At an accident scene on a level road, investigators measure a car’s skid mark (mass of car is M) to be of length d. It was a rainy day and the coefficient of friction was estimated to be μk.
A) Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes.B) Why does the car's mass not matter?1) Since both the change in kinetic energy and the work done by friction are proportional to the mass. The mass cancels out of the equation.2) Since the work done by friction does not depend on mass.3) Since the change in kinetic energy and the work done by friction do not depend on mass.

Answer:

  k = 6,547 N / m

Explanation:

This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is

         w = √ (k / m)

angular velocity and rel period are  related

         w = 2π / T

substitution

         T = 2π √(m / K)

in Experimental measurements give us the following data

  m (g)     A (cm)    t (s)   T (s)

  100        6.5         7.8    0.78

  150        5.5          9.8   0.98

   200      6.0        10.9    1.09

   250       3.5        12.4    1.24

we look for the period that is the time it takes to give a series of oscillations, the results are in the last column

        T = t / 10

To find the spring constant we linearize the equation

        T² = (4π²/K)    m

therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is

         m ’= 4π² / k

where m’ is the slope

           k = 4π² / m'

the equation of the line of the attached graph is

       T² = 0.00603 m + 0.0183

therefore the slope

       m ’= 0.00603  s²/g

    we calculate

         k = 4 π² / 0.00603

          k = 6547 g / s²

we reduce the mass to the SI system

         k = 6547 g / s² (1kg / 1000 g)

         k = 6,547 kg / s² =

         k = 6,547 N / m

let's reduce the uniqueness

         [N / m] = [(kg m / s²) m] = [kg / s²]

The spring-mass system forms a linear graph between the time period and mass. And the value of spring-constant from the given data is 6.46 N/m.

Given data:

Mass suspended by spring is, [tex]m=100 \;\rm g =0.1 \;\rm kg[/tex].

Number of oscillations is, [tex]n =10\;\rm oscillations[/tex].

Time period of oscillation is, [tex]T=7.8 \;\rm s[/tex].

The expression for the angular frequency of spring-mass system is,

[tex]\omega =\drac \sqrt{\dfrac{k}{m} }[/tex]  ......................................................(1)

Here, k is the spring constant.

Angular frequency is also expressed as,

[tex]\omega = 2 \pi f[/tex] .........................................................(2)

here, f is the linear frequency of spring-mass system.

And linear frequency is,

[tex]f=\dfrac{n}{T}\\f=\dfrac{10}{7.81}\\f=1.28 \;\rm cycles/sec[/tex]

Then substitute equation (2) in equation (1) as,

[tex]2 \pi f=\drac \sqrt{\dfrac{k}{m} }\\2 \pi \times 1.28=\drac \sqrt{\dfrac{k}{0.1} }\\(2 \pi \times 1.28)^{2}= \dfrac{k}{0.1}\\k = 6.46 \;\rm N/m[/tex]

Thus, the value of spring constant is 6.46 N/m. And the suitable graph for the spring-mass system is given below.

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Answer:

a) v₀ₓ = v₀ cos θ , b) v_{oy} = v₀ sin θ , c) y = v_{oy}² / 2g,  y = 24.25 m

e) R = 138.46 m

Explanation:

This is a projectile launch exercise

a) let's use trigonometry to find the components of the initial velocity

  cos θ = v₀ₓ / v₀

  v₀ₓ = v₀ cos θ

v₀ₓ = 38 cos 35

v₀ₓ = 31.13 m / s

b) sin θ = [tex]v_{oy}[/tex] / v₀

    v_{oy} = v₀ sin θ

    v_{oy} = 38 sint 35

    v_{oy} = 21 80 m / s

c, d) to find the maximum height, the vertical speed is zero

     v_{y}² = v_{oy}² - 2 g y

     0 = [tex]v_{oy}[/tex]² - 2 gy

     y = v_{oy}² / 2g

let's calculate

     y = 21.80 2 / (2 9.8)

     y = 24.25 m

e) They ask to find the horizontal distance

    for this we can use the expression of reaches

       R = v₀² sin 2θ / g

let's calculate

      R = 38² sin (2 35) / 9.8

       R = 138.46 m

Explanation:

Momentum = mass × velocity

p = mv

p = (1000 kg) (14 m/s)

p = 14000 kg m/s

The momentum of the car as it moves down the highway at the given speed is 14000-kg.m/s

Given the data in the question

Momentum is the product of the mass of a particle and its velocity.

