A glass vessel that can be repeatedly filled with precisely the same volume of liquid is called a pycnometer. A certain pycnometer, when empty and dry, weighed 25.296 g. When filled with water at 25 oC the pycnometer and water weighed 34.914 g. When filled with a liquid of unknown composition the pycnometer and its contents weighed 33.485 g. At 25 oC the density of water is 0.9970 g/ml. What is the density of the unknown liquid?

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:

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

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:

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: 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

Explanation:

Given that,

Weight of the astronaut on the surface of Earth, W = 450 N

We know that the acceleration due to gravity on the earth is, [tex]g=9.8\ m/s^2[/tex]

Weight of an object is given by, W = mg

[tex]m=\dfrac{W}{g}[/tex]

[tex]m=\dfrac{450\ N}{9.8\ m/s^2}[/tex]

m = 45.91 kg

Also, the acceleration due to gravity on the surface of moon is one -sixth of the acceleration due to gravity on the surface of Earth, [tex]g'=1.62\ m/s^2[/tex]

As mass remains constant. So, the weight on the moon is :

W' = mg'

[tex]W'=45.91\times 1.62[/tex]

W' = 74.37 N

Hence, this is the required solution.

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:

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:

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]

The charge on each ball is approximately [tex]\( 8.08 \times 10^{-9} \) C[/tex].

Step 1

To determine the charge on each ball, let's analyze the forces acting on one of the charged balls when they are in equilibrium. The forces are:

1. Gravitational force [tex](\( F_g \))[/tex] downward.

2. Tension [tex](\( T \))[/tex] in the thread.

3. Electrostatic force [tex](\( F_e \))[/tex] between the two charges.

Step 2

Given:

Length of the thread [tex](\( L \)) = 0.26 m[/tex]

Mass of each ball [tex](\( m \)) = \( 7.75 \times 10^{-4} \) kg[/tex]

Angle between the threads = 35°, so each thread makes an angle of 17.5° with the vertical.

Step 3

First, calculate the gravitational force:

[tex]\[ F_g = mg = 7.75 \times 10^{-4} \, \text{kg} \times 9.8 \, \text{m/s}^2 = 7.595 \times 10^{-3} \, \text{N} \][/tex]

The electrostatic force must balance the horizontal component of the tension:

[tex]\[ F_e = T \sin(17.5^\circ) \][/tex]

The vertical component of the tension balances the gravitational force:

[tex]\[ T \cos(17.5^\circ) = F_g \][/tex]

[tex]Solving for \( T \):[/tex]

[tex]\[ T = \frac{F_g}{\cos(17.5^\circ)} = \frac{7.595 \times 10^{-3} \, \text{N}}{\cos(17.5^\circ)} \approx 7.99 \times 10^{-3} \, \text{N} \][/tex]

Step 4

Now calculate the horizontal component of the tension:

[tex]\[ T \sin(17.5^\circ) = 7.99 \times 10^{-3} \, \text{N} \times \sin(17.5^\circ) \approx 2.41 \times 10^{-3} \, \text{N} \][/tex]

Step 5

This is the electrostatic force:

[tex]\[ F_e = \frac{k q^2}{r^2} \][/tex]

where [tex]\( r \)[/tex] is the distance between the balls.

Step 6

To find [tex]\( r \)[/tex]:

[tex]\[ r = 2L \sin(17.5^\circ) = 2 \times 0.26 \, \text{m} \times \sin(17.5^\circ) \approx 0.156 \, \text{m} \][/tex]

Using Coulomb's law:

[tex]\[ 2.41 \times 10^{-3} \, \text{N} = \frac{(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) q^2}{(0.156 \, \text{m})^2} \][/tex]

Step 7

Solving for [tex]\( q \)[/tex]:

[tex]\[ q^2 = \frac{2.41 \times 10^{-3} \, \text{N} \times (0.156 \, \text{m})^2}{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2} \approx 6.53 \times 10^{-17} \, \text{C}^2 \][/tex]

[tex]\[ q \approx \sqrt{6.53 \times 10^{-17}} \approx 8.08 \times 10^{-9} \, \text{C} \][/tex]

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:

[tex]f_{beat}=f_2-f_1=353Hz -350Hz=3Hz[/tex]

Explanation:

When there are two waves with similar frequency, the superposition of these waves create a new wave with a particular beat. The beat frequency from this constructive superposition is equal to the value of the difference in frequency of the two waves.

