A block of mass M is connected by a string and pulley to a hanging mass m.
The coefficient of kinetic friction between block M and the table is 0.2, and also, M = 20 kg, m = 10 kg.
b. Find the acceleration of the system and tensions on the string.
c. How far will block m drop in the first seconds after the system is released?
d. How long will block M move during the above time?
e. At the time, calculate the velocity of block M
f. Find out the deceleration of block M if the connection string is removal by cutting after the first second. Then, calculate the time taken to contact block M and pulley
How far will block m drop in the first seconds after the system is released?

Answers

Answer 1

(b) Use Newton's second law. The net forces on block M are

• ∑ F (horizontal) = T - f = Ma … … … [1]

• ∑ F (vertical) = n - Mg = 0 … … … [2]

where T is the magnitude of the tension, f is the mag. of kinetic friction between block M and the table, a is the acceleration of block M (but since both blocks are moving together, the smaller block m also shares this acceleration), and n is the mag. of the normal force between the block and the table.

Right away, we see n = Mg, and so f = µn = 0.2Mg.

The net force on block m is

• ∑ F = mg - T = ma … … … [3]

You can eliminate T and solve for a by adding [1] to [3] :

(T - 0.2Mg) + (mg - T ) = Ma + ma

(m - 0.2M) g = (M + m) a

a = (10 kg - 0.2 (20 kg)) (9.8 m/s²) / (10 kg + 20 kg)

a = 1.96 m/s²

We can get the tension from [3] :

T = m (g - a)

T = (10 kg) (9.8 m/s² - 1.96 m/s²)

T = 78.4 N

(c/d) No time duration seems to be specified, so I'll just assume some time t before block M reaches the edge of the table (whatever that time might be), after which either block would move the same distance of

1/2 (1.96 m/s²) t

(e) Assuming block M starts from rest, its velocity at time t is

(1.96 m/s²) t

(f) After t = 1 s, block M reaches a speed of 1.96 m/s. When the string is cut, the tension force vanishes and the block slows down due to friction. By Newton's second law, we have

F = -f = Ma

The effect of friction is constant, so that f = 0.2Mg as before, and

-0.2Mg = Ma

a = -0.2g

a = -1.96 m/s²

Then block M slides a distance x such that

0² - (1.96 m/s²) = 2 (-1.96 m/s²) x

x = (1.96 m/s²) /  (2 (1.96 m/s²))

x = 0.5 m

(I don't quite understand what is being asked by the part that says "calculate the time taken to contact block M and pulley" …)

Meanwhile, block m would be in free fall, so after 1 s it would fall a distance

x = 1/2 (-9.8 m/s²) (1 s)

x = 4.9 m


Related Questions

Gsjskebjwkksmndkkwksjdkdkskkskskkehdhjdj

Answers

Answer:

I DON'T UNDERSTAND

Explanation:

GUESS A MISUNDERSTANDING PLZ PUT A UNDERSTANDABLE QUESTION.

i.Name two commonly used thermometric liquids.


ii.State two advantages each of the thermometric liquids mentioned above​

Answers

Answer:

mercury and alcohol

ii) used to test temperatures

Mercury and AlcoholMercury:

i) It is a good conductor of heat and therefore the whole liquid reaches the temperature of the surroundings quickly.

ii) It does not wet (cling to the sides of) the tube.

Alcohol:

i) Alcohol has greater value of temperature coefficient of expansion than mercury.

ii) it's freezing point is below –100°C.

A heat engine exhausts 3 000 J of heat while performing 1 500 J of useful work. What is the efficiency of the engine

Answers

efficiency=work output/work input×100

since it exhausts(use up)3000j of heat that's the work input and the 1500j is the work input

efficiency=1500/3000×100

=50%

What is the total number of moles of products involved in the following reaction?
CaCO3 (s) + 2HCl (aq) - CaCl2 (aq) + CO2 (g) + H20 (g)
O 6
2.
3
5

Answers

Answer:

3

Explanation:

You must first make sure the equation is balanced. This one is. Then, you simply add up the coefficients of each compound on the products side of the equation. When the coefficient is not specified, you can assume it is 1 mole. So, in this equation, there is 1 mole of CaCl₂, 1 mole of CO₂, and 1 mole of H₂O = 3 moles.