Momentum = Mass × Velocity

[tex]P = m \ * \ v[/tex]

We substitute our given values into the equation

[tex]P = 1000kg \ * \ 14m/s\\\\P = 14000kg.m/s[/tex]

Therefore, the momentum of the car as it moves down the highway at the given speed is 14000-kg.m/s

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Answer:

Average velocity v = 21.18 m/s

Average acceleration a = 2 m/s^2

Explanation:

Average speed equals the total distance travelled divided by the total time taken.

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

Average acceleration equals the change in velocity divided by change in time.

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

Where;

v1 and v2 are velocities at time t1 and t2 respectively.

And x1 and x2 are positions at time t1 and t2 respectively.

Given;

t1 = 3.0s

t2 = 20.0s

v1 = 11 m/s

v2 = 45 m/s

x1 = 25 m

x2 = 385 m

Substituting the values;

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

v = (385-25)/(20-3)

v = 21.18 m/s

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

a = (45-11)/(20-3)

a = 2 m/s^2

Answer

The rate at which the magnetic field is changing is  [tex][\frac{dB}{dt} ] = 0.000467 T/s[/tex]

Explanation

From the question we are told that

   The electric field strength is [tex]E = 3.5mV/m = 3.5 *10^{-3} \ V/m[/tex]

    The radius is  [tex]r = 1.5 \ m[/tex]

The rate of change of the  magnetic  field  is mathematically represented as

        [tex]\frac{d \phi }{dt} = \int\limits^{} {E \cdot dl}[/tex]

Where [tex]dl[/tex] is change of a unit length

     [tex]\frac{d \phi}{dt} = A * \frac{dB}{dt}[/tex]

Where A is the area which is mathematically represented as

     [tex]A = \pi r^2[/tex]

    So

    [tex]E \int\limits^{} { dl} = ( \pi r^2) (\frac{dB}{dt} )[/tex]  

  [tex]E L = ( \pi r^2) (\frac{dB}{dt} )[/tex]  

where L is the circumference of the circle which is mathematically represented as

     [tex]L = 2 \pi r[/tex]

So

     [tex]E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ][/tex]

      [tex]E = \frac{r}{2} [\frac{dB}{dt} ][/tex]

       [tex][\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }[/tex]

substituting values

      [tex][\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }[/tex]

      [tex][\frac{dB}{dt} ] = 0.000467 T/s[/tex]    

Complete question is:

A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.

Answer:

(V_A) = 31.32 m/s

Explanation:

We are given;

car's mass, m = 1200 kg

h_A = 100 m

h_B = 150 m

v_B = 0 m/s

From law of conservation of energy,

the distance from point A to B is;

h = 150m - 100 m = 50 m

From Newton's equations of motion;

v² = u² + 2gh

Thus;

(V_B)² = (V_A)² + (-2gh)

(negative next to g because it's going against gravity)

Thus;

(V_B)² = (V_A)² - (2gh)

Plugging in the relevant values;

0² = (V_A)² - 2(9.81 × 50)

(V_A) = √981

(V_A) = 31.32 m/s

Answer:

E = 3.45*10^-19 N/C

Explanation:

a) The electric field between two parallel plates id given by the following formula:

[tex]E=\frac{\sigma}{\epsilon_o}[/tex]           (1)

where:

σ: surface charge density of the plates = 39.0nC/m^2

εo: dielectric permittivity of vacuum = 8.85*10^-12 C/Nm^2

You replace these values in the equation (1):

[tex]E=\frac{39.0*10^{-9}C/m^2}{8.85*10^{-12}C^2/Nm^2}\\\\E=3.45*10^{-19}\frac{N}{C}[/tex]

The electric field in between the parallel plates is 3.45*10^-19 N/C

Answer:

Work is done by friction = -165 J

Explanation:

Given:

Mass of block (m) = 15 kg

Ramp inclined = 28°

Friction force (f) = 30 N

Distance (d) = 5.5 m

Find:

Work is done by friction.

