Which one of the following situations is NOT Possible? A body has zero velocity and non-zero acceleration. A body travels with a northward velocity and a southward acceleration. A body travels with a northward velocity and a northward acceleration. A body travels with a constant velocity and a constant non-zero acceleration. A body travels with a constant acceleration and a time-varying velocity.

Answer:

COMPOSITION

Explanation:

the correct answer is COMPOSITION.

specific heat of the body can be defined as the heat required to raise the temperature of a unit mass of a body by 1 °C.                                  

so, A composition of the material is not affected by the specific heat.

The unit of specific heat is Joule per kelvin (J / K ).

Answer: work all of the above

Explanation: kinetic, electric and  potential are energies so their units must be energy  and the Joule is.

Answer:

work all of the above

Explanation:

Answer:

Explanation:

The two charges are q and Q - q. Let the distance between them is r

Use the formula for coulomb's law for the force between the two charges

[tex]F = \frac{Kq_{1}q_{2}}{r^{2}}[/tex]

So, the force between the charges q and Q - q is given by

[tex]F = \frac{K\left ( Q-q \right )q}}{r^{2}}[/tex]

For maxima and minima, differentiate the force with respect to q.

[tex]\frac{dF}{dq} = \frac{k}{r^{2}}\times \left ( Q - 2q \right )[/tex]

For maxima and minima, the value of dF/dq = 0

So, we get

q = Q /2

Now [tex]\frac{d^{2}F}{dq^{2}} = \frac{-2k}{r^{2}}[/tex]

the double derivate is negative, so the force is maxima when q = Q / 2 .

Answer:

[tex]v_{f} =25m/s[/tex]

Explanation:

Kinematics equation for constant acceleration:

[tex]v_{f}  =v_{o} + at=15+2*5=25m/s[/tex]

Answer: 2.80 N/C

Explanation: In order to calculate the electric firld inside the solid cylinder

non conductor we have to use the Gaussian law,

∫E.ds=Q inside/ε0

E*2πrL=ρ Volume of the Gaussian surface/ε0

E*2πrL= a*r^2 π* r^2* L/ε0

E=a*r^3/(2*ε0)

E=6.2 * (0.002)^3/ (2*8.85*10^-12)= 2.80 N/C

Answer:

diameter = 9.951 × [tex]10^{-6}[/tex] m

Explanation:

given data

NA = 0.1

refractive index = 1.465

wavelength = 1.3 μm

to find out

What is the largest core diameter for which the fiber remains single-mode

solution

we know that for single mode v number is

V ≤ 2.405

and v = [tex]\frac{2*\pi *r}{ wavelength} NA[/tex]

here r is radius    

so we can say

[tex]\frac{2*\pi *r}{ wavelength} NA[/tex]    = 2.405

put here value

[tex]\frac{2*\pi *r}{1.3*10^{-6}} 0.1[/tex]    = 2.405

solve it we get r

r = 4.975979 × [tex]x^{-6}[/tex] m

so diameter is = 2  ×  4.975979 × [tex]10^{-6}[/tex] m

diameter = 9.951 × [tex]10^{-6}[/tex] m

Answer:

11.27 m /s

2.98 m / s.

Explanation:

80 km / h = 22.22 m /s

Tanq = 4 / 100

Sinq = .0399

Deceleration acting on inclined plane = g sinq

= 9.8 x .0399

= .3910

Initial speed u = 22.22 m/s

acceleration = - .3910 ms⁻²

v = u - a t

= 22.22 - .3910 x 28

= 22.22 - 10.95

= 11.27 m /s

b ) v² = u² - 2 a s

v² = ( 22.22) ² - 2 x .3910 x 620

= 493.7284 - 484.84

= 8.8884

v = 2.98 m / s.

Answer:

Speed, u = 29.4 m/s

Explanation:

Given that, A ball thrown straight up climbs for 3.0 sec before falling, t = 3 s

Let u is speed with which the ball is thrown up. When the ball falls, v = 0

Using first equation of motion as :

v = u + at

Here, a = -g

So, u = g × t

[tex]u=9.8\times 3[/tex]

u = 29.4 m/s

So, the speed with which the ball was thrown is 29.4 m/s. Hence, this is the required solution.

The velocity at which the ball was thrown is 29.4 m/s.

To calculate the velocity at which the ball was thrown, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, The velocity at which the ball was thrown is 29.4 m/s.

