ASK YOUR TEACHER A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 51.5-gram mass is attached at the 16.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick

B and D are both true statements. I'm not comfortable saying that either one is better than the other one.

The statement that best describes one way that molecules differ from atoms is a molecule can contain two atoms of the same element, and only a molecule can be broken down into two or more different elements. The correct options are b and d.

According to science, an atom is the smallest component of an element that can exist freely or not. A molecule, on the other hand, is the smallest component of a chemical and is made up of a group of atoms linked together by a bond.

A molecule is the smallest component of a substance that has the chemical properties of the compound.

The term "independent molecule" is not commonly used to refer to atoms and complexes linked by non-covalent interactions such as hydrogen or ionic bonds. Molecules are common constituents of matter.

Therefore, the correct options are b and d.

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Answer:

c. 5.50 km

Explanation:

8 min * 1h/(60 min) = 8/60 = 2/15 h

15 sec* 1 min/60 sec = 1/4 min * 1h/(60 min) =  1/240 h

8 min 15 sec = (2/15+1/240)h

40 km/h *(2/15 +1/240)h =5.50 km

Answer: 5.50 km

Explanation:

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The uncertainty in inverse frequency is  [tex]\Delta [\frac{1}{w} ]= \frac{3}{2000} \ s[/tex]

Explanation:

From the question we are told that

   The value of the proportionality constant is  [tex]k = 5 \frac{Hz }{T}[/tex]

   The strength of the magnetic field is  [tex]B = 20 \ T[/tex]

   The change in this strength of magnetic field is  [tex]\Delta B = 3 \ T[/tex]

The magnetic field is given as

           [tex]B = \frac{k}{\frac{1}{w} }[/tex]

Where [tex]w[/tex] is frequency

The uncertainty or error of the field is given as

         [tex]\Delta B = \frac{k }{[\frac{1}{w}^]^2 } \Delta [\frac{1}{w} ][/tex]

The uncertainty in inverse frequency is given  as

           [tex]\Delta [\frac{1}{w} ] = \frac{\Delta B}{k [\frac{1}{w^2} ]}[/tex]

                    [tex]\Delta [\frac{1}{w} ]= \frac{\Delta B}{k (B)^2 }[/tex]

substituting values

                  [tex]\Delta [\frac{1}{w} ]= \frac{3}{5 (20)^2 }[/tex]

               [tex]\Delta [\frac{1}{w} ]= \frac{3}{2000} \ s[/tex]

Answer:

A) The two potatoes cover the same horizontal distance from the launcher.

B) Mark's potato spends more time in the air than Marvin's potato before hitting the ground.

C) Marvin's potato hits the ground with a greater speed than Mark's potato

Explanation:

A) For projectile motion, the final horizontal distance of the projectile from where it was initially launched (its range) is given as

R = (u² sin 2θ)/g

where

u = initial velocity of the projectile

θ = angle above the horizontal at which the projectile was launched = 30°, 60°

g = acceleration due to gravity on Mars

Since, u and g are the same for Mark and Marvin, sin 2θ would determine which range is higher.

Sin (2×60°) = sin 120°

Sin (2×30°) = sin 60°

Sin 120° = Sin 60°

Hence, the two potatoes cover the same horizontal distance from the launcher.

B) Time spent in the air for a projectile is given as

T = (2u sin θ)/g

Again, since u and g are the same for Mark and Marvin on Mars, sin θ will give the required idea of whose potato spends more time in the air.

Sin 60° = 0.866

Sin 30° = 0.50

Sin 60° > Sin 30°

Hence, Mark's potato spends more time in the air than Marvin's potato.

C) The horizontal velocity for projectile motion is constant all through the motion and is equal to u cos θ

u cos 60° < u cos 30°

And the initial vertical velocity is u sin θ

Final vertical velocity

= (initial vertical velocity) - gt

g = acceleration due to gravity on Mars

T = time of flight

For Mark,

initial vertical velocity = u sin 60°, greater than Marvin's u sin 30°

And Mark's potato's time of flight is greater as established in (B) above.

But for Marvin

initial vertical velocity = u sin 30°, less than Mark's u sin 60°

And Marvin's potato's time of flight is lesser as established in (B) above

So, at the end of the day, the final vertical velocity is almost the same for both Mark's and Marvin's potatoes.

Hence, the horizontal component of the final velocity edges the final speed of the potatoes just before hitting the ground in Marvin's favour.

Hope this Helps!!!