The reactant side of the equation also has three moles:

1 mole of CaCO₃ and 2 moles of HCl.

~~~~NEED HELP ASAP~~~~

Block A slides into block B along a frictionless surface. They are moving in the direction from left o the right.

Block A= 3kg

Block B= 4kg

Block A velocity before collision =30m/s.

Block B velocity before collision = 15 m/s

The velocity of block B after the collision is 20m/s.


a.) What is the velocity of block A after collision?

b.) Is the collision elastic? Show work to explain answer why or why not.

Answers

Answer:

Block A velocity is 23.33 m/s and the collission is not elastic.

Explanation:

a) m1v1 + m2v2 = m1v1' + m2v2'

Plug in givens

90+60=3v1'+80

solve for v1'= 23.33m/s

b) Find the initial and final kinetic energy of Block B

Ki= 1/2(4)(15)^2 + 1/2(3)(30)^2 = 1800 J

Kf= 1/2(4)(20)^2 + 1/2(3)*(23.33)^2= 1616.433J

Since Ki does not equal Kf the collision is not elastic

Una cuerda horizontal tiene una longitud de 5 m y masa de 0,00145 kg. Si sobre esta cuerda se da un pulso generando una longitud de onda de 0,6 m y una frecuencia de 120 Hz. La tensión a la cual está sometida la cuerda es:

a. 1,5 N

b. 15,0 N

c. 3,1 N

d. 5,2 N

Answers

Answer:

Option (A) is correct.

Explanation:

A horizontal rope has a length of 5 m and a mass of 0.00145 kg. If a pulse occurs on this string, generating a wavelength of 0.6 m and a frequency of 120 Hz. The tension to which the string is subjected is

mass of string, m = 0.00145 kg

Frequency, f = 120 Hz

wavelength = 0.6 m

Speed = frequency x wavelength

speed = 120 x 0.6 = 72 m/s

Let the tension is T.

Use the formula

[tex]v =\sqrt\frac{T L}{m}\\\\72 = \sqrt\frac{T\times 5}{0.00145}\\\\T = 1.5 N[/tex]

Option (A) is correct.

A tire is filled with air at 22oC to a gauge pressure of 240 kPa. After driving for some time, if the temperature of air inside the tire is 45oC, what fraction of the original volume of air must be removed to maintain the pressure at 240 kPa?

Answers

Answer:

7.8% of the original volume.

Explanation:

From the given information:

Temperature [tex]T_1[/tex] = 22° C = 273 + 22 = 295° C

Pressure [tex]P_1[/tex] = 240 kPa

Temperature [tex]T_2[/tex] = 45° C

At initial temperature and pressure:

Using the ideal gas equation:

[tex]P_1V_1 =nRT_1[/tex]

making V_1 (initial volume) the subject:

[tex]V_1 = \dfrac{nRT_1}{P_1}[/tex]

[tex]V_1 = \dfrac{nR*295}{240}[/tex]

Provided the pressure maintained its rate at 240 kPa, when the temperature reached 45° C, then:

the final volume [tex]V_2[/tex] can be computed as:

[tex]V_2 = \dfrac{nR*318}{240}[/tex]

Now, the change in the volume ΔV =  V₂ - V₁

[tex]\Delta V = \dfrac{nR*318}{240}- \dfrac{nR*295}{240}[/tex]

[tex]\Delta V = \dfrac{23nR}{240}[/tex]

The required fraction of the volume of air to keep up the pressure at (240) kPa can be computed as:

[tex]= \dfrac{\dfrac{23nR}{240}}{ \dfrac{295nR}{240}}[/tex]

[tex]= {\dfrac{23nR}{240}} \times { \dfrac{240}{295nR}}[/tex]

[tex]= 0.078[/tex]

= 7.8% of the original volume.

Two cars are facing each other. Car A is at rest while car B is moving toward car A with a constant velocity of 20 m/s. When car B is 100 from car A, car A begins to accelerate toward car B with a constant acceleration of 5 m/s/s. Let right be positive.
1) How much time elapses before the two cars meet? 2) How far does car A travel before the two cars meet? 3) What is the velocity of car B when the two cars meet?
4) What is the velocity of car A when the two cars meet?

Answers

Answer:

Let's define t = 0s (the initial time) as the moment when Car A starts moving.

