Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all automobiles are brought to rest.Automobile 1: 500kg, 10m/sAutomobile 2: 2000kg, 5m/sAutomobile 3: 500kg, 20m/sAutomobile 4: 1000kg, 20m/sAutomobile 5: 1000kg, 10m/sAutomobile 6: 4000kg, 5m/sRequired:a. Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest.b. Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest.c. Rank the automobiles based on the magnitude of the force needed to stop them, from largest to smallest.

Answer:

N2 = ¼N1

Explanation:

First of all, let's define the terms;

N1 = number of electric field lines going through the sphere of radius R

N2 = number of electric field lines going through the sphere of radius 2R

Q = the charge enclosed at the centre of concentric spheres

ε_o = a constant known as "permittivity of the free space"

E1 = Electric field in the sphere of radius R.

E2 = Electric field in the sphere of radius 2R.

A1 = Area of sphere of radius R.

A2 = Area of sphere of radius 2R

Now, from Gauss's law, the electric flux through the sphere of radius R is given by;

Φ = Q/ε_o

We also know that;

Φ = EA

Thus;

E1 × A1 = Q/ε_o

E1 = Q/(ε_o × A1)

Where A1 = 4πR²

E1 = Q/(ε_o × 4πR²)

Similarly, for the sphere of radius 2R,we have;

E2 = Q/(ε_o × 4π(2R)²)

Factorizing out to get;

E2 = ¼Q/(ε_o × 4πR²)

Comparing E2 with E1, we arrive at;

E2 = ¼E1

Now, due to the number of lines is proportional to the electric field in the each spheres, we can now write;

N2 = ¼N1

Answer:

From the relation above we can conclude that the  as the distance between the two plate increases the electric field strength decreases

Explanation:

I cannot  find any attached photo, but we can proceed anyways theoretically.

The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point

i.e

[tex]E=\frac{F}{Q}[/tex]

But the force F

[tex]F= \frac{kQ1Q2}{r^2}[/tex]

But the electric field intensity due to a point charge Q at a distance r meters away is given by

[tex]E= \frac{\frac{kQ1Q2}{r^2}}{Q} \\\\\E= \frac{Q1}{4\pi er^2 }[/tex]

From the relation above we can conclude that the  as the distance between the two plate increases the electric field strength decreases

Explanation:

The velocity of light in a medium of refractive index [tex]n[tex] is given by,

[tex]v=\frac{c}{n}[/tex]

[tex]v \text { is the velocity of light in the medium }[/tex]

[tex]c \text { is speed of light in vacuum }[/tex]

The exact value of speed of light in vacuum is [tex]299792458 \mathrm{m} / \mathrm{s}[/tex].

For Cherenkov radiation to be emitted, the velocity of the charged particle traversing the medium must be greater than this velocity. Thus, the threshold velocity of for creating Cherenkov radiation is,

[tex]v_{\text {Cherenkov }} \geq \frac{c}{n}[/tex]

[tex]v_{\text {threshod }}=\frac{c}{n}[/tex]

For water [tex]n=1.33,[tex] thus the threshold velocity for producing Cherenkov radiation in water is,

[tex]v_{\text {threatold }(\text { water })} &=\frac{299792458 \mathrm{m} / \mathrm{s}}{1.33}[/tex]

[tex]=225407863 \mathrm{m} / \mathrm{s}[/tex]

[tex]=2.254 \times 10^{8} \mathrm{m} / \mathrm{s}[/tex]

For ethanol [tex]n=1.36[tex], thus the threshold velocity for producing Cherenkov radiation in water is,

[tex]v_{\text {threstold }( \text { ettanol) } } &=\frac{299792458 \mathrm{m} / \mathrm{s}}{1.36}[/tex]

[tex]=220435630 \mathrm{m} / \mathrm{s}[/tex]

[tex]=2.204 \times 10^{8} \mathrm{m} / \mathrm{s}[/tex]

Answer:

The answer is "2.2 × [tex]\bold{10^8}[/tex]".