Computation:

Work is done by friction = -Fd

Work is done by friction = -(30)(5.5)

Work is done by friction = -165 J

Answer:

Maximum height of jump on Rhea is 37.16 times of that on Earth, i.e 37.16h

Maximum range of jump on Rhea is 37.16 of times that on Earth, i.e 37.16R

Explanation:

The acceleration due to gravity on Rhea = 0.264 m/s^2

Acceleration due to gravity on earth here = 9.81 m/s^2

this means that the acceleration due to gravity g on earth is 9.81/0.264 = 37.16 times that on Rhea.

maximum height that can be achieved by the froghopper is given by the equation;

h = [tex]\frac{u^{2}sin^{2} \alpha}{2g}[/tex]

let us put all the numerator of the equation as k, since the velocity of take off is the same for Earth and Rhea. The equation is simplified to

h = [tex]\frac{k}{2g}[/tex]

for earth,

h =  [tex]\frac{k}{2*9.81}[/tex] =  [tex]\frac{k}{19.62}[/tex]

for Rhea,

h  =  [tex]\frac{k}{2*0.264}[/tex] =  [tex]\frac{k}{0.528}[/tex]

therefore,

h on Rhea is [tex]\frac{k}{0.528}[/tex] ÷ [tex]\frac{k}{19.62}[/tex] = 37.16 times of that on Earth, i.e 37.16h

Equation for range R is given as

R =  [tex]\frac{u^{2}sin 2\alpha}{g}[/tex]

following the same approach as before,

R on Rhea will be [tex]\frac{k}{0.264}[/tex] ÷ [tex]\frac{k}{9.81}[/tex] = 37.16 of times that on Earth, i.e 37.16R

Answer:

Q = - 5264 W = - 5.26 KW

Here, negative sign indicates the outflow of heat

Explanation:

Fourier's Law of heat conduction, gives the following formula:

Q = - KAΔT/t

where,

Q = Rate of Heat Conduction out of window = ?

K = Thermal Conductivity of Glass = 0.84 W/m.°C

A =Surface Area of window = 2.82 m²

ΔT = Difference in Temperature of both sides of surface

ΔT = Inner Surface Temperature - Outer Surface Temperature= 5°C - (- 10°C)

ΔT = 15°C

t = thickness of window = 0.675 cm = 0.00675 m

Therefore,

Q = - (0.84 W/m.°C)(2.82 m²)(15°C)/0.00675 m

Q = - 5264 W = - 5.26 KW

Here, negative sign indicates the outflow of heat.

Answer:

4.8 × 10^15 N

Explanation:

Electric Field is defined as Force per unit Charge.

This is expressed mathematically as;

E= F/Q

Where E- Electric Field

F- Force

Q- charge

From the expression above by change of subject of formula for F, we have;

F=E×Q

= 1.2 * 10^6 ×4.0 * 10^9

= 4.8 × 10^15 N

Answer:

i) angle of incidence;i = 29.43°

ii) δm = 38.92°

Explanation:

Prism is equilateral so angle of prism (A) = 60°

Refractive index of glass; n_glass = 1.52

A) Let's assume the incident angle = i and Critical angle = θc

We know that, sin θc = 1/n

Thus;

sin θc = 1/n_glass

θc = sin^(-1) (1/n_glass)

θc = sin^(-1) (1/1.52)

θc = 41.14°

Now, the angle of prism will be the sum of external angle that is critical angle and reflected angle.