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Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×[tex]10^{-4}[/tex] m

dynamic viscosity = 1.75 ×[tex]10^{-5}[/tex] Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ [tex]\frac{du}{dy}[/tex]

so

= µ  [tex]\frac{v}{h}[/tex]   ............1

put here value

= 1.75×[tex]10^{-5}[/tex] × [tex]\frac{v}{10^{-4}}[/tex]

= 0.175 v

and

area between air and puck is given by

Area = [tex]\frac{\pi }{4} d^{2}[/tex]

area  =  [tex]\frac{\pi }{4} 0.1^{2}[/tex]

area = 7.85 × [tex]\frac{v}{10^{-3}}[/tex] m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × [tex]10^{-3}[/tex]

force = 1.374 × [tex]10^{-3}[/tex] v    

and now apply newton second law

force = mass × acceleration

- force = [tex]mass \frac{dv}{dt}[/tex]

- 1.374 × [tex]10^{-3}[/tex] v = [tex]0.03 \frac{0.9v - v }{t}[/tex]

t =  [tex] \frac{0.1 v * 0.03}{1.37*10^{-3} v}[/tex]

time = 2.18

so time required after impact for a puck is 2.18 seconds

To calculate the time required for the puck to lose 10% of its initial speed, you can use the equation for deceleration. First, find the final velocity of the puck using the given 10% decrease. Then, calculate the acceleration of the puck using the equation for acceleration. Finally, substitute the acceleration back into the equation for time to find the answer.

To calculate the time required for the puck to lose 10% of its initial speed, we need to find the deceleration of the puck. We can use the equation for deceleration, which is a = (v_f - v_i) / t, where a is the acceleration, v_f is the final velocity, v_i is the initial velocity, and t is the time. In this case, since the puck is losing speed, we can use the negative value of the acceleration.

Given that the initial speed of the puck is v_i = (2 * distance) / t, we can calculate the final velocity v_f = 0.9 * v_i, where 0.9 represents the 10% decrease. Substituting these values into the deceleration equation, we can solve for t as follows:

t = (v_f - v_i) / a = (0.9 * v_i - v_i) / (-a) = 0.1 * v_i / a.

Now we can find the acceleration by using the equation a = (6 * π * η * r) / (m * v_i), where η is the dynamic viscosity of air, r is the radius of the puck (which is half the diameter), m is the mass of the puck, and v_i is the initial velocity. Substituting the given values, we can calculate the acceleration. Finally, substituting the acceleration back into the equation for t, we can calculate the time required for the puck to lose 10% of its initial speed.

Answer:

E=[8.1X-9.63Y]*10^{3}N/m

Explanation:

Field in the point is the sum of the point charge electric field and the field of the infinite line.

First, we calculate the point charge field:

[tex]E_{Charge}=\frac{1}{4\pi \epsilon_0}  *\frac{Q}{||r_p -r||^2} *Unitary vector\\||r_p -r||^2=(x_p-x)^2+(y_p -y)^2=1.25 m^2\\Unitary Vector=\frac{(r_p -r)}{||r_p -r||}=\frac{(x_p-x)X+(y_p -y)Y}{||r_p -r||}\\=\frac{2\sqrt{5} }{5}X- \frac{\sqrt{5} }{5}Y \\E_{Charge}=K*\frac{3\mu C}{1.25m^2}*(\frac{2\sqrt{5} }{5}X- \frac{\sqrt{5} }{5}Y)[/tex]

It is vectorial, where X and Y represent unitary vectors in X and Y. we recall the Coulomb constant k=[tex]\frac{1}{4\pi \epsilon_0}[/tex] and not replace it yet. Now we compute the line field as follows:

[tex]E_{Line}=\frac{\lambda}{2\pi \epsilon_0 distance} *Unitary Vector\\Unitary Vector=X[/tex] (The field is only in the perpendicular direction to the wire, which is X)

[tex]E_{Line}=\frac{-3\mu C/m*2}{2*2\pi \epsilon_0 5*m}X=K*\frac{6\mu C/m}{ 5*m}(-X)[/tex]

We multiplied by 2/2 in order to obtain Coulomb constant and express it that way. Finally, we proceed to sum the fields.

[tex]E=K*\frac{3\mu C}{1.25m^2}*(\frac{2\sqrt{5} }{5}X- \frac{\sqrt{5} }{5}Y)+K*\frac{6\mu C}{ 5*m^2}(-X)\\E=K*[2.15-1.2]X-K*[1.07]Y \mu N/m\\E=K*[0.9X-1.07Y] \mu C/m^2\\E=[8.1X-9.63Y]*10^{3}N/m[/tex]

Answer:

The speed is 33.5 m/s.