[tex]f_{beat}=f_2-f_1=353Hz -350Hz=3Hz[/tex]

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:

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.

Answer:

The column compressed is 57.142 Pa.

Explanation:

Given that,

Height = 3.50 m

Diameter = 50.0 cm

Radius = 25.0 cm

Load [tex]F=5.00\times10^{5}\ kg[/tex]

We need to calculate the column compressed

Using formula of compressed

[tex]P = \dfrac{F}{A}[/tex]

[tex]P=\dfrac{F}{l\times r}[/tex]

Where, F = load

l = length

r = radius

Put the value into the formula

[tex]P=\dfrac{5.00\times10^{5}}{3.50\times25\times10^{2}}[/tex]

[tex]P=57.142\ Pa[/tex]

Hence, The column compressed is 57.142 Pa.

Answer:

Part a)

[tex]a = 4.86 m/s^2[/tex]

Part b)

[tex]d = 155.52 m[/tex]

Part c)

[tex]v_f = 48.6 m/s[/tex]

Explanation:

As we know that car start from rest and reach to final speed of 87 mph

so we have

[tex]v_f = 87 mph = 38.88 m/s[/tex]

now we have

Part a)

acceleration is rate of change in velocity

[tex]a = \frac{v_f - v_i}{t}[/tex]

[tex]a = \frac{38.88 - 0}{8}[/tex]

[tex]a = 4.86 m/s^2[/tex]

Part b)

distance moved by car with uniform acceleration is given as

[tex]d = \frac{v_f + v_i}{2} t[/tex]

[tex]d = \frac{38.88 + 0}{2} 8[/tex]

[tex]d = 155.52 m[/tex]

Part c)

As we know that the car start from rest

so final speed after t = 10 s

[tex]v_f = v_i + at[/tex]

[tex]v_f = 0 + (4.86)10[/tex]

[tex]v_f = 48.6 m/s[/tex]

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:

A)Ep=-81.3N/C  :Electric field at the point x = +0.2

Ep Magnitude =81.3N/C  

Direction of the electric field ( Ep): -x

B)Graphic attached

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m

Graphic attached

The attached graph shows the field due to the charges:

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

E₁: Electric Field at point  Xp=0.2 m due to charge q₁. As the charge q1 is positive (q₁+) ,the field leaves the charge.

E₁: Electric Field at point  Xp=0.2 m due to charge q₂. As the charge q1 is positive (q₂+) ,the field leaves the charge

E₃: Electric Field at point Xp=0.2 m due to charge q₃. As the charge q₃ is negative (q₃-), the field enters the charge.

Equivalence

+1.0μC=1* 10⁶C

Data

q₁=+1.0μC=1* 10⁶C

q₂=+1.0μC=1* 10⁶C

q₃=-2.0μC=-2* 10⁶C

Xp=0.2m

x₁=0

x₂=0.01 m

x₃=0.02m

Calculation of the distances of the charges to the point P

d= Xp-x

d₁=Xp-x₁= 0.2-0= 0.2m

d₂=Xp-x₂=0.2-0.01= 0.19m

d₃=Xp-x₃=0.2-002= 0.18m

Calculation of electric fields due to charges q1, q2 and q3 at point P

E₁=k*q₁/d₁²=9*10⁹*1*10⁻⁶/0.2²=225*10³N/C

E₂=k*q₂/d₂²=9*10⁹*1*10⁻⁶/0.19²=249.3*10³N/C

E₃=-k*q₃/d₃²=9*10⁹*2*10⁻⁶/0.18²=-555.6*10³N/C

Calculation of electric field at point P due to charges q₁, q₂ and q₃

To calculate Ep, the electric fields E₁,E₂ and E₃ are added algebraically:

Ep=E₁+E₂ +E₃

Ep=(225*10³+249.3*10³ -555.6*10³)N/C

Ep=-81.3N/C  

Ep Magnitude =81.3N/C  in -x direction

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:

Explanation:

Force due to charges 1.75 and 5 nC is given below

F =K Q₁Q₂ / d²

F₁ = [tex]\frac{9\times10^9\times1.75\times10^{-9}\times5\times 10^{-9}}{(2.7\times10^{-2})^2}[/tex]

F₁ = 10.8 X 10⁻⁵ N . It will at in x direction.

