2. Ms. Evelyn lined up her students at recess on the soccer field. She told each student to kick a soccer ball as hard as they could. Then she measured how far each student kicked the soccer ball and marked the distance with a flag. After the activity she brought the class back inside and created a graph from the data. Did Ms. Evelyn's class do an experiment? A. Yes, because they did several tests with the soccer balls and recorded their observations. B. Yes, because every student kicked the same soccer ball. C. No, because they did not start with a hypothesis or a control group. This was an investigation. D. No, because all experiments require the use of a laboratory and scientists.​

Answer:

   x_total = 3.07 m

Explanation:

Let's analyze the situation presented, in this case we have the data to find the angle of incidence and with the law of refraction we can find the angle of refraction.

Let's start looking for the angle with which the laser pointer reaches the water, let's use trigonometry

          tan θ₁ = 1.0 /2.0

          θ₁ = tan⁻¹ 0.50

          θ₁ = 26.57º

now let's use the law of refraction to find the angle of refraction (in water)

          n₁ sin θ₁ = n₂ sin θ₂

the refractive index of air is n₁ = 1 and that of water n₂ = 1.33

         sin θ₂ = [tex]\frac{n_1}{n_2} \ sin \theta_1[/tex]

         sin θ₂ = [tex]\frac{1}{1.33} \ sin \ 26.57[/tex]

         θ₂ = sin⁻¹ 0.3363

         θ₂ = 19.65º

now we can find the distance from the entry point to the water to the lenses

         tan θ₂ = x₂ / 3

         x₂ = 3 tan θ₂

         x₂ = 3 tan 19.65

         x₂ = 1.07 m

the total distance from the edge of the pool is

        x_total = 2 + x₂

        x_total = 2 + 1.07

        x_total = 3.07 m

Answer:

[tex](75.0 \times 30) + (60.0 \times 6.0) = (75.0 \times V) + (60.0 \times 0) \\ 2250 + 360 = 75V \\ 75V = 2610 \\ V = 34.8 \: m {s}^{ - 1} [/tex]

Answer:

O The environment did work on an object

Explanation:

Answer:

Tenemos dos problemas a resolver acá:

Primero, debemos encontrar la velocidad con la que la pelota impacta el suelo.

Acá podemos usar la conservación de la energía.

E = U + K

U = energía potencial = m*g*H

m = masa

g = aceleración gravitatoria = 9.8m/s^2

H = altura

K = energía cinética = (m/2)*V^2

donde V es la velocidad.

Inicialmente, cuando la pelota es soltada, su velocidad es cero, entonces solo tenemos energía potencial:

Ei = U = m*(9.8m/s^2)*2.5m

Al final, cuando la pelota esta por impactar el suelo, la altura tiende a cero, entonces ya no hay energía potencial, solo hay energía cinética:

Ef = (m/2)*V^2

Y como la energía se conserva, la energía final es igual a la inicial, entonces:

m*(9.8m/s^2)*2.5m = (m/2)*V^2

Podemos resolver esto para V, y asi obtener la velocidad con la que la pelota impacta el suelo.

V = √(2*(9.8m/s^2)*2.5m) = 7m/s

Ahora respondamos la segunda parte.

Una vez la pelota rebota, su aceleración va a estar dada solamente por la aceleración gravitatoria, entonces tenemos:

A(t) = -9.8m/s^2

Para obtener su velocidad integramos:

V(t) = (-9.8m/s^2)*t + V0

donde V0 es la velocidad con la que la pelota reboto, que sabemos que es 3/4 de 7m/s

V0 = (3/4)*7m/s = (21/4) m/s

Así, la ecuación de la velocidad es:

V(t) = (-9.8m/s^2)*t + (21/4) m/s

Sabemos que la altura máxima se da cuando la velocidad es igual a cero, entonces primero calculemos el valor de t tal que esto ocurra:

V(t) = 0 = (-9.8m/s^2)*t + (21/4) m/s

         t =  (21/4) m/s/9.8m/s^2 = 0.54 s

Ahora debemos encontrar la ecuación de la posición y evaluarlo en este tiempo.

