In quantum mechanics, the fundamental constant called Planck's constant, h, has dimensions of [ML^2T^-1 ]. Construct a quantity with the dimensions of length using h, a mass m, and c, the speed of light.

Answer:

h = 0.204 m

Explanation:

given data:

radius r = 0.131 m

mass m = 9.86 kg

density of copper = 8960 kg/m3

we knwo that density is given as

[tex]\rho = \frac{mass}{volume}[/tex]

[tex]volume = \pi * r^2 h[/tex]

[tex]density  =  \frac{ mass}{\pi * r^2 h}[/tex]

[tex]h = \frac{ mass}{\pi * r^2 * density}[/tex]

putting all value to get thickness value

[tex]h = \frac{ 98.6}{ \pi 0.131^2*8960}[/tex]

h = 0.204 m

Using Ohm's law and properties of series and parallel circuits, it's possible to find the resistances. In series, resistances are directly added and for parallel, the reciprocal of total resistance is the sum of reciprocals of individual resistances. Applying these principles with given current and voltage, one can solve for resistances.

This question pertains to electricity and specifically the characteristics of resistors when they are connected in series or parallel. Using Ohm's Law, we know that the voltage (V) is the product of the current (I) and the resistance (R). Therefore, when the resistors are connected in series, the combined resistance (Rtotal) is the sum of the individual resistances, while the current remains the same. This gives us Rtotal = V/I = 12.0V / 1.12A.

When in parallel, however, the total resistance can be found differently. In a parallel circuit, the total resistance is given by 1/Rtotal = 1/R+ 1/R. As per the problem, we know that the total current of the circuit connected in parallel is 9.39 A, so we can use the equation Itotal = V/ Rtotal to find the total resistance in the parallel circuit.

Combining the information from both the circuits would allow us to solve two simultaneous equations to get the values of R and R.

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The two resistances connected in series have a total resistance of 10.71 Ω, and connected in parallel have a total resistance of 1.28 Ω. Solving the equations, we find the resistances to be 1.43 Ω and 9.28 Ω.

Let's solve this step-by-step:

Step 1: Series Connection

When two resistors are connected in series, the total resistance, [tex]R_{total[/tex], is the sum of the resistances:

[tex]R_{total[/tex] = R₁ + R₂

Using Ohm's Law (V = IR), we can find the total resistance in series:

V = 12.0 V;

I = 1.12 A;

[tex]R_{total[/tex] = V/I

[tex]R_{total[/tex] = 12.0 V / 1.12 A = 10.71 Ω

Step 2: Parallel Connection

When the same resistors are connected in parallel, the total resistance, [tex]R_{parallel[/tex], can be found using the formula:

1/[tex]R_{parallel[/tex] = 1/R₁ + 1/R₂

Again, using Ohm's Law, we first find [tex]R_{parallel[/tex]:

V = 12.0 V; [tex]I_{total[/tex] = 9.39 A;

[tex]R_{parallel[/tex] = V/[tex]I_{total[/tex] = 12.0 V / 9.39 A = 1.28 Ω

Step 3: Solving the Equations

We now have two equations:

R₁ + R₂ = 10.71 Ω

(R₁ * R₂) / (R₁ + R₂) = 1.28 Ω (since 1/[tex]R_{parallel[/tex] = 1/R₁ + 1/R₂ )

Let's solve these equations:

Substitute R₂ = 10.71 - R₁ into the parallel equation:

(R₁ * (10.71 - R₁)) / 10.71 = 1.28

R₁ * (10.71 - R₁) = 1.28 * 10.71

10.71R₁ - R₁² = 13.69

R₁² - 10.71R₁ + 13.69 = 0

Solving the quadratic equation using the quadratic formula:

R₁ = [10.71 ± √((10.71)² - 4*1*13.69)] / 2

Solving this, we get R₁ ≈ 1.43 Ω or R₁ ≈ 9.28 Ω

Then, R₂ = 10.71 - 1.43 = 9.28 Ω ,

R₂ = 10.71 - 9.28 = 1.43 Ω

Answer:

time to fall is 3.914 seconds

Explanation:

given data

half distance time = 1.50 s

to find out

find the total time of its fall

solution

we consider here s is total distance

so equation of motion for distance

s = ut + 0.5 × at²   .........1

here s is distance and u is initial speed that is 0 and a is acceleration due to gravity = 9.8 and t is time

so for last 1.5 sec distance is 0.5 of its distance so equation will be

0.5 s = 0 + 0.5 × (9.8) × ( t - 1.5)²     ........................1

and

velocity will be

v = u + at

so velocity v = 0+ 9.8(t-1.5)    ..................2

so first we find time

0.5 × (9.8) × ( t - 1.5)² = 9.8(t-1.5)  + 0.5 ( 9.8)

solve and we get t

t = 3.37 s

so time to fall is 3.914 seconds

Answer:

acceleration, a = [tex]20 m/s^{2}[/tex]

distance, d = 80 m

Given:

Initial velocity of object, v = 20 m/s

Final velocity of object, v' = 60 m/s

Time interval, [tex]\Delta t = 2.0 s[/tex]

Solution:

(a) Acceleration is the rate at which velocity changes and constant acceleration is when the velocity changes by equal amount in equal intervals of time.