Let's find the movement equations of each car.

A:

We know that Car A accelerations with a constant acceleration of 5m/s^2

Then the acceleration equation is:

[tex]A_a(t) = 5m/s^2[/tex]

To get the velocity, we integrate over time:

[tex]V_a(t) = (5m/s^2)*t + V_0[/tex]

Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:

[tex]V_a(t) = (5m/s^2)*t[/tex]

To get the position equation we integrate again over time:

[tex]P_a(t) = 0.5*(5m/s^2)*t^2 + P_0[/tex]

Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:

[tex]P_a(t) = 0.5*(5m/s^2)*t^2[/tex]

Now let's find the equations for car B.

We know that Car B does not accelerate, then it has a constant velocity given by:

[tex]V_b(t) =20m/s[/tex]

To get the position equation, we can integrate:

[tex]P_b(t) = (20m/s)*t + P_0[/tex]

This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:

[tex]P_b(t) = (20m/s)*t + 100m[/tex]

Now we can answer this:

1) The two cars will meet when their position equations are equal, so we must have:

[tex]P_a(t) = P_b(t)[/tex]

We can solve this for t.

[tex]0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0[/tex]

This is a quadratic equation, the solutions are given by the Bhaskara's formula:

[tex]t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)} }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}[/tex]

We only care for the positive solution, which is:

[tex]t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s[/tex]

Car A reaches Car B after 11.48 seconds.

2) How far does car A travel before the two cars meet?

Here we only need to evaluate the position equation for Car A in t = 11.48s:

[tex]P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m[/tex]

3) What is the velocity of car B when the two cars meet?

Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s

4)  What is the velocity of car A when the two cars meet?

Here we need to evaluate the velocity equation for Car A at t = 11.48s

[tex]V_a(t) = (5m/s^2)*11.48s = 57.4 m/s[/tex]

What is the biggest planet in the solar system

Answers

Answer:

Jupiter

Explanation:

Answer:

The answer is Jupiter.

Explanation:

Jupiter is an orange/yellow colored planet.

In a
DC source, which has more cuwent?
(i)R L Circuit
(ii)RC Circuit (series)
(iii)LC Cirenit (series)
(iv)RLC Circuit (series)

Answers

Answer:

Answer is LC Cirenit (seres)

A transverse sine wave with an amplitude of 2.50 mm and a wavelength of 1.80 m travels, from left to right along a long, horizontal stretched string with a speed of 36.0 m s. I Take the origin at the left end of the undisturbed string. At time t = 0 the left end of the string has its maximum upward displacement,
(a) What is the frequency of the wave?
(b) What is the angular frequency of the wave?
(c) What is the wave number of the wave?
(d) What is the function y(x,t) that describes the wave?
(e) What is y(t) for a particle at the left end of the string?
(f) What is y(t) for a particle 1.35 m to the right of the origin?
(g) What is the maximum magnitude of transverse velocity of any particle of the string?
(h) Find the transverse displacement of a particle 1.35 m to the right of the origin at time t = 0.0625 s.
(i) Find the transverse velocity of a particle 1.35 m to the right of the origin at time t = 0.0625 s.

Answers

Explanation:

Given that,

Amplitude, A = 2.5 nm

Wavelength,[tex]\lambda=1.8\ m[/tex]

The speed of the wave, v = 36 m/s

At time t = 0 the left end of the string has its maximum upward displacement.

(a) Let f is the frequency. So,

[tex]f=\dfrac{v}{\lambda}\\\\f=\dfrac{36}{1.8}\\\\f=20\ Hz[/tex]

