is 2 elements behind Beryllium. ThBeryllium wants to lose two electrons. When it does that, Beryllium will have a positive chargeof two, and it will be stated as B-e two plus.

The Beryllium (Be) has an atomic number of 4 and belongs to Group-2 elements. The Beryllium will form a divalent cation (+2). Thus, option C is correct.

In an atom, the number of electrons equals the number of protons. If the electrons are removed from the atom or the electrons are added to the atom, the atom has an excessive positive or negative charge.

This excessive of electrons or lack of electrons forms Ions. The excess of electrons has a negative charge or anions and the lack of electrons has a positive charge or cations.

Beryllium has 4 electrons. Two electrons are occupied in the valence shell of beryllium. Group 2 elements always form the positive ions or cations, to become stable ions.

The outermost shell of beryllium has two electrons. In order to form a stable ion, beryllium should lose its two electrons or gain six electrons. Beryllium belongs to the Group-2 element, it always loses two electrons and forms Be²⁺, to form a stable ion.

Hence, Beryllium forms an ion with a +2 charge. Thus, the correct option is C.

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Answer:

the situation in which the force or any kinds of things rhat are done by force is called work done.ex.typed,wrote

Answer:

WORK ONLY HAS MAGNITUDE AND NO DIRECTION.

hope it helps

have a nice day

Answer:

Explanation:

The mass of that science book...wow. In pounds that would be 35.2! Yikes!

Anyway, we need final velocity here, and the mass of the book has nothing to do with how fast it falls. Everything is pulled by the same gravity. A feather falls at 9.8 m/s/s and so does an elephant. Mass is useless information. The equation we will use is

[tex]v^2=v_0^2+2a[/tex]Δx  where

v is the final velocity, our unknown,

v₀ is the initial velocity which is 0 since someone had to be holding the book before dropping it,

a is the pull of gravity which is always -9.8 m/s/s, and

Δx = -120 which is the displacement (it's negative because the book falls below the point from which it was dropped). Filling in:

[tex]v^2=0^2+2(-9.8)(-120)[/tex] so

[tex]v=\sqrt{2(-9.8)(-120)}[/tex] and

v = 48 m/s

As far as how far above the bottom of the cliff the object is when it is moving at 12 m/s we will use the same equation, but the velocity will be 12:

[tex]12^2=0^2+2(-9.8)[/tex]Δx and

144 = -19.6Δx so

Δx = -7.3 m. That's how far from the top of the cliff it is. We subtract then t find out how far it is from the bottom:

120 - 7.3 = 112.7 m off the ground.

Answer: Members of the Flat Earth Society claim to believe the Earth is flat. Walking around on the planet's surface, it looks and feels flat, so they deem all evidence to the contrary, such as satellite photos of Earth as a sphere, to be fabrications of a "round Earth conspiracy" orchestrated by NASA and other government agencies.

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Answer:

The efficiency of Carnot's heat engine is 26.8 %.

Explanation:

Temperature of hot reservoir, TH = 100 degree C = 373 K

temperature of cold reservoir, Tc = 0 degree C = 273 K

The efficiency of Carnot's heat engine is

[tex]\eta = 1-\frac{Tc}{T_H}\\\\\eta = 1 -\frac{273}{373}\\\\\eta = 0.268 =26.8 %[/tex]

The efficiency of Carnot's heat engine is 26.8 %.

Answer:

336.96m/s

Explanation:

answer is in photo above

The speed of sound on that day is  336.96 m/s.

Speed is distance travelled by the object per unit time. Due to having no direction and only having magnitude, speed is a scalar quantity With SI unit meter/second.

given that:

frequency of the sound = 312 Hz.

2nd resonant length = 81.0 cm.

If the wavelength of sound is λ; the 2nd resonant length of  plastic pipe with an adjustable closed end = 3λ/4

Hence, wavelength of sound is = 81×(4/3) cm = 0.81 ×(4/3) m.

So,  the speed of sound on that day is = frequency × wavelength

= 312 Hz ×  0.81 ×(4/3) m.

= 336.96 m/s.

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Answer:

C

Explanation:

horizintal speed stays same

only vertical speed changes

so H speed will stay 30 m/s

Complete question is;

A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:

(a) the work done by the force of gravity on the projectile,

(b) the change in kinetic energy of the projectile since it was fired, and

(c) the final kinetic energy of the projectile.

(d) Are any of the answers changed if the initial angle is changed?