Thus

acceleration, a = [tex]\frac{v' - v}{\Delta t}[/tex]

a = [tex]\frac{60 - 20}{2.0} = 20 m/s^{2}[/tex]

(b) Now, the distance covered is  given by:

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

[tex]d = 20\times 2.0 + \frac{1}{2}20\times 2^{2} = 80 m[/tex]

Answer:

option (C)

Explanation:

Speed of car = 30 m/s

Let the time taken by the police car to catch the speeding car is t

The distance traveled by the speeding car in t + 1 second is equal to the distance traveled by the police car in time t

Distance traveled by the police car in time t

[tex]s=ut + 0.5 at^{2}[/tex]    .... (1)

Distance traveled by the speeding car in t + 1 second

s = 30 (t + 1) = 300

t + 1 = 10

t = 9 s

Put the value of t in equation (1), we get

300 = 0 + 0.5 x a x 9 x 9

a = 7.41 m/s^2

C. 7.41 meters per square second.

In this question, the car is travelling at constant speed, whereas the police car accelerates uniformly after some delay to catch the car, the respective kinematic formulas are shown below:

Car

[tex]x_{C} = x_{o} + v_{C}\cdot t[/tex] (1)

Police car

[tex]x_{P} = x_{o} + \frac{1}{2}\cdot a_{P}\cdot (t-t')^{2}[/tex] (2)

Where:

If we know that [tex]x_{o} = 0\,m[/tex], [tex]x_{C} = x_{P} = 300\,m[/tex], [tex]t' = 1\,s[/tex] and [tex]v_{C} = 30\,\frac{m}{s}[/tex], then we have the following system of equations:

[tex]300 = 30\cdot t[/tex] (1)

[tex]300 = \frac{1}{2}\cdot a_{P}\cdot (t-1)^{2}[/tex] (2)

By (1):

[tex]t = 10[/tex]

Then we find that acceleration of the police car must be:

[tex]300 = \frac{1}{2}\cdot a_{P}\cdot (10-1)^{2}[/tex]

[tex]a_{P} = 7.407\,\frac{m}{s^{2}}[/tex]

Therefore, the correct choice is C.

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Answer:

[tex]y=-4.9x10^{-16}m[/tex]

Explanation:

From the exercise we have initial velocity on the x-axis, the final x distance and acceleration of gravity.

[tex]v_{ox}=4.3x10^{7}m/s[/tex]

[tex]x=43cm=0.43m\\g=9.8m/s^{2}[/tex]

From the equation on moving particles we can find how long does it take the electron beam to strike the screen

[tex]x=x_{o}+v_{ox}t+\frac{1}{2}at^{2}[/tex]

Since [tex]x_{o}=0[/tex] and [tex]a_{x}=0[/tex]

[tex]0.43m=(4.3x10^{7}m/s)t[/tex]

Solving for t

[tex]t=1x10^{-8} s[/tex]

Now, from the equation of free-falling objects we can find how far does the electron beam fell

[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]

[tex]y=-\frac{1}{2}(9.8m/s^{2})(1x10^{-8} s)=-4.9x10^{-16}m[/tex]

The negative sign means that the electron beam fell from its initial point.

The magnitude of the velocity with which the ball hits the crater floor is approximately 401.24 m/s.

Let's convert the height to meters:

Height = 23 km

= 23,000 m

Now we can use the kinematic equation to find the time it takes for the ball to reach the ground:

[tex]h=\frac{1}{2}gt^2[/tex]

Where: h = height (23,000 m)

g = acceleration due to gravity (3.5 m/s²)  

t = time

Solving for t:

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

Plug in the values:

[tex]t=\sqrt{\frac{2 \times 23000}{3.5}}[/tex]

t=114.64 s

So, the time for the ball to reach the crater floor is approximately 114.64 seconds.

Now calculate the magnitude of the velocity with which the ball hits the crater floor.