(b) Angular frequency of the wave,

[tex]\omega=2\pi f\\\\=2\pi \times 20\\\\=125.7\ rad/s[/tex]

(c) The wave number of the wave[tex]=\dfrac{1}{\lambda}[/tex]

[tex]=\dfrac{1}{1.8}\\\\=0.56\ m^{-1}[/tex]

Strategies for good health management involve:
A Avoiding stressful situations that may cause depression or moodiness insomnia, or lack motivation.
B) Denying, ignoring, or repressing feelings or problems, so that you don't have to face them.
Eating your favorite foods, imagining yourself working out (mind is power), sleeping a few hours a day, so as to make
the most of party time.
D Eating healthy, maintaining and ideal weight, resting, exercising, and establishing healthy relationships.

Answers

Answer:

D

Explanation:

This is a great way to manage health.

A would be avoiding everything which isnt good.

B. would be emotionally draining and damaging to bottle feelings and ignore them.

C. is unhealthy to not exercise and eat food while doing nothing.

why do you like the full moon ?

Answers

Answer:

The Moon brings perspective. Observing the Moon, and I mean really looking – sitting comfortably, or lying down on a patch of grass and letting her light fill your eyes, it's easy to be reminded of how ancient and everlasting the celestial bodies are. When I do this, it always puts my life into perspective.

Answer:

because it look more impressive than empty dark sky .

An electric eel can generate a 180-V, 0.1-A shock for stunning its prey. What is the eel's power output

Answers

Power output = volts x amps

Power output = 170 volts x 0.1 amps

Power output = 18 watts

What is the value of the charge that experiences a force of 2.4×10^-3N in an electric field of 6.8×10^-5N/C

Answers

Hi there!

[tex]\large\boxed{\approx 35.29 C}[/tex]

Use the following formula:

E = F / C, where:

E = electric field (N/C)

F = force (N)

C = Charge (C)

Thus:

6.8 × 10⁻⁵ = 2.4 × 10⁻³ / C

Isolate for C:

C = 2.4 × 10⁻³  / 6.8 × 10⁻⁵

Solve:

≈ 35.29 C

if a body covers 100m in 5 second from rest find the acceleration produced by a body in 10 second​

Answers

Answer:

a=10m/s^2

Explanation:

acceleration= final velocity+ initial velocity/time taken

v-u/t=a

100-0/5=a

100/5=a

a=20m/s^2

case2

100-0/10=a

100/10=a

a=10m/s^2

Don't forget to write the units.

Hope this helps

please mark me as brainliest.

1. Compare and contrast the SI and the English systems of measurement.

Answers

Answer:The SI system is based on the number 10 as well as multiples and products of 10. This makes it much easier to use, and so it has been the accepted system in scientific and technical applications. The English system is more complicated as relationships between units of the same quantity aren't uniform.

Explanation:

Answer:

The metric system is an internationally agreed decimal system of measurement while The International System of Units (SI) is the official system of measurement in almost every country in the world

The north pole of magnet A will __?____ the south pole of magnet B

Answers

Answer:

A will attract

B will repare

Solids diffuse because the particles cannot move.
A. Can
B. Not enough info
C. Cannot
D. Sometimes will

Answers

Solids cannot diffuse.

Answer: C. Cannot
They don’t space to move.

trình bày nguyên lý Đa lăm be

Answers

I don’t understand this question

Diffuse reflection occurs when parallel light waves strike which surface? a mirror a rippling fountain a polished silver plate a still pond

Answers

Answer: a rippling fountain

Explanation: diffuse reflection happens on rough surfaces, so using the process of elimination, that leaves us with b, a rippling fountain (I also just took this test I'm pretty sure I'm right)

SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.7 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building

Answers

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

Cell phone conversations are transmitted by high-frequency radio waves. Suppose the signal has wavelength 35 cm while traveling through air. What are the
(a) frequency and
(b) wavelength as the signal travels through 3-mm-thick window glass into your room?

Answers

Answer:

(a) 8.57 x 10^8 Hz

(b) 23.3 cm

Explanation:

Wavelength = 35 cm = 0.35 m

speed =3 x10^8 m/s

Let the frequency is f.

(a) The relation is

speed  = frequency x wavelength

3 x 10^8 = 0.35 x f

f = 8.57 x 10^8 Hz

(b) refractive index of glass  is 1.5

The relation for the refractive index and the wavelength is

wavelength in glass= wavelength in air/ refractive index.

Wavelength in glass= 35/1.5 = 23.3 cm

A uniformly dense solid disk with a mass of 4 kg and a radius of 4 m is free to rotate around an axis that passes through the center of the disk and perpendicular to the plane of the disk. The rotational kinetic energy of the disk is increasing at 21 J/s. If the disk starts from rest through what angular displacement (in rad) will it have rotated after 3.3 s?

Answers

Answer:

3.44 rad

Explanation:

The rotational kinetic energy change of the disk is given by ΔK = 1/2I(ω² - ω₀²) where I = rotational inertia of solid sphere = MR²/2 where m = mass of solid disk = 4 kg and R = radius of solid disk = 4 m, ω₀ = initial angular speed of disk = 0 rad/s (since it starts from rest) and ω = final angular speed of disk

Since the kinetic energy is increasing at a rate of 21 J/s, the increase in kinetic energy in 3.3 s is  ΔK = 21 J/s × 3.3 s = 69.3 J

So, ΔK = 1/2I(ω² - ω₀²)

Since ω₀ = 0 rad/s

ΔK = 1/2I(ω² - 0)

ΔK = 1/2Iω²

ΔK = 1/2(MR²/2)ω²

ΔK = MR²ω²/4

ω² = (4ΔK/MR²)

ω = √(4ΔK/MR²)

ω = 2√(ΔK/MR²)

Substituting the values of the variables into the equation, we have

ω = 2√(ΔK/MR²)

ω = 2√(69.3 J/( 4 kg × (4 m)²))

ω = 2√(69.3 J/[ 4 kg × 16 m²])

ω = 2√(69.3 J/64 kgm²)

ω = 2√(1.083 J/kgm²)

ω = 2 × 1.041 rad/s

ω = 2.082 rad/s

The angular displacement θ is gotten from

θ = ω₀t + 1/2αt² where ω₀ = initial angular speed = 0 rad/s (since it starts from rest), t = time of rotation = 3.3 s and α = angular acceleration = (ω - ω₀)/t = (2.082 rad/s - 0 rad/s)/3.3 s = 2.082 rad/s ÷ 3.3 s = 0.631 rad/s²

Substituting the values of the variables into the equation, we have

θ = ω₀t + 1/2αt²

θ = 0 rad/s × 3.3 s + 1/2 × 0.631 rad/s² (3.3 s)²

θ = 0 rad + 1/2 × 0.631 rad/s² × 10.89 s²

θ = 1/2 × 6.87159 rad

θ = 3.436 rad

θ ≅ 3.44 rad

Starting from rest, a wheel undergoes constant angular acceleration for a period of time T. At which of the following times does the average angular acceleration equal the instantaneous angular acceleration?

a. 0.50 T
b. 0.67 T
c. 0.71 T
d. all of the above

Answers

I think it would be a

Question 9 of 10
According to the law of conservation of momentum, the total initial
momentum equals the total final momentum in a(n)
A. Interacting system
B. System interacting with one other system
C. Isolated system
D. System of balanced forces

Answers

Answer:

The answer is C. Isolated System

Answer:

C. Isolated system

Explanation :

∵According to law of  conservation of momentum ,In an isolated system ,the total momentum remains conserved.

A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?​

Answers

Answer:

The correct answer is "[tex]4.49\times 10^{10} \ joules[/tex]".

