A rigid, insulated tank that is initially evacuated is connected through a valve to a supply line that carries steam at 1 MPa and 300 ºC. Now the valve is opened and steam is allowed to flow slowly into the tank until the pressure reaches 1 MPa, at which point the valve is closed. Determine the final temperature of the steam in the tank.

Answer:

Height, H = 25.04 meters

Explanation:

Initially the ball is at rest, u = 0

Time taken to fall to the ground, t = 2.261 s

Let H is the height from which the ball is released. It can be calculated using the second equation of motion as :

[tex]H=ut+\dfrac{1}{2}at^2[/tex]

Here, a = g

[tex]H=\dfrac{1}{2}gt^2[/tex]            

[tex]H=\dfrac{1}{2}\times 9.8\times (2.261)^2[/tex]

H = 25.04 meters

So, the ball is released form a height of 25.04 meters. Hence, this is the required solution.

Final answer:

Using the equation of motion under gravity, it was calculated that a ball dropped and taking 2.261 s to hit the ground was released from a height of approximately 25.07 meters.

Explanation:

To find out from what height H a ball was released if it takes 2.261 s to fall to the ground, we use the equation of motion under gravity, which is H = 0.5 * g * t^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2 on the surface of the Earth) and t is the time in seconds.

Substituting the given values in the equation, we get: H = 0.5 * 9.8 * (2.261)^2. After calculating, we find that the height H approximately equals 25.07 meters.

Therefore, the ball was released from a height of approximately 25.07 meters.

Final answer:

To estimate the electric field near a uniformly charged wire, one must determine the linear charge density and integrate the contributions from infinitesimal charge elements along the wire, considering the symmetry that leads to the cancellation of horizontal components.

Explanation:

To estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire, we use the principle of superposition and integrate the contributions of the infinitesimally small charge elements along the wire.

First, we determine the linear charge density λ by dividing the total charge Q by the length L of the wire:

λ = Q / L

Then, we apply the formula for the electric field due to an infinitesimal charge element dQ at a distance r:

dE = (k * dQ) / r²

Where k is the Coulomb's constant (approximately 8.99 × 10⁹ Nm²/C²). Since we need to integrate over the length of the wire to find the total electric field, we use the linear charge density λ and a differential element of the wire's length dl:

dQ = λ * dl

The resulting integral, when evaluated, will give us the magnitude of the electric field E at the specified distance from the wire.

It is important to note that due to the symmetry of the problem, the horizontal components of the electric field due to charge elements on the wire will cancel out, leaving only the vertical components to contribute to the total electric field at the point of interest

Answer:

Depends on the specific relation of acceleration as function of time, but it would always look like decresing or increasing negatively.

Explanation:

Acceleration is the derivative of velocity with respect to time. That means that it is the change. The fact that is negative means that it is decreasing, or increasing in the negative direction (i.e. going backwards faster).

Attached is the graph with constant negative acceleration, using kinematics relations, it would be a linear equation with negative slope.

Answer:a straight line with negative slope

Explanation:

Answer:  b) TRUE and d) TRUE

Explanation:  a) FALSE the electric field lines are used to represent the charges postives and negatives.

b) TRUE it is the definition of  electric field lines , they are tangent to the electric field vector.

c) FALSE  the electric field lines only for conductor are perpendicular to the surface in other any situation there is a tangencial electric field components.

d) TRUE since it is a way to describe and imagine the effect of vectorial fields.

e) FALSE

Answer:

[tex]F=39,68N[/tex]

Explanation:

Data:

Mass [tex]m=25 Kg[/tex]

Coefficient of kinetic friction [tex]\mu=0.12[/tex]

Angle = [tex]29^{0}[/tex]

Acceleration = [tex]0.12 \frac{m}{s^{2} }[/tex]

Solution:

By Newton's first law we know that for the x-axis:

[tex]F_{rope_x}-F_f=F_R[/tex] Where [tex]F_R[/tex] is the resulting force, and [tex]F_f[/tex] is the friction force.

And for the y-axis:

[tex]F_{rope_y}+N=W[/tex], where N is the normal force, and W is the weight of the sled.  

We know that the resulting force's acceleration is [tex]0.12 \frac{m}{s^{2} }[/tex], and by using Newton's second law, we obtain:

[tex]F=m.a[/tex]

[tex]F_R=25Kg. 0.12\frac{m}{s^2} \\ F_R=3N[/tex] .

