A certain car takes 30m to stop when it is traveling at 25m/s. If a pedestrian is 28m in front of this car when the driver starts braking (starting at 25m/s), how long does the pedestrian have to get out of the way?

Answer:

R = 668.19 ft  

Explanation:

given,

speed of the ball thrown by the pitcher = 100 mph

to travel maximum distance θ = 45°

distance traveled by the ball = ?

using formula

1 mph = 0.44704 m/s

100 mph = 44.704 m/s

[tex]R = \dfrac{u^2sin2\theta}{g}[/tex]

[tex]R = \dfrac{44.704^2sin2\times 45}{9.81}[/tex]

R = 203.71 m

1 m  = 3.28 ft

R = 203.71 × 3.28

R = 668.19 ft  

hence, ball will go at a distance of 668.19 ft   when pitcher throw it at 100 mph.

Final answer:

The weight of a 2.5-kg pumpkin is calculated by multiplying the mass by the acceleration due to gravity, resulting in 24.5 newtons (N) when rounded to one decimal place.

Explanation:

To find out the weight of a 2.5-kg pumpkin, you would need to multiply its mass by the acceleration due to gravity. In most physics problems, this is approximately 9.8 meters per second squared (m/s²). Therefore, the weight W can be calculated using the formula W = m×g, where m is the mass in kilograms and g is the acceleration due to gravity.

So, for a 2.5-kg pumpkin, W = 2.5 kg × 9.8 m/s² = 24.5 kg·m/s². Since weight is a force, it is measured in newtons (N) in the International System of Units (SI). Therefore, the pumpkin weighs 24.5 N.

Final answer:

The weight of a 2.5-kg pumpkin is calculated by multiplying its mass by the gravitational acceleration, which results in a weight of 24.5 Newtons (N) on Earth.

Explanation:

The weight of a pumpkin is the force with which it is pulled towards the Earth due to gravity. To calculate the weight, you can use the formula:

Weight (W) = mass (m) × gravitational acceleration (g)

Where the gravitational acceleration on Earth is typically 9.8 meters per second squared (9.8 m/s²). For a 2.5-kg pumpkin, the calculation would be:

Weight (W) = 2.5 kg × 9.8 m/s²

Weight (W) = 24.5 kg·m/s²

Since weight is measured in Newtons (N), and 1 kg·m/s² is equivalent to 1 Newton, the weight of the pumpkin is 24.5 Newtons (N). When converting measurements and rounding, it's important to consider significant figures and the least precise measurement. In this calculation, both 2.5 kg and 9.8 m/s² are exact enough, so our final answer in Newtons does not need additional rounding.

Answer:

a) 12.56 rad per minute

b)62.8 inches per minute

Explanation:

Given:

a)

Let [tex]\omega[/tex] be the angular velocity of the blade which is given by

[tex]\omega =2\pi f\\\omega=2\times 3.14 \times 2\\\omega=12.56\ \rm rad/min[/tex]

b)

Let v be the linear speed of the tip of the blade which is given by

[tex]v=\omega \times R\\v=12.56\times 5\\v=62.8\ \rm inches/min[/tex]

Hence we have calculated the angular velocity and linear speed.

Final answer:

To find the distance the cat strikes from the edge of the piano, we can break the motion into horizontal and vertical components, and solve for the time of flight and horizontal distance traveled. By substituting the given values into relevant equations and solving, we can determine that the cat will strike the floor approximately 1.71 meters from the edge of the piano.

Explanation:

To solve this problem, we can break the motion into horizontal and vertical components. The horizontal component of the initial velocity will not change, so the cat will travel a horizontal distance of 3m/s *cos(37°) * t, where t is the time of flight. To find t, we can use the equation h = v0y * t + (1/2) * g * [tex]t^2[/tex], where h is the vertical displacement, v0y is the vertical component of the initial velocity, and g is the acceleration due to gravity.

Substituting the given values into the equation, we get 1.3m = 3m/s * sin(37°) * t - (1/2) * 9.8m/[tex]s^2[/tex] * [tex]t^2[/tex]. Solving this quadratic equation, we find two possible values for t: approximately 0.364s and 0.203s. Since the cat jumps off the piano, we only consider the positive value of t. So the cat will strike the floor approximately 3m/s * cos(37°) * 0.364s = 1.71 meters from the edge of the piano.

