An object on a level surface experiences a horizontal force of 12.7 N due to kinetic friction. The coefficient of kinetic friction is 0.42.
What is the mass of the object? (Express your answer to two significant figures)kg

Answer:

Resistance of each bulb = 360 ohms

Explanation:

Let each bulb have a resistance r .

Since, even after removing one of the bulbs, the circuit is closed and the other bulbs glow. Therfore, the bulbs are connected in Parallel connection.

[tex] \frac{1}{r(equivalent)} = \frac{1}{r1} + \frac{1}{r2} + + + + \frac{1}{r15} [/tex]

[tex] \frac{1}{r(equivalent)} = \frac{15}{r} [/tex]

R(equivalent) = r/15

Now, As per Ohms Law :

V = I * R(equivalent)

120 V = 5 A * r/15

r = 360 ohms

Answer:

Explanation:

Using the Continuity equation

v X A = v' xA'

so if A is 1/2of A' then A velocity must be 2 times the A'

after-contraction v = 2 x 5.0m/s = 10m/s

Using the Bernoulli equation

p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂

, the "h" terms cancel

3.5 x 10^ 5Pa + ½ x 1000kg/m³x (5.0m/s)² = p₂ + ½ x 1000kg/m³ x (10m/s)²

p₂ = 342500pa

Answer:

0.2812

Explanation:

Given that

mass of skater 1, m1 = 45 kg

mass of skater 2, m2 = 60 kg

speed of skater 1, v1 = 0.375 m/s

To attempt this question, we would be using the Law of conservation of momentum That says the momentum is constant, before and after the movement.

Thus, momentum p = mv

Law of conservation of momentum infers that,

m1v1 = m2v2

Now we proceed to substitute our values into the formula.

45 * 0.375 = 60 * v2

v2 = 16.875 / 60

v2 = 0.2812 m/s

Therefore the speed of the second skater has to be 0.2812 m/s

Answer:

1) designed to measure the difference in speed of light in different directions , 1887

Explanation:

1) This experiment was designed to measure the difference in speed of light in different directions and therefore find the speed of the ether.

2) was made in 1887

3) At that time it was assumed that it was the medium in which light traveled and it is everywhere

4) the speed of the wave depends on the characteristics of the medium where it travels,

for the one in a string depends on the tension and density

for an electromagnetic wave of the permittivity and permeability of the vacuum

5) In this type of interferometer the beam is divided into two rays

6) In his interrupter, he had to accurately measure the displacement of the fringes in a telescope, for which he had to minimize vibrations, he had problems in the movement of one of the arms, changes in temperature

7) In Michelsom's second experiment, the apparatus could measure 0.01 fringes by increasing the length of the arms by 11 m

8) The new interferometer floated on a bed of mercury

9) Couldn't measure any difference in speed of light in different directions

10) Physics was forced to eliminate the concept of ETHER

11) One of the principles of relativities that the speed of light is constant in all inertial efficiency systems

12) Michelson in 1907

13) It seems that Einstein did not know the results of this experiment

Answer:

2/5 m/s

Explanation:

There are two vectors  v and w . Let θ be angle b/w the two vector.

[tex]cos\theta =\frac{\overleftarrow{v}\cdot \overleftarrow{w}}{\left | v \right |\left | w \right |}\\=\frac{6-4}{\sqrt(2^2+1^2)\sqrt(3^2+4^2)} =\frac{2}{5\sqrt(5)}[/tex]

velocity of the ball in direction of the the wind

[tex]\left | vcos\theta \right |\\\left | v \right |cos\theta\\\sqrt(2^2+1^2)\frac{2}{5\sqrt(5)} = \frac{2}{5}[/tex]

The size of the velocity of the ball in the direction of the wind is 2/5 ms.

Since there are two vectors v and w

Also, here we assume θ be angle b/w the two vector.

So

Cos θ = 6-4 / √(2^2 + 1^2) √(3^2 + 4^2)

= 2/5√5

Now the velocity of the ball should be

= √(2^2 + 1^2) 2 ÷ 5√(5)

= 2 /5

hence, The size of the velocity of the ball in the direction of the wind is 2/5 ms.