Thus;

A = r + θc

r = A - θc

So;

r = 60° - 41. 14°

r = 18.86°

From, Snell's law. If we apply it to this question, we will have;

(sin i)/(sin r) = n_glass

Where;

i is angle of incidence and r is angle of reflection.

Let's make i the subject;

i = sin^(-1) (n_glass × sin r)

i = sin^(-1) (1.52 × sin 18.86)

i = sin^(-1) 0.4914

i = 29.43°

B) The formula to calculate minimum deviation would be from;

μ = [sin ((A + δm)/2)]/(sin A/2)

Where;

μ is Refractive index

δm is minimum angle of deviation

A is angle of prism

Now Refractive index is given by a formula; μ = (sin i)/(sin r)

So; μ = (sin 29.43)/(sin 18.86)

μ = 1.52

Thus;

1.52 = [sin ((60 + δm)/2)]/(sin 60/2)

1.52 * sin 30 = sin ((60 + δm)/2)

0.76 = sin ((60 + δm)/2)

sin^(-1) 0.76 = ((60 + δm)/2)

49.46 × 2 = (60 + δm)

98.92 - 60 = δm

δm = 38.92°

Answer:

Distance = 16.9 m

Explanation:

We are given;

Power; P = 70 W

Intensity; I = 0.0195 W/m²

Now, for a spherical sound wave, the intensity in the radial direction is expressed as a function of distance r from the center of the sphere and is given by the expression;

I = Power/Unit area = P/(4πr²)

where;

P is the sound power

r is the distance.

Thus;

Making r the subject, we have;

r² = P/4πI

r = √(P/4πI)

r = √(70/(4π*0.0195))

r = √285.6627

r = 16.9 m

Answer:

16.9 m

Explanation:

Answer:

9.05 m/s ,   -14.72°  (respect to x axis)

Explanation:

To find the final velocity of the bowling ball you take into account the conservation of the momentum for both x and y component of the total momentum. Then, you have:

[tex]p_{xi}=p_{xf}\\\\p_{yi}=p_{yf}\\\\[/tex]

[tex]m_1v_{1xi}+m_2v_{2xi}=m_1v_1cos\theta+m_2v_{2}cos\phi\\\\0=m_1v_1sin\theta-m_2v_2sin\phi[/tex]

m1: mass of the bowling ball = 5.50 kg

m2: mass of the bowling pin = 0.850 kg

v1xi: initial velocity of the bowling ball = 9.0 m/s

v2xi: initial velocity of bowling pin = 0m/s

v1: final velocity of bowling ball = ?

v2: final velocity of bowling pin = 15.0 m/s

θ: angle of the scattered bowling pin = ?

Φ: angle of the scattered bowling ball = 85.0°

Where you have used that before the bowling ball hits the pin, the y component of the total momentum is zero.

First you solve for v1cosθ in the equation for the x component of the momentum:

[tex]v_1cos\theta=\frac{m_1v_{1xi}-m_2v_2cos\phi}{m_1}\\\\v_1cos\theta=\frac{(5.50kg)(9.0m/s)-(0.850kg)(15.0m/s)cos85.0\°}{5.50kg}\\\\v_1cos\theta=8.79m/s[/tex]

and also you solve for v1sinθ in the equation for the y component of the momentum:

[tex]v_1sin\theta=\frac{(0.850kg)(15.0m/s)sin(85.0\°)}{5.50kg}\\\\v_1sin\theta=2.3m/s[/tex]

Next, you divide v1cosθ and v1sinθ:

[tex]\frac{v_1sin\theta}{v_1cos\theta}=tan\theta=\frac{2.3}{8.79}=0.26\\\\\theta=tan^{-1}(0.26)=14.72[/tex]

the direction of the bawling ball is -14.72° respect to the x axis

The final velocity of the bawling ball is:

[tex]v_1=\frac{2.3m/s}{sin\theta}=\frac{2.3}{sin(14.72\°)}=9.05\frac{m}{s}[/tex]

hence, the final velocity of the bawling ball is 9.05 m/s

Answer:

rt

Explanation:

Answer:

100 J

Explanation:

Work = change in energy

W = ΔKE

W = ½ mv² − ½ mv₀²

W = ½ m (v² − v₀²)

W = ½ (10 kg) ((6 m/s)² − (4 m/s)²)

W = 100 J

Answer:

1.25 m/s^2

Explanation:

F = m*a ...... force = mass * acceleration

force = 100 N, mass = 80 kg

100 = 80 * a

100/80 = a = 1.25 m/s^2

Answer:

The acceleration is 1.25m/s².

Explanation:

You have to apply Newton's Second Law which is F = m×a where F represents force, m is mass and a is acceleratipn. Then you have to substitute the following values into the formula :

[tex]f = m \times a[/tex]

Let F = 100,

Let m = 80,

[tex]100 = 80 \times a[/tex]

[tex]100 = 80a[/tex]

[tex]a = 100 \div 80[/tex]

[tex]a = 10 \div 8[/tex]

[tex]a = 1.25[/tex]

Answer:

5.06*10^6 m/s

Explanation:

Given that

Initial velocity, u = 4*10^6 m/s

Acceleration, a = 6*10^12 m/s²

Distance traveled, s = 80 cm

Final velocity, v = ?

We can find the final velocity by using one of the equations of motion.

v² = u² + 2as

On substituting the values, we have

v² = (4*10^6)² + 2 * 6*10^12 * 0.8

v² = 2.56*10^13

v = √2.56*10^13

v = 5.06*10^6 m/s

Therefore, the final velocity of the proton is adjudged to be 5.06*10^6 m/s

The final velocity of the proton over the given distance is [tex]5.06 \times 10^6 \ m/s[/tex].

The given parameters;

The final velocity of the proton over the given distance is calculated as follows;

[tex]v^2 = u^2 + 2as\\\\v^2 = (4\times 10^6)^2 \ + \ 2(6.0 \times 10^{12})(0.8)\\\\v^2 = 2.56 \times 10^{13} \\\\v = \sqrt{2.56 \times 10^{13} } \\\\v = 5.06 \times 10^6 \ m/s[/tex]

Thus, the final velocity of the proton over the given distance is [tex]5.06 \times 10^6 \ m/s[/tex]

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_____________________________

Solution,

Work=50 Joule

Time=10 seconds

Power=?

Now,

Power=Work/time

= 50/10

= 5 Watt.

So the right answer is 5 W

Hope it helps..

Good luck on your assignment

__________________________

Answer:

A. we will see that the notion [tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex] which denotes Stephen Curry  illustrates that Stephen Curry minigolf ball shot is closer

B.  Lebron James hits at an angle of 17.48° North -East.

The direction of Stephen is   = 20° due to East of North

Explanation:

Let [tex]r ^ {\to[/tex] represent the position vector of the hole;

Also; using  the origin as starting point. Let the east direction be along the positive x axis and the North direction  be + y axis

Thus:

[tex]r ^ {\to[/tex]  = [tex]8.2 \ sin 20^0 \hat i + 8.2 \ cos 20 \hat j[/tex]

[tex]r ^ {\to[/tex]  = [tex](2.8046 \hat i + 7.7055 \hat j ) m[/tex]

Let [tex]r_1 ^ \to[/tex] be the position vector for Lebron James's first shot

So;

[tex]r_1 ^ \to[/tex] = [tex](8.6 \ sin \ 35.2 )^0 \hat i + 8.6 \ cos \ ( 35.2)^0 \hat j[/tex]

[tex]r^ \to = (4.9573 \hat i + 7.02745 \hat j) m[/tex]

Let [tex]r_2 ^ \to[/tex] be the position vector for Stephen Curry's shot

[tex]r_2 ^ \to[/tex]  [tex]=6.1 \ sin 20^0 \hat i + 6.1 \ cos \ 20 \hat j[/tex]