Answer:

63.57 kg

Explanation:

weight = 140 lbs

Let the mass  is m.

1 lbs = 4.45 N

The weight of an object is defined as the force with which our earth attracts the body towards its centre.

Weight is the product of mass of the body and the acceleration due to gravity of that planet.

W = m x g

On earth surface g = 9.8 m/s^2

Now convert lbs in newton

So, 140 lbs = 140 x 4.45 = 623 N

So, m x 9.8 = 623

m = 63.57 kg

Thus, the mass is 63.57 kg.

Answer:

The force will be zero

Explanation:

Due to the symmetric location of the +2μC charges the forces the excert over the +5μC charge will cancel each other resulting in a net force with a magnitude of zero.However in this case it would be an unstable equilibrium, very vulnerable to a kind of bucking. If the central charge is not perfectly centered on the vertical axis the forces will have components in that axis that will add together instead of canceling each other.

The weight of the astronaut's life support backpack on the moon would be one-sixth of its weight on Earth, which equates to 50lb.

The acceleration due to gravity on the moon is one-sixth that on Earth. Therefore, the weight of any object on the moon will be one-sixth of its weight on Earth. Since weight is a product of mass and gravity, if gravity is reduced, weight is also reduced accordingly.

To calculate the weight of an astronaut's life support backpack on the moon, you need to divide its Earth weight (300lb) by 6 (the relative gravity of the moon compared to Earth). So, on the moon, the backpack would weigh 300/6 = 50lb.

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Answer:

E = 2.88*10^6 N/C

v = 0.742 *10^6m/s

Explanation:

The two disks form a parallel-plate capacitor, which will cause an electric field equal to:

[tex]E = \frac{Q}{e_0 A}[/tex]

Where Q is the charge of the disks, A is the area of the disks and e_0 is the vacuum permisivity equal to 8.85*10^-12 C^2/Nm^2:

[tex]E = \frac{8*10^{-9} C}{(\frac{1}{4}\pi(0.02m)^2)*8.85*10^{-12}C^2/Nm^2}= 2.88*10^6 N/C[/tex]

Now, for the second part of the problem, we can use conservation of energy. The addition of potential and kinetic energy at launch point should be equal to the addition at the positive disk. Because the proton has positive charge, the potential energy of the proton will increase as its distance to the negative disk increases too. This is because the proton will be attracted towards the negative disk. The potential energy is given by:

[tex]E_p = V*q[/tex]

Where V is the difference in potential (voltage) between the disks. In a parallel-plate capacitor:

[tex]V = E*d[/tex], where d is the difference in position with the frame of reference. Our frame of reference will be the negative disk.

q is the charge of the proton.

The kinetic energy is given by:

[tex]E_k = \frac{1}{2}mv^2[/tex]

Then:

[tex]E_p_1+E_k_1 = E_p_2+E_k_2\\V_1*q+\frac{1}{2}mv_1^2= V_2*q+\frac{1}{2}mv_2^2 |v_2=0, E_p_1 = 0\\\frac{1}{2} mv_1^2 = E*d*q \\v = \sqrt{\frac{2Edq}{m}} = \sqrt{\frac{2*2.88*10^6N/m*0.001m*1.6*10^{-19}C}{1.67*10^{-27}}} = 0.742 *10^6m/s[/tex]

(a) The electric field strength between the disks is 2.88 N/C

(b) The launch speed of the proton to reach the positive disk is 7.43 x 10⁵ m/s.

The given parameters;

The electric field strength between the disks is calculated as follows;

[tex]E = \frac{Q}{\varepsilon _o A} \\\\E = \frac{Q}{\varepsilon _o \pi r^2} \\\\E = \frac{8 \times 10^{-9} }{8.85\times 10^{-12} \times \pi \times (0.01)^2} \\\\E = 2.88 \times 10^ 6 \ N/C[/tex]

The launch speed of the proton to reach the positive disk is calculated as follows;

[tex]K.E = W\\\\\frac{1}{2} mv^2 = Fd\\\\\frac{1}{2} mv^2 = Eqd\\\\mv^2 = 2Eqd\\\\v = \sqrt{\frac{2Eqd}{m} } \\\\v = \sqrt{\frac{2\times 2.88 \times 10^6 \times 1.6\times 10^{-19} \times 0.001 }{1.67 \times 10^{-27}} }\\\\v = 7.43 \times 10^5 \ m/s[/tex]

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Answer:

(a) 5.45 m

(b) no change

Explanation:

refractive index of water, n = 1.333

h = 2.4 m

Let the radius of circle is r.