Para ello integramos de vuelta:

P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t + P0

donde P0 es la posición inicial, como la pelota rebota en el suelo, la posición inicial es el suelo, el cual representamos con 0, entonces la ecuación de la posición es:

P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t  

La altura máxima estará dada por esta ecuación evaluada en t = 0.54 s

P(0.54s) =  (1/2)(-9.8m/s^2)*(0.54s)^2 + (21/4 m/s)*0.54s = 1.81 m

La altura máxima es 1.81 metros.

Y entre que rebota y llega a esta altura máxima, transcurren 0.54 segundos.

Answer:

The power is 465.44 W.

Explanation:

mass, m = 70 kg

number of steps,  n = 1576

height of each step, h = 0.241 m

time taken, t = 9.33 min= 9.33 x 60 s

The power is given by the rate of doing work.

W = n m g h

W = 1576 x 70 x 9.8 x 0.241

W = 260553.776 J

The power is given by

[tex]P = \frac{W}{t}\\\\P = \frac{260553.776}{9.33\times 60}\\\\P = 465.44 W[/tex]  

The statement that can be explained using Kepler's Second Law is "Mars moves faster in its orbit when it is closer to the Sun than when it is farther from the Sun."

According to Encyclopedia Britannica, Kepler's second law states that; "a planet moves in its ellipse so that the line between it and the Sun placed at a focus sweeps out equal areas in equal times."

This law is sometimes referred to as the law of equal areas. It describes the fact that a planet moves faster when it is closer to the sun and slower when it farther from the sun.

An implication of this law is that mars moves faster in its orbit when it is closer to the Sun than when it is farther from the Sun.

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Answer:

(a) t = 1.14 s

(b) h = 0.82 m

(c) vf = 7.17 m/s

Explanation:

(b)

Considering the upward motion, we apply the third equation of motion:

[tex]2gh = v_f^2 - v_i^2[/tex]

where,

g = - 9.8 m/s² (-ve sign for upward motion)

h = max height reached = ?

vf = final speed = 0 m/s

vi = initial speed = 4 m/s

Therefore,

[tex](2)(9.8\ m/s^2)h = (0\ m/s)^2-(4\ m/s)^2\\[/tex]

h = 0.82 m

Now, for the time in air during upward motion we use first equation of motion:

[tex]v_f = v_i + gt_1\\0\ m/s = 4\ m/s + (-9.8\ m/s^2)t_1\\t_1 = 0.41\ s[/tex]

(c)

Now we will consider the downward motion and use the third equation of motion:

[tex]2gh = v_f^2-v_i^2[/tex]

where,

h = total height = 0.82 m + 1.8 m = 2.62 m

vi = initial speed = 0 m/s

g = 9.8 m/s²

vf = final speed = ?

Therefore,

[tex]2(9.8\ m/s^2)(2.62\ m) = v_f^2 - (0\ m/s)^2\\[/tex]

vf = 7.17 m/s

Now, for the time in air during downward motion we use the first equation of motion:

[tex]v_f = v_i + gt_1\\7.17\ m/s = 0\ m/s + (9.8\ m/s^2)t_2\\t_2 = 0.73\ s[/tex]

(a)

Total Time of Flight = t = t₁ + t₂

t = 0.41 s + 0.73 s

t = 1.14 s

Answer:

wait

Explanation:

Answer:

D. Non- polar solvant

Explanation:

l think that's it

Answer:

I think the answer is D polar solvent

Answer:

A

Explanation:

If the object is moving at a constant speed, the object isn't accelerating as the velocity doesn't change.

Answer: C.

Explanation:  plato users

Answer:

Mechanical energy

Explanation:

If an object does not have any momentum, then it doesn't have mechicanl energy.

Answer:

it is energy resulting in charged particles

Electricity a form of energy resulting from the existence of charged particles (such as electrons or protons), either statically as an accumulation of charge or dynamically as a current

Answer:

M' = μ₀n₁n₂πr₂²

Explanation:

Since r₂ < r₁ the mutual inductance M = N₂Ф₂₁/i₁ where N₂ = number of turns of solenoid 2 = n₂l where n₂ = number of turns per unit length of solenoid 2 and l = length of solenoid, Ф₂₁ = flux in solenoid 2 due to magnetic field in solenoid 1 = B₁A₂ where B₁ = magnetic field due to solenoid 1 = μ₀n₁i₁ where μ₀ = permeability of free space, n₁ = number of turns per unit length of solenoid 1 and i₁ = current in solenoid 1. A₂ = area of solenoid 2 = πr₂² where r₂ = radius of solenoid 2.