[tex]r_2 ^ \to[/tex]  = [tex](2.0863 \hat i + 5.7321 \hat j )m[/tex]

However;

[tex]r ^ \to - r_1 ^\to = (-2.1527 \hat i + 0.67805 \hat j) m[/tex]

[tex]\mathbf{|r ^ \to - r_1 ^\to| =2.25696 \ m }[/tex]

Also;

[tex]r ^ \to - r_2 ^\to = (0.71013 \hat i - 1.9734 \hat j) m[/tex]

[tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex]

Thus; from above ; we will see that the notion [tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex] which denotes Stephen Curry  illustrates that Stephen Curry minigolf ball shot is closer

B .

For Lebron James ;

The angle can be determine using the trigonometric function:

[tex]tan \theta = ( \dfrac{0.67805}{-2.1527}) \\ \\ tan \theta = -0.131498 \\ \\ \theta = tan ^{-1} ( -0.31498) \\ \\ \mathbf{\theta = -17.48^0}[/tex]

Thus  Lebron James hits at an angle of 17.48° North -East.

For Stephen Curry;

[tex]tan \theta = ( \dfrac{-1.9734}{0.7183}) \\ \\ tan \theta = -2.74732 \\ \\ \theta = tan ^{-1} ( -2.74732) \\ \\ \mathbf{\theta = -70.0^0}[/tex]

The direction of Stephen is  = 90° - 70° = 20° due to East of North

Answer:

Explanation:

capacitance of air capacitor

C = ε₀ A /  d

ε₀ is permittivity of medium , A is plate area , d is distance between plate .

C = 8.85 x 10⁻¹² x 25 x 10⁻⁴ / 1.3 x 10⁻²

= 170.19 x 10⁻¹⁴ F .

charge on the capacitor when it is charged to  potential of 255 V

= CV , C is capacitance and V is potential

= 170.19 x 10⁻¹⁴  x 255

= 4.34 x 10⁻¹⁰ C .

After it is disconnected from the source , and it is immersed in water , charge on it remains the same .

So its charge when immersed in water will be constant at 4.34 x 10⁻¹⁰ C.

b )

When it is immersed in water its capacity increases  k times where k is dielectric constant of water which is 80 .

capacitance of capacitor in water = 80 x 170.19 x 10⁻¹⁴  F

= 13615.2  x 10⁻¹⁴ F .

= 1.36 x 10⁻¹⁰ F

potential difference = charge / capacitance

= 4.34 x 10⁻¹⁰ / 1.36 x 10⁻¹⁰

= 3.2 V

c )

Energy of capacitor = 1/2 C V²

Initial energy = 1/2 x 170.19 x 255² x 10⁻¹⁴

=  55.33 x 10⁻⁹ J

Final energy = 1/2 x 1.36 x 10⁻¹⁰ x 3.2²

= .7  x 10⁻⁹ J .

decrease of energy = 54.63 x 10⁻⁹ J .

Answer:

The apparent weight of the person as she pass the highest point is  [tex]N = 458.8 \ N[/tex]

Explanation:

From the question we are told that

   The radius of the Ferris wheel is [tex]r = 5.0 \ m[/tex]

    The period of revolution is [tex]T = 8.0 \ s[/tex]

     The weight of the person is  [tex]W = 670 \ N[/tex]

Generally the speed of the wheel is mathematically represented as

      [tex]v = \frac{2 \pi r}{T }[/tex]

substituting values

      [tex]v = \frac{2 * 3.142 * 5}{8 }[/tex]

       [tex]v = 3.9 3 \ m/s[/tex]

The apparent weight (the normal force exerted on her by the bench) at the highest point is mathematically evaluated as

          [tex]N = mg - \frac{mv^2}{r}[/tex]

Where m is the mass of the person which is mathematically evaluated as

     [tex]m = \frac{W}{g}[/tex]

substituting values

    [tex]m = \frac{670}{9.8}[/tex]