Here we use the concept of total internal reflection.

Let i be the angle of incidence.

By using Snell's law

Refractive index of water with respect to air is the ratio of Sin of angle of incidence to the Sin of angle of refraction.

here refraction takes place from denser medium to rarer medium

So, [tex]\frac{1}{n}=\frac{Sin i}{Sin r}[/tex]

Here, angle r is 90 for total internal reflection

[tex]\frac{1}{1.33}=\frac{Sin i}{Sin 90}[/tex]

Sin i = 0.75

i = 48.6°

According to the triangle ΔOAB

[tex]tan i = \frac{AB}{OA}[/tex]

[tex]tan 48.6 = \frac{r}{2.4}[/tex]

r = 2.73 m

Diameter = 2 r = 2 x 2.73  = 5.45 m

Thus, the diameter of the circle is 5.45 m.

(b) As the value of r depends on the angle of incidence i and angle of incidence depends on the value of refractive index. As refractive index is constant, so the value of diameter remains same.

Due to refraction, a fish sees the world outside through a circle on the water's surface. The diameter of this circle, around 3.6 meters for a fish 2.4 m deep, decreases as the fish descends due to the narrowing range of incident light angles.

The question revolves around the concept of refraction, a phenomenon that occurs when light changes the medium, therefore, also changing direction. This comes into play when light travels from water to air (or vice versa), like how a fish observes the world outside the water body.

In response to the student's question part (a): When the fish is 2.4 m below the water's surface, it sees through a circle on the surface due to refraction. Knowing that the index of refraction of water is 1.333, the apparent depth of the fish to an observer on the surface would be the real depth divided by the refractive index, or around 1.8 meters. Therefore, the light from objects outside the lake would reach the fish within a circle having a diameter of about 3.6 meters (twice the apparent depth).

For part (b) of the question: If the fish were to dive deeper, the diameter of the circle through which it could observe the world outside would diminish. This is because as the actual depth increases, the range of incident angles for light rays entering the water from air, for which total internal reflection doesn’t occur, gets narrower. Therefore, the fish's field of view on the outside world decreases.

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Answer:

1.21 cm.

Explanation:

Given that the wavelength of light is 550 nm.

And the distance between the slits is 0.10 mm.

And the distance to the screen is 2.2 m.

As we know that the lateral separation can be defined by the formula.

[tex]LS=\dfrac{m\lambda D}{d}[/tex]

Here, m is the order of maxima, [tex]\lambda[/tex] is the wavelength , d is the distance between the slits and D is the distance to the screen.

For first maxima m=1, and for second maxima m=2.

Difference in the lateral maxima will be,

[tex]\Delta LS=\dfrac{2\lambda D}{d}-\dfrac{1\lambda D}{d}\\\Delta LS=\dfrac{\lambda D}{d}[/tex]

Substitute all the variables in the above equation.

[tex]\Delta LS=\dfrac{550 nm\times 2.2m}{0.10mm}\\\Delta LS=\dfrac{550\times 10^{-9} nm\times 2.2m}{0.10\times 10^{-3} }\\ \Delta LS=121\times 10^{-4}m\\ \Delta LS=1.21 cm[/tex]

Therefore, the lateral separation difference  between between first order and second order maxima is 1.21 cm.

Answer:

The distance is 3.1 m

Explanation:

The position vector of the fly relative to the corner of the wall is

r = (3.1, 0.5).

The distance of the fly from the corner will be calculated as the magnitude of the vector "r"

magnitude of vector  [tex] r = \sqrt{(3.1 m)^{2} + (0.5 m)^{2}} = 3.1 m[/tex]

Since the numbers to be added have only one decimal place 3.1 and 0.5, the result of the sum will have to have one decimal place. The result of the square root will also have one decimal place.