Force due to other charge placed at origin

F₂ = [tex]\frac{9\times10^9\times1.75\times10^{-9}\times5\times 10^{-9}}{22.5\times10^{-4}}[/tex]

F₂ = 3.5 x 10⁻⁵ N.

Its x component

= F₂ Cos θ

= 3.5 x 10⁻⁵ x 3.9/ 4.74

= 2.88 x 10⁻⁵ N

Its y component

F₂ sin θ

= 3.5 x 10⁻⁵ x 2.7/4.743

= 1.99 x 10⁻⁵ N

Total x  component

=  10.8 X 10⁻⁵ +2.88 x 10⁻⁵

= 13.68 x 10⁻⁵ N.

Magnitude of total force  F

F²  = (13.68 x 10⁻⁵)² + (1.99 x 10⁻⁵ )²

F = 13.82 X 10⁻⁵ N

Direction θ with x axis .

Tanθ = 1.99/ 13.68

θ = 8 °

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:

Explanation:

Given

Velocity of women relative to the car is u

Let v be the velocity of car after women jump off

Therefore women velocity relative to the ground is v-u

Conserving momentum

Mv+Nm(v-u)=0

Mv+Nmv=Nmu

[tex]v=\frac{Nm}{M+Nm}[/tex]

(b)Let [tex]v_r[/tex] be the velocity of car after r women has jumped and [tex]v_N[/tex] be the final velocity of the car after N women have jumped.

After r women jumped and N-r on cart momentum is given by

[tex]P_r=\left [ M+\left ( N-r\right )m\right ]v_r[/tex]

After next woman jumps car has velocity of [tex]v_{r+1}[/tex] while the woman has velocity is [tex]v_{r+1}-u[/tex](relative to the ground)

Total momentum

[tex]P_{r+1}=\left [ M+\left ( N-r-1\right )m\right ]v_{r+1}+m\left ( v_{r+1}-u\right )[/tex]

Since total momentum in horizontal direction is conserved then

[tex]P_{r+1}=P_r[/tex]

[tex]v_{r+1}-v_r=\frac{mu}{\left ( M+\left ( N-r\right )m\right )}[/tex]

Summing the above expression from r=0 to r=N-1 we get

[tex]\sum_{j=1}^{j=N}\frac{mu}{\left ( M+jm\right )}[/tex]

(c)Answer in part b is greater because j\leq N[/tex]

Thus Velocity in part b is greater.

Intuitively if you jump after an another person you will impart extra momentum to the cart compared to when all persons jumped off simultaneously.

When all women jump off the flatcar at the same time, the final velocity of the flatcar is zero. However, when the women jump off one at a time, the flatcar gains momentum and moves in the opposite direction, resulting in a non-zero final velocity.

To answer the given question, we can apply the principle of conservation of momentum. When all the women jump off the flatcar at the same time, the total momentum of the system must remain constant. Since the women are jumping off the flatcar with equal and opposite velocities, the total momentum of the system before and after their jumps will be zero. Therefore, the final velocity of the flatcar will also be zero.

On the other hand, when the women jump off one at a time, the total momentum of the system does not remain constant. Initially, the flatcar and the first woman have a total momentum of zero. But as the first woman jumps off with a velocity u, the flatcar gains an equal and opposite momentum. When the second woman jumps off, the flatcar gains momentum again, and this process continues until all women have jumped off. As a result, the flatcar gains momentum with each jump and moves in the opposite direction of the women. Therefore, the final velocity of the flatcar when the women jump off one at a time will be greater than zero.