So, M = N₂Ф₂₁/i₁

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

M = N₂Ф₂₁/i₁

M = N₂B₁A₂/i₁

M = n₂lμ₀n₁i₁πr₂²/i₁

M = lμ₀n₁n₂πr₂²

So, the mutual inductance per unit length is M' = M/l = μ₀n₁n₂πr₂²

M' = μ₀n₁n₂πr₂²

(a) Attached to the response as Figure 1.

(b) 35.0Ω

(c) Across 5.0Ω = 1.3V

   Across 10.0Ω = 2.6Ω

   Across 20.0Ω = 5.2Ω

(a) The labelled circuit using the correct symbols (for the resistors and battery) has been attached to this response.

(b) Since the resistors are hooked up in series, their equivalent resistance R, is found by adding the individual resistances of the resistors (R₁, R₂ and R₃). i.e

R = R₁ + R₂ + R₃               -------------------(i)

Where;

R₁ = 5.0 Ω

R₂ = 10.0 Ω

R₃ = 20.0 Ω

Substitute these values into equation (i) as follows;

∴ R = 5.0 Ω + 10.0 Ω + 20.0 Ω

∴ R = 35.0 Ω

Therefore, the equivalent resistance is ∴ R = 35.0Ω

(c) When resistors are connected in series, the same current passes through them. To get the current through each resistor;

i. First, replace the resistors by their equivalent resistor as calculated above. The diagram has been attached to this response.

ii. As seen in the diagram, the current flowing through the equivalent resistor can be calculated using Ohm's law as follows;

V = I R              ------------------(ii)

Where;

V = Voltage supplied to the circuit = 9.0V

I = Current through the circuit

R = Resistance of the equivalent resistor = 35.0Ω

Substitute these values into equation (ii)

9.0 = I x 35.0

I = [tex]\frac{9.0}{35.0}[/tex]

I = 0.26A

This is also the current flowing through each of the resistors separately.

iii. Calculate the voltage drop across

1. 5.0 Ω resistor

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 5.0Ω resistor

I = current through the 5.0Ω resistor = 0.26A

R = resistance of the 5.0Ω resistor = 5.0Ω

=> V = 0.26 x 5.0

=> V = 1.3V

2. 10.0 Ω resistor

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 10.0Ω resistor

I = current through the 10.0Ω resistor = 0.26A

R = resistance of the 10.0Ω resistor = 10.0Ω

=> V = 0.26 x 10.0

=> V = 2.6V

3. 20.0 Ω resistor

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 20.0Ω resistor

I = current through the 20.0Ω resistor = 0.26A

R = resistance of the 20.0Ω resistor = 10.0Ω

=> V = 0.26 x 20.0

=> V = 5.2V

Answer:

0.53 quart

Explanation:

The volume expansion of the coolant is gotten from ΔV = VγΔθ where ΔV   = change in volume of the coolant, V = initial volume of coolant = 15 quart, γ = coefficient of volume expansion of coolant = 410 × 10⁻⁶ /°C and Δθ = temperature change = θ₂ - θ₁ where θ₁ = initial temperature of coolant = 6 °C and θ₂ = final temperature of coolant = 92 °C. So, Δθ = θ₂ - θ₁ = 92 °C - 6 °C = 86 °C

Since, ΔV = VγΔθ

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

ΔV = VγΔθ

ΔV = 15 × 410 × 10⁻⁶ /°C × 86 °C

ΔV = 528900 × 10⁻⁶ quart

ΔV = 0.528900 quart

ΔV ≅ 0.53 quart

Since the change in volume of the coolant equals the spill over volume, thus the overflow from the radiator will spill into the reservoir when the coolant reaches its operating temperature of 92 °C is 0.53 quart.

Answer:

y = 5 cos 2πt

Explanation:

We will use the formula for simple harmonic motion curve where;

y = a cos ωt

Where;

a is amplitude

t is period

ω is angular frequency with the formula; ω = 2π/t

We are told that when the spring is compressed, the mass is located 5 cm above its rest position.

Thus;

a = 5 cm

it's highest point is 5 cm, but we are told that after 1/2 second of being released, it reaches its lowest point.

Since highest point is 5, then lowest point will be -5.

The difference in time between the highest and lowest point is ½ s. Which is half of the period.