Explanation:

In the given question the value of n is missing which can be defined as follows:

n= 1.36

The velocity value of the threshold(ethanol) for a generation the Cerenkov light from the charged particle by travel through ethanol as:

know we will have to use an equation as follows:

Formula:      

(ethanol) or the vthreshold = [tex]\frac{c}{n}[/tex]

                                         [tex]= \frac{3\times 10^8} {1.36} \\\\= 2.2 \times 10^8[/tex]

The water in vthreshold:

[tex]= 2.2 \times 10^8 \ \ \frac{m}{ s} \\\\[/tex]

Express the value in c, that is multiple, so, the value of vthreshold(water) is:

=(0.735) c

Complete question is;

A rope, under a tension of 153 N and fixed at both ends, oscillates in a second harmonic standing wave pattern. The displacement of the rope is given by

y = (0.15 m) sin[πx/3] sin[12π t].

where x = 0 at one end of the rope, x is in meters, and t is in seconds. What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c)the mass of the rope? (d) If the rope oscillates in a third - harmonic standing wave pattern, what will be the period of oscillation?

Answer:

A) Length of rope = 4 m

B) v = 24 m/s

C) m = 1.0625 kg

D) T = 0.11 s

Explanation:

We are given;

T = 153 N

y = (0.15 m) sin[πx/3] sin[12πt]

Comparing this displacement equation with general waveform equation, we have;

k = 2π/λ = π/2 rad/m

ω = 2πf = 12π rad/s

Since, 2π/λ = π/2

Thus,wavelength; λ = 4 m

Since, 2πf = 12π

Frequency;f = 6 Hz

A) We are told the rope oscillates in a second-harmonic standing wave pattern. So, we will use the equation;

λ = 2L/n

Since second harmonic, n = 2 and λ = L = 4 m

Length of rope = 4 m

B) speed is given by the equation;

v = fλ = 6 × 4

v = 24 m/s

C) To calculate the mass, we will use;

v = √T/μ

Where μ = mass(m)/4

Thus;

v = √(T/(m/4))

Making m the subject;

m = 4T/v²

m = (4 × 153)/24²

m = 1.0625 kg

D) Now, the rope oscillates in a third harmonic.

So n = 3.

Using the formula f = 1/T = nv/2L

T = 2L/nv

T = (2 × 4)/(3 × 24)

T = 0.11 s

Answer:

In a tuning fork, two basic qualities of sound are considered, they are

1) The pitch of the waveform: This pitch depends on the frequency of the wave generated by hitting the tuning fork.

2) The loudness of the waveform: This loudness depends on the intensity of the wave generated by hitting the tuning fork.

Hitting the tuning fork harder will make it vibrate faster, increasing the number of vibrations per second. The number of vibration per second is proportional to the frequency, so hitting the tuning fork harder increase the frequency. From the explanation on the frequency above, we can say that by increasing the frequency the pitch of the tuning fork also increases.

Also, hitting the tuning fork harder also increases the intensity of the wave generated, since the fork now vibrates faster. This increases the loudness of the tuning fork.

Answer:

In an electromagnetic wave in free space, the ratio of the magnitudes of electric and magnetic field vectors E and B is equal:  speed of light(c)

Explanation:

Generally the ratio of the E(electric field ) and  the B(magnetic field ) is  equal to the speed of the electromagnetic wave i.e the speed of  light (c) the value is

    [tex]c = 3.0 *10^{8} \ m/s[/tex]

Answer:

Explanation:

[tex]Force = 25N\\Area = 0.05m^2\\\\Pressure = \frac{Force}{Area}\\ \\Pressure = \frac{25}{0.05}\\\\ Pressure = 500 Pascals[/tex]

Answer:

the weight of the wire + 25kg

Explanation:

Answer:

The new volume will be 0.918 x 10^5 L

Explanation:

initial volume  [tex]V_{1}[/tex] = 1.21 x 10^5 L

Initial temperature [tex]T_{1}[/tex] = 265 K

Final volume [tex]V_{2}[/tex] = ?

Final temperature [tex]T_{2}[/tex] = 201 K

Th gas is an ideal gas.