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Answer:

Explanation:

[tex]Power = IV[/tex]

From ohms law we know that

[tex]V= IR\\\\I= \frac{V}{R} \\\\Power= \frac{V}{R}*V\\\\Power= \frac{V^2}{R}[/tex]

Given data

P1 = 0.5 Watt

P2 = ?

V1= 3 Volts

V2= 1 Volt

Thus we can solve for the power dissipated as follows

[tex]P1= \frac{V1^2}{R1}\\\\P2= \frac{V2^2}{R2}[/tex]

[tex]\frac{P1}{P2} = \frac{V1^2}{V2^2}\\\\ P2=\frac{ V2^2}{ V1^2} *P1\\\\ P2=\frac{ 1^2}{ 3^2} *0.5= 0.055= 0.056 W[/tex]

The  resistor will dissipate 0.056 Watt

Answer:D

Explanation:I got it right

Answer:

They needed to be able to unite even though they spoke different languages.

Explanation:

Answer:

[tex]\theta=10.20^{\circ}[/tex]  

[tex]\Delta l=0.10 ft[/tex]    

Explanation:

First of all, we analyze the system of blocks before starting to move.

[tex]\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0[/tex]  

[tex]\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0[/tex]

[tex]11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0[/tex]

[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]

[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]  

[tex]16sin(\theta)-2.91cos(\theta)=0[/tex]  

[tex]tan(\theta)=0.18[/tex]  

[tex]\theta=arctan(0.18)[/tex]  

[tex]\theta=10.20^{\circ}[/tex]  

Hence, the incline angle θ for which both blocks begin to slide is 10.20°.

Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.  

[tex]P_{A}sin(\theta)-F_{fA}-F_{spring}=0[/tex]

Where:

[tex]F_{spring} = k\Delta l=2.1\Delta l[/tex]

[tex]P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0[/tex]

[tex]\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}[/tex]

[tex]\Delta l=0.10 ft[/tex]    

Therefore, the required stretch or compression in the connecting spring is 0.10 ft.

I hope it helps you!

(a) The inclined angle for which both blocks begin to slide is 10.3⁰.

(b) The compression of the spring is 0.22 ft.

The given parameters;

The normal force on block A and B:

[tex]F_n_A = m_Agcos \ \theta\\\\F_n_B = m_Bgcos \ \theta[/tex]

The frictional force on block A and B:

[tex]F_f_A = \mu_s_AF_n_A \\\\F_f_B = \mu_s_BF_n_A[/tex]

The net force on the blocks when they starts sliding;

[tex](m_Ag sin \theta+ m_Bgsin\theta) - (F_f_A + F_f_B) = 0\\\\m_Ag sin \theta+ m_Bgsin\theta = F_f_A + F_f_B\\\\m_Ag sin \theta+ m_Bgsin\theta = \mu_Am_Agcos\theta \ + \ \mu_Bm_Bgcos\theta\\\\gsin\theta(m_A + m_B) = gcos\theta (\mu_Am_A + \mu_Bm_B)\\\\\frac{sin\theta}{cos \theta} = \frac{\mu_Am_A\ + \ \mu_Bm_B}{m_A\ + \ m_B} \\\\tan\theta = \frac{(0.16\times 11) \ + \ (0.23 \times 5)}{11 + 5} \\\\tan\theta = 0.1819\\\\\theta = tan^{-1}(0.1819)\\\\\theta = 10.3 \ ^0[/tex]

The change in the energy of the blocks is the work done in compressing the spring;

[tex]\Delta E = W\\\\F_A (sin \theta )d- \mu F_n d= \frac{1}{2} kd^2\\\\F_A sin\theta \ - \ \mu F_A cos\theta = \frac{1}{2} kd\\\\d = \frac{2F_A(sin\theta - \mu cos \theta) }{k} \\\\d = \frac{2\times 11(sin \ 10.3\ - \ 0.16\times cos \ 10.3) }{2.1} \\\\d = 0.22 \ ft[/tex]

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Answer:

[tex]\frac{di}{dt} = 7.31 \ A/s[/tex]

Explanation:

From the question we are told that  

     The  number of turns is  [tex]N = 450 \ turns[/tex]

      The  radius is  [tex]r = 1.17 \ cm = 0.0117 \ m[/tex]