    [tex]m = 68.37 \ kg[/tex]

So

    [tex]N = 68.37 * 9.8 - \frac{68.37 * {3.93}^2}{5}[/tex]

    [tex]N = 458.8 \ N[/tex]

Answer:

a bale

Explanation:

a bale is a group of turtles

Answer:

A bale or nest

Explanation:

Answer:

A₁/A₂ = 0.44

Explanation:

The emissive power of the bulb is given by the formula:

P = σεAT⁴

where,

P = Emissive Power

σ = Stefan-Boltzman constant

ε = Emissivity

A = Surface Area

T = Absolute Temperature of Surface

FOR BULB 1:

Since, emissivity and emissive power are constant.

Therefore,

P = σεA₁T₁⁴   ----------- equation 1

where,

A₁ = Surface Area of Bulb 1

T₁ = Temperature of Bulb 1 = 3000 k

FOR BULB 2:

Since, emissivity and emissive power are constant.

Therefore,

P = σεA₂T₂⁴   ----------- equation 2

where,

A₂ = Surface Area of Bulb 2

T₂ = Temperature of Bulb 1 = 2000 k

Dividing equation 1 by equation 2, we get:

P/P = σεA₁T₁⁴/σεA₂T₂⁴

1 = A₁(3000)²/A₂(2000)²

A₁/A₂ = (2000)²/(3000)²

A₁/A₂ = 0.44

Answer:

Plutonium–238

Explanation:

The stability of isotopes is largely dependent on their half-life.

Half life of an isotope is the time taken for the initial mass of the isotope to be halfed or we can say that the half-life of an isotope is the time taken for the mass of the isotope to become half the initial mass.

From the above definition, we discovered that if the time taken for the mass of the isotope to become half its initial mass is long, then the isotope must be very stable. On the other hand, if the time taken to become half its initial mass is short, then the isotope is unstable because.

We can thus say that, the longer the half-life the more stable the isotope and the shorter the half-life, the less stable the isotope will be.

Considering the table given in the question above and with the ideas we have obtained from the explanation above, we can see that Plutonium–238 has the longest half-life. Therefore Plutonium–238 will be more stable.

Answer:

Explanation:

Given that:

width b=100mm

depth h=150 mm

length L=2 m =200mm

point load P =500 N

Calculate moment of inertia

[tex]I=\frac{bh^3}{12} \\\\=\frac{100 \times 150^3}{12} \\\\=28125000\ m m^4[/tex]

Point C is subjected to bending moment

Calculate the bending moment of point C

M = P x 1.5

= 500 x 1.5

= 750 N.m

M = 750 × 10³ N.mm

Calculate bending stress at point C

[tex]\sigma=\frac{M.y}{I} \\\\=\frac{(750\times10^3)(25)}{28125000} \\\\=0.0667 \ MPa\\\\ \sigma =666.67\ kPa[/tex]

Calculate the first moment of area below point C

[tex]Q=A \bar y\\\\=(50 \times 100)(25 +\frac{50}{2} )\\\\Q=250000\ mm[/tex]

Now calculate shear stress at point C

[tex]=\frac{FQ}{It}[/tex]

[tex]=\frac{500*250000}{28125000*100} \\\\=0.0444\ MPa\\\\=44.4\ KPa[/tex]

Calculate the principal stress at point C

[tex]\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2} \pm\sqrt{(\frac{\sigma_x-\sigma_y}{2} ) + (\tau)^2} \\\\=\frac{666.67+0}{2} \pm\sqrt{(\frac{666.67-0}{2} )^2 \pm(44.44)^2} \ [ \sigma_y=0]\\\\=333.33\pm336.28\\\\ \sigma_1=333.33+336.28\\=669.61KPa\\\\\sigma_2=333.33-336.28\\=-2.95KPa[/tex]

Calculate the maximum shear stress at piont C

[tex]\tau=\frac{\sigma_1-\sigma_2}{2}\\\\=\frac{669.61-(-2.95)}{2} \\\\=336.28KPa[/tex]

Answer:

2025m

Explanation:

Since all materials of the sphere is made to a cylindrical wire, it implies the volume of the sphere material is same as that of the cylinder. This is expressed mathematically thus.