Answer:

A) W = mgh

B) ΔKE = mgh

C) K2 = mgh + ½mv_o²

D) No they wouldn't change

Explanation:

We are expressing in terms of m, v0​, h, and g. They are;

m is mass

v0 is initial velocity

h is height of projectile fired

g is acceleration due to gravity

A) Now, the formula for workdone by force of gravity on projectile is;

W = F × h

Now, Force(F) can be expressed as mg since it is force of gravity.

Thus; W = mgh

Now, there is no mention of any angles of being fired because we are just told it was fired horizontally.

Therefore, even if the angle is changed, workdone will not change because the equation doesn't depend on the angle.

B) Change in kinetic energy is simply;

ΔKE = K2 - K1

Where K2 is final kinetic energy and K1 is initial kinetic energy.

However, from conservation of energy, we now that change in kinetic energy = change in potential energy.

Thus;

ΔKE = ΔPE

ΔPE = U2 - U1

U2 is final potential energy = mgh

U1 is initial potential energy = mg(0) = 0. 0 was used as h because at initial point no height had been covered.

Thus;

ΔKE = ΔPE = mgh

Again like a above, the change in kinetic energy will not change because the equation doesn't depend on the angle.

C) As seen in B above,

ΔKE = ΔPE

Thus;

½mv² - ½mv_o² = mgh

Where final kinetic energy, K2 = ½mv²

And initial kinetic energy = ½mv_o²

Thus;

K2 = mgh + ½mv_o²

Similar to a and B above, this will not change even if initial angle is changed

D) All of the answers wouldn't change because their equations don't depend on the angle.

Answer:

Explanation:

Since the path an airplane travels is pretty much horizontal, then acceleration in this dimension is 0 and the equation we use is strictly the d = rt one. Here we use it and solve for t:

[tex]t=\frac{d}{r}[/tex] and filling in:

[tex]t=\frac{2.00}{940}[/tex] so

t = .002 hrs which is the same as 7.2 seconds

Answer:

1210 ohm

Explanation:

Given :

P=40 W

V=220 V

Now,

[tex]P=\frac{V^{2} }{R} \\40=\frac{(220)^{2} }{R} \\40R=48400\\R=\frac{48400}{40} \\R=1210 ohm[/tex]

Therefore, resistance of bulb will be 1210 ohm

Answer:

v₀ₓ = 24.24 m / s

Explanation:

This is a projectile launching exercise, where the windshield comes out with a horizontal initial velocity.

Y axis  

initial vertical velocity is zero

          y = y₀ + v_{oy} t - ½ g t²

when it reaches the ground its height is zero and the initial height is y₀=1.2m

          0 = y₀ + 0 - ½ g t²

          t = [tex]\sqrt{2 y_o/g}[/tex]

          t = [tex]\sqrt{2 \ 1.2 / 9.8}[/tex]

          t = 0.495 s

X axis

          x = v₀ₓ t

          v₀ₓ = x / t

          v₀ₓ = 12 / 0.495

          v₀ₓ = 24.24 m / s

Answer:

A bridge suspended by cables

Explanation:

Both objects represent a contact force (in this case, normal force) acting on each other. The force occurs since both objects are in direct physical contact.

Answer:

the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1

Explanation:

Given the data in the question;

Hank and Harry are two ice skaters, since both are on top of ice, we assume that friction is negligible.

We know that from Newton's Second Law;

Force = mass × Acceleration

F = ma

Since they hold on to opposite ends of the same rope. They have the same magnitude of force |F|, which is the same as the tension in the rope.

Now,

Mass[tex]_{Hank[/tex] × Acceleration[tex]_{Hank[/tex] = Mass[tex]_{Henry[/tex] × Acceleration[tex]_{Henry[/tex]

so

Mass[tex]_{Hank[/tex] /  Mass[tex]_{Henry[/tex] = Acceleration[tex]_{Henry[/tex] / Acceleration[tex]_{Hank[/tex]

given that; magnitude of Hank's acceleration is 1.26 times greater than the magnitude of Harry's acceleration,

Mass[tex]_{Hank[/tex] /  Mass[tex]_{Henry[/tex] = 1 / 1.26

Mass[tex]_{Hank[/tex] /  Mass[tex]_{Henry[/tex] = 0.7937 or [ 0.7937 : 1 ]

Therefore, the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1 ]

Answer:

v₀ = 2.67 m / s

Explanation:

This problem can be solved using the Kinetic Enemy Work Theorem

         W = ΔK

Work is defined by the relation

         W = fr. d

The bold letters indicate vectors, in this case the blow is in the direction of the slope of the ramp and the displacement is also in the direction of the ramp, therefore the angle between the force and the displacement is zero.

the friction force opposes the displacement therefore its angle is 180º

          W = - fr d

Let's use Newton's second law, we define a reference frame with the horizontal axis parallel to the plane

Y axis  

           N- Wy = 0

           N - W cos tea = 0

the friction force has the expression

          fr = μ N

          fr = μ W cos θ

we substitute

           W = - μ W cos θ d

let's look for kinetic energy

the minimum velocity at the highest point is zero

           K_f = 0

the initial kinetic energy is

            K₀ = ½ m v₀²

we substitute energy in the work relationship

         - μ W cos θ d = 0 - ½ m v₀²

           v₀² = - μ W cos θ  2d / m

Let's use trigonometry to find distance d

         sin θ=  y / d

         d = y /sin θ

         d = 3.50 / sin 30

         d = 7 m

let's calculate

           v₀² = (6 10⁻² 2.50 9.8 cos 30) 2 7 / 2.50

           v₀ = √7.129

           v₀ = 2.67 m / s

Answer:

less than the weight of the block.

Explanation:

From the free body diagram, we get.

The normal force is N = Mg cosθ

The tension in the string is T = Mg sinθ

Wight of the block when the block is static, W = Mg

Now since the magnitude of cosθ is in the range of : 0 < cosθ < 1,

therefore, the normal force is less than the weight of the static block.

Answer:

C) 300 Ohm.

Explanation:

In a series circuit, total resistance is just adding all the resistance together. So R (total) = 75 +75 +75+ 75 = 300 ohms

Parallel circuit are different because you add the inverses of resistance and you flip the final answer.

You can confirm your answers using the tools below:

https://www.omnicalculator.com/physics/series-resistor

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Answer:

[tex]{ \tt{n _{w} = \frac{real \: depth}{aparent \: depth} }} \\ \\ \frac{4}{3} = \frac{4}{d} \\ { \tt{d = 3 \: meters}}[/tex]

A boy observes a fish at the depth of 4 meters under the water's surface. The refractive index of the water is 4/3 then the apparent depth of the fish as seen by the boy is 3 meters.

Refractive index is also known as the refraction, is a measure of the way a light beam bends when it goes through different media. The refractive index n is computed by dividing the sine of the angle of incidence by the sine of the angle of refraction, i.e., n = sin i/sin r, if I will be the angle of incidence of a light source in a void (angle between the inbound ray and the normal, or perpendicular to the surface of a medium).

The Index of refraction is also equal to c/v, or the ratio of a wavelength of light's velocity in a substance to its velocity in a void.

As per the given data in the question,

n(w) = real depth/apparent depth

4/3 = 4/d

d = 3 meters.

Therefore, the apparent depth will be equal to 3 meters.

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Answer:

If air resistance is taken as negligible, then the ball is in freefall the moment it is thrown so gravity is the only force acting on the object. If air resistance is not negligible then gravity will be the greatest force acting on the ball while it is going up and coming down, because Fair has to be less than gravity at all times otherwise the atmosphere would wither away.

Answer:

[tex]\mu=0.185[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=17kg[/tex]

Force [tex]F=33N[/tex]

Velocity [tex]v=1.6m/s[/tex]

Distance [tex]d= 9.8m[/tex]

Generally the equation for Work done is mathematically given by

 [tex]W=\triangle K.E+\triangle P.E[/tex]

Where

 [tex]\triangle K.E=(F-F_f)*2[/tex]

 [tex]F_f=F+\frac{\triangle K.E}{d}[/tex]

 [tex]F_f=33+\frac{0.5*17*1.6^2}{9.8}[/tex]

 [tex]F_f=30.8N[/tex]

Since

 [tex]f = \mu*m*g[/tex]

 [tex]\mu= 30.8/(m*g)[/tex]

 [tex]\mu= 30.8/(17*9.81)[/tex]

 [tex]\mu=0.185[/tex]

Explanation:

1. First, let's find the total resistance of the circuit. We begin by combining [tex]R_{4}[/tex], [tex]R_{5}[/tex] and [tex]R_{6}[/tex]:

[tex]R_{456}=R_{4} + \dfrac{R_{5}R_{6}}{R_{5} + R_{6}}[/tex]