       The  position from the center consider is  x =  3.45 cm  =  0.0345 m

       The  induced emf is  [tex]e = 8.20 *10^{-6} \ V/m[/tex]

Generally according to Gauss law

        [tex]\int\limits { e } \, dl = \mu_o * N * \frac{di}{dt } * A[/tex]

=>    [tex]e * 2\pi x = \mu_o * N * \frac{d i }{dt } * A[/tex]

Where A is the  cross-sectional area of the solenoid which is mathematically represented as

                [tex]A = \pi r ^2[/tex]

=>      [tex]e * 2\pi x = \mu_o * N * \frac{d i }{dt } * \pi r^2[/tex]

=>       [tex]\frac{di}{dt} = \frac{2e * x }{\mu_o * N * r^2}[/tex]ggl;

Here  [tex]\mu_o[/tex] is the permeability of free space with value

          [tex]\mu_o = 4\pi * 10^{-7} \ N/A^2[/tex]

=>     [tex]\frac{di}{dt} = \frac{2 * 8.20*10^{-6} * 0.0345 }{ 4\pi * 10^{-7} * 450 * (0.0117)^2}[/tex]

=>      [tex]\frac{di}{dt} = 7.31 \ A/s[/tex]

The value of di/dt from the given values of the solenoid electric field is;

di/dt = 7.415 A/s

We are given;

Number of turns; N = 450 per m

Radius; r = 1.17 cm = 0.0117 m

Electric Field; E = 8.2 × 10⁻⁶ V/m

Position of electric field; r' = 3.45 cm = 0.0345 m

According to Gauss's law of electric field;

∫| E*dl | = |-d∅/dt |

Now, ∅ = BA = μ₀niA

where;

n is number of turns

i is current

A is Area

μ₀ = 4π × 10⁻⁷ H/m

Thus;

E(2πr') = (d/dt)(μ₀niA)  (negative sign is gone from the right hand side because we are dealing with magnitude)

Since we are looking for di/dt, then we have;

E(2πr') = (di/dt)(μ₀nA)

Making di/dt the subject of the formula gives;

di/dt = E(2πr')/(μ₀nA)

Plugging in the relevant values gives us;

di/dt = (8.2 × 10⁻⁶ × 2 × π × 0.0345)/(4π × 10⁻⁷ × 450 × π × 0.0117²)

di/dt = 7.415 A/s

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Answer:

300 W

Explanation:

power of each bulb P = 75 W

voltage in the circuit = 120 V

we know that electrical power P = IV    ....1

and V = IR

we can also say that I = V/R

substituting for I in  equation 1, we have

P = [tex]V^{2}/R[/tex]    ....2

The total total power in the circuit = 75 x 2 = 150 W

from equation 2, we have

150 = [tex]120^{2} /R[/tex]

R = [tex]120^{2}/150[/tex] = 96 Ω    this is the resistance of the whole circuit.

This resistance is due to the two light bulbs, for each light bulb since they are arranged in series

R = 96/2 = 48 Ω

From P =  [tex]V^{2}/R[/tex]  

for each light bulb, power is

P = [tex]120^{2} /48[/tex] = 300 W

Answer:

E. None of these statements are true.

Explanation:

We can't say the exact or approximate amount of tension on the rope, since we do know for sure from the statement who is winning.

for A, the tension on the rope does not depend on if both teams pull are in equilibrium.

for B, the 49ers would be pulling with a force more than 5000 N, if they were winning. The problem is that we can't say with all confidence that they'd be winning.

for C, we don't know how much tension exists on the rope, and its direction, so we can't work out how much tension the 49ers are pulling the rope with.

for D,  just as for C above, we can't work out how much tension there is on the rope, since we do not know how much force the 49ers are pulling with.

we go with option E.

