Help needed for science homework

This question can be solved using the equations of motion. There are two scenarios where the equations of motion can be used. The first scenario is the free-fall motion of the piece of bread. The second scenario is the uniformly accelerated motion of the bird.

The acceleration of the bird is  "a = 26.13 m/s²".

First, we will calculate the time taken by the bread to fall 3 m. Using the second equation of motion for this free-fall motion:

[tex]h = v_it + \frac{1}{2}gt^2[/tex]

where,

h = height fall = 3 m

vi = initial velocity = 0 m/s

g = acceleration due to gravity = 9.8 m/s²

t = time taken = ?

Therefore,

[tex]3\ m = (0\ m/s)t+\frac{1}{2}(9.8\ m/s^2)t^2\\t = \sqrt{\frac{(3\ m)(2)}{9.8\ m/s^2}}\\\\t = 0.78\ s[/tex]

The bird took the same time to catch the bread. Now applying the second equation of motion to the bird's motion:

[tex]s = v_it + \frac{1}{2}at^2[/tex]

where,

s = distance covered by the bird = 5 m + 3 m = 8 m

vi = initial velocity of the bird = 0 m/s

a = acceleration of the bird = ?

t = time taken = 0.78 s

Therefore, using these values we get:

[tex]8\ m = (0\ m/s)(0.78\ s)+\frac{1}{2}a(0.78\ s)^2\\\\a = \frac{16\ m}{(0.78\ s)^2}[/tex]

a = 26.13 m/s²

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Part A - The apparent displacement of the mark is 4 cm.

The refractive index of a material, n = real depth, D/apparent depth, L.

So, n = D/L

Since the refractive index of the triangular glass prism is 1.5, n = 1.5. Also, the triangular glass prism is 12 cm thick and placed on a mark on a piece of paper resting on a horizontal bench. So, the real depth of the mark on the piece of paper through the triangular prism is D = 12 cm and its apparent depth is L.

Since n = D/L,

making L, subject of the formula, we have

L = D/n

Substituting the values of the variables into the equation, we have

L = D/n

L = 12 cm/1.5

L = 8 cm

so, the apparent depth of the mark is 8 cm.

So, the apparent displacement of the mark is d = D - L = 12 cm - 8 cm = 4 cm

Thus the apparent displacement of the mark is 4 cm.

Part B - The refractive index of water, n = 1.33.

The refractive index of a material, n = real depth, D/apparent depth, L.

So, n = D/L

Since the depth of water in the jar is 24 cm, the real depth, D = 24 cm.

Also, the bottom of the jar appears to be raised by 6 cm. So, the apparent depth, L = 24 cm - 6 cm = 18 cm

Since n = D/L, the refractive index of water n = D/L

Substituting the values of the variables into the equation, we have

n = D/L

= 24 cm/18 cm

= 1.33

So, the refractive index of water, n = 1.33.

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Answer:

option b is correct..................

Answer:

4.097 Joules

Explanation:

Applying,

P.E = mgh............... Equation 1

Where P.E = Potential energy, m = lemming mass, h = height, g = acceleration due to gravity

From the question,

Given: m = 0.0780 kg, h = 5.36 m

Constant: g = 9.8 m/s²

Substitute these values into equation 1

P.E = 0.0780(9.8)(5.36)

P.E = 4.097 Joules.

Hence the potential energy is 4.097 Joules

Answer:

4.097 J

Explanation:

Answer:

Computer hard drives use magnetism to store the data on a rotating disk. More complex applications include: televisions, radios, microwave ovens, telephone systems, and computers. An industrial application of magnetic force is an electromagnetic crane that is used for lifting metal objects.

Answer:

Examples of magnetic force is a compass, a motor, the magnets that hold stuff on the refrigerator, train tracks, and new roller coasters. All moving charges give rise to a magnetic field and the charges that move through its regions, experience a force.

I Hope this will help you if not then sorry :)

Answer:

The Reference standard with which we carry out the measurement of any physical quantity of the same kind is know as unit .

I HOPE I HELP YOU.

Answer:

The standard quantity which is used for the comparison with an unknown quantity is called unit.

Unit is a known or standard quantity in terms of which other physical quantities are measured.

for examples;kilogram,metre,second,etc.