I have to say, i love this kind of problems.

So, we got the linear acceleration, as we all know, linear the acceleration its, in dimensional units:

[tex][a]=[\frac{distance}{time^2}][/tex].

Now, we got the radius

[tex][r] = [distance][/tex]

the angular frequency

[tex][\omega] = \frac{1}{s}[/tex]

and the mass

[tex][m]=[mass][/tex].

Now, the acceleration doesn't have units of mass, so it can't depend on the mass of the particle.

The distance in the acceleration has exponent 1, and so does in the radius. As the radius its the only parameter that has units of distance, this means that the radius must appear with exponent 1. Lets write

[tex]a \propto r[/tex].

The time in the acceleration has exponent -2 As the angular frequency its the only parameter that has units of time, this means that the angular frequency must appear, but, the angular frequency has an exponent of -1, this means it must be squared

[tex]a \propto r \omega^2[/tex].

We are almost there. If this were any other problem, we would write:

[tex]a = A r \omega^2[/tex]

where A its an dimensionless constant. Its common for this constants to appears if we need an conversion factor. If we wanted the acceleration in cm/s^2, for example. Luckily for us, the problem states that there is no dimensionless constant involved, so:

[tex]a = r \omega^2[/tex]

Final answer:

The linear acceleration a of a particle in circular motion depends on the radius r of the circle and the square of the angular frequency ω, which is expressed by the formula a = rω².

Explanation:

To determine how the linear acceleration a in m/s² of a particle traveling in a circle depends on the properties such as the radius r of the circle, the angular frequency ω in s⁻¹, and the mass m of the particle, we can use dimensional analysis and known relationships from physics.

Firstly, we recognize that mass m does not directly affect a, because a depends on the force per unit mass. By considering the relationship between linear and angular variables, we know that linear velocity v is related to ω by v = rω, and the centripetal acceleration of an object moving in a circle is ac = v²/r. Substituting v = rω into the expression for centripetal acceleration, we get ac = (rω)²/r = rω².

Therefore, the linear acceleration a of a particle undergoing circular motion is directly proportional to the radius r and the square of the angular frequency ω. This relationship can be expressed as a = rω², where a is the linear acceleration, r is the radius of the circle, and ω is the angular frequency.

Answer:

Explanation:

let the ball is thrown vertically downwards with velocity u.

So, initial velocity, = - u (downwards)

acceleration = - g (downwards)

let the velocity is v after time t.

use first equation of motion

v = u + at

- v = - u - gt

v = u + gt

So, it is a straight line having slope g and y intersept is u.

The graph I shows the velocity - time graph.

Now the value of acceleration remains constant and it is equal to - 9.8 m/s^2.

So, acceleration time graph is a starigh line parallel to time axis having slope zero.

the graph II shows the acceleration - time graph.

Use III equation of motion to find the final speed in terms height.

[tex]v^{2}=u^{2}+2gh[/tex]

And the time is

v = u + gt

[tex]t=\frac{v-u}{g}[/tex]

Answer:

50 km is equivalent to 50,000 metres

Answer:

The electric flux through the car's bottom is [tex]1.75\times 10^{4} Wb[/tex]

Solution:

As per the question:

Magnitude of vertical Electric field, [tex]E_{v} = 1.85\times 10^{4} N/C[/tex]

The area of the rectangular surface of the car, [tex]A_{bottom} = 6\times 3 = 18 m^{2}[/tex]

Downward slope at an angle, [tex]\angle\theta = 19.0^{\circ}[/tex]

Now, the electric flux, [tex]\phi_{E}[/tex] is given by:

[tex]\phi_{E} = \vec{E_{v}}.\vec{A_{bottom}} = E_{v}A_{bottom}cos\theta[/tex]

Now, substituting the appropriate values in the above formula:

[tex]\phi_{E} = 1.85\times 10^{4}\times 18cos19.0^{\circ}[/tex]

[tex]\phi_{E} = 1.75\times 10^{4} Wb[/tex]

Answer:

[tex]t=d/(v_{o}*cos(\alpha ))[/tex]

[tex]y =d*tan(\alpha )- 1/2*g^{2}d^{2}/(v_{o}^{2}*(cos(\alpha ))^2)[/tex]

Explanation:

Kinematics equation in the axis X:

[tex]x=v_{o}*cos(\alpha )*t[/tex]

The projectile strikes the building at time t:

[tex]d=v_{o}*cos(\alpha )*t[/tex]

[tex]t=d/(v_{o}*cos(\alpha ))[/tex]  (1)

Kinematics equation in the axis Y:

[tex]y =v_{o}*sin(\alpha )*t - 1/2*gt^{2}[/tex]  (2)

We replace (1) in (2):

[tex]y =v_{o}*sin(\alpha )*d/(v_{o}*cos(\alpha )) - 1/2*g(d/(v_{o}*cos(\alpha )))^{2}[/tex]  

[tex]y =d*tan(\alpha )- 1/2*g^{2}d^{2}/(v_{o}^{2}*(cos(\alpha ))^2)[/tex]  (2)

Answer:

The total time to reach ground is 24.89 seconds

Explanation:

Since the height of the helicopter is given by

[tex]h(t)=3.30t^{3}[/tex] thus at time t = 2.30 seconds the height of the helicopter is

[tex]h(2.30)=3.30\times (2.30)^{3}=40.151m[/tex]

The velocity of helicopter upwards at time t = 2.30 is given by

[tex]v=\frac{dh(t)}{dt}\\\\v=\frac{d}{dt}(3.30t^{3})\\\\v(t)=9.90t^{2}\\\\\therefore v(2.30)=9.90\times (2.30)^2=120.45m/s[/tex]

Now the time after which it becomes zero can be obtained using the equations of kinematics as

1) Time taken by the mailbag to reach highest point equals

[tex]v=u+gt\\\\0=120.45-9.81\times t\\\\\therefore t_{1}=\frac{120.45}{9.81}=12.28s[/tex]

2) Time taken by the mailbag to reach ground from a height of 40.151 meters equals

[tex]s=ut+\frac{1}{2}gt^{2}\\\\40.151=120.45t+4.9t^{2}[/tex]

Solving for t we get[tex]t_{2}=0.3289secs[/tex]

Now the total time of the journey is

[tex]\\\\2\times t_{1}+t_{2}\\\\=2\times 12.28+0.3289=24.89secs[/tex]

Answer: a) 23 m/s

Explanation: In order to solve this problem we have to consider the circular movement  where the friction force  change de direcion of the velocty to keep a circular trajectory.

By using Second Newton law, we have:

F=m*acentripeta

μ*N= m*v^2/r  where N  ( equal to mg) is the normal force and μ is the coefficient of friction. r is the redius of the trajectory.

so

the maximun speed permited to keep a circular trajectory is calculated as:

v= (μ*r*g)^1/2=(0.65*80m*9.8 m/s^2)^1/2=22.57 m/s

Answer:

[tex]W = 462.5 keV[/tex]

Explanation:

As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron

So here we have to find the work done by electric field on moving electron

So we have

[tex]F = qE[/tex]

[tex]F = (1.6 \times 10^{-19})(1.85 \times 10^6)[/tex]

[tex]F = 2.96 \times 10^{-13} N[/tex]

now the distance moved by the electron is given as

[tex]d = 0.25 m[/tex]

so we have

[tex]W = F.d[/tex]

[tex]W = (1.6 \times 10^{-19})(1.85 \times 10^6)(0.25)[/tex]

[tex]W = 7.4 \times 10^{-14} J[/tex]

now we have to convert it into keV units

so we have

[tex]1 keV = 1.6 \times 10^{-16} J[/tex]

[tex]W = 462.5 keV[/tex]

Answer:

210.3 degrees

Explanation:

The net force exerted on charge A = 59.5 N

Use the x and y coordinates of net force to get the direction

arctan (y/x)

Answer:

F = -51.357i -29.958j

abs(F) = 59.45 N

Explanation:

To solve the problem we use coulomb's law with vectorial notation, F = q1*q2/(4*pi*eo*r^2) where q1 and q2 are the charges and r is the distance between them:

Point B exerts a force on A in '-i' direction  

Point C exerts a force on A in '-j' direction  

Fba = 4*7/(4*pi*eo*0.07^2) = 51.357N

Fca = 4*3/(4*pi*eo*0.06^2) = 29.958N  

F = -51.357i -29.958j

abd(F) = 59.45 N

Answer:

Explanation:

In Both Physics and Math

y=mx+b is plotted as straight line where

m=slope of line

b=intercept on Y-axis

whereas Equation of parabola is something like this

[tex]y^2=4ax[/tex]

or

[tex]x^2=4ay[/tex]

Math is a tool to solve Physics problems so equations are same in math and physics

Answer:

it take for the proton to reach the negative plate is 0.1089 ns

Explanation:

given data

area = 21 cm² = 21 × [tex]10^{-4}[/tex] m²

separated d = 3 mm = 3 × [tex]10^{-3}[/tex] m

charge Q = 9.8 nC = 9.8 × [tex]10^{-9}[/tex] C

to find out

How long does it take for the proton to reach the negative plate

solution

we know that electric field between the plate is express as

electric field E = [tex]\frac{Q}{A*\epsilon}[/tex]   ..................1

here Q is charge and A is area and ∈ is permittivity of free space

put here value

electric field E = [tex]\frac{9.8 ×10^{-9}}{21×10^{-4}*8.85×10^{-12}}[/tex]

electric filed E = 527306.968 N/C

and

we know force on proton by electric field is

Force = electric filed × charge on proton

and force according to newton law

force = mass × acceleration

so

mass × acceleration = electric filed × charge on proton

here put all value and find acceleration

1.67 × [tex]10^{-27}[/tex] × acceleration = 527306.968 × 1.6 × [tex]10^{-19}[/tex]

acceleration = 5.05 × [tex]10^{13}[/tex] m/s²

and

final velocity of proton by equation of motion

v² - u² = 2as   ..........2

here u is zero and v is  final velocity and s is distance and a is acceleration

v² - 0 = 2(5.05 × [tex]10^{13}[/tex]) (3 × [tex]10^{-3}[/tex] )

velocity = 5.50 × [tex]10^{5}[/tex] m/s

so time is

time = [tex]\frac{velocity}{acceleration}[/tex]

time = [tex]\frac{5.50*10^{5}}{5.05*10^{13}}[/tex]

time = 0.1089 × [tex]10^{-9}[/tex] s

so  it take for the proton to reach the negative plate is 0.1089 ns

In a parallel plate, the capacitor plate is parallel while the plates are separated by distance. The time required for the proton to reach the negative plate will be 0.1089×10⁻⁹ second.

What is a parallel plate capacitor?

It is a type of capacitor is in which two metal plates arranged in such a way so that they are connected in parallel and have some distance between them.

A dielectric medium is a must in between these plates helps to stop the flow of electric current through it due to its non-conductive nature.

The following date given in the problem,

A id the area of plates  = 21 cm² = 21 × 10m²

d is the distance separated= 3 mm = 3 ×10⁻³m

q is the charge on capicitor = 9.8 nC = 9.8 ×10⁻⁹C

The formula of the electric field is as

[tex]\rm E = \frac{q}{A\varepsilon } \\\\ \rm E = \frac{9.8\times10^{-9}}{ 21\times10^{-4}\times8.85\times10^{-12}} \\\\\rm E = 527306\; N/C}[/tex]

The electric force is equal to the product of the charge and the electric field. Here the electric force is balanced by mechanical force.

[tex]\rm F= qE\\\\\rm{ F= ma}\\\\qE =ma\\\\ \rm{ a= \frac{qE}{m} }\\\\\rm{ a= \frac{1.6\times10^{-9}\times527306}{1.2\times10^{-10}} }\\\\ \rm a = 5.05\times10^{-31}\;m/sec^2[/tex]

From newton's third equation of motion

As the initial velocity is zero u=0

[tex]\rm{v^2= u^2-2as} \\\\ \rm{v^2= 2\times5.05\times10^-3\times10^{-3}} \\\\ \rm{v= \sqrt{2\times5.05\times10^-3\times10^{-3}}\\\\[/tex]

[tex]\rm v = 5.50\times10^5 \;m/sec.[/tex]

Time is defined as the ratio of velocity and acceleration.

[tex]\rm t= \frac{v}{a} \\\\ \rm t= \frac{5.50\times10^5}{5.05\times10^{-3}} \\\\ \rm t= 0.1089\times 10^{-9}}\;sec[/tex]

Hence the time required for the proton to reach the negative plate will be 0.1089×10⁻⁹ second.