Answer:

It does not violate the first law because the total energy taken is what is used 100J = 25J + 75J

But violates 2nd lawbecause the engine has a higher energy after doing work than the initial for e.g A cold object in contact with a hot one never gets colder, transferring heat to the hot object and making it hotter confirming the second law

Answer:

t = 134834.31 years

Explanation:

First we find the speed of the ship:

v = 0.890 c

where,

v = speed of the ship = ?

c = speed of light = 3 x 10⁸ m/s

Therefore, using the values, we get:

v = (0.89)(3 x 10⁸ m/s)

v = 2.67 x 10⁸ m/s

Now, we find the distance in meters:

Distance = s = (1.2 x 10⁵ light years)(9.461 x 10¹⁵/1 light year)

s = 11.35 x 10²⁰ m

Now, for the time we use the following equation:

s = vt

t = s/v

t = (11.35 x 10²⁰ m)/(2.67 x 10⁸ m/s)

t = (4.25 x 10¹² s)(1 h/3600 s)(1 day/24 h)(1 year/365 days)

t = 134834.31 years

Answer:

  K = 1.6 10⁻¹⁵ J

Explanation:

In an electron microscope, electrons are used to form images, these electrons are accelerated in electric fields so that they have a kinetic energy that allows obtaining a good amplification with the microscope.

electrical potential energy is converted to kinetic energy

                U = K

                e V = ½ m v²

                v = √2eV /m

the wavelength of these electrons we obtain from the de Broglie equation

                λ = h / p

p = mv

               λ = h / mv

               λ = h / mra 2eV / m

               λ = h / ra 2eVm

where we can see that as the potential energy increases, it electrifies the shorter the wavelength of the electrons and consequently the greater the magnification of the microscope

in general these microscopes use from 10000X onwards therefore for this saponification

                K = e V

               K = 1.6 10⁻¹⁹ 10000

               K = 1.6 10⁻¹⁵ J

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:

The planet that is farther from the star must have a time period greater.

Explanation:

We can determine the ratio of the period's planet with the radius of the circular orbit in the star's equatorial plane:

[tex] T = 2\pi*\sqrt{\frac{r^{3}}{GM}} [/tex]     (1)

Where:

r: is the radius of the circular orbit of the planet and the star

T: is the period

G: is the gravitational constant

M: is the mass of the planet

From equation (1) we have:

[tex] T = 2\pi*\sqrt{\frac{r^{3}}{GM}} = k*r^{3/2} [/tex]   (2)          

Where k is a constant

From equation (2) we have that of the two planets, the planet that is farther from the star must have a time period greater.

I hope it helps you!

Explanation:

When an ideal gas undergoes a slow isothermal expansion, following phenomenon occur

1. Work done bu the gas = Energy absorbed as heat.

2. Work done by environment = Energy absorbed as heat.

3. Increase in internal energy= Heat absorbed= work done by gas = work done by environment.

Hence all option are correct.

Increase in internal energy is equal to the heat absorbed or work done by gas or environment. All the statements are correct.

If an ideal gas undergoes a slow isothermal expansion,

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Answer:

Will be equal to alpha x r; less than UsN

Answer:

20.96 m/s^2 (or 21)

Explanation:

Using the formula (final velocity - initial velocity)/time = acceleration, we can plug in values and manipulate the problem to give us the answer.

At first, we know a car is going 8 m/s, that is its initial velocity.

Then, we know the acceleration, which is 1.8 m/s/s

We also know the time, 7.2 second.

Plugging all of these values in shows us that we need to solve for final velocity. We can do so by manipulating the formula.

(final velocity - initial velocity) = time * acceleration

final velocity = time*acceleration + initial velocity

After plugging the found values in, we get 20.96 m/s/s, or 21 m/s

Answer:

A) V_air = 1.295 L

B) Volume is not reasonable

Explanation:

A) Let;

m be total mass of the man

m_p be the mass of the man that pulled out of the water because of the buoyant force that pulled out of the lung

m_3 be the mass above the water with the empty lung

m_5 be the mass above the water with full lung

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

V_a be the volume of air inside man's lungs

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

Now, we are given;

m = 78.5 kg

m_3 = 3.2% × 78.5 = 2.512 kg

m_5 = 4.85% × 78.5 = 3.80725 kg

Now, m_p = m_5 - m_3

m_p = 3.80725 - 2.512

m_p = 1.29525 kg

From archimedes principle, we have the formula for buoyant force as;

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

Where ρ_water is density of water = 1000 kg/m³

Thus;

F_b = w_p = 1.29525 × 9.81

F_b = 12.7064 N

As earlier said,

F_b = (ρ_water × V_air × g)

Thus;

V_air = F_b/(ρ_water × × g)

V_air = 12.7064/(1000 × 9.81)

V_air = 1.295 × 10^(-3) m³

We want to convert to litres;

1 m³ = 1000 L

Thus;

V_air = 1.295 × 10^(-3) × 1000

V_air = 1.295 L

B) From research, the average lung capacity of an adult human being is 6 litres of air.