We can use the following kinematic equation:

v=gt

Where:

v = final velocity

g = acceleration due to gravity (3.5 m/s²)

t = time (114.64 s)

v=3.5×114.64

v=401.24 m/s

Hence, the magnitude of the velocity with which the ball hits the crater floor is approximately 401.24 m/s.

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The time it would take for the ball to hit the crater floor would be approximately 360.8 seconds, and the velocity it would hit the floor with would be approximately 1262.8 m/s.

To calculate the time it takes for the ball to reach the crater floor, we use the equation of motion d = 1/2gt^2, where 'd' represents the distance, 'g' represents the acceleration due to gravity, and 't' represents time. Here, the distance is 23,000m (converted from km to m), and the acceleration due to gravity is 3.5 m/s^2. By solving for 't' in this equation, we get t = sqrt(2d/g) = sqrt((2*23000)/3.5) = approx. 360.8s.

To calculate the magnitude of the final velocity as the ball hits the crater floor, we use the equation v = gt, where 'v' is the final velocity and 't' is the amount of time it has fallen. Plugging in the values we have for 'g' and 't', we get v = 3.5 * 360.8s = approx. 1262.8 m/s

Therefore, the time it would take for the ball to hit the crater floor would be approximately 360.8 seconds, and the velocity it would hit the floor with would be approximately 1262.8 m/s.

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Answer:

a) [tex]x=26m[/tex]

b) [tex]v_{y}=-21.5m/s[/tex]

Explanation:

From the exercise we know initial velocity, initial height

[tex]y_{o}=20m[/tex]

[tex]v_{o} =12m/s[/tex] [tex]\beta =45[/tex]º

a) The range of the stone is defined by how far does it goes. From the theory of free falling objects, we have:

[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]

The stone strike the water at y=0

[tex]0=20+12sin(45)t-\frac{1}{2}(9.8)t^{2}[/tex]

Solving for t, using the quadratic formula

[tex]t=-1.33s[/tex] or [tex]t=3.06s[/tex]

Since time can't be negative, the answer is t=3.06s

Now, we can calculate the range of the stone

[tex]x=v_{o}t=(12cos(45))m/s(3.06s)=26m[/tex]

b) We can calculate the velocity were the stone strike the water using the following formula  

[tex]v_{y}=v_{oy}+gt=12sin(45)m/s-(9.8m/s^{2})(3.06s)=-21.5m/s[/tex]

The negative sign indicates that the stone is going down

Answer:

(a) x= 26m

(b) vf= 23.15m/s  

Explanation:

Given data

h=20m

Ɵ=45°

to find

(a) range of stone=x=?

(b) velocity=vf=?

Solution

For part (a)

You need to solve for time first using

yf = yi + visinƟt + 1/2gt^2

0 = 20m + 12sin45t + 1/2(-9.8)t^2

and use the quadratic equation to solve for t

t = 3.064 sec

To solve for the distance traveled use

x = xi + vicosƟt + 1/2at^2 there is no acceleration in the x direction so that cancels

x = 12cos(45)(3.064)

x= 26m

For part(b)

For b I'm not sure if you what direction you want the final velocity in the x, y, or the direction its traveling so I'll just give all 3.

Theres no change in the velocity in the x direction so its just vfx = vixcosƟ = 12cos45 = 8.49m/s

For the y direction its vfy^2 = viy^2 + 2g(Δy)

vfy = sqrt((12sin(45))^2 + 2(-9.8)(0-20m)) = 21.54m/s

The velocity the direction the stone is traveling is vf = sqrt(vx^2 + vy^2) = sqrt(8.49^2 + 21.54^2)

vf= 23.15m/s

Answer:

Three linear operators A,B, and C will satisfy the condition [tex][[A, B],C] + [[B,C), A] + [[C, A], B] = 0[/tex].

Explanation:

According to the question we have to prove.

[tex][[A, B],C] + [[B,C), A] + [[C, A], B] = 0[/tex]

Now taking Left hand side of the equation and solve.

[tex][[A, B],C] + [[B,C), A] + [[C, A], B][/tex]

Now use commutator property on it as,

[tex]=[A,B] C-C[A,B]+[B,C]A-A[B,C]+[C,A]B-B[C,A]\\=(AB-BA)C-C(AB-BA)+(BC-CA)A-A(BC-CB)+(CA-AC)B-B(CA-AC)\\=ABC-BAC-CAB+CBA+BCA-CAB-ABC+ACB+CAB-ACB-BCA+BAC\\=0[/tex]

Therefore, it is proved that [tex][[A, B],C] + [[B,C), A] + [[C, A], B] = 0[/tex].