Explanation:

According to the question,

The work will be:

⇒ [tex]Work=-\frac{kQq}{R}[/tex]

              [tex]=-\frac{1}{4 \pi \varepsilon \times (18-30)\times 3\times 0.2}[/tex]

              [tex]=-\frac{1}{4 \pi \varepsilon \times (-12)\times 3\times 0.2}[/tex]

              [tex]=\frac{0.3978}{\varepsilon }[/tex]

              [tex]=4.49\times 10^{10} \ joules[/tex]

Thus the above is the correct answer.    

We have that the workdone  is mathematically given as

W=4.49*10e10 J

From the question we are told

A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?​

Workdone

Generally the equation for the workdone   is mathematically given as

W=-kQq/R

Therefore

0.3978/ε0 =-1/(4πε0*(18-30)*3*0.2

Hence

W=4.49*10e10 J

For more information on Charge visit

https://brainly.com/question/9383604

If the source moves, the wavelength of the sound in front of the direction of motion is____than the wavelength behind the direction of motion.
a. the same.
b. smaller than.
c. unrealted to.
d. larger then.

Answers

Answer:

B. Smaller than

Explanation:

This question is from the Doppler effect. As the object which is in motion goes off from the other, there's a reduction in the frequency. This is due to the fact that successive soundwave get to be longer. So that the pitch will then be lowered. When the person observing moves towards what is making the sound, each soundwave that follows gets faster than the previous.

A basketball of mass 0.608 kg is dropped from rest from a height of 1.37 m. It rebounds to a height of 0.626 m.
(a) How much mechanical energy was lost during the collision with the floor?
(b) A basketball player dribbles the ball from a height of 1.37 m by exerting a constant downward force on it for a distance of 0.132 m. In dribbling, the player compensates for the mechanical energy lost during each bounce. If the ball now returns to a height of 1.37 m, what is the magnitude of the force?

Answers

Answer:

a)[tex]|\Delta E|=4.58\: J[/tex]  

b)[tex]F=61.90\: N[/tex]

Explanation:

a)

We can use conservation of energy between these heights.

[tex]\Delta E=mgh_{2}-mgh_{1}=mg(h_{2}-h_{1})[/tex]  

[tex]\Delta E=0.608*9.81(0.6026-1.37)[/tex]

Therefore, the lost energy is:

[tex]|\Delta E|=4.58\: J[/tex]  

b)

The force acting along the distance create a work, these work is equal to the potential energy.

[tex]W=\Delta E[/tex]

[tex]F*d=mgh[/tex]

Let's solve it for F.

[tex]F=\frac{mgh}{d}[/tex]

[tex]F=\frac{0.608*9.81*1.37}{0.132}[/tex]

Therefore, the force is:

[tex]F=61.90\: N[/tex]

I hope is helps you!

Mary needs to row her boat across a 160 m-wide river that is flowing to the east at a speed of 1.5 m/s. Mary can row with a speed of 3.6 m/s. If Mary points her boat due north, how far from her intended landing spot will she be when she reaches the opposite shore? What is her speed with respect to the shore?

Answers

Answer: 66.67 m, 44.44 s

Explanation:

Given

Velocity of flow is [tex]u=1.5\ m/s[/tex]

Mary can row with speed [tex]v=3.6\ m/s[/tex]

Width of the river [tex]y=160\ m[/tex]

Flow will drift the Mary towards east, while Mary boat will cause it to travel in North direction

time taken to cross river

[tex]\Rightarrow t=\dfrac{160}{3.6}\\\\\Rightarrow t=\dfrac{400}{9}\ s[/tex]

Flow will drift Mary by

[tex]\Rightarrow x=ut\\\\\Rightarrow x=1.5\times \dfrac{400}{9}\\\\\Rightarrow x=66.67\ m[/tex]

Velocity w.r.t shore is

[tex]\Rightarrow v_{net}=\sqrt{3.6^2+1.5^2}\\\Rightarrow v_{net}=\sqrt{15.21}\\\Rightarrow v_{net}=3.9\ m/s[/tex]

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