Thus;

t/2 = ½

Thus, t = 1 s

Now, we know that;

t = 1/f = 2π/ω

Since t = 1, then 1 = 1/f

f = 1

Thus;

2π/ω = 1

Thus, ω = 2π

Thus, the equation is;

y = 5 cos 2πt

The equation that describes the motion of the mass is y = 5 cos 2πt.

The given parameters;

The general equation of the wave is given as;

[tex]y = A\ cos\ \omega t[/tex]

where;

The angular speed of the mass is calculated as;

[tex]\omega = 2\pi f\\\\[/tex]

The period of the oscillation is calculated as;

[tex]T = 2t \\\\T = 2(0.5 s) = 1 \ s[/tex]

The frequency of the wave is calculated as;

[tex]f = \frac{1}{T} \\\\f = \frac{1}{1} \\\\f = 1\ Hz[/tex]

The equation that describes the motion of the mass is calculated as;

[tex]y = A\ cos \ \omega t\\\\y = A\ cos \ 2\pi ft\\\\y = 5\ cos \ 2\pi (1) t\\\\y = 5 \ cos \ 2\pi t[/tex]

Thus, the equation that describes the motion of the mass is y = 5 cos 2πt.

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Answer:

O Distant objects are blurry. describes farsightedness.

Explanation:

Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry. The degree of your farsightedness influences your focusing ability.Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry.

Answer:

144 meters

Explanation:

the ball is thrown with a speed of 24 meters per second right so if the ball reaches the ground in 6 seconds. the hight of the cliff must be S=v.t

S (height cliff)=24m/s×6s=144

Answer:

d) and e)

Explanation:

Answer:

The correct solution is "11.51 mA".

Explanation:

Given:

Time average power,

[tex]P_{avg}=0.05 \ W/m^2[/tex]

n = 377

As we now,

⇒ [tex]P_{avg}=\frac{E_0^2}{n}[/tex]

or,

⇒ [tex]E_0^2=0.05\times 377[/tex]

⇒       [tex]=4.341 \ V[/tex]

hence,

⇒ [tex]H_0=\frac{E_0}{n}[/tex]

By putting the values, we get

         [tex]=\frac{4.341}{377}[/tex]

         [tex]=11.51 \ mA[/tex]

Answer:

8 N

Explanation:

Applying,

(F'+F) = ma............... Equation 1

Where F' = Amhed's force, F = Rashid's force, m = mass of the box, a = acceleration of the box.

From the question,

Given: F = 12 N, m = 4 kg, a = 5 m/s²

Substitute these values into equation 1

(F'+12) = 4×5

(F'+12) = 20

F' = 20-12

F' = 8 N.

Hence Ahmed's force is 8 N

Answer:

a) B = 3.11 km.  θ= 54.7º E of S

b) B = 3.11 km  θ= 54.7º W of S

Explanation:

a)

        [tex]C=\sqrt{(2.2km)^{2} + B^{2} } = 3.81 Km (1)[/tex]

       [tex]B=\sqrt{(3.81km)^{2} -(2.2km)^{2} } = 3.11 Km (2)[/tex]

        [tex]\theta = arc tg (\frac{B}{A} )= arc tg ( \frac{3.11}{2.2} )= arctg (1.414) = 54.7 deg (3)[/tex]

b)

      [tex]\theta = arc tg (\frac{-B}{A} )= arc tg ( \frac{-3.11}{2.2} )= arctg (-1.414) = -54.7 deg (4)[/tex]

Answer:

a. The disk

b. Because it has the smallest rotational inertia

Explanation:

a. Which object do you expect to reach the bottom of the inclined plan first?

I would expect the disk to reach the bottom first.

b. Why?

This is because the disk has the smallest rotational inertia.

The rotational inertial of the hollow sphere, disk and ring are 2/3MR², 1/2MR² and MR² respectively.

Since the three objects are rolling from the same height, they have the same mechanical energy.

But, since the disk has the smallest rotational inertia, it would have the smallest rotational kinetic energy and largest translational kinetic energy.  The disk's smaller rotational kinetic energy will cause  to rotate less but translate more than the other objects and thus reach the bottom first.

The object which is expected to reach the bottom of the inclined plan first is the disk, as it has the lowest rotational inertia.

Moment of inertia is the force which acts in the opposite direction of the force of angular acceleration acting on the body.

There are three objects, hollow sphere, disk and ring.