Answer:

- 3.46 m

Explanation:

initial position, xi = 3.37 m

displacement, Δx = - 6.83 m

Let the final position is xf.

So, displacement = final position - initial position

Δx = xf - xi

- 6.83 = xf - 3.37

xf = 3.37- 6.83

xf = - 3.46 m

Thus, the final position of the squirrel is - 3.46 m.

Answer:

Explanation:

When saw slices wood by exerting a force on the wood , wood also exerts a reaction force on the saw in opposite direction which is equal to the force of action that is 104 N.

So torque exerted by wood on the blade

= force x perpendicular distance from the axis of rotation

= 104 x .128

=13.312 Nm.

Since this torque opposes the movement of blade , it turns the blade slower.

Answer:

carrier power is 7.8 kW

Explanation:

given data

power = 10 kW

modulation percentage = 75 %

to find out

carrier power

solution

we will use here power transmitted equation that is

power = [tex]carrier power * ( 1+  \frac{modulation}{2})[/tex]   .................1

put here value in equation 1  we get carrier power

10 = [tex]carrier power *(1+ \frac{0.75}{2})[/tex]

carrier power = 7.8

so carrier power is 7.8 kW

Answer:

Explanation:

Moles of helium ( n ) = 156.7 / 4 = 39.175

Temperature T₁ = 35.73 +273 = 308.73 K

Volume V₁ = V

Pressure P₁ = 3.55 atm

V₂ =1.75 V

a ) For isothermal change

P₁ V₁ = P₂V₂

P₂ = P₁ V₁ / V₂

= 3.55 X V / 1.75 V

= 2.03 atm.

b ) Work done by the gas = nRT ln(V₂/V₁)

= 39.175 X 8.321 X 308.73 X ln 1.75

= 56318.8 J

Work done on the gas = - 56318.8 J

c ) Since there is no change in temperature , internal energy of gas is constant

Q = ΔE + W

ΔE  = 0

Q = W

Work done by gas = heat absorbed

heat absorbed = 56318.8 J

d ) Change in the internal energy of gas is zero because temperature is constant.

Answer:

force is 96 N

Explanation:

given data

mass = 40 kg

acceleration = 2.4 m/s²

to find out

force

solution

we know force is mass time acceleration so

we will apply here force formula that is express as

force = m × a   ..............1

here m is mass and a is acceleration so

put here value in equation 1 we get force

force = 40 × 2.4

force = 96

so force will be 96 N

The speed of the ball when it leaves Sarah's hand is 8.2 m/s.

Given that;

The ball leaves Sarah's hand at a distance of 1.5 meters above the ground, reaches a maximum height of 8 m above the ground, and takes 1.619 s to get directly over Julie's head.

For the speed of the ball when it leaves Sarah's hand, use the equations of motion and consider the vertical motion of the ball.

Since the ball is thrown vertically upward and then comes back down, the time taken to reach the maximum height is half of the total time of flight.

Therefore, the time to reach the maximum height is,

t/2 = 1.619 s / 2.

So, the time to reach the maximum height is 0.8095 s.

Now, let's find the initial vertical velocity of the ball using the equation:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

At the maximum height, the final vertical velocity is 0 m/s because the ball momentarily stops.

The acceleration due to gravity, a, is -9.8 m/s² (negative because it acts downward).

Using the equation, we have:

0 m/s = u + (-9.8 m/s²) × 0.8095 s

Simplifying the equation:

u = 7.93 m/s

So, the initial vertical velocity of the ball is 7.93 m/s.

Since the ball travels in a parabolic path, the time taken to reach Julie's horizontal position is the same as the time taken to reach the maximum height, which is 0.8095 s.

Now, let's calculate the initial horizontal velocity of the ball, using the equation:

s = ut

where s is the horizontal distance travelled, u is the initial horizontal velocity, and t is the time.

The horizontal distance travelled is equal to the horizontal distance between Sarah and Julie, which we don't have.

However, since we are only interested in the initial horizontal velocity, we can assume that the horizontal distance travelled is equal to the distance between Sarah and Julie.

Therefore, s = 1.5 m.

Using the equation, we have:

1.5 m = u × 0.8095 s

u = 1.5 m / 0.8095 s

Calculating u, we find:

u ≈ 1.853 m/s

So, the initial horizontal velocity of the ball is 1.853 m/s.

Finally, the speed of the ball when it leaves Sarah's hand by combining the horizontal and vertical components of velocity using the Pythagorean theorem:

speed = √(horizontal velocity² + vertical velocity²)

speed = √(1.853 m/s)² + (7.951 m/s)²

Calculating the speed, we find:

speed ≈ 8.2 m/s

Therefore, the speed of the ball when it leaves Sarah's hand is 8.2 m/s.