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Answer:

Discussed

Explanation:

Percentage error is the measure of how far is your measured value from the actual value. It give the magnitude of error in form of percentage.

it is calculated as ratio of  difference of calculated and the actual value to the actual value multiplied by 100.

If you are repeating the experiment the percent difference in your results will give the measure of how much variation there is in your own techniques, and therefore will give the precision of your experimental procedure.

Answer:

[tex]I(a,T)=\frac{2hc^2}{a^5}\frac{1}{e^{\frac{hc}{akT}}-1}[/tex]

Explanation:

The intensity of the radiation emitted by a black body with a certain temperature T and frequency [tex]\nu[/tex], is given by Planck's law:

[tex]I(\nu,T)=\frac{2h\nu^3}{c^2}(\frac{1}{e^{\frac{h\nu}{kT}}-1})[/tex]

Considering the frequency range between [tex]\nu[/tex] and [tex]\nu + \delta \nu[/tex] and [tex]dI[/tex] the amount of energy emitted in this frequency range. Since an increase in frequency means a decrease in wavelength. Then:

[tex]I(a,T)da=-I(\nu,T)d\nu\\I(a,T)=-\frac{d\nu}{da}I(\nu,T)[/tex]

Now recall that [tex]\nu=\frac{c}{a}[/tex], differentiate both sides:

[tex]d\nu=-\frac{c}{a^2}da\\\frac{d\nu}{da}=-\frac{c}{a^2}[/tex]

Replacing this in previous equation:

[tex]I(a,T)=\frac{c}{a^2}I(\nu,T)\\I(a,T)=\frac{c}{a^2}(\frac{2h\nu^3}{c^2}\frac{1}{e^{\frac{h\nu}{kT}}-1})[/tex]

Rewriting [tex]\nu^3[/tex] as [tex]\frac{c^3}{a^3}[/tex] and [tex]\nu[/tex] as [tex]\frac{c}{a}[/tex]

[tex]I(a,T)=\frac{c}{a^2}(\frac{2hc^3}{a^3c^2}\frac{1}{e^{\frac{hc}{akT}}-1})\\I(a,T)=\frac{2hc^2}{a^5}\frac{1}{e^{\frac{hc}{akT}}-1}[/tex]

Finally, we obtain Planck's radiation law in terms of wavelength

Answer:

Volume of Tank 2, V' = [tex]2.17 m^{3}[/tex]

Equilibrium Pressure, [tex]P_{eq} = 278.82 kPa[/tex]

Given:

Volume of Tank 1, V = 1 [tex]m^{3}[/tex]

Temperature of Tank 1, T = [tex]25^{\circ}C[/tex] = 298 K

Pressure of Tank 1, P = 500 kPa

Mass of air in Tank 2, m = 5 kg

Temperature of tank 2, T' = [tex]35^{\circ}C[/tex] = 303 K

Pressure of Tank 2, P' = 200 kPa

Equilibrium temperature, [tex]20^{\circ}C[/tex] = 293 K

Solution:

For Tank 1, mass of air in tank can be calculated by:

PV = m'RT

[tex]m' = \frac{PV}{RT}[/tex]

[tex]m' = \frac{500\times 1}{0.287\times 298} = 5.85 kg[/tex]

Also, from the eqn:

PV' = mRT

V' = volume of Tank 2

Thus

V' = [tex]\frac{mRT}{P}[/tex]

V' = [tex]\frac{5\times 0.287\times 303}{200} = 2.17 m^{3}[/tex]

Now,

Total Volume, V'' = V + V' = 1 + 2.17 = 3.17[tex]m^{3}[/tex]

Total air mass, m'' = m + m' = 5 + 5.85 = 10.85 kg

Final equilibrium pressure, P'' is given by:

[tex]P_{eq}V'' = m''RT_{eq}[/tex]

[tex]P_{eq} = \frac{m''RT_{eq}}{V''}[/tex]

[tex]P_{eq} = \frac{10.85\times 0.87\times 293}{3.17} = 287.82 kPa[/tex]

[tex]P_{eq} = 287.82 kPa[/tex]

Answer:

the buoyant force acting on the  ice cube is 0.119 N

Explanation:

given,

volume of ice cube = 15 cm³ = 15 × 10⁻⁶ m³

density of ice cube = 917 kg/m³

density of ethyl alcohol = 811 kg/m³

buoyant force = ?