For ideal gases, the equation [tex]V_{1}[/tex]/[tex]T_{1}[/tex] = [tex]V_{2}[/tex]/[tex]T_{2}[/tex] = constant

substituting value, we have

(1.21 x 10^5)/265 = [tex]V_{2}[/tex]/201

[tex]V_{2}[/tex] =  24321000/265 = 91777.4 L

= 0.918 x 10^5 L

Answer:

[tex]\huge \boxed{\mathrm{Option \ D}}[/tex]

Explanation:

Two forces are acting on the object.

Subtracting 2 N from both forces.

2 N → Object ← 5 N

- 2 N                 - 2N

0 N → Object ← 3 N

The force 3 N is pushing the object to the left side.

The mass of the object is 10 kg.

Applying formula for acceleration (Newton’s Second Law of Motion).

a = F/m

a = 3/10

a = 0.3

Explanation:

Hey there!

According to the question;

A person can lift mass of 60 kg on earth.

mass(m1) = 60kg

acceleration due to gravity on earth (a) = 9.8m/s²

Now;

force (f) = m.a

= 60*9.8

= 588 N

Since, there is application of same magnitude of force on moon,

mass(m) =?

acceleration due to gravity on moon (a) = 1.67m/s²

Now;

force (f) = m.a

588 = m*1.67

m = 352.09 kg

Therefore, the person who can lift the mass of 60 kg on earth can lift mass of 352 kg on moon.

Hope it helps!

The value of n in the relationship between F₁ and F₂, (F₂ = n·F₁), is n = 2

The reason for the above value is as follows;

The given parameters of the first pair of books are;

The masses of the books = m₁, and m₂

The distance between the two m₁ and m₂ = r

The gravitational force between the masses = F₁

The given parameters of the second pair of books are;

The masses of the books second pair = 2·m₁, and 4·m₂

The distance between the two masses in second pair = 2·r

The gravitational force between the masses in second pair = F₂

The relationship between the two forces is F₂ = n·F₁

The required parameter;

The value of n in F₂ = n·F₁

According to Newton's law of universal gravitation, we have;

[tex]\mathbf{F = G \times \dfrac{m_1 \times m_2}{r^2}}[/tex]

Therefore, we get;

[tex]F_1 = G \times \dfrac{m_1 \times m_2}{r^2}[/tex]

[tex]F_2 = G \times \dfrac{2\cdot m_1 \times 4 \cdot m_2}{(2 \cdot r)^2} = G \times \dfrac{8 \times m_1 \times m_2}{4 \times r^2} = 2 \times G \times \dfrac{ m_1 \times m_2}{r^2} = 2 \times F_1[/tex]

Therefore;

F₂ = 2·F₁

The value of n in F₂ = n·F₁ is n = 2

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Answer:

hello the needed diagram is missing attached below is the diagram and the detailed solution

The tension in BC = [tex]\frac{\sqrt{2} }{4} mg[/tex]

Explanation:

ATTACHED BELOW IS THE DETAILED SOLUTION T THE GIVEN PROBLEM

Ma = mg - T/ [tex]\sqrt{2}[/tex]  equation 1

Ma = 3T / [tex]\sqrt{2}[/tex]   equation 2

equate both equations to determine the tension on BC

Answer:

-0.73mA

Explanation:

Using amphere's Law

ε =−dΦB/ dt

=−(2.6T)·(7.30·10−4 m2)/ 1.00 s

=−1.9 mV

Using ohms law

ε=V =IR

I = ε/ R =−1.9mV/ 2.60Ω =−0.73mA

Answer:

Unbalenced

Explanation:

when balenced forces are applied to an object there is no motion. When you apply unbalenced force the object you are applying the force to will move in the opposite direction of the force.

Answer:

im pretty sure it unbalenced

Explanation:

i just am

Answer:

400m

Explanation:

Answer:

Explanation:i think this would help u

Answer:

Fluoroscopy

Explanation:

A Fluoroscopy is an imaging technique that uses X-rays to obtain real-time moving images of the interior of an object. In its primary application of medical imaging, a fluoroscope allows a physician to see the internal structure and function of a patient, so that the pumping action of the heart or the motion of swallowing, for example, can be watched.