Now, the horizontal component of the force in the rope will be given by

[tex]F_{rope_x}=F_{rope}.cos(29^0)=F_R+F_f[/tex], since the resulting force is completely on the x-axis, and the friction opposes to the speed of the sled.

To obtain the friction force, we must know the normal force:

[tex]F_f=\mu. N[/tex]

Clearing N in the y-axis equation:

[tex]N=W-F_{rope_y}=W-F_{rope}.sin(29^0)[/tex]

So we can express the x-axis equation as follows:

[tex]F.cos(29^0)=F_R+\mu.(W-F_{rope}.sin(29^0))[/tex]

Finally, solving for F we get

[tex]F = (F_R + \mu. m.g) / (cos (29^0) + \mu.sin (29^0))[/tex]

[tex]F=39,68N[/tex]

Answer:

Explanation:

Shoveling gravel into the water will increase the buoyancy of the barge, which will make it float higher. Without data it is hard to tell if it will raise the barge enough to be counterproductive, but in any case throwing away the payload is not a good idea. Adding some weights would make the barge float lower, and then maybe the plie can make it under the bridge.

Answer:

The horizontal distance is 478.38 m

Solution:

As per the question:

Initial Speed of the bullet in horizontal direction, [tex]v_{x} = 670 m/s[/tex]

Initial vertical velocity of the bullet, [tex]v_{y} = 0 m/s[/tex]

Vertical distance, y = 0.025 m

Now, for the horizontal distance, 'x':

We first calculate time, t:

[tex]y = v_{y}t - \frac{1}{2}gt^{2}[/tex]

(since, motion is vertically downwards under the action of 'g')

[tex]0.025 = 0 - \frac{1}{2}\times 9.8t^{2}[/tex]

[tex]t = \sqrt{0.05}{9.8} = 0.0714 s[/tex]

Now, the horizontal distance, x:

[tex]x = v_{x}t + \frac{1}{2}a_{x}t^{2}[/tex]

[tex]x = v_{x}t + \frac{1}{2}0.t^{2}[/tex]

(since, the horizontal acceleration will always be 0)

[tex]x = 670\times 0.714 = 478.38 m[/tex]

Answer:

18.86° , it will bend towards normal.

Explanation:

For refraction,

Using Snell's law as:

[tex]n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}[/tex]

Where,  

[tex]{\theta_i}[/tex]  is the angle of incidence  ( 25.0° )

[tex]{\theta_r}[/tex] is the angle of refraction  ( ? )

[tex]{n_r}[/tex] is the refractive index of the refraction medium  (n=1.7)

[tex]{n_i}[/tex] is the refractive index of the incidence medium ( n=1.3)

Hence,  

[tex]1.3\times {sin\ 25.0^0}={1.7}\times{sin\theta_r}[/tex]

Angle of refraction = [tex]sin^{-1}0.3232[/tex] = 18.86°

Since, the light ray is travelling from a material with a refractive index of 1.3 into a material with a refractive index of 1.7 or lighter to denser medium, it will bend towards normal.

The diagram is shown below:

Answer:

b. precise

Explanation:

Accuracy means how close a measurement is to the true value (in our example the true value is score a goal )

Precision refers how close two or more measurements to each other.

So if the playing soccer hit always the left goal post, he is not accurate (e doesn't score any goals ) but he is very precise

Answer:

a) time taken = 2.66 s

b) v = 28.34

Explanation:

given,

rate of descending = 2.3 m/s

height of camera above ground = 41 m              

using equation of motion                      

[tex]h = u t + \dfrac{1}{2}gt^2[/tex]              

[tex]41 =2.3t + \dfrac{1}{2}\times 9.8\times t^2[/tex]

4.9 t² + 2.3 t - 41 =0                      

t = 2.66 ,-3.13                    

time taken = 2.66 s

b) v² = u² + 2 g h

v² = 0 + 2× 9.8 × 41

v = 28.34                            

Answer:

33 seconds.

Explanation:

The equation for speed with constant acceleration at time t its:

[tex]V(t) \ = \ V_0 \ + \ a \ t[/tex]

where [tex]V_0[/tex] is the initial speed, and  a its the acceleration.

Starting at rest, the initial speed will be zero, so

[tex]V_0 = 0[/tex]

the final speed is

[tex]V(t_{f1}) = 23 \frac{m}{s}[/tex]

and the acceleration is

[tex]a = 2.3 \frac{m}{s^2}[/tex].