Explanation:

Given that,

Mass = 0.064 kg

Wavelength [tex]\lambda= 3.09\times10^{-34}\ m[/tex]

We need to calculate the speed

Using formula of he de Broglie wavelength

[tex]\lambda=\dfrac{h}{mv}[/tex]

[tex]v=\dfrac{h}{m\lambda}[/tex]

Where, h = Planck constant

m = mass

[tex]\lambda[/tex] = wavelength

Put the value into the formula

[tex]v = \dfrac{6.63\times10^{-34}}{0.064\times 3.09\times10^{-34}}[/tex]

[tex]v=33.5\ m/s[/tex]

Hence, The speed is 33.5 m/s.

Answer:

Explanation:

If we assume there is a sharp boundary between the two masses of air, there will be a refraction. The refractive index of each medium will depend on the relative speeds of light.

n = c / v

If light travels faster in warmer air, it will have a lower refractive index

nh < nc

Snell's law of refraction relates angles of incidence and refracted with the indexes of refraction:

n1 * sin(θ1) = n2 * sin(θ2)

sin(θ2) = sin(θ1) * n1/n2

If blue light from the sky passing through the hot air will cross to the cold air, then

n1 = nh

n2 = nc

Then:

n1 < n2

So:

n1/n2 < 1

The refracted light will come into the cold air at angle θ2 wich will be smaller than θ1, so the light is bent upwards, creating the appearance of water in the distance, which is actually a mirror image of the sky.

Explanation:

Given that,

Radius a= 5.0 mm

Radial distance r= 0, 2.9 mm, 5.0 mm

Current density at the center of the wire is given by

[tex]J_{0}=420\ A/m^2[/tex]

Given relation between current density and radial distance

[tex]J=\dfrac{J_{0}r}{a}[/tex]

We know that,

When the current passing through the wire changes with radial distance,

then the magnetic field is induced in the wire.

The induced magnetic field is

[tex]B=\dfrac{\mu_{0}i_{ind}}{2\pi r}[/tex]...(I)

We need to calculate the induced current

Using formula of induced current

[tex]i_{ind}=\int_{0}^{r}{J(r)dA}[/tex]

[tex]i_{ind}= \int_{0}^{r}{\dfrac{J_{0}r}{a}2\pi r}[/tex]

[tex]i_{ind}={\dfrac{2\pi J_{0}}{a}}\int_{0}^{r}{r^2}[/tex]

[tex]i_{ind}={\dfrac{2\pi J_{0}}{a}}{\dfrac{r^3}{3}}[/tex]

We need to calculate the magnetic field

Put the value of induced current in equation (I)

[tex]B=\dfrac{\mu_{0}{\dfrac{2\pi J_{0}}{a}}{\dfrac{r^3}{3}}}{2\pi r}[/tex]

[tex]B=\dfrac{\mu_{0}J_{0}r^2}{3a}[/tex]

(a). The  magnetic field at a distance r = 0

Put the value into the formula

[tex]B=\dfrac{4\pi\times10^{-7}\times420\times0}{3\times5.0\times10^{-3}}[/tex]

[tex]B = 0[/tex]

The  magnetic field at a distance 0 is zero.

(b). The  magnetic field at a distance r = 2.9 mm

[tex]B=\dfrac{4\pi\times10^{-7}\times420\times(2.9\times10^{-3})^2}{3\times5.0\times10^{-3}}[/tex]

[tex]B = 2.95\times10^{-7}\ T[/tex]

The  magnetic field at a distance 2.9 mm is [tex]2.95\times10^{-7}\ T[/tex]

(c). The  magnetic field at a distance r = 5.0 mm

[tex]B=\dfrac{4\pi\times10^{-7}\times420\times(5.0\times10^{-3})^2}{3\times5.0\times10^{-3}}[/tex]

[tex]B = 8.79\times10^{-7}\ T[/tex]

The  magnetic field at a distance 5.0 mm is [tex]8.79\times10^{-7}\ T[/tex]

Hence, This is the required solution.

Answer:

The image height is 0.00283 m or 0.283 cm and is inverted due to the negative sign

Explanation:

u = Object distance =  9 m

v = Image distance = 1.7 cm (as the image is forming on the retina)

[tex]h_u[/tex]= Object height = 1.5 m

Magnification

[tex]m=-\frac{v}{u}=\frac{h_v}{h_u}\\\Rightarrow -\frac{v}{u}=\frac{h_v}{h_u}\\\Rightarrow -\frac{0.017}{9}=\frac{h_v}{1.5}\\\Rightarrow h_v=-0.00283\ m[/tex]

The image height is 0.00283 m or 0.283 cm and is inverted due to the negative sign

The image formed on the retina is 1.8 cm behind the lens.