Answer:

The work done by the force is  5.76 J

Explanation:

Given;

mass of canister , m = 3.2 kg

magnitude of force, f = 6.7 N

initial velocity of the canister on x-axis,  [tex]v_i[/tex]= 3.3i m/s

final velocity of the canister on y- axis, [tex]v_f[/tex] = 6.9j m/s

The work done on the canister = change in the kinetic energy of the canister

[tex]W = K.E_f - K.E_i[/tex]

where;

K.Ei is the initial kinetic energy

K.Ef is the final kinetic energy

The initial kinetic energy:

[tex]K.E_i = \frac{1}{2} *m\sqrt{i^2 +j^2+z^2}\\\\K.E_i = \frac{1}{2} *3.2\sqrt{3.3^2 +0^2+0^2}\\\\K.E_i = 5.28 \ J[/tex]

The final kinetic energy:

[tex]K.E_f = \frac{1}{2} *m\sqrt{i^2 +j^2+z^2}\\\\K.E_f = \frac{1}{2} *3.2\sqrt{0^2 +6.9^2+0^2}\\\\K.E_f = 11.04 \ J\\[/tex]

W = 11.04 - 5.28

W = 5.76 J

Therefore, work done on the canister by the 6.7 N force during this time is 5.76 J

Answer:

7.2 m/s

Explanation:

The maximum speed is the amplitude times the frequency.

v = Aω

v = (0.76 m / 2) (180 rev × 2π rad/rev / 60 s)

v = 7.2 m/s

Answer:

About 3.06 seconds

Explanation:

[tex]v_f=v_o+at[/tex]

Since at the peak of its trajectory, the ball will have no velocity, you can write the following equation:

[tex]0=30+(-9.81)t\\\\-30=-9.81t\\\\t\approx 3.06s[/tex]

Hope this helps!

Answer:

Explanation:

Let the tension in the string be T . At the top of the circle , total force acting on them = T + mg . This will provide centripetal force

T + mg = m v² / r

4 + .25 x 9.8 = .25 x v² / .62

6.45 = .25 v² / .62

v² = 16

v = 4 m /s .

Answer:

Ke- = Ke+ = 0.294MeV

Explanation:

To fins the kinetic energy of both electron and positron you use the following formula, for the case of annihilation of one electron an positron:

2[tex]E_p=2E_o+K_{e^-}+K_{e^+}[/tex]   (1)

Ep: photon energy = 0.804MeV

Eo: rest energy of one electron (and positron) = 0.51MeV

Ke-: kinetic energy of electron

Ke+: kinetic energy of positron

You replace the values of Ep and Eo in the equation (1):

[tex]K_{e^-}+K_{e^+}=2E_p-2E_o=2(0.804MeV-0.51MeV)=0.588MeV[/tex]

Iy you assume both positron and electron have the same speed, then, the kinetic energy of them are equal, and the kinetic energy of each one is:

[tex]K_{e^-}=K_{e^+}=\frac{0.588MeV}{2}=0.294MeV[/tex]

Answer:

The volume of the marble is [tex]523.33\ mm^2[/tex].

Explanation:

Marble is spherical in shape. The diameter of marble is 10 mm. It radius will be 5 mm.

The volume of spherical shaped object is given by :

[tex]V=\dfrac{4}{3}\pi r^3[/tex]

Plugging all the values, we get :

[tex]V=\dfrac{4}{3}\times 3.14\times (5)^3\\\\V=523.33\ mm^2[/tex]

So, the volume of the marble is [tex]523.33\ mm^2[/tex].

Answer:

a = 5.34 m/s²

T = 37.86 N

Explanation:

This is the case where two masses are hanging vertically on sides of the pulley. In such case, the formula for acceleration of objects is derived to be:

a = g(m₂ - m₁)/(m₂ + m₁)

where,

a = acceleration of both masses = ?

g = 9.8 m/s²

m₂ = heavier mass = 8.5 kg

m₁ = lighter mass = 2.5 kg

Therefore,

a = (9.8 m/s²)(8.5 kg - 2.5 kg)/(8.5 kg + 2.5 kg)

a = (9.8 m/s²)(6 kg)/(11 kg)

a = 5.34 m/s²

The formula for tension in cable is derived to be:

T = 2m₁m₂g/(m₁ + m₂)

T = (2)(2.5 kg)(8.5 kg)(9.8 m/s²)/(2.5 kg + 8.5 kg)

T = 37.86 N

Answer:

 v = 126 m / s

Explanation:

Let's analyze this exercise a little, they give us the thrust that is the applied force and the time that it lasts, and they ask us for the final speed, so we can use the Impulse ratio and the variation of the amount of movement

       I = F t = Dp

       F t = pf -p₀

      Now let's use Newton's second law to find the net thrust

        F = E - fr

the friction force has the formula

       fr = μ N

let's write Newton's second law on the y-axis

       N-W = 0

       N = W

we substitute

      fr = μ mg

we look for the net out

       F = 200 - μ mg

With the skater starting from rest, the initial speed is zero (vo = 0)

we substitute

      (200 - very m g) t = m v

       v = (200 µm - very g) t

let's calculate

     v = (200/75 - 0.10 9.8) 75

      v = 126 m / s

Answer:

Air masses move with the global pattern of winds. In most of the United States, air masses generally move from west to east. They may move along with the jet stream in more complex and changing patterns. 

