3) if you have a convex (converging) lens (f#12 cm) that produced an image with a
magnification of -2.5, how far must the object be placed from the lens?
I
b) Draw a ray-tracing diagram of this situation below (label all points in cm)

Answer:

Explanation:

magnitude: 180-50=130N

Direction: in the direction same as the second horizontal force

Answer:

the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

Explanation:

Given the data in the question;

To determine the maximum intensity of an electromagnetic wave, we use the formula;

[tex]I[/tex] = [tex]\frac{1}{2}[/tex]ε₀cE[tex]_{max[/tex]²

where ε₀ is permittivity of free space ( 8.85 × 10⁻¹² C²/N.m² )

c is the speed of light ( 3 × 10⁸ m/s )

E[tex]_{max[/tex] is the maximum magnitude of the electric field

first we calculate the maximum magnitude of the electric field ( E[tex]_{max[/tex]  )

E[tex]_{max[/tex] = 350/f kV/m

given that frequency of 60 Hz, we substitute

E[tex]_{max[/tex] = 350/60 kV/m

E[tex]_{max[/tex] = 5.83333 kV/m

E[tex]_{max[/tex] = 5.83333 kV/m × ( [tex]\frac{1000 V/m}{1 kV/m}[/tex] )

E[tex]_{max[/tex] = 5833.33 N/C

so we substitute all our values into the formula for  intensity of an electromagnetic wave;

[tex]I[/tex] = [tex]\frac{1}{2}[/tex]ε₀cE[tex]_{max[/tex]²

[tex]I[/tex] = [tex]\frac{1}{2}[/tex] × ( 8.85 × 10⁻¹² C²/N.m² ) × ( 3 × 10⁸ m/s ) × ( 5833.33 N/C )²

[tex]I[/tex] = 45 × 10³ W/m²

[tex]I[/tex] = 45 × 10³ W/m² × ( [tex]\frac{1 kW/m^2}{10^3W/m^2}[/tex] )

[tex]I[/tex] = 45 kW/m²

Therefore, the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

Answer:

160 watts.

Explanation:

Remark

Power = Work / Time

Work = F * d

Note: Since he is running up stairs he is doing work against gravity.

Givens

Formula

Solution

Answer:

Explanation:

Suppose m₂ is greater than m₁ and it is going down . m₁ will be going up.

Let tension in string be T₁ and T₂  . Let common acceleration of system be a

For motion of m₁

T₁-m₁g = m₁a ----- (1)

For motion of m₂

m₂g- T₂ = m₂a ------- (2)

For motion of pulley

(T₂-T₁ )R represents net torque

(T₂-T₁ )R = I x α where I is moment of inertia of disc and α is angular acceleration of disc

(T₂-T₁ )R = 1/2 M R² x a / R

(T₂-T₁ ) = M a /2

Adding (1) and (2)

(m₂-m₁)g = (m₂+m₁)a + (T₂-T₁ )

(m₂-m₁)g = (m₂+m₁)a + Ma/2

(m₂-m₁)g = (m₂+m₁+ 0.5M)a

a = (m₂-m₁)g / ( (m₂+m₁+ 0.5M)

The acceleration of the masses is [tex]\dfrac{g(m_2+m_1)}{(m_2-m_1-0.5M)}[/tex].

Given to us

Masses = m₁, m₂

The radius of the pulley = R

Mass of the pulley = M

Assumption

We know the acceleration due to gravity is denoted by g,

[tex]T_1 = m_1(a+g)[/tex]......... equation 1

[tex]T_2 = m_2(a-g)[/tex]......... equation 2

The inertia of the pulley,  [tex]I = \dfrac{1}{2}MR^2[/tex]

Acceleration of the pulley, [tex]a = {\alpha }\times {R}[/tex]

[tex](T_2-T_1)R = I \times \alpha \\\\[/tex]

Substitute the values we get,

[tex][m_2(a-g)-m_1(a+g)]R = \dfrac{1}{2} MR^2 \times \dfrac{a}{R}\\\\[/tex]

[tex][m_2a-m_2g-m_1a-m_1g]R= \dfrac{1}{2} MR \times a[/tex]

[tex]m_2a-m_1a-m_2g-m_1g= \dfrac{1}{2} M \times a[/tex]

[tex]a(m_2-m_1)-g(m_2+m_1)= 0.5M \times a[/tex]

[tex]a(m_2-m_1)-0.5Ma=g(m_2+m_1)\\\\a(m_2-m_1-0.5M) = g(m_2+m_1)\\\\a = \dfrac{g(m_2+m_1)}{(m_2-m_1-0.5M)}[/tex]

Hence, the acceleration of the masses is [tex]\dfrac{g(m_2+m_1)}{(m_2-m_1-0.5M)}[/tex].