Explanation:

hope it helps you

The kinematic energy of the positive charge is 2 10⁻⁸ J

This electrostatics exercise must be done in parts, the first part: let's start by finding the charge of the capacitor, the capacitance is defined by

        C = [tex]\frac{Q}{\Delta V}[/tex]

        C = ε₀ [tex]\frac{A}{d}[/tex]

we solve for the charge (Q)

        [tex]\frac{Q}{\Delta V} = \epsilon_o \frac{A}{d}[/tex]

indicates that for the initial point d₁ = 3 mm = 0.003 m and the voltage is DV₁ = 12

         Q = [tex]\epsilon_o \ \frac{A \ \Delta V_1 }{d_1}[/tex]

Now the voltage source is disconnected so the charge remains constant across the ideal capacitor.

For the second part, the condenser is separated at d₂ = 5mm = 0.005 m

         Q = \epsilon_o \  \frac{A \ \Delta V_2 }{d_2}

we match the expressions of the charge and look for the voltage

          [tex]\frac{\Delta V_1}{d_1} = \frac{\Delta V_2}{d_2}[/tex]

          ΔV₂ = [tex]\frac{d_2}{d_1 } \ \Delta V_1[/tex]

The third part we use the concepts of conservation of energy

starting point. With the test load (q = 1 nC = 1 10⁻⁹ C) next to the left plate

          Em₀ = U = q DV₂

          Em₀ = q  \frac{d_2}{d_1 } \ \Delta V_1

final point. Proof load on the right plate

         Em_f = K

energy is conserved

         Em₀ = em_f

         q  \frac{d_2}{d_1 } \ \Delta V_1 = K

we calculate

         K = 1 10⁻⁹  12  [tex]\frac{0.005}{0.003}[/tex]  

         K = 20 10⁻⁹ J

In this exercise, as the conditions at two different points of separation give, the area of ​​the condenser is not necessary and with conservation of energy we find the final kinetic energy of 2 10⁻⁸ J

Answer:

Đáp án là vị trí để đặt q3 tại vị trí cân bằng là cách A một đoạn 0,0411 m

và nằm trong AB

Explanation:

Do q1 và q2 trái dấu nên q3 nằm trong

gọi r1 là vị trí từ A đến q3 suy ra r2=0.08-r1

Vị trí cân bằng nên [tex]\frac{q1}{r1^{2} }[/tex] = [tex]\frac{q2}{(0.08-r1)^{2} }[/tex]

Answer:

d

Explanation:

I'm not perfectly sure so check again

Answer:

Obeying to weights & measurement regulations in both national and international metrology legislation, standards and test procedures is a requirement to participate in any market because it's aimed to safeguard the consumers and promote fair competition, which provides efficiency and saves unnecessary costs to U.S. businesses and stakeholders.

Explanation:

In the U.S., the National Institute of Standards and Technology (NIST) has an Office of Weights and Measures (OWM) that represents the country in the International Organization of Legal Metrology (OIML)

Answer:

F1=26N and F2=09N ..this is from the two simultaneously equations

Answer:

Explanation:

As,

In Option,

A:

There are two same positive ions so they will move away from each other.

B:

There is one negative and positive ion so they will move towards each other.

C:

Again there is one negative and positive ion so they will move towards each other.

D:

Here, there is neutral ions so they will not move and its the correct option.

Answer:

No,

Explanation:

Efficiency of a machine is always less than 100% as output energy is always less than the input energy, because some energy is lost to overcome friction.

Explanation:

light travel slower in daimond

Answer:

nature of the material

hope this help you

Answer:

DU = 120 Joules

Explanation:

Given the following data;

Quantity of energy = 200 J

Work = 80 J

To find the change in internal energy;

Mathematically, the change in internal energy of a system is given by the formula;

DU = Q - W

Where;

DU is the change in internal energy.

Q is the quantity of energy.

W is the work done.

Substituting into the formula, we have;

DU = 200 - 80

DU = 120 Joules

Answer:

Un diafragma es una pieza de silicona o de goma reutilizable que cubre el cuello del útero. El diafragma se coloca dentro de la vagina con espermicida antes de tener relaciones sexuales para prevenir un embarazo. El diafragma es un dispositivo anticonceptivo que impide que los espermatozoides ingresen en el útero.