Volume of Sphere= volume of cylinder

4/3 ×π×R^3= π× r2× L

4/3 ×R^3= r^2×L

Hence

L = 3/4 × R^3/ r^2

But R = 6.0/2 = 3.0cm{ Diameter is twice raduis}

r= 0.2/2 = 0.1mm=>0.01cm{ Diameter is twice raduis and unit converted by dividing by 10 since 10mm = 1cm}

Substituting R and r into the expression for L, we have :

L = 3/4 × 3^3/ 0.01^2= 0.75 ×27/0.0001 = 202500cm

202500/100= 2025m{ we divide by 100 because 100cm=1m}

Answer:

ΔL = 1.011 mm

Explanation:

Let's begin by listing out the given information:

Length (L) = 600 mm = 0.6 m,

Diameter (D) = 40 mm = 0.04 m ⇒ Radius (r) = 20 mm = 0.2 m,

Area (cross sectional) = πr² = 3.14 x .02² = 0.001256 m²,

Modulus of Elasticity (E) = 85 GN/m²,

Compressive load (F) = 180 KN

Using the formula, Stress = Load ÷ Area

Mathematically,

σ = F ÷ A = 180 x 10³ ÷ 0.001256

σ = 143312.1 KN/m²

Modulus of elasticity = stress ÷ strain

E = σ ÷ ε

ε = ΔL/L

85 x 10⁹ = 143312.1 x 10³ ÷ (ΔL/L)

ΔL = 143312.1 x 10³ ÷ 85 X 10⁹ = 1686.02 * 10⁻⁶

ΔL = L x 1686.02 * 10⁻⁶

ΔL = 0.6 * 1686.02 * 10⁻⁶ = 1011.61 x 10⁻⁶

ΔL = 1.011 x 10⁻³ m

ΔL = 1.011 mm

∴The bar contracts by 1.011 mm

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

[tex]K=qV[/tex]          (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

[tex]K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J[/tex]

The kinetic energy of the particle is also:

[tex]K=\frac{1}{2}mv^2[/tex]       (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}[/tex]

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

[tex]|F_B|=qvBsin(\theta)[/tex]

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

[tex]|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N[/tex]

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

[tex]r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm[/tex]

the radius of the trajectory of the electron is 10.1 cm

The speed, magnetic force and radius are respectively; 10.75 * 10⁶ m/s; 7.57 * 10⁻¹² N; 0.101 m

A) We know that the formula for kinetic energy can be expressed as;

K = qV

where;

q is charge of the particle = 2e = 2(1.6 × 10⁻¹⁹ C) = 3.2 × 10⁻¹⁹ C

V is potential difference = 1.2 × 10⁶ V

K = 3.2 × 10⁻¹⁹ *  1.2 × 10⁶

K = 3.84 × 10⁻¹³ J

Also, formula for kinetic energy is;

K = ¹/₂mv²

where v is speed

Thus;

v = √(2K/m)

v = √(2 * 3.84 × 10⁻¹³)/(6.64 * 10⁻²⁷)

v = 10.75 * 10⁶ m/s

B) The magnetic force is given by the formula;

F_b = qvB

F_b = (3.2 × 10⁻¹⁹ * 10.75 * 10⁶ * 2.2)

F_b = 7.57 * 10⁻¹² N

C) The formula to find the radius is;

r = mv/qB

r = (6.64 * 10⁻²⁷ * 10.75 * 10⁶)/(1.6 × 10⁻¹⁹ * 2.2)

r = 0.101 m

Read more about magnetic field at; https://brainly.com/question/7802337

Answer:

Dear Kaleb

Answer to your query is provided below

Acceleration of the vehicle is 12m/s^2

Explanation:

Explanation for the same is attached in image