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Answer:

1) A. .33 hr

2) B. 6ft

3) A. 58mi/hr

Answer:

    vₐ₀ = 29.56 m / s

Explanation:

In this exercise the initial velocity of car A is asked, to solve it we must work in parts

* The first with the conservation of the moment

* the second using energy conservation

let's start with the second part

we must use the relationship between work and kinetic energy

             W = ΔK                             (1)

for this part the mass is

             M = mₐ + m_b

the final velocity is zero, the initial velocity is v

friction force work is

              W = - fr x

the negative sign e because the friction forces always oppose the movement

we write Newton's second law for the y-axis

              N -W = 0

              N = W = Mg

friction forces have the expression

              fr =μ N

              fr = μ M g

we substitute in 1

               -μ M g x = 0 - ½ M v²

             v² = 2 μ g x

let's calculate

              v² = 2  0.750  9.8  6.00

              v = ra 88.5

              v = 9.39 m / s

Now we can work on the conservation of the moment, for this part we define a system formed by the two cars, so that the forces during the collision are internal and therefore the tsunami is preserved.

Initial instant. Before the crash

         p₀ = + mₐ vₐ₀ - m_b v_{bo}

instant fianl. Right after the crash, but the cars are still not moving

         p_f = (mₐ + m_b) v

         p₀ = p_f

         + mₐ vₐ₀ - m_b v_{bo} = (mₐ + m_b) v

         mₐ vₐ₀ = (mₐ + m_b) v + m_b v_{bo}

let's reduce to the SI system

          v_{bo} = 64.0 km / h (1000m / 1km) (1h / 3600s) = 17.778 m / s

let's calculate

         660 vₐ₀ = (660 +490) 9.39 + 490 17.778

         vₐ₀ = 19509.72 / 660

         vₐ₀ = 29.56 m / s

we can see that car A goes much faster than vehicle B

Answer:

25.1

Explanation:

Answer:

Explanation:

In order to answer this question, we simply have to refer to the laws of the equations of gravitational mechanics.

The equation given by Newton tells us that  

[tex]F = \frac{Gm_{1} m_{2} }{r^{2} }[/tex]

In the case where we compare a specific place where the Force of Gravity is greater or lesser, we focus on the term assigned to the Planet's Radius.

In the case of [tex]G, m_{1} ,m_{2}[/tex], we understand that they are constant.

We can easily notice that the more the Radius (Height seen from a viewer on the ground), the lower the force will be.

In other words, the smaller the radius in which the measurement is made with respect to the center of the earth, the greater the gravitational force.

In that order of ideas the smallest radio has South Pole, which is about 6356 km from the center of the Earth on the Equator line

Answer:

14,432 g

Explanation:

PE=mh

360,800 J = (m)(25)

[tex]\frac{360,800 J}{25 m}[/tex] = m

14,432 g = m

Answer:

yes its c

Explanation:

took it already

The greatest electric field between them is Option B.

An electric field is the physical subject that surrounds electrically charged particles and exerts a force on all other charged debris inside the area, either attracting or repelling them. It also refers back to the bodily field of a machine of charged particles.

Reasoning:-

Electric field = Kq/r²

Since the distance is inversely proportional to the square root of the distance between them. therefore decreasing the distance electric field will be the greatest. Hence Option B.

The electrical field is defined mathematically as a vector discipline that can be related to each point in space, the force in step with unit charge exerted on a fine take a look at the price at rest at that factor. the electric field is generated by the electrical rate or through time-varying magnetic fields. electric fields provide us with the pushing force we need to set off the contemporary float.