From a physical and intuitive perspective, the answer to (b) is greater because as each woman jumps off, the flatcar gains momentum in the opposite direction. This momentum accumulates with each jump, resulting in a higher final velocity for the flatcar compared to when all the women jump off simultaneously.

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Answer:

(a) 10 s

(b) 236.5 m

(c) Kathy's speed = 47.3 m/s

    Stan's speed = 42.9 m/s

Explanation:

Given:

Assumptions:

Part (a):

 Kathy will catch Stan only if the distances traveled by each of them are equal at the same instant.

[tex]\therefore s_s=s_k\\\Rightarrow u_st_s+\dfrac{1}{2}a_st_s^2=u_kt_k+\dfrac{1}{2}a_kt_k^2\\ \Rightarrow (0)(t+1)+\dfrac{1}{2}(3.9)(t+1)^2=(0)(t)+\dfrac{1}{2}(4.73)t^2\\ \Rightarrow \dfrac{1}{2}(3.9)(t+1)^2=\dfrac{1}{2}(4.73)t^2\\\Rightarrow (3.9)(t+1)^2=(4.73)t^2\\\Rightarrow \dfrac{(t+1)^2}{t^2}=\dfrac{4.73}{3.9}\\\textrm{Taking square root in both sides}\\\dfrac{t+1}{t}= 1.1\\\Rightarrow t+1=1.1t\\\Rightarrow 0.1t = 1\\\Rightarrow t = 10\\[/tex]

Hence, Kathy catches Stan after 11 s from the Stan's starting times.

Part (b):

Distance traveled by Kathy to catch Stan will be distance the distance traveled by her in 10 s.

[tex]s_s = u_kt_k+\dfrac{1}{2}a_kt_k^2\\\Rightarrow s_s= (0)(t)+\dfrac{1}{2}(4.73)t^2\\\Rightarrow s_s= \dfrac{1}{2}(4.73)(10)^2\\\Rightarrow s_s= 236.5[/tex]

Hence, Kathy traveled a distance of 236.5 m to overtake Stan.

Part (c):

[tex]v_k = u_k+a_kt_k\\\Rightarrow v_k = 0+(4.73)(t)\\\Rightarrow v_k = (4.73)(10)\\\Rightarrow v_k =47.3[/tex]

The speed of Kathy at the instant she catches Stan is 47.3 m/s.

[tex]v_s = u_s+a_st_s\\\Rightarrow v_s = 0+(3.9)(t+1)\\\Rightarrow v_s = (3.9)(10+1)\\\Rightarrow v_s =42.9[/tex]

The speed of Stan at the instant Kathy catches him is 42.9 m/s.

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:

option (c) 4.71 seconds

Explanation:

Given:

Distance to be covered = 1 km = 1000 m

Acceleration, a = 90 m/s²

From Newton's equation of motion

[tex]s=ut+\frac{1}{2}at^2[/tex]

where,  

s is the distance

u is the initial speed  = 0 m/s ( since it starts from rest )

a is the acceleration

t is the time

on substituting the respective values, we get

[tex]1000=0\times t+\frac{1}{2}\times90\times t^2[/tex]

or

45t² = 1000

or

t² = 22.22

or

t = 4.71 seconds

Hence, the correct answer is option (c) 4.71

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:

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:

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:

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:

1.875 cm from 60 microcoulomb charge.

Explanation:

Let the third charge be Q. Let it be put at x distance from 60 micro coulomb charge for balance.

Force on this charge due to first charge

= [tex]\frac{k\times60\times10^{-6}Q}{x^2}[/tex]

Force on this charge due to second charge

= [tex]\frac{k\times6.66\times10^{-6}Q}{(2.5-x)^2}[/tex]

Since both these forces are equal [tex]\frac{k\times60\times10^{-6}Q}{x^2}=\frac{k\times6.66\times10^{-6}Q}{(2.5-x)^2}[/tex]

[tex]\frac{60}{6.66} = \frac{x^2}{(2.5-x)^2}[/tex][tex]\frac{x}{2.5 -x} = \frac{3}{1}[/tex]

x = 1.875

1.875 cm from 60 microcoulomb charge.