Answer:

E) True. The girl has a larger tangential acceleration than the boy.

Explanation:

In this exercise they do not ask us to say which statement is correct, for this we propose the solution to the problem.

Angular and linear quantities are related

          v = w r

          a = α r

the boy's radius is r₁ = 1.2m the girl's radius is r₂ = 1.8m

as the merry-go-round rotates at a constant angular velocity this is the same for both, but the tangential velocity is different

          v₁ = w 1,2 (boy)

          v₂ = w 1.8 (girl)

whereby

          v₂> v₁

reviewing the claims we have

          a₁ = α 1,2

          a₂ = α 1.8

          a₂> a₁

A) False. Tangential velocity is different from zero

B) False angular acceleration is the same for both

C) False. It is the opposite, according to the previous analysis

D) False. Angular acceleration is equal

E) True. You agree with the analysis above,

Answer:

a)188.65m

b)154.35m

c)243.7m

Explanation:

Given data:

[tex]t_A=0.55s[/tex]

[tex]t_B=0.45s[/tex]

(a) The distance from the kicker to each of the 2 spectators is given by:

[tex]d_A=v \times t_A[/tex]

where,

v= speed of sound

[tex]t_A[/tex]=time taken for the sound waves to reach the ears

[tex]d_A=343\times 0.55=188.65[/tex]m

(b)[tex]d_B=v \times t_B[/tex]

where,

v= speed of sound

[tex]t_B[/tex]=time taken for the sound waves to reach the ears

[tex]d_B=343\times 0.45=154.35m[/tex]

(c)As the angle b/w slight lines  from the two spectators to the player is right angle,

hypotenuse=the distance b/w 2 spectators

and, the slight lines are the other 2 lines

[tex]D^2=d_A^2+d_B^2\\D=\sqrt{188.65^2+154.35^2} \\D= 243.7m[/tex]

Answer:

(a) 0.0885 Hz

(b) 0.21 m/s

Explanation:

(a) Frequency: This can  be defined as the number of cycle completed in one seconds.

From the question,

Note: 2 crest = one cycle,

If four crest = 22.6 s,

Then two crest = (22.6/2) s

= 11.3 s.

T = 11.3 s

But,

F = 1/T

F = 1/11.3

F = 0.0885 Hz.

(b)

Using,

V = λF...................... Equation 1

Where V = speed of wave, F = Frequency of wave, λ = wave length.

Given: F = 0.0885 Hz, λ = 2.37 m.

Substitute these values into equation 1

V = 2.37(0.0885)

V = 0.21 m/s.

Answer:

I = 4.189 mA    V = 0.338 V

Explanation:

In order to do this, we need to apply the following expression:

I = Is[exp^(qV/kT) - 1]   (1)

However, as the junction of the diode is illuminated, the above expression changes to:

I = Iopt + Is[exp^(qV/kT) - 1]   (2)

Now, as the shunt resistance becomes infinite while the current becomes zero, we can say that the leakage current is small, and so:

I ≅ Iopt

Therefore:

I ≅ I₀Aλq / hc  (3)

Where:

I₀A = Area of diode (radiation)

λ: wavelength

q: electron charge (1.6x10⁻¹⁹ C)

h: Planck constant (6.62x10⁻³⁴ m² kg/s)

c: speed of light (3x10⁸ m/s)

Replacing all these values, we can get the current:

I = (8x10⁻³) * (0.65x10⁻⁶) * (1.6x10⁻¹⁹) / (6.62x10⁻³⁴) * (3x10⁸)

I = 4.189x10⁻³ A or 4.189 mA

Now that we have the current, we just need to replace this value into the expression (2) and solve for the voltage:

I = Is[exp^(qV/kT) - 1]

k: boltzman constant (1.38x10⁻²³ J/K)

4.189x10⁻³ = 9x10⁻⁹ [exp(1.6x10⁻¹⁹ V / 1.38x10⁻²³ * 300) - 1]

4.189x10⁻³ / 9x10⁻⁹ = [exp(38.65V) - 1]

465,444.44 + 1  = exp(38.65V)

ln(465,445.44) = 38.65V

13.0508 = 38.65V

V = 0.338 V

Answer:

Coefficient of kinetic friction = 0.235

Explanation:

Given:

Mass of crate = 330 kg

1st force = 430 N

2nd force = 330 N

Find:

Coefficient of kinetic friction.