The image formed on the retina can be calculated using the lens formula: 1/f = 1/do + 1/di, where f is the focal length of the lens, do is the object distance from the lens, and di is the image distance.

In this case, the lens-to-retina distance is given as 2.00 cm, so di = -2.00 cm. The object distance do can be calculated as the difference between the person's height and the distance at which they are standing: do = 9.0 m - 1.5 m = 7.5 m. Substituting the values into the lens formula, we get 1/2.00 cm = 1/7.5 m + 1/di. Solving for di, we find that the image is formed 1.8 cm behind the lens.

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Answer:

Mass of each objects, m = 185.69 kg

Explanation:

The gravitational force between two equal masses is, [tex]F=2.3\times 10^{-8}\ N[/tex]

Separation between the masses, d = 10 m

Let m is the mass of each object. The gravitational force is given by :

[tex]F=G\dfrac{m^2}{d^2}[/tex]

[tex]m=\sqrt{\dfrac{F.d^2}{G}}[/tex]

[tex]m=\sqrt{\dfrac{2.3\times 10^{-8}\times (10)^2}{6.67\times 10^{-11}}}[/tex]

m = 185.69 kg

So, the mass of each objects is 185.69 kg. Hence, this is the required solution.

To find the mass of each object, we can use Newton's law of gravitation and solve an equation.

To find the mass of each object, we can use Newton's law of gravitation, which states that the gravitational force between two objects is given by the equation: F = G * M₁ * M₂ / R². Rearranging the equation to solve for the mass, we have: M = F * R² / (G * M₂), where M is the mass of one object.

Using the given values, the gravitational force is 2.30 x 10^‐8 N and the distance between the objects is 10.0 m. Plugging these values into the equation, we have: M = (2.30 x 10^‐8 N) * (10.0 m)² / (6.67 × 10-¹¹ Nm²/kg² * M₂).

Since the masses of the two objects are equal, we can assume M₁ = M₂. Therefore, the mass of each object would be the same and can be calculated by solving the equation.

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Answer:

q₃ is located at X= - 0.144m

Explanation:

Coulomb law:

Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

F=K*q₁*q₂/d² Formula (1)  

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)  

d: distance between the charges in meters(m)

Equivalence  

1uC= 10⁻⁶C

Data

F=0.66 N

K=8.99x10⁹N*m²/C²

q₁ =  +3.00 μC = +3.00  *10⁻⁶C

q₂=  -5.00  μC =  -5.00  *10⁻⁶C

q₃= -8.00 μC= -8.00 *10⁻⁶C

d₁₂= 0.2m: distance between q₁ and q₂

Fn=7 N: Net force on q₁

Calculation of the magnitude of the forces exerted on q₁

F₂₁ = K*q₁*q₂/d₁₂² = (8.99 * 10⁹ * 3.00 * 10⁻⁶ * 5.00 * 10⁻⁶)/(0.2)² = 3.37 N

F₃₁ = K*q₁*q₃/d₁₃² = (8.99 * 10⁹ * 3.00 * 10⁻⁶ * 8.00 * 10⁻⁶)/(d₁₃)² = 0.21576/(d₁₃)² N

Calculation of the distance d₁₃ between q₁ and q₃

In order for the net force to be negative on the x axis, the charge q₃ must be located on the x axis to the left of the charge q₁

F₂₁: It is a force of attraction and goes to the right(+)

F₃₁: It is a force of attraction and goes to the left(-)

-Fn₁ = F₂₁ - F₃₁

-7 = 3.37 - 0.21576/(d₁₃)²

3.37 + 7 = 0.21576/(d₁₃)²

10.37*(d₁₃)²=0.21576

(d₁₃)² = (0.215769 )/ (10.37)

(d₁₃)² =  20.8 * 10⁻³

[tex]d_{13} = \sqrt{20.8 * 10^{-3}} = 0.144 m[/tex]

d₁₃=0.144m

q₃ is located at X= - 0.144m

Answer:

The area of oil slick is calculate as [tex]785.7 m^{2}[/tex]

Solution:

Volume of oil slick, [tex]V_{o} = 1.1 m^{3}[/tex]

The thickness of one molecule on a side, w = 1.4 mm = [tex]1.4\times 10^{- 3}[/tex]

Now, in order to determine the area of oil slick, [tex]A_{o}[/tex]:

Volume, V = [tex]Area\times thickness[/tex]

Thus

[tex]Area,\ A_{o} = \frac{V_{o}}{w}[/tex]