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Final answer:

Using the kinematic equations for vertical motion, the initial speed of the ball Sarah threw is calculated to be approximately 7.94 m/s, based on the given maximum height and time to reach Julie's horizontal position.

Explanation:

To find the initial speed of the ball when it leaves Sarah's hand, we need to use the information about the ball's motion under gravity. We know it reaches a maximum height of 8 m and takes 1.619 s to get directly over Julie's head, starting from a height of 1.5 m.

The motion of the ball can be divided into two segments: ascending and descending. During the ascending part, the ball slows down due to gravity until it reaches its maximum height. In the descending part, it accelerates back down. Since the motion is symmetrical, the time to reach the maximum height is half of the total time, which is 1.619 s / 2 = 0.8095 s.

To find the initial velocity (v_i), we can use the kinematic equation for vertical motion:
v_i = v_f - g*t
where v_f is the final velocity (0 m/s at the maximum height), g is the acceleration due to gravity (9.81 m/[tex]s^2[/tex]), and t is the time to reach maximum height.

Plugging in the values, we get:
v_i = 0 m/s - (-9.81 m/[tex]s^2[/tex] * 0.8095 s) = 7.94 m/s
Therefore, the initial speed of the ball when it leaves Sarah's hand is about 7.94 m/s.

Answer:

76 degree  south of west.

Explanation:

We shall represent velocities in vector form , considering east as  x axes and west as Y axes.

V₁ = 4.4 j

V₂ = 3.8, 30 degree  west of south

V₂ = -  3.8 sin 30 i - 3.8 cos 30 j

= - 1.9 i - 3.29 j

Change in velocity

= V₂ - V₁

= - 1.9 i - 3.29 j - 4.4 j

= - 1.9 i - 7.69 j

Acceleration

= change in velocity / time

(- 1.9 i - 7.69 j  ) / 60 ms⁻² .

Direction of acceleration θ

Tan θ = 7.69 / 1.9 =4.047

θ = 76 degree  south of west.

Answer:

(a) 0.9 s

(b) 7.78 m/s

Explanation:

height, h = 4 m

Horizontal distance, d = 7 m

Let it takes time t to reach the ground and u be the initial velocity of the jet.

(a) Use second equation of motion in vertical direction

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

In vertical direction, uy = 0 m/s, a = g = - 9.8 m/s^2, h = - 4 m

By substituting the values, we get

[tex]-4 = 0 - \frac{1}{2}\times 9.8\times t^{2}[/tex]

t = 0.9 second

Thus, the time taken by water jet in air is 0.9 second.

(b) Use

Horizontal distance = horizontal velocity x time

d = u t

7 = u x 0.9

u = 7.78 m/s

Thus, the initial velocity of water jet is 7.78 m/s.

The duration the water is in the air is found using the formula for the motion under gravity, which depends on the vertical distance and the acceleration due to gravity. After finding the time, the initial velocity of the water is calculated using the horizontal distance and the fraction of time the water was in motion.

To determine how long the water is in the air (a), and the initial velocity of the water (b) when a water gun is fired horizontally from a hill, we can use the principles of projectile motion. The time a projectile is in the air is solely determined by its vertical motion. Since the water gun is fired horizontally, it has an initial vertical velocity of 0 m/s.

For part (a), the time (t) it takes for the water to reach the ground can be calculated using the formula for the motion under gravity, which is y = 0.5 * g * t2, where y is the vertical distance (4 meters in this case) and g is the acceleration due to gravity (approximated to 9.81 m/s2). Solving for t gives us the time the water is in the air.

For part (b), once we have the time, we can use the horizontal distance (7 meters) to find the initial velocity (v0) using the formula x = v0 * t. This provides the initial horizontal velocity of the water gun's jet. The overall process involves solving for time first and then using that time to find the initial velocity.

Answer:

[tex]a=2.36\ m/s^2[/tex]

T=157.06 N

Explanation:

Given that

Mass of first block = 21.1 kg

Mass of second block = 12.9 kg

First mass is heavier than first that is why mass second first will go downward and mass second will go upward.

Given that pulley and string is mass  less that is why both mass will have same acceleration.So lets take their acceleration is 'a'.

So now from force equation

[tex]m_1g-m_2g=(m_1+m_2)a[/tex]

21.1 x 9.81 - 12.9 x 9.81 =(21.1+12.9) a

[tex]a=2.36\ m/s^2[/tex]

Lets tension in string is T

[tex]m_1g-T=m_1a[/tex]

[tex]T=m_1(g-a)[/tex]

T=21.1(9.81-2.36) N

T=157.06 N

The magnitude of their acceleration is 2.36m/s² while the tension in the rope is 157.06N.