Computation:

We know that, velocity is constant.

So, acceleration (a) = 0

So, net force (f) = 430 N + 330 N

Net force (f) = 760 N

F = μmg

μ = f / mg                                   [∵ g = 9.8]

μ = 760 / [330 × 9.8]

μ = 760 / [3,234]

μ = 0.235

Coefficient of kinetic friction = 0.235

Answer:

1.25kg

Explanation:

Simply multiply volume and density together

Explanation:

Complete the first and second sentences, choosing the correct answer from the given ones.

1. T = 100 K

[tex]^{\circ}C=K-273[/tex]

Put T = 100 K

[tex]T=100-273=-173^{\circ} C[/tex]

A temperature of 100 K corresponds on a Celsius scale to (-173 °C)

2. T = 50 °C

[tex]K=^{\circ}C+273\\\\K=50+273\\\\T=323\ K[/tex]

So, At 50 °C, it corresponds to a Kelvin scale of 323 K.

Answer:

Explanation:

The equivalent resistance of resistor connected parallel in the circuit is [tex]66.66 ohm[/tex]

The equivalent resistance is the total resistance measured in a parallel or series circuit. If several resistors are connected together and connected to a battery, the current supplied by the battery depends on the equivalent resistance of the circuit.

Resistors are in series whenever the current flows through the resistors sequentially. It is given by

[tex]R_{eq} = R_{1} + R_{2} + ....[/tex]

Resistors are in parallel when one end of all the resistors are connected by a continuous wire and the other end of all the resistors are also connected to one another through a continuous wire.

The equivalent resistance is the total resistance measured in a parallel. It is given by

[tex]\frac{1}{R_{eq} } = \frac{1}{R_{1} } + \frac{1}{R_{2} }+ ....[/tex]

Given:

Resistor, [tex]R_{1} = 100 ohm[/tex]

Resistor, [tex]R_{2} = 200 ohm[/tex]

Voltage, [tex]V = 10 Volt[/tex]

Since, resistors are connected in parallel, the equivalent resistor is given by,

[tex]\frac{1}{R_{eq} } = \frac{1}{R_{1} } + \frac{1}{R_{2} }[/tex]

[tex]\frac{1}{R_{eq} } = \frac{1}{100 } + \frac{1}{200 }[/tex]

[tex]R_{eq} = \frac{100*200}{100+200} \\R_{eq} = 66.66 ohm[/tex]

Hence, the equivalent resistor is [tex]66.66 ohm[/tex].

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Answer: A and D

a. Many objects move in arcs of circles or complete circles at times, and polar coordinates allowthe motion of such objects to be comprehended more easily. Some of Newton's laws in certaincases, such as calculation ofg from first principles, are much easier to calculate using polar coordinates.

b. "These coordinates are just used to confuse students.

c. ""Physics teachers are helping math teachers by getting students to practice theirtrigonometry.

d. ""In some cases, such as addition of forces, where a force magnitude is specified, it is simpler todescribe the forces in polar coordinates and be able to convert to the xy representation.

Explanation:

(A). Any force possessing a fixed magnitude and direction, it's always important to describe the for exactly as it is and also be able to convert it from one coordinate representation to another.

(D). Anything which involves radial motion(this is a motion along a radius) or motion along an arc of a circle or ellipse, this kind of motion is best explained and easily understood when in polar coordinates.

Explanation:

The answer is B because Nuclear family mean a family with two kids and Mrs. Parker have two kids

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is  B

Explanation:

The system is best considered a closed system because looking at process we can see that there was no exchange of matter between the system and the surrounding,(as the was no escape of matter from the system to the surrounding )

Secondly we can  deduce that there is  a variation in the volume. from [tex]1.0 m^3[/tex] to [tex]0.5 m^3[/tex]

Answer:

89.11kg

Explanation:

Note an object weighs less when in a fluid and the weight of the volume of the fluid displaced is known as the upthrust.

Now, the person is going to displace the volume 89/1025 =0.087m3 { from density D = mass(M)/volume(V)}

The weight of the fluid displaced is the density of the fluid × volume of fluid displaced.

The weight of the fluid=0.087m3× 1kg/me = 0.087kg

Now the weight of the fluid displaced is referred to as the upthrust.