The density of ice is more than ethyl alcohol hence it will sink.

buoyant  force acting on the ice cube = ρ V g

                                                                = 811 × 15 × 10⁻⁶ × 9.81

                                                                =0.119 N

so, the buoyant force acting on the  ice cube is 0.119 N

Answer: D) 936 K

Explanation:

Given:

Initial temperature of the gas, [tex]T = 450\ K[/tex]

Initial Pressure of the gas, [tex]P=4\ atm[/tex]

initial volume of the gas, [tex]V=10\ L[/tex]

It it given that the process is adiabatic, so for a adiabatic process we have

Let [tex]T_f \ \ and\ \ V_f [/tex] be the final temperature and volume of the gas.

[tex]T_iV_i^{\gamma -1}=T_fV_f^{\gamma -1}[/tex]

For monotomic gas [tex]\gamma=1.67[/tex]

[tex]450\times V_i^{1.67 -1} =T_f\left (\dfrac{V_i}{3} \right )^{1.67-1}\\T_f=936 K[/tex]

Hence the final temperature of the gas is 936 K. So option D is correct

Answer:

acceleration = 120 km/h²

so correct option is e 120 km/h²

Explanation:

given data

speed va = 30 km/h

speed vb = 60 km/h

time = 15 minutes

to find out

acceleration

solution

we know here car speed increases from 30 km/h to  60 km/h

sop change in speed is = vb - va

change in speed = 60 - 30 = 30  km/h

and

change in time = 15 minute = 0.25 hour

so acceleration is express as

acceleration = [tex]\frac{change in speed}{change in time}[/tex]  ........1

put here value

acceleration = [tex]\frac{30}{0.25}[/tex]

acceleration = 120 km/h²

so correct option is e 120 km/h²

Answer:51.44 units

Explanation:

Given

x component of vector is [tex]-27.3\hat{i}[/tex]

y component of vector is [tex]43.6\hat{j}[/tex]

so position vector is

[tex]r=-27.3\hat{i}+43.6\hat{j}[/tex]

Magnitude of vector is

[tex]|r|=\sqrt{27.3^2+43.6^2}[/tex]

[tex]|r|=\sqrt{2646.25}[/tex]

|r|=51.44 units

Direction

[tex]tan\theta =\frac{43.6}{-27.3}=-1.597[/tex]

vector is in 2nd quadrant thus

[tex]180-\theta =57.94[/tex]

[tex]\theta =122.06^{\circ}[/tex]

Answer:

[tex]V_{3.01}=-93.2m/s[/tex]

[tex]V_{3.005}=-93.1m/s[/tex]

[tex]V_{3.002}=-93.04m/s[/tex]

[tex]V_{3.001}=-93.02m/s[/tex]

[tex]V_{3}=-93m/s[/tex]

Explanation:

To calculate average velocity we need the position for both instants t0 and t1.

Now we will proceed to calculate all the positions we need:

[tex]Y_{3}=-99m/s[/tex]

[tex]Y_{3.01}=-99.932m/s[/tex]

[tex]Y_{3.005}=-99.4655m/s[/tex]

[tex]Y_{3.002}=-99.18608m/s[/tex]

[tex]Y_{3.001}=-99.09302m/s[/tex]

Replacing these values into the formula for average velocity:

[tex]V_{3-3.01}=\frac{Y_{3.01}-Y_{3}}{3.01-3}=-93.2m/s[/tex]

[tex]V_{3-3.005}=\frac{Y_{3.005}-Y_{3}}{3.005-3}=-93.1m/s[/tex]

[tex]V_{3-3.002}=\frac{Y_{3.002}-Y_{3}}{3.005-3}=-93.04m/s[/tex]

[tex]V_{3-3.001}=\frac{Y_{3.001}-Y_{3}}{3.001-3}=-93.02m/s[/tex]

To know the actual velocity, we derive the position and we get:

[tex]V=27-40t = -93m/s[/tex]