Taking all this together, we got

[tex]V(t_{f1}) = 23 \frac{m}{s} = 0 + 2.3 \ \frac{m}{s^2} t_{f1}[/tex]

[tex]23 \frac{m}{s} = 2.3 \ \frac{m}{s^2} t_{f1}[/tex]

[tex]\frac{23 \frac{m}{s}}{2.3 \ \frac{m}{s^2}} =  t_{f1}[/tex]

[tex]10 s =  t_{f1}[/tex]

So, for the first half of the problem we got a time of 10 seconds.

Now, the initial speed will be

[tex]V_0 = 23 \frac{m}{s}[/tex],

the acceleration

[tex]a=-1.0 \frac{m}{s^2}[/tex],

with a minus sign cause its slowing down, the final speed will be

[tex]V(t_{f2}) = 0[/tex]

Taking all together:

[tex]V(t_{f2}) = 0 = 23 \frac{m}{s} -  1.0 \frac{m}{s^2} t_{f2}[/tex]

[tex] 23 \frac{m}{s} =  1.0 \frac{m}{s^2} t_{f2}[/tex]

[tex] \frac{23 \frac{m}{s}}{1.0 \frac{m}{s^2}} = t_{f2}[/tex]

[tex] 23 s = t_{f2}[/tex]

So, for the first half of the problem we got a time of 23 seconds.

[tex]t_total = t_{f1} + t_{f2} = 33  \ s[/tex]

Answer:

Gravitational potential energy

Explanation:

Hydroelectric dams are the power plants which generates electricity by using the energy of falling water from a great height.

A the water is stored in big reservoirs and at a great height, it contain lot of potential energy due to the height. As it falls downwards, the potential energy is converted into kinetic energy of the water. this kinetic energy of the falling water is used to run the turbine, and then the electric energy is generated.

So, in hydroelectric power stations, the potential energy of water is converted into the electric energy.

Answer:

(a) 4.04 ohm

(b) 6.93 A

(c) 8.53°

Explanation:

f = 12 kHz = 12000 Hz

Vo = 28 V

R = 4 ohm

L = 30 micro Henry = 30 x 10^-6 H

C = 8 micro Farad = 8 x 10^-6 F

(a) Let Z be the impedance

[tex]X_{L} = 2\pi fL=2\times3.14\times12000\times30\times10^{-6}= 2.26 ohm[/tex]

[tex]X_{c} = \frac{1}{2\pi fC}=\frac{1}{2\times3.14\times12000\times8\times10^{-6}}= 1.66 ohm[/tex]

[tex]Z = \sqrt{R^{2}+(X_{L}-X_{C})^{2}}=\sqrt{4^{2}+\left ( 2.26-1.66 \right )^{2}}[/tex]

Z = 4.04 Ohm

(b) Let Io be the amplitude of current

[tex]I_{o}=\frac{V_{o}}{Z}[/tex]

[tex]I_{o}=\frac{28}{4.04}[/tex]

Io = 6.93 A

(c) Let the phase difference is Ф

[tex]tan\phi = \frac{X_{L}-X_{C}}{R}[/tex]

[tex]tan\phi = \frac{2.26-1.66}{4}[/tex]

tan Ф =0.15

Ф = 8.53°

Answer:

8717 meters.

Explanation:

We need to know the conversion factors. We know that:

1 mile = 1609.34 meters

1 yard = 0.9144 meters

This means that:

[tex]\frac{1609.34 meters}{1 mile}=1[/tex]

[tex]\frac{0.9144 meters}{1 yard}=1[/tex]

It is convenient to leave the units we want at the end in the numerator so the ones in the denominator cancel out with the ones we want to remove, as will be seen in the next step.

We will convert first the miles, then the yards, and add them up.

[tex]5miles=5miles\frac{1609.34 meters}{1 mile}=8046.7meters[/tex]

[tex]733yards=733yards\frac{0.9144 meters}{1 yard}=670.2552meters[/tex]

So total distance is the sum of these,  8717 meters.

Answer: The two places altitudes are: 16.17 m and 40.67 m

Explanation:

Hi!