[tex]Area,\ A_{o} = \frac{1.1}{1.4\times 10^{- 3}} = 785.7 m^{2}[/tex]

Answer:

t = 8.45 sec

car distance d = 132.09  m

bike distance d = 157.08 m

Explanation:

GIVEN :

motorcycle is 25 m behind the car , therefore distance need to covered by bike to overtake car is 25+ d, when car reache distance d at time t

for car

by equation of motion

[tex]d  = ut + \frac{1}{2}at^2[/tex]

u = 0 starting from rest

[tex]d = \frac{1}{2}at^2[/tex]

[tex]t^2 = \frac{2d}{a}[/tex]

for bike

[tex]d+25 = 0 + \frac{1}{2}*4.40t^2[/tex]

[tex]t^2= \frac{d+25}{2.20}[/tex]

equating time of both

[tex]\frac{2d}{a} = \frac{d+25}{2.20}[/tex]

solving for d we get

d = 132 m

therefore t is[tex] = \sqrt{\frac{2d}{a}}[/tex]

[tex]t =  \sqrt{\frac{2*132}{3.70}}[/tex]

t = 8.45 sec

each travelled in time 8.45 sec as

for car

[tex]d = \frac{1}{2}*3.70 *8.45^2[/tex]

d = 132.09  m

fro bike

[tex]d = \frac{1}{2}*4.40 *8.45^2[/tex]

d = 157.08 m

To find the time when the motorcycle overtakes the car, equate the distances they have traveled given their individual accelerations and solve for time. Once the time is known, calculate the distance each has traveled using the equations of motion for uniform acceleration.

To determine when the motorcycle (MC) overtakes the car, we need to calculate the time at which both have traveled the same distance, considering the initial 25.0 m advantage of the car. We can use the equation of motion d = ut + (1/2)at2 where d is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.

The car starts from rest, so its initial velocity is 0, and it accelerates at 3.70 m/s2. The motorcycle also starts from rest, with an acceleration of 4.40 m/s2, but it needs to cover an additional 25.0 m to catch up with the car.

We can set the equations equal to each other to find the time t when the distances are equal:

To find the time when MC overtakes the car, we equate Car_d to MC_d and solve for t:

(1/2)(3.70)t2 = 25.0 + (1/2)(4.40)t2

After solving for t, we can calculate the distance each has traveled using the original equations of motion for uniform acceleration.

Answer:

Electric Field at a distance d from one end of the wire is [tex]E=\dfrac{Q}{4\pi \epsilon_0(L+d)d}[/tex]

Electric Field when d is much grater than length of the wire =[tex]\dfrac{Q}{4\pi \epsilon_0\ d^2}[/tex]

Explanation:

Given:

Let dE be the Electric Field due to the small elemental charge on the wire at a distance x from the one end of the wire and let [tex]\lambda[/tex] be the charge density of the wire

[tex]E=\dfrac{dq}{4\pi \epsilon_0x^2}[/tex]

Now integrating it over the entire length varying x from x=d to x=d+L we have and replacing [tex]\lambda=\dfrac{Q}{L}[/tex] we have

[tex]E=\int\dfrac{\lambda dx}{4\pi \epsilon_0x^2}\\E=\dfrac{Q}{4\pi \epsilon_0 (L+d)(d)}[/tex]

When d is much greater than the length of the wire then we have

1+\dfrac{L]{d}≈1

So the Magnitude of the Electric Field at point P = [tex]\dfrac{Q}{4\pi \epsilon_0\ d^2}[/tex]

Answer:

Th magnitude of each Force will be [tex]=62.35\times10^{-3}\ \rm N[/tex]

Explanation:

Given:

Since all the charges are identical and distance between them is same so magnitude of the force between each charge is equal.

Let F be the force between the particles. According to Coulombs Law we have

[tex]F=\dfrac{kQ^2}{L^2}\\=\dfrac{9\times10^9\times (2\times10^{-6})^2}{1^2}\\F=36\times10^{-3}\ \rm N[/tex]

Now the the force on any charge by other two charges will be F and the angle between the two force is [tex]60^\circ[/tex]

Let [tex]F_{resultant}[/tex] be the force on nay charge by other two

By using vector Law of addition we have

[tex]F_{resultant}=\sqrt{(F^2+F^2+2F\times F \times cos60^\circ)}\\=\sqrt{3}F\\=\sqrt{3}\times36\times10^{-3}\ \rm N\\=62.35\times10^{-3}\ \rm N[/tex]

The angle made by the resultant vector will be

[tex]\tan\beta=\dfrac{F\sin60^\circ}{F+F\cos60^\circ}\\\beta=30^\circ[/tex]