Answer:

3 to 7 seconds.

Explanation:

Answer:

11760 J

Explanation:

cuz potential engery is PE = MHG

SO 80×15×9.8= 11760 J

Answer:

11769j

Explanation:

here,

mass(m)=80 kg

height(h)=15m

acceleration due to gravity(g)=9.8m/s^2

now,

potential energy = m×g×h

= 80×9.8×15

= 11760j

The frequency of the other car is 669HZ

Let start off by writing out the parameters given in the question

frequency= 595 HZ

Speed of sound(v)= 343m/s

Speed of the car(vs)= 20m/s

Speed of the second car(vo) = 20m/s

The frequency of the other car can be calculated as follows

(v+vo/v-vs) f

= (343+20/343-20)595

= (363/323)595

= 1.1238(595)

= 669 HZ

Hence the frequency of the other car is 669 HZ

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Answer:

Amplitude.

Explanation:

A wave can be defined as a disturbance in a medium that progressively transports energy from a source location to another location without the transportation of matter.

In Science, there are two (2) types of wave and these include;

I. Electromagnetic waves: it doesn't require a medium for its propagation and as such can travel through an empty space or vacuum. An example of an electromagnetic wave is light.

II. Mechanical waves: it requires a medium for its propagation and as such can't travel through an empty space or vacuum. An example of a mechanical wave is sound.

An amplitude can be defined as a waveform that's measured from the center line (its origin or equilibrium position) to the bottom of a trough or top of a crest.

Hence, an amplitude is a word that describes the maximum displacement a point moves from its rest position when a wave passes.

On a graph, the vertical axis (y-axis) is the amplitude of a waveform and this simply means that, it's measured vertically.

Mathematically, the amplitude of a wave is given by the formula;

x = Asin(ωt + ϕ)

Where;

x is displacement of the wave measured in meters.

A is the amplitude.

ω is the angular frequency measured in rad/s.

t is the time period measured in seconds.

ϕ is the phase angle.

Chlorine has the smallest atomic radius since the atomic radius decreases as you travel to the right and up

Answer:

When a tree is vigorously shaken, the branches of the tree come in motion but the leaves tend to continue in their state of rest due to inertia of rest. As a result of this, leaves get separated from the branches of the tree and hence fall down.

Answer:

When a tree is vigorously shaken, the branches of the tree come in motion moving left and right and shake the leaves off, that's why some of the tree got detached when we vigorously shake its branch.

Explanation:

Like I did in the answer section.

Answer:

Torque; τ = 4.712 × 10^(-3) J

Magnetic moment; M = 0.0248 J/T

Explanation:

Torque is gotten from the formula;

τ = BIA

Where;

B is magnetic field

I is current

A is area

We are given;

B = 0.19T

I = 6.2A

Rectangle dimensions = 5cm by 8cm = 0.05m by 0.08m

Thus;

Area; A = 0.05m × 0.08m = 0.004 m²

Thus;

τ = 0.19 × 6.2 × 0.004

τ = 4.712 × 10^(-3) J

Formula for the magnetic moment is given by;

M = IA

M = 6.2 × 0.004

M = 0.0248 J/T

Answer:

when it comes to machines being Rusty oil is like the root loosener

Answer:

The three primaryfactors of evaporation are heat, atmospheric pressure (which determines the percentage of moisture) and the movementofair. At the molecular level, there is no strict boundary between the liquid state and the vapor state

Answer:

I. Vs = 8.0 Volts.

II. Is = 4.5 Amperes.

Explanation:

Given the following data;

Np = 1.5Ns = [tex] \frac {N_{P}}{N_{S}} = 1.5 [/tex]  ..... equation 1

Ip = 3.0 A

Vp = 12 V

To find the voltage and current on the secondary coil;

I. For the voltage in the secondary coil (Vs), we would use the following formula;

[tex] \frac {V_{P}}{V_{S}} = \frac {N_{P}}{N_{S}} [/tex]  ...... equation 2.