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The energy that a substance contains is known as heat. Thus, the correct option for this question is B.

Entropy may be characterized as the measurement of the degree of randomness or in other words, it is the increase in the disorganization within a system. It is the measure of a system's thermal energy per unit temperature that is unavailable for doing useful work.

Equilibrium is a state of balance between opposing forces or actions that is either static or dynamic. It symbolizes the circumstance of being balanced in nature.

Thermal energy is a type of energy that is significantly contained within a system that is responsible for its temperature. Heat is the flow of thermal energy.

Therefore, the energy that a substance contains is known as heat. Thus, the correct option for this question is B.

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Answer:

Sledgehammer A has more momentum

Explanation:

Given:

Mass of Sledgehammer A = 3 Kg

Swing speed = 1.5 m/s

Mass of Sledgehammer B = 4 Kg

Swing speed = 0.9 m/s

Find:

More momentum

Computation:

Momentum = mv

Momentum sledgehammer A = 3 x 1.5

Momentum sledgehammer A = 4.5 kg⋅m/s

Momentum sledgehammer B = 4 x 0.9

Momentum sledgehammer B = 3.6 kg⋅m/s

Sledgehammer A has more momentum

Answer:

a) 10.23°C

b) - 0.8795 kJ / k

Explanation:

Given data :

mass of ice = 100 gram = 0.1 kg

mass of water in calorimeter = 100 grams = 0.1kg

temperature of water in calorimeter = 373K

a) Determine the Final temperature of the water once thermal equilibrium is established

Heat lost by water  = heat gained by Ice

Mwater * CPwater * Δt = Mice * Lice + Mice * CPwater * Δt

0.1 * (4.2 * 10^3 ) * ( 100 - t ) = ( 0.1 * 334 * 10^3 ) +  0.1 * (4.2 *1063 ) * ( T - 0 )

42000 = 33400 + 0.2 * 4200 * T

∴ equilibrium temp ( T ) = 10.23°C

B) Determine the entropy change of the universe of the process

Entropy change of ice = 33400/ 273 + 0.1 * 42 * 10^3 * In 283.23 / 273

                                     = 276.84 J/K

Entropy change of water = 0.1 * 42* 10^3 * In 283.23 / 273

                                          = -1156.34 J/K

Hence Entropy change of variance = 276.84 J/K + ( - 1156.34 J/K )

= - 0.8795 kJ / k

Answer:

The magnitude of the lift generated by the wings of the airplane is 130,340 N.

Explanation:

Given;

mass of the airplane, m = 13,300 kg

speed of the airplane, v = 560 km/h = 155.56 m/s

The magnitude of the lift generated by the wings of the airplane is calculated as;

[tex]F_l = mg\\\\where;\\\\F_l \ is \ the \ magnitude \ of \ the \ lift \ generated\\g \ is \ acceleration \ due \ to \ gravity = 9.8 \ m/s^2\\\\F_l = 13,300 \times \ 9.8\\\\F_l = 130,340 \ N[/tex]

Therefore, the magnitude of the lift generated by the wings of the airplane is 130,340 N.

Answer:

C?

Explanation:

Yep,It's C all right.

Answer:

Yep,It's C all right.

Explanation:

Answer:

B = 5.23 T

Explanation:

Given that,

Electric field, E = 550 V/m

The speed of the proton, v = 105 m/s

We need to find the magnitude of the magnetic field such that the beam of protons continues along its straight-line trajectory.

To move in a straight line, the magnitude of the electric force from the field and the magnetic field must be equal i.e.

[tex]qE=qvB\\\\B=\dfrac{E}{v}\\\\B=\dfrac{550}{105}\\\\B=5.23\ T[/tex]

So, the magnitude of the magnetic field is equal to 5.23 T.

Answer:

PE=mgh

M= Mass (kg)

G= Gravitational field strength (N/kg)

H= Hight (m)

PE= Gravitational Potential Energy (J)

Explanation:

Gravitational Potential Energy is the energy stored in a object due to its position above the Earth's surface.