Answer:

1.89 seconds

Explanation:

t = Time taken

u = Initial velocity = 9.6 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

[tex]v=u+at\\\Rightarrow 0=9.6-9.8\times t\\\Rightarrow \frac{-9.6}{-9.8}=t\\\Rightarrow t=0.97 \s[/tex]

Time taken to reach maximum height is 0.97 seconds

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=9.6\times 0.97+\frac{1}{2}\times -9.8\times 0.97^2\\\Rightarrow s=4.7\ m[/tex]

So, the stone would travel 4.7 m up

So, total height ball would fall is 4.7+12.8 = 17.5 m

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 17.5=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{17.5\times 2}{9.8}}\\\Rightarrow t=1.89\ s[/tex]

Time taken by the stone to travel 17.5 m is 1.89 seconds

Answer:

No. of laps of Hannah are 7 (approx).

Solution:

According to the question:

The total distance to be covered, D = 5000 m

The distance for each lap, x = 400 m

Time taken by Kara, [tex]t_{K} = 17.9 min = 17.9\times 60 = 1074 s[/tex]

Time taken by Hannah, [tex]t_{H} = 15.3 min = 15.3\times 60 = 918 s[/tex]

Now, the speed of Kara and Hannah can be calculated respectively as:

[tex]v_{K} = \frac{D}{t_{K}} = \frac{5000}{1074} = 4.65 m/s[/tex]

[tex]v_{H} = \frac{D}{t_{H}} = \frac{5000}{918} = 5.45 m/s[/tex]

Time taken in each lap is given by:

[tex](v_{H} - v_{K})t = x[/tex]

[tex](5.45 - 4.65)\times t = 400[/tex]

[tex]t = \frac{400}{0.8}[/tex]

t = 500 s

So, Distance covered by Hannah in 't' sec is given by:

[tex]d_{H} = v_{H}\times t[/tex]

[tex]d_{H} = 5.45\times 500 = 2725 m[/tex]

No. of laps taken by Hannah when she passes Kara:

[tex]n_{H} = \frac{d_{H}}{x}[/tex]

[tex]n_{H} = \frac{2725}{400} = 6.8[/tex] ≈ 7 laps

The number of laps that Hannah has run when she passes Kara is 7 laps.

The speed of each athlete is calculated as follows;

Kara = (5000) / (17.9 x 60) = 4.66 m/s

Hannah = (5000) / (15.3 x 60) = 5.47 m/s

The time taken in each lap if Hannah passes kara is calculated as follow;

(5.47 - 4.66)t = 400

0.81t = 400

t = 493.83 s

Distance covered by Hannah when she passes kara;

d = 493.83 x 5.47 = 2,701.25 m

n = 2,701.25/400

n = 6.8 ≈ 7 laps.

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Answer:

it increases-

Explanation:

When the mass of a rocket decreases as it burns through its fuel and the force ( thrust) is constant then by newtons second law of motion

F= ma  here F is constant this means that   ma= constant

⇒ m= F /a    this implies that mass is inversely proportional to  acceleration.

its means when the mass decreases the acceleration must increase. hence the acceleration increases

Answer:

a)the ball will leave the bat at an angle of  61.3°  .

b) the velocity at which it will hit the ground will be v = 27.1 m/s

Explanation:

Given,

v = 47.24 m

h = 0.42 m

t = 5.73 s

R = 130 m

a)We know that

R = v cosθ × t

cosθ = [tex]\dfrac{R}{v t } = \dfrac{130}{47.24\times 5.73 } =0.4803[/tex]

θ = 61.3°  

the ball will leave the bat at an angle of  61.3°  .

b)Vx = v cos(θ) = 47.24 x cos 61.3 = 22.7 m/s

v = u + at

Vy = 47.24 x sin 61.3 - 9.81 x 5.73

    = -14.8 m/s

v = [tex]\sqrt{v_x^2 + v_y^2)}[/tex]

v = [tex]\sqrt{22.7^2 + -14.8^2}[/tex]

v = 27.1 m/s

the velocity at which it will hit the ground will be v = 27.1 m/s

Answer:

Explanation:

Given

Horizontal bar rises with 300 mm/s

Let us take the horizontal component of P be

[tex]P_x=rcos\theta [/tex]

[tex]P_y=rsin\theta [/tex]

where [tex]\theta [/tex]is angle made by horizontal bar with x axis

Velocity at y=150 mm

[tex]150=300sin\theta [/tex]

thus [tex]\theta =30^{\circ}[/tex]

position of[tex]P_x=rcos\theta =300\cdot cos30=300\times \frac{\sqrt{3}}{2}[/tex]