[tex]= 6\:Ω + \dfrac{(3\:Ω)(5\:Ω)}{3\:Ω+5\:Ω} = 7.9\:Ω[/tex]

Now time to combine [tex]R_{2}[/tex] and [tex]R_{3}[/tex] and they are connected in series so

[tex]R_{23} =R_{2} + R_{3} = 17\:Ω[/tex]

Note that [tex]R_{23}[/tex] and [tex]R_{456}[/tex] are connected in parallel so

[tex]R_{23456} = \dfrac{R_{23}R_{456}}{R_{23}+R_{456}}=5.4\:Ω[/tex]

Finally, [tex]R_{23456}[/tex] is connected in series with [tex]R_{1}[/tex] so the total resistance [tex]R_{T}[/tex] is

[tex]R_{T} = R_{1} + R_{23456} = 10\:Ω + 5.4\:Ω = 15.4\:Ω[/tex]

2. The total current in the circuit is

[tex]I_{T} = \dfrac{V}{R_{T}} = \dfrac{20\:V}{15.4\:Ω} = 1.3\:A[/tex]

3. The voltage drop across [tex]R_{1},\:V_{1}[/tex] is

[tex]V_{1} = I_{T}R_{1} = (1.3\:A)(10\:Ω) = 13\:V[/tex]

4. We can see that [tex]I_{T} = I_{1} + I_{2}[/tex]. To solve for [tex]I_{1}[/tex], we need [tex]V_{23}[/tex], which is just [tex]V_{T} - V_{1} = 20\:V - 13\:V = 7\:V[/tex] , which gives us

[tex]I_{1} = \dfrac{V_{23}}{R_{23}} = \dfrac{7\:V}{17\:Ω} = 0.4\:A[/tex]

5. From #2 & #4, [tex]I_{2} = 1.3\:A - 0.4\:A = 0.9\:A[/tex] and we also know that the voltage drop across [tex]R_{456}[/tex] is 7 V, the same as that of [tex]R_{23}[/tex]. The voltage drop across [tex]R_{4}[/tex] is

[tex]V_{4} = I_{2}R_{4} =(0.9\:A)(6\:Ω) = 5.4\:V[/tex]

This means that the voltage drop across [tex]R_{6}[/tex] is 7 V - 5.4 V = 1.6 V. Knowing this, the current through [tex]R_{6}[/tex] is

[tex]I_{6} = \dfrac{1.6\:V}{5\:Ω} = 0.3\:A[/tex]

Answer:

The second statement.

Answer:

21.73 kg

Explanation:

Applying,

T = 2π√(m/k)............... Equation 1

Where T = period, m = mass on the spring, k = spring constant, π = pie.

make m the subject of the equation

m = T²k/4π²................. Equation 2

From the question,

Given: T = 3.45 s, k = 72.0 N/m, π = 3.14

Substitute these values into equation 2

m = (3.45²×72)/(4×3.14²)

m = 21.73 kg.

Hence the mass should be 21.73 kg

Answer:

The wave speed is 2.8 m.

Explanation:

Wavelength = 1.4 m

frequency, f = 2 Hz

the wave speed is given by

wave speed = wavelength x frequency

wave speed = 1.4 x 2 = 2.8 m

option (D) is correct.  

Explanation:

Given that,

The radius of rain drop, r = 0.12 cm = 0.0012 m

The viscocity of air is, [tex]\eta=18\times 10^{-5}\ poise[/tex]

Let the viscous force is, [tex]F = 0.010173\ N[/tex]

The viscous force is given by :

[tex]F=6\pi \eta rv\\\\v=\dfrac{F}{6\pi \eta r}[/tex]

Put all the values,

[tex]v=\dfrac{0.010173}{6\pi 18\times 10^{-5}\times 0.0012 }\\\\v=2498.58\ m/s[/tex]

Answer:

The heat source for our planet is the sun. Energy from the sun is transferred through space and through the earth's atmosphere to the earth's surface. Since this energy warms the earth's surface and atmosphere, some of it is or becomes heat energy.

Answer:

the linear speed of the car is 28.83 m/s

Explanation:

Given;

radius of the car, r = 0.33 m

angular speed of each tire, ω = 13.9 rev/s = 13.9 x 2π = 87.35 rad/s

The linear speed of the car is calculated as;

V = ωr

V = 87.35 rad/s x 0.33 m

V = 28.83 m/s

Therefore, the linear speed of the car is 28.83 m/s