The potential energy of the ball is 29.4 joules

The parameters given are;

mass= 3 kg

height= 1 meter

g= 9.8 m/s²

P.E= mgh

= 3 × 9.8 × 1

= 29.4

Hence the potential energy is 29.4 joules

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Answer:

The force they will exert on each other is 1.6*10⁻¹⁰ N

Explanation:

The electromagnetic force is the interaction that occurs between bodies that have an electric charge. When the charges are at rest, the interaction between them is called the electrostatic force. Depending on the sign of the interacting charges, the electrostatic force can be attractive or repulsive. The electrostatic interaction between charges of the same sign is repulsive, while the interaction between charges of the opposite sign is attractive.

Coulomb's law is used to calculate the electric force acting between two charges at rest. This force depends on the distance "r" between the electrons and the charge of both.

Coulomb's law is represented by:

[tex]F=k*\frac{q1*q2}{r^{2} }[/tex]

where:

In this case:

Replacing:

[tex]F=9*10^{9} \frac{N*m^{2} }{C^{2} }*\frac{1.602*10^{-19} C*1.602*10^{-19} C}{(1.2*10^{-9} )^{2} }[/tex]

and solving you get:

F=1.6*10⁻¹⁰ N

The force they will exert on each other is 1.6*10⁻¹⁰ N

Answer:

the effect on energy is due to the force applied to the frisbee to throw it back in the tree.

Explanation:

Hello,

✔ We have: KE = PE (potential energy)

PE = m x g x h

The potential energy that the pebble of mass 1 has is called PE1 and the potential energy that the pebble of mass 2 has is called PE2  

PE1 = PE2 ⇔ PE1/PE2 = 1

[tex]\frac{m_1\times g\times h}{m_2\times g\times 4h} = 1 \\ \\ \frac{m_1}{m_2\times 4} = 1 \\ \\ \frac{m_1}{m_2} = 4[/tex]

The mass m1 is therefore 4 times greater than that of the stone of mass m2.

Answer:

I believe its a and c but my notes are all kinds of messed up so im sorry if its wrong

Explanation:

None of the  regions can the gravitational field strength due to the two planets be zero.

What is gravitational field?

A gravitational field is a model used in physics to explain the effects that a large thing has on the area surrounding it, exerting a force on smaller, less massive bodies. Consequently, a gravitational field, which is measured in newtons per kilogram, is employed to describe gravitational processes.

Field due to given masses can not be zero in the given region they do not cancel each other.

None of the  regions can the gravitational field strength due to the two planets be zero.

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Answer:

hello your question has some missing parts below are the missing parts

A Circular, 10-turn coil has a radius of 10.7 cm and is oriented with its plane perpendicular to a 0.2-T magnetic field.

answer : 1 volt

Explanation:

Determine the Average induced EMF during this mishap

A' = A/2  ( for a semi circle )

where A = [tex]\frac{\pi r^2}{2}[/tex]

To determine the Average induced EMF apply the relation below

| E | =  η * [tex]\frac{\beta A}{T}[/tex]   ----- ( 1 )

Replace A in equation 1 with  A = [tex]\frac{\pi r^2}{2}[/tex]

hence equation becomes : | E | =  η * βπr^2 / 2T'

where : T' = 0.0365 ,  β = 0.2 , η = 10 , r = 0.107

∴| E |  = 0.999 ≈ 1volts

Explanation:

W=F×s

w=20N×6m=120J

W=120J

Explanation:

The diameter of a record, d = 0.3 cm

Radius, r = 0.15 cm

It rotates 33.5 times per minute.

(a) Frequency, f = 33.5 rotation/minute

= (33.5/60) rotation/second

= 0.55 rotation/second

(b) Time period,

T = 1/f

So,

[tex]T=\dfrac{1}{0.55}\\\\T=1.81\ s[/tex]

(c) Linear speed of a point on its rim,

[tex]v=r\omega\\\\v=0.15\times 10^{-2}\times 2\pi \times 0.55\\\\v=5.18\times 10^{-3}\ m/s[/tex]

(d) Centripetal acceleration,

[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{(5.18\times 10^{-3})^2}{0.15\times 10^{-2}}\\\\a=0.017\ m/s^2[/tex]

Hence, this is the required solution.