       [tex]I=\dfrac{2}{3}mr^2[/tex]

        [tex]I=mr^2[/tex]

Here, (m) is the mass and (r) is the radius of the object.

These three objects are going to race by releasing from rest at the top of an inclined plane to the bottom of the plane.

As moment of inertia is the force which acts in the opposite direction of the force of angular acceleration acting on the body.

Thus the less the value of inertia will result in less the time required to reach at the bottom of the inclined plane.

Hence, the object which is expected to reach the bottom of the inclined plan first is the disk, as it has the lowest rotational inertia.

Learn more about the force of inertia here;

https://brainly.com/question/10454047

The question is incomplete, the complete question is;

In addition to absorption of a photon, energy can be transferred to an atom by collision. Consider a hydrogen atom in its ground state. Incident on the atom are electrons having a kinetic energies of 10.5 eV. What is a possible result?

A) The atom moves to a state of lower energy

B) The atom is ionized

C) One of the electrons leaves the atom

D) The atom can be excited to a higher energy state

Answer:

The atom can be excited to a higher energy state

Explanation:

According to the Bohr model of the atom, electrons in an atom can be excited from a lower to a higher energy level when energy is absorbed by the atom.

If electrons having an energy of 10.5ev are incident on a hydrogen atom, this energy is transferred to the atom by collision. Since the energy transferred is less than the ionization energy of hydrogen atom in its ground state(13.6ev), the atom is not ionized.

Rather, the atom is excited from ground state to a higher energy level.

Answer:

  a₃ = -1.08 m/s²,    K = 1.42 J

Explanation:

The particle is in a periodic motion, so the general expression is

           x = A cos (wt + Ф)

let's compare the terms with the expression they give us

           x = 2 cos (t - 2)

the amplitude of motion is A = 2 m, the angular velocity w = 1 rad / s, and the phase is Ф = - 2.

to find the acceleration we use its definition

          v = dx / dt

          a = dv / dt

          a = [tex]\frac{ d^2x}{dt^2}[/tex]

let's perform the derivative

          v = - A w sin (wt + Ф)

          a = - A w² cos wt + Ф)

substituting the values

          a = - 2 1² cos (t-2)

for t = 3 s

          a₃ = 2 cos (3-2)

remember angles are in radians

          a₃ = -1.08 m/s²

To calculate kinetic energy, let's find the velocity for t = 3 s

          v = - 2 sin (t-2)

          v = -2 sin (3-2)

          v = - 1.683 m / s

body mass is m = 1 kg

we calculate

          K = ½ m v²

          K = ½ 1 (-1.683) ²

          K = 1.42 J

a. The force exerted by the ground on each of the front wheels is 4918.16 Newton.

b. The force exerted by the ground on each of the back wheels is 2627.84 Newton.

Given the following data:

a. To determine the force exerted by the ground on each of the front wheels:

First of all, we would take moment about the rear wheels.

[tex]F(3.13) - 1540(9.8) \times (3.13 - 1.09) = 0\\\\3.13F - 15092 \times 2.04 =0\\\\3.13F -30787.68=0\\\\F=\frac{30787.68}{3.13}[/tex]

Force, F = 9836.32 Newton

For each front wheel:

[tex]Force = \frac{9836.32}{2}[/tex]

Force = 4918.16 Newton.

b. To determine the force exerted by the ground on each of the back wheels:

We would determine the sum of the vertical forces acting on the wheels.

[tex]9836.32 + B - 1540(9.8) = 0\\\\9836.32 + B - 15092 = 0\\\\B=15092-9836.32[/tex]

B = 5255.68 Newton.

For each back wheel:

[tex]Force = \frac{5255.68}{2}[/tex]

Force = 2627.84 Newton.

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Answer:

As it is the most efficient heat engine, it's efficiency is [T1 - T2]/T1. It can be measured for every Carnot cycle. From the formula and diagram, we can understand that the efficiency of an ideal heat engine also depends on the difference between the hot and cold reservoirs.

Explanation:

Answer: When it comes to the movement of air, friction is greater near the ground surface.

Explanation:

A resistance in motion observed by an object while on another object is called friction.

For example, a vehicle moving on road will have friction between its tires and the road.

Friction is more near the ground surface rather than away from the ground surface.

Thus, we can conclude that when it comes to the movement of air, friction is greater near the ground surface.