Answer:

Argument in favor of less total energy consumption if the store is kept at a low temperature

Explanation:

Have in mind that if the store has numerous refrigerators and freezers, the energy consumption of those machines have to be included into the analysis.

Recall that the efficiency (or Coefficient Of Performance - COP) of a frezzer or refrigerator is inversely proportional to the temperature difference between the inside of th machine and the environment where it is operation, therefore the smaller the difference, the highest their efficiency. Therefore, the cooler the environment (the temperature at which the store is kept) the better performance of the running refrigerators and freezers.

Answer:

towards the west

One can conclude that the magnetic field must have a component towards the east. The correct option is c.

A magnetic field is a vector field that describes the magnetic influence on moving charges, currents, and magnetic materials.

A moving charge in a magnetic field is subjected to a force that is perpendicular to both its own velocity and the magnetic field.

Every magnet possesses a north as well as a south pole. The opposite poles attract each other, whereas the same poles repel each other.

When you rub a piece of iron against a magnet, the atoms' north-seeking poles line up in the same direction.

Moving north, a proton enters a magnetic field. The proton curves downward as a result of this field. We can conclude that the magnetic field must have an eastward component.

Thus, the correct option is c.

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#SPJ2

The resistance R = 20 Ω

The capacitance C = 106.1 μF

The current, I is 2.773 A at 56.31°.

The phase angle of the between the current and the voltage is 56.31° leading.

Since the impedance Z = 20 - j30 Ω, the resistance, R is the real part of the impedance. So R = ReZ = 20 Ω

So, the resistance R = 20 Ω

To find the capacitance, we need first to find the reactance of the capacitor X. Since the impedance Z = 20 - j30, the reactance of the capacitor X. is the imaginary part of the impedance. So X = ImZ = 30 Ω.

Now the reactance of the capacitor X = 1/ωC where ω = angular frequency of the circuit = 2πf where f = frequency of the circuit = 50 Hz and C = capacitance  

So, C = 1/ωX = 1/2πfX

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

C = 1/2πfX

C = 1/(2π × 50 Hz × 30 Ω)

C = 1/3000π

C = 1/9424.778

C = 1.061 × 10⁻⁴ F

C = 106.1 × 10⁻⁶ F

C = 106.1 μF

So, the capacitance is 106.1 μF

The current I = V/Z where V = voltage = 100 V at 0° and Z = impedance.

The magnitude of Z = √(20² + (-30)²)

= √(400 + 900)

= √1300

= 36.06 Ω

and its angle Φ = tan⁻¹(ImZ/ReZ)

= tan⁻¹(-30/20)

= tan⁻¹(-1.5) = -56.31°

So, V = 100 ∠ 0° and Z = 36.06 ∠ -56.31°

So, the current, I = V/Z =  (100 ∠ 0°)/36.06 ∠ -56.31°

= 100/36.06 ∠(0° - (-56.31° ))

= 2.773 ∠ 56.31° A

So, the current is 2.773 A at 56.31°.

Since the current is 2.773 A at 56.31°, the phase angle of the between the current and the voltage is 56.31° leading.

So, the phase angle of the between the current and the voltage is 56.31° leading.

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Answer:

[tex]q = -461532.5 \ C[/tex]

Explanation:

From the question we are told that

     The  electric filed is  [tex]E = 102 \ N/C[/tex]  

Generally according to Gauss law

=>   [tex]E A = \frac{q}{\epsilon_o }[/tex]

Given that  the electric field is pointing downward  , the equation become

    [tex]- E A = \frac{q}{\epsilon_o }[/tex]

Here   [tex]q[/tex] is the excess charge on the surface of the earth

          [tex]A[/tex] is the surface  area of the of the earth which is mathematically represented as

     [tex]A = 4\pi r^2[/tex]

Where r is the radius of the earth which has a value [tex]r = 6.3781*10^6 m[/tex]

 substituting values

    [tex]A = 4 * 3.142 * (6.3781*10^6 \ m)^2[/tex]