Substituting eqn 1 into eqn 2, we have;

[tex] \frac {V_{P}}{V_{S}} = 1.5 [/tex]

[tex] \frac {12}{V_{S}} = 1.5 [/tex]

Cross-multiplying, we have;

[tex] V_{S} * 1.5 = 12 [/tex]

[tex] V_{S} = \frac {12}{1.5} [/tex]

Vs = 8.0 V

II. For the current in the secondary coil (Is), we would use the following formula;

[tex] \frac {I_{S}}{I_{P}} = \frac {N_{P}}{N_{S}} [/tex]  .... equation 3

Substituting eqn 1 into eqn 3, we have;

[tex] \frac {I_{S}}{I_{P}} = 1.5 [/tex]

[tex] \frac {I_{S}}{3.0} = 1.5 [/tex]

Cross-multiplying, we have;

[tex] I_{S} = 1.5 * 3.0 [/tex]

Is = 4.5 A

Work = force x distance

Work = 5500 x 1.8

Work = 9900 N

The work measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement.

Work = force x distance

Work = 5500 x 1.8

Work = 9900 N

therefore, work does 9900 N

Work is force applied over distance. Examples of work include lifting an object against the Earth's gravitation, driving a car up a hill, and pulling down a captive helium balloon. Work is a mechanical manifestation of energy. The standard unit of work is the joule (J), equivalent to a newton - meter (N · m).

Work is defined as transferring energy into an object so that there is some displacement. Energy is defined as the ability to do work. Work done is always the same. Energy can be of different types such as kinetic and potential energy.

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Answer:

La fuerza resultante sobre q₃ es  -1.2245 × 10⁻¹⁵ i  + -0.24 × 10⁻¹⁵ j

La magnitud de la fuerza resultante sobre q₃ es aproximadamente 1.25 × 10⁻¹⁵ N

Explanation:

q₁ = -45 μC = -45 × 10⁻⁶ C

r₁₂ = 30 mm = 30 × 10⁻³ m

q₂ = 25 μC = 25 × 10⁻⁶ C

r₂₃ = 50 mm = 50 × 10⁻³ m

q₃ = 20 μC = 20 × 10⁻⁶ C

k = 9×10⁻⁹ N·m²/C²

Por lo tanto;

r₁₃ = √(50² + 30²) = 10·√(34)

F₁₂ = 9×10⁻⁹ × (-45 × 10⁻⁶)×(25 × 10⁻⁶)/(30 × 10⁻³)² = -1.125 × 10⁻¹⁴

F₁₂ = -1.125 × 10⁻¹⁴ N

F₂₃ = 9×10⁻⁹ × (20 × 10⁻⁶)×(25 × 10⁻⁶)/(50 × 10⁻³)² = 1.8 × 10⁻¹⁵ j

F₁₃ = 9×10⁻⁹ × (-45 × 10⁻⁶)×(20 × 10⁻⁶)/(10·√34 × 10⁻³)² = -2.38× 10⁻¹⁵

Los componentes de F₁₃ son;

-2,38 × 10⁻¹⁵ × cos (arctan (30/50)) = -2,04 × 10⁻¹⁵ j

-2,38 × 10⁻¹⁵ × sin (arctan (30/50)) = -1,2245 × 10⁻¹⁵ i

La fuerza resultante sobre la carga q₃, [tex]\left | \underset {F_3} \rightarrow \right |[/tex] = [tex]\underset{F_{13}}{\rightarrow}[/tex] + [tex]\underset{F_{23}}{\rightarrow}[/tex]

∴ [tex]\left | \underset {F_3} \rightarrow \right |[/tex] = 1.8 × 10⁻¹⁵ j + -1.2245 × 10⁻¹⁵ i + -2.04 × 10⁻¹⁵ j

La fuerza resultante sobre q₃ es [tex]\left | \underset {F_3} \rightarrow \right |[/tex]  = -1.2245 × 10⁻¹⁵ i  + -0.24 × 10⁻¹⁵ j

La magnitud de la fuerza resultante sobre q₃,

[tex]\left | F_3 \right |[/tex] = √((-1.2245 × 10⁻¹⁵)² + (-0.24 × 10⁻¹⁵)²) ≈ 1.25 × 10⁻¹⁵

La magnitud de la fuerza resultante sobre q₃, [tex]\left | F_3 \right |[/tex] ≈ 1.25 × 10⁻¹⁵ N.