Now the real weight - the apparent weight = the upthrust.

Hence the apparent weight = real weight - upthrust

Apparent weight = 89.2-0.087 = 89.11kg

Answer:

[tex]B_T=2.0*10^-5[-\hat{i}+\hat{j}]T[/tex]

Explanation:

To find the magnitude of the magnetic field, you use the following formula for the calculation of the magnetic field generated by a current in a wire:

[tex]B=\frac{\mu_oI}{2\pi r}[/tex]

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

I: current = 6.0 A

r: distance to the wire in which magnetic field is measured

In this case, you have four wires at corners of a square of length 9.0cm = 0.09m

You calculate the magnetic field in one corner. Then, you have to sum the contribution of all magnetic field generated by the other three wires, in the other corners. Furthermore, you have to take into account the direction of such magnetic fields. The direction of the magnetic field is given by the right-hand side rule.

If you assume that the magnetic field is measured in the up-right corner of the square, the wire to the left generates a magnetic field (in the corner in which you measure B) with direction upward (+ j), the wire down (down-right) generates a magnetic field with direction to the left (- i)  and the third wire generates a magnetic field with a direction that is 45° over the horizontal in the left direction (you can notice that in the image attached below). The total magnetic field will be:

[tex]B_T=B_1+B_2+B_3\\\\B_{T}=\frac{\mu_o I_1}{2\pi r_1}\hat{j}-\frac{\mu_o I_2}{2\pi r_2}\hat{i}+\frac{\mu_o I_3}{2\pi r_3}[-cos45\hat{i}+sin45\hat{j}][/tex]

I1 = I2 = I3 = 6.0A

r1 = 0.09m

r2 = 0.09m

[tex]r_3=\sqrt{(0.09)^2+(0.09)^2}m=0.127m[/tex]

Then you have:

[tex]B_T=\frac{\mu_o I}{2\pi}[(-\frac{1}{r_2}-\frac{cos45}{r_3})\hat{i}+(\frac{1}{r_1}+\frac{sin45}{r_3})\hat{j}}]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[(-\frac{1}{0.09m}-\frac{cos45}{0.127m})\hat{i}+(\frac{1}{0.09m}+\frac{sin45}{0.127m})]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[-16.67\hat{i}+16.67\hat{j}]\\\\B_T=2.0*10^-5[-\hat{i}+\hat{j}]T[/tex]

Answer:

Explanation:

If there is only 1 force, the body can never be in equilibrium, providing that the force is not zero (and that would hardly be a force. Zero is possible in math and it means something. It is debatable in physics).

You cannot think of a condition where something is stationary on planet earth and there are not 2 forces or more forces involved.

Think of something like a block of wood sitting on a table. It is not moving, we'll say. Gravity is holding it down, but what is pushing up on it?

The table is. There are 2 forces and they are equal in magnitude, but opposite in direction. That matters.

Answer:

the force we will feel is F

Explanation:

 According to the Gauss law, electric field due to very large sheet of charge is as follows.

[tex]E = \frac{\sigma}{2 \times \epsilon_{o}}[/tex]

where,        

[tex]\sigma[/tex] =  charge per unit area

Since, it is given that there are two sheets of equal and opposite charge. Therefore, electric field between the plates will be as follows.   [tex]E = \frac{\sigma}{2 \times \epsilon_{o}} + \frac{\sigma}{2 \times \epsilon_{o}}[/tex]

Also, we know that relation between force and electric field is as follows.

                      F = qE

Hence, force felt by the charge present inside the plates will be as follows.

                 [tex]F = q \times \frac{\sigma}{2 \times \epsilon_{o}}[/tex]

This depicts that force is not dependent on the distance and the charge is kept from one of the plate. Therefore,  force F felt by the charge is same when it is placed at a distance d/2 and at a distance d/4 from one of the plate.

Answer:

Explanation:

According to snell's law which states that the ratio of the sin of incidence (i) to the angle of refraction(n) is a constant for a given pair of media.

sini/sinr = n

n is the constant = refractive index

Since the diver shines light up to the surface of a flat glass-bottomed boat, the refractive index n = nw/ng

nw is the refractive index of water and ng is that of glass

sini/sinr = nw/ng

given i = 30°, nw = 1.33, ng = 1.5, r = angle the light leave the glass

On substitution;

sin 30/sinr = 1.33/1.5

1.5sin30 = 1.33sinr

sinr = 1.5sin30/1.33

sinr = 0.75/1.33

sinr = 0.5639

r = arcsin0.5639

r ≈35°

angle the light leave the glass is 35°