Lets call z to the vertical direction (z= is ground) . Then the positions of the balloon and the pellet, using the values of the velocities we are given, are:

[tex]z_b =\text{balloon position}\\z_p=\text{pellet position}\\z_b=(7\frac{m}{s})t\\z_p=30\frac{m}{s}(t-t_0)-\frac{g}{2}(t-t_0)^2\\g=9.8\frac{m}{s^2}[/tex]

How do we know the value of t₀? This is the time when the pellet is fired. At this time the pellet position is zero: its initial position. To calculate it we know that the pellet is fired when the ballon is in z = 12m. Then:

[tex]t_0=\frac{12}{7}s[/tex]

We need to know the when the z values of balloon and pellet is the same:

[tex]z_b=z_p\\(7\frac{m}{s})t =30\frac{m}{s}(t-\frac{12}{7}s)-\frac{g}{2}(t-\frac{12}{7}s)^2[/tex]

We need to find the roots of the quadratic equation. They are:

[tex]t_1=2.31s\\t_2=5.81[/tex]

To know the altitude where the to objects meet, we replace the time values:

[tex]z_1=16,17m\\z_2=40,67m[/tex]

Answer:

The initial velocity of the ball should be 50 m/s.

Explanation:

Since the trip of the ball shall consist of upward ascend and the downward descend and since the ascend and the descend of the ball is symmetrical we infer that the upward ascend of the ball shall last for a time of 5 seconds.

Now since the motion of ball is uniformly accelerated we can find the initial speed of the ball using first equation of kinematics as

[tex]v=u+gt[/tex]

where,

'v' is final velocity of the ball

'u' is initial velocity of the ball

'g' is acceleration due to gravity

't' is the time of motion

Now we know that the ball will continue to ascend until it's velocity becomes zero hence to using the above equation we can write

[tex]0=u-9.81\times 5\\\\\therefore u=9.81\times 5=49.05m/s\approx 50m/s[/tex]

Answer:

h = 50.49 m

Explanation:

Data provided:

Speed of skier, u = 2.0 m/s

Maximum safe speed of the skier, v = 30.0 m/s

Mass of the skier, m = 85.0

Total work = 4000 J

Height from the starting gate = h

Now, from the law of conservation of energy

Total energy at the gate = total energy at the time maximum speed is reached

[tex]\frac{1}{2}mu^2+mgh=4000J+\frac{1}{2}mv^2[/tex]

where, g is the acceleration due to the gravity

on substituting the values, we get

[tex]\frac{1}{2}\times85\times2.0^2+85\times9.81\times h=4000J+\frac{1}{2}\times85\times30^2[/tex]

or

170 + 833.85 × h = 4000 + 38250

or

h = 50.49 m

Answer:

[tex]F = 1361.1 N[/tex]

Explanation:

As we know that a charge is split into two parts

so we have

[tex]q_1 + q_2 = 7\mu C[/tex]

now we have

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

here we know

[tex]F = \frac{kq_1(7\mu C - q_1)}{r^2}[/tex]

here we have

r = 9 mm

now to obtain the maximum value of the force between two charges

[tex]\frac{dF}{dq_1} = 0[/tex]

so we have

[tex]7 \mu C - q_1 - q_1 = 0[/tex]

so we have

[tex]q_1 = q_2 = 3.5 \mu C[/tex]

so the maximum force is given as

[tex]F = \frac{(9\times 10^9)(3.5 \mu C)(3.5 \mu C)}{0.009^2}[/tex]

[tex]F = 1361.1 N[/tex]

Answer:

Ff= 8.9N:Force of friction acting on the block.

Explanation

Box kinetics:  in x₁-y₁ :We apply  the second law of Newton

∑Fx₁=ma : second law of Newton  Formula (1)

Where:

∑Fx₁:algebraic sum of forces in the direction of x1, positive in the direction of movement of the block, which is down and negative in the direction opposite to the movement of the block

m: is the mass of the block

The x₁ axis coincides with the plane of sliding of the block, that is, the x₁ axis forms 45 degrees with the horizontal

The total weight (W) of the block is in the vertical direction and the tip of the vector down and its magnitude is calculated as follows:

W=m*g

Where:

m: is the mass of the block

g: is the acceleration due to gravity

Calculation of the weight in the direction x₁

Wx₁=Wcos45°= m*g*cos 45 Equation( 1)

Data

m=1.8kg

g=9.8 m/s²

a=2m/s

Wx₁=m*g*cos 45°= 1.8*9.8 *[tex]\frac{\sqrt{2} }{2}[/tex]= 12.47 N

Friction force (Ff ) calculation

We apply formula (1)