From the information given, the following can be depicted:

The force equation can be used to calculate the magnitude of the acceleration which will go thus:

(21.1 × 9.81) - (12.9 × 9.81) = (21.1 + 12.9)a

(8.2 × 9.81) = 34a

a = (8.2 × 9.81) / 34

a = 2.36m/s².

Therefore, the tension will be:

T = m(g - a)

T = 21.1(9.81 - 2.36)N

T = 157.06N

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Answer:

18.96 m

Explanation:

Height from person jump, h = 10 m

Let it takes time t to reach to the car.

Use second equation of motion

[tex]s = ut +0.5at^2[/tex]

Here, a  g = 10 m/s^2 , u = 0, h = 10 m  

By substituting the values, we get

10 = 0 + 0.5 x 10 x t^2

t = 1.414 s

The speed of car, v = 30 mi/h = 13.41 m/s

Distance traveled by the car in time t , d = v x t = 13.41 x 1.414 = 18.96 m

So, the distance of car from the bridge is 18.96 m as the man jumps.

Answer:

120 m/s^2

(a) 6000 m

(b) 1200 m/s

Explanation:

mass, m = 10 kg

initial velocity, u = 0

Force, F = 1200 N

time, t = 10 s

Let a be the acceleration of the block.

By use of Newton,s second law

Force = mass x acceleration

1200 = 10 x a

a = 120 m/s^2

(a) Let the block travels by a distance s.

Use second equation of motion

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

s = 0 + 0.5 x 120  x 10 x 10

s = 6000 m

(b) Let v be the final velocity of the block

Use first equation of motion

v = u + at

v = 0 + 120 x 10

v = 1200 m/s

Answer:

We should only agree with the first statement of the first student but not with the second part of his statement. While as the statement of the other student is completely wrong.

Explanation:

The earth and the moon are locked in a process known as tidal locking. Tidal locking occurs when any object which revolves around other object takes the same amount of time to rotate around it's own axis as it takes to revolve around the planet.

This is exactly the case with moon and the earth system ,the moon is tidally locked with earth thus we cannot see the other side of the moon but this side is not dark as claimed by the first student but this side of the moon is also illuminated by the sunlight as the face of moon that we are able to see.

The second student is wrong as we cannot see the other side of moon from earth.

Final answer:

The Moon is in synchronous rotation with Earth, which means the same side, the near side, always faces Earth. The far side, incorrectly called the 'dark side,' is not perpetually dark but simply not visible from Earth and receives sunlight just like the near side.

Explanation:

The student who claimed that we always see only one side of the Moon is correct; however, their assertion that the other side is the "dark side" and always dark is incorrect. This is because the Moon is in synchronous rotation with Earth—it rotates on its axis in the same amount of time it takes to orbit Earth. Because of this, only one hemisphere, the near side, is visible from Earth, while the other hemisphere, the far side, remains out of view.

The idea that the far side is the "dark side" is a misconception; it receives sunlight just as the near side does, only at different times during the Moon's orbit. The term "dark side" only means that it's the side not visible from Earth. As the Moon revolves around Earth, different parts experience day and night, similar to how the Sun rises and sets on Earth. Therefore, the "dark side" is not perpetually dark; it simply refers to the lunar face we do not see that is also subject to changing illumination by the Sun.

In summary, the correct understanding is that the same side of the Moon always faces Earth, not because it doesn't rotate, but because it has a synchronous orbit with Earth, and the use of the term "dark side" is a misnomer since all parts of the Moon experience both sunlight and darkness at different times.

Answer:t=0.54 s

Explanation:

Given

Jacob is traveling 5 m/s in North direction

Jacob throw a ball with a in south direction with a velocity of 5 m/s

Ball is thrown in opposite direction of motion of car therefore it seems as if it is dropped from car as its net horizontal velocity is 5-5=0

Time taken by ball to reach ground

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

[tex]1.45=0+\frac{9.81\times t^2}{2}[/tex]

[tex]t^2=frac{2\times 1.45}{9.81}[/tex]

t=0.54 s

Motion of ball will be straight line

Answer:

The speed of the train is 7.75 m/s towards station.

The speed of the train is 8.12 m/s away from the station.