[tex]P_x=259.80 mm[/tex]

[tex]P=259.80\hat{i}+150\hat{j}[/tex]

Velocity at this instant

[tex]u_x=-rsin\theta =300\times sin30=-150 mm/s[/tex]

[tex]u_y=rcos\theta =300\times cos30=259.80 mm/s[/tex]

Answer:

a) 0.018 kg

b) 262 kPa

Explanation:

The volume of the concentric cylinders would be:

V = π/4 * h * (D^2 - d^2)

V = π/4 * 13 * (52^2 - 33^2) = 16500 cm^3 = 0.0165 m^3

The state equation of gases:

p * V = m * R * T

Rearranging:

m = (p * V) / (R * T)

R is 287 J/(kg * K) for air

25 C = 298 K

m0 = 202000 * 0.0165 / (287 * 298) = 0.039 kg

After pumping more air the volume remains about the same, but temperature and pressure change.

30 C = 303 K

m1 = 303000 * 0.0165 / (287 * 303) = 0.057 kg

The mass that was added is

m1 - m0 = 0.057 - 0.039 = 0.018 kg

If that air is cooled to 0 C

0 C  is 273 K

p = m * R * T / V

p = 0.057 * 278 * 273 / 0.0165 = 262000 Pa = 262 kPa

Answer:

Slope of position time is velocity

Explanation:

The position time graph means a relation between the position of the object and the time which is represent on a graph.

The graph line shows that how the position of an object changes with respect to time.

The slope of the position time graph shows the rate of change of position of the object with respect to time.

The rate of change of position with respect to time is called velocity.

thus, the slope of position time graph gives the velocity of the object.

Answer:

So the car displacement after 3.7 sec is 0.030 km

Explanation:

We have given initial velocity u = 6.64 m/sec

Acceleration [tex]a=0.85m/sec^2[/tex]

Time t = 3.7 sec

Final velocity v = 9.8 m/sec

We have to find the displacement after that time

From second equation of motion we know that [tex]s=ut+\frac{1}{2}at^2[/tex], here s is displacement, u is initial velocity, t is time , and a is acceleration

So displacement [tex]s=ut+\frac{1}{2}at^2=6.64\times 3.7+\frac{1}{2}\times 0.85\times 3.7^2=30.386m[/tex]

We know that 1 km = 1000 m

So 30.386 m = 0.030 km

Answer:

The  displacement of car after that time is 30.56 m.

Explanation:

Given that,

Initial velocity = 6.64 m/s

Acceleration = 0.85 m/s²

Time = 3.7 s

Final velocity = 9.8 m/s

We need to calculate the displacement

Using equation of motion

[tex]v^2=u^2+2as[/tex]

[tex]s=\dfrac{v^2-u^2}{2a}[/tex]

Put the value into the formula

[tex]s=\dfrac{9.8^2-6.64^2}{2\times0.85}[/tex]

[tex]s =30.56\ m[/tex]

Hence, The  displacement of car after that time is 30.56 m.

Explanation:

Given that,

Distance = 402000 km

Speed = 2.23 m/s

Angle = 150

(a). We need to calculate the eccentricity of the trajectory

Using formula of eccentricity

[tex]\epsilon=\dfrac{v^2}{2}-\dfrac{\mu}{r}[/tex]

Put the value into the formula

[tex]\epsilon=\dfrac{2.23^2}{2}-\dfrac{398600}{402000}[/tex]

[tex]\epsilon=1.4949\ km^2/s^2[/tex]

We need to calculate the angular momentum

Using formula of  the angular momentum

[tex]h^2=-\dfrac{1}{2}\dfrac{\mu^2}{\epsilon}(1-e^2)[/tex]

[tex]h^2=-\dfrac{1}{2}\dfrac{(398600)^2}{1.4949}(1-e^2)[/tex]

[tex]h^2=-5.3141\times10^{10}(1-e^2)[/tex]...(I)

The orbit equation is

[tex]h^2=\mu r(1+e\cos\theta)[/tex]

[tex]h^2=398600\times402000(1-+\cos150)[/tex]

[tex]h^2=16.02372\times10^{10}(1-e0.8660)[/tex]

[tex]h^2=16.02372\times10^{10}-e13.877\times10^{10}[/tex]....(II)

Equating the value of h²

[tex]-5.3141\times10^{10}(1-e^2)=16.02372\times10^{10}-e13.877\times10^{10}[/tex]

[tex]-5.3141+5.3141e^2=13.877e+16.02372[/tex]