Thus, the calculated lung volume is not reasonable

Answer:

A) 3.13 m/s

B) 5.34 N

C) W = 26.9 J

Explanation:

We are told that the position as a function of time is given by;

x(t) = αt² + βt³

Where;

α = 0.210 m/s² and β = 2.04×10^(−2) m/s³ = 0.0204 m/s³

Thus;

x(t) = 0.21t² + 0.0204t³

A) Velocity is gotten from the derivative of the displacement.

Thus;

v(t) = x'(t) = 2(0.21t) + 3(0.0204t²)

v(t) = 0.42t + 0.0612t²

v(4.5) = 0.42(4.5) + 0.0612(4.5)²

v(4.5) = 3.1293 m/s ≈ 3.13 m/s

B) acceleration is gotten from the derivative of the velocity

a(t) = v'(t) = 0.42 + 2(0.0612t)

a(4.5) = 0.42 + 2(0.0612 × 4.5)

a(4.5) = 0.9708 m/s²

Force = ma = 5.5 × 0.9708

F = 5.3394 N ≈ 5.34 N

C) Since no friction, work done is kinetic energy.

Thus;

W = ½mv²

W = ½ × 5.5 × 3.1293²

W = 26.9 J

Answer: I0*0.853

Explanation:

Ok, the Malus's law says that:

If you have light polarized along a given line with an intensity I0, and it passes through a polaroid which axis of polarization forms an angle θ with respect to the polarization of the light, then the intensity of the resulting beam is:

I(θ) = I0*cos^2(θ)

For example, if the axis of the polaroid is exactly the same as the axis of polarization of the light beam that will impact it, then we have θ = 0°, and the equation above says that the intensity of the beam will not change.

In this particular case, we have that the intensity of the light is I0, and the angle is θ = 22.5°

Then:

I(22.5°) = I0*cos^2(22.5°) = I0*0.853

Answer:

Explanation:

dipole moment = qs = q x s

= charge x charge separation

charge = q

separation between charge = s

half separation l = s / 2

dipole has two charges + q and - q separated by distance s .

Potential at distance x along x axis due to + q

[tex]v_1=\frac{1}{4\pi \epsilon } \times\frac{q}{x-l}[/tex]

Potential at distance x along x axis due to - q

[tex]v_2=\frac{1}{4\pi \epsilon } \times\frac{-q}{x+l}[/tex]

Total potential

v = v₁ + v₂

[tex]v=\frac{1}{4\pi \epsilon } \times( \frac{q}{x-l}-\frac{q}{x+l})[/tex]

[tex]v=\frac{1}{4\pi \epsilon } \times\frac{2ql}{x^2-l^2}[/tex]

[tex]v=\frac{1}{4\pi \epsilon } \times\frac{qs}{x^2-(\frac{s}{2}) ^2}[/tex]

Potential at distance y along y axis due to + q

[tex]v_1=\frac{1}{4\pi \epsilon } \times\frac{qs}{(y^2+\frac{s^2}{4})^\frac{1}{2} }[/tex]

Potential at distance y along y axis due to - q

[tex]v_1=\frac{1}{4\pi \epsilon } \times\frac{-qs}{(y^2+\frac{s^2}{4})^\frac{1}{2} }[/tex]

Total potential

v = v₁ + v₂

[tex]v= 0[/tex]

Answer:

Change in kinetic energy (ΔKE) = 12.8 × 10⁴ J

Explanation:

Given:

Mass of truck(m) = 2,100 kg

Initial speed(v1) = 38 km/h = 38,000 / 3600 = 10.56 m/s

Final speed(v2) = 55 km/h = 55,000 / 3600 = 15.28 m/s

Find:

Change in kinetic energy (ΔKE)

Computation:

Change in kinetic energy (ΔKE) = 1/2(m)[v2² - v1²]

Change in kinetic energy (ΔKE) = 1/2(2100)[15.28² - 10.56²]

Change in kinetic energy (ΔKE) = 1,050[233.4784 - 111.5136]

Change in kinetic energy (ΔKE) = 1,050[121.9648]

Change in kinetic energy (ΔKE) = 128063.04

Change in kinetic energy (ΔKE) = 12.8 × 10⁴ J

Answer:

The direction of the magnetic field on point P, equidistant from both wires, and having equal magnitude of current flowing through them will be pointed perpendicularly away from the direction of the wires.