Explanation:

Given that,

Frequency of the whistles f= 175 Hz

Beat frequency [tex]\Delta f= 4.05 Hz[/tex]

Speed of observer = 0

We need to calculate the frequency

Using formula of beat frequency

[tex]\Delta f=f'-f[/tex]

[tex]f'=\Delta f+f[/tex]

[tex]f'=4.05+175[/tex]

[tex]f'=179.05\ Hz[/tex]

When the train moving towards station, then the frequency heard is more than the actual

Using Doppler effect

[tex]f'=f(\dfrac{v-v_{o}}{v-v_{s}})[/tex]

[tex]v=v-\dfrac{vf}{f'}[/tex]

Put the value into the formula

[tex]v=343-\dfrac{343\times175}{179.05}[/tex]

[tex]v=7.75\ m/s[/tex]

The speed of the train is 7.75 m/s towards station.

When the train moving away form the station

Again beat frequency

[tex]\Delta f=f-f'[/tex]

[tex]f'=f-\Delta [/tex]

[tex]f'=175-4.05[/tex]

[tex]f'=170.95\ Hz[/tex]

We need to calculate the speed

Using Doppler effect

[tex]f'=f(\dfrac{v-v_{o}}{v+v_{s}})[/tex]

[tex]v=\dfrac{vf}{f'}-v[/tex]

Put the value into the formula

[tex]v=\dfrac{343\times175}{170.95}-343[/tex]

[tex]v=8.12\ m/s[/tex]

The speed of the train is 8.12 m/s away from the station.

Hence, This is the required solution.

Answer:

t = 2.96 s

Explanation:

Since the two stones hit the water at same instant of time

so we will have

[tex]d =vt + \frac{1}{2}gt^2[/tex]

here we know that

d = 47 m

v = 1.4 m/s

[tex]g = 9.81 m/s^2[/tex]

[tex]d = 1.40 t + \frac{1}{2}(9.81) t^2[/tex]

now by solving above equation for  d = 47 m

t = 2.96 s

Answer:1.26 m/s

Explanation:

Given

translation speed of ball =3.5 m/s

Moment of inertia of ball about com [tex]I=\frac{2}{5}mr^2[/tex]

Initial Energy

[tex]E_i=\frac{1}{2}mu^2+\frac{1}{2}I\omega _i^2(\omega =\frac{u}{r})[/tex]

Final Energy

[tex]E_f=\frac{1}{2}mv^2+\frac{1}{2}I\omega _f^2+mgh[/tex]

Equating energy as no energy loss take place

[tex]E_i=E_f[/tex]

[tex]\frac{1}{2}mu^2+\frac{1}{2}I\omega _i^2=\frac{1}{2}mv^2+\frac{1}{2}I\omega _f^2+mgh[/tex]

[tex]\frac{1}{2}mu^2+\frac{1}{2}\times \frac{2}{5}mr^2\times \left ( \frac{u}{r}\right )^2=\frac{1}{2}mv^2+\frac{1}{2}\times \frac{2}{5}mr^2\times \left ( \frac{v}{r}\right )^2+mgh[/tex]

m term get cancel

[tex]\left ( \frac{u^2}{2}\right )+\left ( \frac{2u^2}{10}\right )=\left ( \frac{v^2}{2}\right )+\left ( \frac{2v^2}{10}\right )+gh[/tex]

[tex]\frac{7}{10}u^2=\frac{7}{10}v^2+gh[/tex]

[tex]v^2=3.5^2-\frac{10}{7}\times 9.81\times 0.76[/tex]

[tex]v=\sqrt{1.6}=1.26 m/s[/tex]

Answer:

1.93 m

Explanation:

Initial velocity of fish u = 11 m /s

Angle of jump = 34 degree.

Vertical component of its velocity = u sin34 = 11 x.5592 = 6.15 m /s

Considering its motion in vertical direction ,

u = 6.15 m/s

g = - 9.8 m /s

maximum height attained = h

From the formula

v² = u² - 2gh

0 = 6.15x 6.15 - 2 x 9.8 h

h = (6.15 x 6.15 )/ 2 x 9.8

= 1.93 m .

Answer:

A) The maximum height of the blue ball is 29.7 m above the ground.

B) The height of the blue ball after 1.8 s of throwing the red ball is 22.3 m

C) The balls are at the same height 1.41 s after the red ball is thrown

Explanation:

A) At maximum height, the velocity of the blue ball is 0 because for that instant, the ball does not go up nor down.

The equation for velocity for an accelerated object moving in a straight line is:

v = v0 + a*t

where

v = velocity.

v0 = initial velocity.

a = acceleration, in this case, it is the acceleration of gravity, g, 9.81 m/s².

t = time.