[tex]5.3141e^2+13.877e-21.33782=0[/tex]

[tex]e = 0, 1.086[/tex]

(b). We need to calculate the altitude at closest approach

Put the value of e in equation (I)

[tex]h^2=16.02372\times10^{10}-1.086\times13.877\times10^{10}[/tex]

[tex]h^2=9.53298\times10^{9}\ km^4/s^2[/tex]

Now, using the formula of the altitude at closest

[tex]r_{perigee}=\dfrac{h^2}{\mu}\dfrac{1}{1+e}[/tex]

[tex]r_{perigee}=\dfrac{9.53298\times10^{9}}{398600}\dfrac{1}{1+1.086}[/tex]

[tex]r_{perigee}=11465\ km[/tex]

So, The altitude is

[tex]z_{perigee}=r_{perigee}-r_{earth}[/tex]

[tex]z_{perigee}=11465-6378[/tex]

[tex]z_{perigee}=5087\ km[/tex]

(c). We need to calculate the  speed at the closest approach.

Using formula of speed

[tex]v_{perigee}=\dfrac{h}{r_{perigee}}[/tex]

[tex]v_{perigee}=\dfrac{\sqrt{9.53298\times10^{9}}}{11465}[/tex]

[tex]v_{perigee}=8.516\ km/s[/tex]

Hence, This is the required solution.

(a) The eccentricity of the meteoroid's trajectory is approximately -0.226. This value indicates the deviation of the orbit from a perfect circle, with negative eccentricity suggesting an elliptical orbit.

(b) At its closest approach, or perigee, the meteoroid is at an altitude of about 595,261 kilometers from the Earth's center. This is the point in its trajectory where it is closest to Earth.

(c) The speed of the meteoroid at its closest approach is approximately 3.69 kilometers per second. This velocity is characteristic of its orbital motion when it is nearest to Earth and is determined by the specific orbital energy and gravitational influences.

To calculate the eccentricity (e), altitude at closest approach (perigee), and speed at the closest approach, we can use the vis-viva equation and orbital mechanics. The vis-viva equation relates the specific orbital energy (ε) to the semi-major axis (a), eccentricity (e), and velocity (v) of an object in orbit:

ε = v²/2 - μ/a

Where:

ε = specific orbital energy

v = velocity of the object

μ = standard gravitational parameter of Earth (approximately 3.986 x 10^5 km³/s²)

a = semi-major axis of the trajectory

We are given the speed of the meteoroid at a distance of 402,000 km from the Earth's center, which we'll use as v. We'll also use μ for Earth's gravitational parameter.

First, let's find the semi-major axis (a):

ε = v²/2 - μ/a

Rearrange to solve for a:

a = μ / (2μ/v² - 1/v²)

a = (3.986 x 10^5 km³/s²) / (2 * (3.986 x 10^5 km³/s²) / (2.23 km/s)² - (1 / (2.23 km/s)²))

a ≈ 491,453 km

(a) The semi-major axis (a) is approximately 491,453 km.

Next, let's calculate the eccentricity (e) using the true anomaly (ν) and the semi-major axis:

e = cos(ν) - r/a

e = cos(150°) - (402,000 km) / (491,453 km)

e ≈ -0.226

(b) The eccentricity (e) is approximately -0.226.

Now, to find the altitude at closest approach (perigee), we need to calculate the perigee distance (rp) using the semi-major axis and eccentricity:

rp = a(1 - e)

rp = (491,453 km) * (1 - (-0.226))

rp ≈ 595,261 km

(c) The perigee distance is approximately 595,261 km.

To find the speed at closest approach (closest approach velocity), we can use the vis-viva equation again:

v = sqrt(μ * (2/r - 1/a))

Where r is the distance from the center of the Earth (rp at perigee).

v = sqrt((3.986 x 10^5 km³/s²) * (2 / (595,261 km) - 1 / (491,453 km)))

v ≈ 3.69 km/s

(c) The speed at closest approach is approximately 3.69 km/s.

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Answer:

0.224 psi

Explanation:

The pressure using a differential manometer is calculated with the delta H.

Delta H = 293 - 275 = 18 mm

The formula for the pressure is:

P = rho * g * h,

where rho : density of the fluid inside the manometer

g : gravitational acceleration

h : delta H inside the manometer.

It is importar the use of units.

8.75 g/cm3 = 8750 kg/m3

g = 9.8 m/s2

h = 18 mm = 0.018 m

P = 1543,5 Pa ;  1 psi = 6894.8 Pa

P = 1543,5/6894,8 = 0.224 psi