Explanation:

Using the right hand grip, the direction of the magnet field on the wire M is counterclockwise, and the direction of the magnetic field on wire N is clockwise. Using this ideas, we can see that the magnetic flux of both field due to the currents of the same magnitude through both wires, acting on a particle P equidistant from both wires will act in a direction perpendicularly away from both wires.

Answer:

4.245s

Explanation:

Given that,

Hypothetical value of speed of light in a vacuum is 18 m/s

Speed of the car, 14 m/s

Time given is 6.76 s, and we're asked to find the observed time, T

The relationship between the two times can be given as

T = t / √[1 - (v²/c²)]

The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject

t = T / √[1 - (v²/c²)]

And now, we substitute the values and insert into the equation

t = 6.76 * √[1 - (14²/18²)]

t = 6.76 * √[1 - (196/324)]

t = 6.76 * √(1 - 0.605)

t = 6.76 * √0.395

t = 6.76 * 0.628

t = 4.245 s

Therefore, the time the driver measures for the trip is 4.245s

Answer:

Explanation:

Concave mirrors is otherwise known as converging mirrors: These are mirrors that are caved inwards (reflecting surface is on the outside curved part). It is called a converging mirror due to the fact that light converges to a point when it strikes and reflects from the surface of the mirror. This type of mirror is used to focus light; parallel rays that are directed towards it will be concentrated to a point.

For a concave mirror to reflect light with properties that are the same as a spotlight (directed light rays parallel to each other), one has to consider its property to gather light to a point after reflecting. Meaning that, we can achieve the spotlight by locatng the point where the rays will be parallel, this point is called the focal point.

Therefore, the light bulb should be placed at the focal point of the mirror.

Answer:

Star A has a higher surface temperature than star B.

Explanation:

The effective temperature of a star can be determined by means of its spectrum and Wien's displacement law:

[tex]T = \frac{2.898x10^{-3} m. K}{\lambda max}[/tex] (1)

Where T is the effective temperature of the star and [tex]\lambda_{max}[/tex] is the maximum peak of emission.  

A body that is hot enough emits light as a consequence of its temperature. For example, if an iron bar is put in contact with fire, it will start to change colors as the temperature increase, until it gets to a blue color, that scenario is known as Wien's displacement law. Which establishes that the peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase and higher wavelengths as the temperature decreases.

Therefore, star A has a higher surface temperature than star B, as it is shown in equation 1 since T and [tex]\lambda max[/tex] are inversely proportional.

Answer:

F = 4212 N

Explanation:

Given that,

Mass of a car, m = 1300 kg

Speed of car on the road is 9 m/s

Radius of curve, r = 25 m

We need to find the magnitude of the unbalanced force that steers the car out of its natural straight-  line path. The force is called centripetal force. It can be given by :

[tex]F=\dfrac{mv^2}{r}\\\\F=\dfrac{1300\times 9^2}{25}\\\\F=4212\ N[/tex]

So, the force has a magnitude of 4212 N

Answer:

At 3 meter distance, the per-second count is 222.22 and at a 10 meter distance, the per-second count is 20.

Explanation:

The number of particles (N)  counts are inversely proportional to the distance between the source and the detector.  

By using the below formula we can find the number of counts.

[tex]N2 = \frac{(D1)^2}{(D2)^2} \times N1 \\N1 = 2000 \\D 1 = 1 \ meter \\D2 = 3 \\[/tex]

The number of count per second, when the distance is 3 meters.

[tex]= \frac{1}{3^2} \times 2000 \\= 222.22[/tex]

Number of count per second when the distance is 10 meters.

[tex]= \frac{1}{10^2} \times 2000 \\= 20[/tex]