∑Fx₁=m*a

Wx₁ - Ff= m*a

Ff=Wx₁ - m*a

We replace data

Ff= 12.47 - 1.8*2 =8.87

Ff= 8.9N

Answer:

runway use is 3307.8 feet

Explanation:

given data

velocity = 140 kts = 140 × 0.5144 m/s = 72.016 m/s

time = 28 seconds

weight = 28000 lbs

to find out

How many feet of runway was used

solution

we will use here first equation of motion for find acceleration

v = u + at     ..............1

here v is velocity given and u is initial velocity that is 0 and a is acceleration and t is time

put here value in equation 1

72.016 = 0 + a(28)

a = 2.572 m/s²

and

now apply third equation of motion

s = ut + 0.5×a×t²    .......................2

here s is distance and u is initial speed and t is time and a is acceleration

put here all value in equation 2

s = 0 + 0.5×2.572×28²  

s = 1008.24 m = 3307.8 ft

so  runway use is 3307.8 feet

To find the runway distance used by an aircraft accelerating to 140 knots in 28 seconds, we convert knots to feet per second, find the acceleration, and then apply the kinematic formula for distance to obtain approximately 3308 feet.

To calculate how many feet of runway an aircraft used to accelerate to 140 knots in 28 seconds, we need to convert the speed to consistent units and use the kinematic equations for uniformly accelerated motion. Knots need to be converted to feet per second (since the answer is required in feet). Since 1 knot = 1.68781 feet per second, 140 knots is equivalent to 236.293 feet per second. Now, we can use the formula for distance d when given initial velocity vi, final velocity vf, and acceleration a:

d = (vi + vf) / 2 * t

Here, as the plane is starting from a standstill, vi is 0, vf is 236.293 feet/second, and t is 28 seconds. We first need to find the acceleration:

a = (vf - vi) / t = 236.293 / 28 = 8.439 feet/second2

Now we can calculate the distance:

d = (0 + 236.293) / 2 * 28 = 3308.1 feet

The aircraft used approximately 3308 feet of runway.

Answer:

Its dissipated by the brake.

Explanation:

Traditional brakes use friction to stop the wheels (or axis). This friction its dissipated in the form of heat. There are others mechanism to brake that don't dissipated the energy, they stored it. In electric cars (or hybrids), there are regenerative brakes, that store the kinetic energy as electrical energy.

As the driver slams the brakes, the kinetic energy of the car gets converted into other forms of energy until the car comes to a stop, at which point it becomes zero.

When the driver of a car slams on her brakes to avoid a collision with a deer crossing the highway, the kinetic energy of the car decreases until the car comes to rest. This is due to the principle of energy conservation. Initially, when the car is moving, it has kinetic energy. However, when the brakes are applied, the kinetic energy gets converted into other forms of energy such as heat energy (due to friction between the brakes and the wheels) and potential energy (if the car is moving uphill). The kinetic energy keeps decreasing until the car comes to a complete stop. At that point, the kinetic energy of the car is zero because kinetic energy is associated with motion and the car is no longer moving.

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(a) Initial velocity (v₀) = 0 m/s,  Final velocity (v) = 15.0 m/s.

(b) The dolphin rises approximately 11.5 meters above the water.

(c) The dolphin is in the air for approximately 1.5 seconds.

(a)

Initial velocity (v₀) = 0 m/s (since the dolphin jumps straight up from the water, its initial velocity is zero)

Final velocity (v) = 15.0 m/s (given)

(b)

To solve for the height, we can use the kinematic equation that relates the final velocity (v), initial velocity (v₀), acceleration (a), and displacement (Δy):

v² = v₀² + 2aΔy

v² = v₀² + 2gh

v² = 2gh

Solving for h:

h = (v²) / (2g)

h = (15.0)² / (2 × 9.8)

h ≈ 11.5 m

Therefore, the dolphin rises approximately 11.5 meters above the water.

(c) To determine the time the dolphin is in the air, we can use the equation for time (t) derived from the kinematic equation:

Δy = v₀t + (1/2)at²

Δy = (1/2)at²

t = √((2Δy) / a)

t = √((2 × 11.5) / -9.8)

t ≈ 1.5 s

Therefore, the dolphin is in the air for approximately 1.5 seconds.

To know more about the velocity:

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The dolphin rises 8.64 meters above the water and is in the air for 1.33 seconds.