Then:

0 = v0 + g * t  (if the origin of the reference system is the ground, then g is negative)

0 = 23.8 m/s - 9.81 m/s² * t

-23.8 m/s / -9.81 m/s² = t

t = 2.43 s

With this time, we can calculate the position of the blue ball. The equation for position is:

y = y0 + v0 * t + 1/2 * g * t²

y = 0.8 m + 23.8 m/s * 2.43 s - 1/2 * 9.81 m/s² * (2.43 s)²

y = 29.7 m

the maximum height of the blue ball is 29.7 m above the ground.

B) 1.8 s after throwing the red ball, the blue ball was in the air for (1.8 s - 0.6) 1.2 s. Then, using the equation for the position of the blue ball:

y = 0.8 m + 23.8 m/s * 1.2 s - 1/2 * 9.81 m/s² * (1.2 s)² = 22.3 m

The height of the blue ball after 1.8 s of throwing the red ball is 22.3 m

C) Now, we have to find the time at which both positions are equal. Notice that the time of the blue ball is not the same as the time for the red ball. The time for the blue ball is the time of the red ball minus 0.6 s:

t blue = t red - 0.6 s

Then:

position red ball = position blue ball

 y0 + v0 * t + 1/2 * g * t² = y0 + v0 * (t- 0.6) + 1/2 * g * (t-0.6s)²

25 m + 1.2 m/s * t -1/2 * 9.81 m/s² * t² = 0.8 m + 23.8 m/s * (t-0.6 s) - 1/2 * 9.81 m/s² * (t-0.6 s)²

24.2 m + 1.2 m/s * t -4.91 m/s² * t² = 23.8 m/s * t - 14.28 m - 4.91 m/s² * (t-0.6 s)²

38.5 m - 22.6 m/s * t - 4.91 m/s² * t² = -4.91 m/s² (t² - 1.2 s * t + 0.36 s²)

38.5 m - 22.6 m/s * t - 4.91 m/s² * t² = -4.91 m/s² * t² + 5.89 m/s * t - 1.77 m

40.3 m - 28.5 m/s * t = 0

t = -40.3 m / -28.5 m/s

t = 1.41 s

The balls are at the same height 1.41 s after the red ball is thrown and 0.81 s after the blue ball is thrown.

Answer:1 s

Explanation:

Given

Initial velocity in Y-direction [tex]v_1[/tex]=+4 m/s

Final Velocity in Y-direction [tex]v_2=-5.8 m/s[/tex]

Acceleration in Y-direction is 9.81 [tex]m/s^2[/tex]

Using equation of motion

v=u+at

[tex]-5.8=4+9.81\times t[/tex]

[tex]t=0.998 \approx 1 s[/tex]

Answer:

a) d=65.7 m : The automobile overtake passes the truck at a distance of 65.7 m  from the traffic signal

b) vₐ=16.94m/s : Automobile  speed for t = 7.7s

Explanation:

When the car catches the truck, the two will have passed the same time (t) and will have traveled the same distance (d):

Automobile kinematics:

car moves with uniformly accelerated movement:

d = v₀*t + (1/2)a*t²

v₀ = 0 : initial speed  

d = (1/2)*a*t² Equation (1)

Truck kinematics:

Truck moves with constant speed:

d = v*t Equation (2)

Data:

a=2.20 m/s² : automobile acceleration

v= 8.50 m/s  ; truck speed

Equation (1) =Equation (2)

(1/2)*a*t²= v*t

(1/2)*2.2*t²= 8.5*t   We divide both sides of the equation by t:

1.1*t=8.5

t=8.5÷1.1

t=7.7 s

We replace t=7.7 s in the equation (2)

d=8.5*7.7

d=65.7 m

Automobile  speed for t = 7.7s

vₐ=v₀₊a*t

vₐ=o₊2,2*7.7

vₐ=16.94m/s

The speed of the ball when passing its original height on its way down is 20 m/s. The time it takes to reach the ground level is 3.19 seconds. The ball travels a total distance of 70 meters.

The motion of the tennis ball can be analyzed by using the principles of Physics, particularly the laws of motion and the concept of gravity.

(a) The ball will have the same speed when it passes its original height on its way down, which is 20 m/s. This is based on the principle of conservation of energy. Since air resistance is ignored, the speed when it passes the original height on its way down should be the same as its initial speed.

(b) Calculating the total time it takes the ball to reach the ground involves the formula for time in free fall, which is t = sqrt(2h/g), wherein h is the height and g is the gravity. Thus plug in the values: t = sqrt((2*50)/9.8) = 3.19 seconds.

(c) The total distance traveled by the ball includes it going up and falling back down. The ball rises to a height of 20 m/s before falling from that height plus an additional height corresponding to the height of the cliff, total distance is 70 meters.

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