    [tex]A =5.1128 *10^{14} \ m^2[/tex]

So

   [tex]q = -E * A * \epsilon _o[/tex]

Here [tex]\epsilon_o[/tex] s the permitivity of free space with value

          [tex]\epsilon_o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

substituting values

     [tex]q = -102 * 5.1128 *10^{14} * 8.85 *10^{-12}[/tex]

     [tex]q = -461532.5 \ C[/tex]

Answer:

a)  λ = 555.5 m,  λ = 187.5 nm

Explanation:

The velocity of a wave is given by the relation

           c = λ f

           λ = c /f

a) length of the radii AM

            λ = 3 10⁸/540 10⁻⁷

            λ= 5.555 102 m

            λ = 555.5 m

             f = 1600 kHz

            λ = 3 108/1600 103

             λ = 1.875 102 m

            lam  = 187.5 nm

b) light = 760 Thz = 760 10-12 Hz

         λ=  c/f

            λ = 3 108/760 10-12

            λ = 3.947 10-9

           λ= 30000 Thz

           λ= c/f

      λ = 3 10⁸/ 30000 10-12

       λ = 1  m

Answer:

The working principle of a water pump mainly depends upon the positive displacement principle as well as kinetic energy to push the water.

Explanation:

it mainly depends upon the positive displacement principle and also kinetic energy to push water. hope this hepls!

Answer:

Explanation:

A) Vair = 1.3 L

B) Volume is not reasonable

Explanation:

A)

Assume

m to be total mass of the man

mp be the mass of the man that pulled out of the water

m1 be the mass above the water with the empty lung

m2 be the mass above the water with full lung

wp be the weight that the buoyant force opposes as a result of the air.

Va be the volume of air inside man's lungs

Fb be the buoyant force due to the air in the lung

given;

m = 78.5 kg

m1 = 3.2% × 78.5 = 2.5 kg

m2 = 4.85% × 78.5 = 3.8kg

But, mp = m2- m1

mp = 3.8 - 2.5

mp = 1.3kg

So using

Archimedes principle, the relation for formula for buoyant force as;

Fb = (m_displaced water)g = (ρ_water × V_air × g)

Where ρ_water is density of water = 1000 kg/m³

Thus;

Fb = wp = 1.3× 9.81

Fb = 12.7N

But

Fb = (ρ_water × V_air × g)

So

Vair = Fb/(ρ_water × × g)

Vair = 12.7/(1000 × 9.81)

V_air = 1.3 × 10^(-3) m³

convert to litres

1 m³ = 1000 L

Thus;

V_air = 1.3× 10^(-3) × 1000

V_air = 1.3 L

But since the average lung capacity of an adult human being is about 6-7litres of air.

Thus, the calculated lung volume is not reasonable

Explanation:

Explanation:

It is given that,

Distance covered by the cart to the right is 2 m

Distance covered by the cart to the left is 1 m

We need to find the total distance rolled by the cart. Total distance is equal to the sum of the distances covered by an object. It does depend on the direction.

Total distance = 2 m + 1 m

D = 3 m

The cart rolled to a total distance of 3 m.

Answer:

The cross-sectional area of the larger piston is 392 cm²

Explanation:

Given;

output mass of the piston, m₀ = 2000 kg

input force of the piston, F₁ = 500 N

input area of the piston, A₁ = 10 cm² = 0.001 m²

The output force is given by;

F₀ = m₀g

F₀ = 2000 x 9.8

F₀ = 19600 N

The cross-sectional area of the larger piston or output area of the piston will be calculated by applying the following equations;

[tex]\frac{F_i}{A_i} = \frac{F_o}{A_o} \\\\A_o= \frac{F_o A_i}{F_i} \\\\A_o = \frac{19600*0.001}{500} \\\\A_o = 0.0392 \ m^2\\\\A_o = 392 \ cm^2[/tex]

Therefore, the cross-sectional area of the larger piston is 392 cm²

Cold water kept in an open container acquires heat from the warmer surrounding becomes warm like the air around it due to the transfer of thermal energy

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