(a) The knowns in this problem are:

(b) To find the height above the water, we need to find the time taken by the dolphin to reach that height and use the equation:

Final velocity = Initial velocity + Acceleration × Time

Solving for time, we have:

Time = (Final velocity - Initial velocity) / Acceleration

Substituting the known values into the equation, we get:

Time = (0 m/s - 13.0 m/s) / -9.8 m/s² = 1.33 s

Next, we can use the formula for height:

Height = Initial velocity × Time + 0.5 × Acceleration × Time²

Substituting the known values, we have:

Height = 13.0 m/s × 1.33 s + 0.5 × -9.8 m/s² × (1.33 s)² = 8.64 m

(c) The time the dolphin is in the air is equal to the time calculated above, which is 1.33 seconds.

Answer:

He knocks her. V = 7.97m/s

Explanation:

Let A be Tarzan's starting position and B Jane's position (Tarzan's final position).

Since there is no air resistance, energy at position A must be equal to energy at B.

At position A, Tarzan's speed is zero and since the 45° of the vine is greater than the final 30°, Tarzan will have potential energy at point A.

[tex]E_{A}=m_{T}*g*L*(cos(30)-cos(45))[/tex]

At point B, it is the lowest point (between A and B), so it has no potential energy. It will only have kinetic energy (ideally zero, for Jane's sake, but we don't know).

[tex]E_{B}=\frac{m_{T}*V^{2}}{2}[/tex]

Because of energy conservation, we know that Ea=Eb, so:

[tex]m_{T}*g*L*(cos(30)-cos(45))=\frac{m_{T}*V^{2}}{2}[/tex]  Solving for V:

[tex]V=\sqrt{2*g*L*(cos(30)-cos(45))}=7.97m/s[/tex]

Answer:

2.82 s

Explanation:

The ball will be subject to the acceleration of gravity which can be considered constant. Therefore we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

Y0 is the starting position, 2.3 m in this case.

Vy0 is the starting speed, 13 m/s.

a will be the acceleration of gravity, -9.81 m/s^2, negative because it points down.

Y(t) = 2.3 + 13 * t - 1/2 * 9.81 * t^2

It will reach the ground when Y(t) = 0

0 = 2.3 + 13 * t - 1/2 * 9.81 * t^2

-4.9 * t^2 + 13 * t + 2.3 = 0

Solving this equation electronically gives two results:

t1 = 2.82 s

t2 = -0.17 s

We disregard the negative solution. The ball spends 2.82 seconds in the air.

Answer:

μ= 0.408 : coefficient of kinetic friction

Explanation:

Kinematic equation for the box:

[tex]a=\frac{v_{f} -v_{i} }{t}[/tex] Formula( 1)

a= acceleration

v_i= initial speed =0

v_f= final speed= 20 m/s

t= time= 5 s

We replace data in the formula (1):

[tex]a=\frac{0-20}{5}[/tex]

[tex]a= -\frac{20}{5}[/tex]

a= - 4m/s²

Box kinetics: We apply Newton's laws in x-y:

∑Fx=ma : second law of Newton

-Ff= ma Equation (1)

Ff is the friction force

Ff=μ*N Equation (2)

μ is the coefficient of kinetic friction

N is the normal force

Normal force calculation

∑Fy=0  : Newton's first law

N-W=0   W is the weight of the box

N=W= m*g  : m is the mass of the box and g is the acceleration due to gravity

N=9.8*m

We replace N=9.8m in the equation (2)

Ff=μ*9.8*m

Coefficient of kinetic friction ( μ) calculation

We replace Ff=μ*9.8*m and a= -4m/s² in the equation (1):

-μ*9.8*m: -m*4 : We divide by -m on both sides of the equation

9.8*μ=4

μ=4 ÷ 9.8

μ= 0.408

Answer:

longest length is 13304.05 m

Explanation:

given data

diameter = 25 mm

so radius r = 12.5 × [tex]10^{-3}[/tex] m

density = 5000 kg/m³

tension = 320000 N

to find out

longest length of of this rope

solution

we know that here tension is express as

T = m × g

and m is mass and g is  acceleration due to gravity

here mass = density × volume

and volume = π×r²×l

here r is radius and l is length

so T = density c volume  × g

put here value

320000 = 5000 × π×12.5² × [tex]10^{-3}[/tex] × l × 9.8

solve it we get length l

l = 13304.05

so longest length is 13304.05 m

Answer:

a) 3.39 × 10²³ atoms

b) 6.04 × 10⁻²¹ J

c) 1349.35 m/s

Explanation:

Given:

Diameter of the balloon, d = 29.6 cm = 0.296 m

Temperature, T = 19.0° C = 19 + 273 = 292 K

Pressure, P = 1.00 atm = 1.013 × 10⁵ Pa

Volume of the balloon = [tex]\frac{4}{3}\pi(\frac{d}{2})^3[/tex]

or

Volume of the balloon = [tex]\frac{4}{3}\pi(\frac{0.296}{2})^3[/tex]

or

Volume of the balloon, V = 0.0135 m³

Now,

From the relation,

PV = nRT

where,

n is the number of moles

R is the ideal gas constant = 8.314  kg⋅m²/s²⋅K⋅mol

on substituting the respective values, we get

1.013 × 10⁵ × 0.0135 = n × 8.314 × 292

or

n = 0.563

1 mol = 6.022 × 10²³ atoms

Thus,

0.563 moles will have = 0.563 × 6.022 × 10²³ atoms = 3.39 × 10²³ atoms

b) Average kinetic energy = [tex]\frac{3}{2}\times K_BT[/tex]

where,

Boltzmann constant, [tex]K_B=1.3807\times10^{-23}J/K[/tex]

Average kinetic energy = [tex]\frac{3}{2}\times1.3807\times10^{-23}\times292[/tex]

or

Average kinetic energy = 6.04 × 10⁻²¹ J

c) rms speed = [tex]\frac{3RT}{m}[/tex]

where, m is the molar mass of the Helium = 0.004 Kg

or

rms speed = [tex]\frac{3\times8.314\times292}{0.004}[/tex]

or

rms speed = 1349.35 m/s

The straight-line distance from city A to city C is approximately 458.80 km. The direction from city A to city C is 14.9° north of west. The approximation is due to ignoring Earth's curvature.

To solve this problem, we will use the laws of vector addition as airplanes movements are vector quantities as it involves both magnitude (distance) and direction. Initially, the airplane flies 200 km west. Then, it changes its direction 26.0° north of west and flies an additional 275 km.

When we decompose the second segment of the flight into its westward and northward components, we can use trigonometric principles. The horizontal (westward) distance is 275 km * cos(26.0°) = 245.56 km. Adding that to the 200 km the airplane flew initially gives us a total westward distance of 445.56 km. The vertical (northward) distance is 275 km * sin(26.0°) = 118.40 km.

To find the straight-line distance between City A and City C, we apply the Pythagorean theorem: √[(445.56 km)² + (118.4 km)²] = 458.80 km. To find the direction, we take the arctan of the northward distance over the westward distance: arctan(118.4 km / 445.56 km) = 14.9° north of west.

It's only approximately correct because we ignore the curvature of the Earth, which would slightly modify the distances and angles involved.

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The straight-line distance from city A to city C is approximately 460 km and the direction is 15° north of west. The answer is only an approximation because we've considered a flat plane, ignoring the spherical nature of the Earth.

The question is asking for the linear distance and direction from city A to city C, which can be solved through vector addition and trigonometric calculations in physics. The displacement from A to B is 200 km west. The displacement from B to C is 275 km, 26.0° north of west. These can be seen as components of a right triangle, and by applying the Pythagorean theorem, the straight-line distance, or the hypotenuse, can be calculated.

First, we derive the north and west components of the 275 km using cosine and sine respectively as the motion is angled. The westward motion is 275cos(26) = 244 km. Add this to the 200 km westward motion to get a total westward motion of 444 km. The northward motion is 275sin(26) = 121 km.

Then, we apply Pythagorean theorem. √(444^2 + 121^2) ≈ 460 km. This is the straight-line distance from city A to city C.

For the direction, we find the angle, using tan^(-1)(121/444) ≈ 15° north of west.

The answer is only approximately correct because in real situation we have to take into account the spherical nature of the Earth, but here we've considered a flat plane for simplicity.

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Answer:

Option (c) [tex]2\times 10^{-3}\ m=2\ millimeters[/tex]

Explanation:

Here, it is required to describe the given length in a particular unit. Firstly, we need to see the following conversions as :

1 meter = 0.001 kilometers

1 meter = 10⁻⁶ megameters

1 meter = 1000 millimeters

1 meter = 1000000 micrometers

From the given option, the correct one is (c) because, 1 meter = 1000 millimeters

So, [tex]2\times 10^{-3}\ m=2\ millimeters[/tex]

Hence, the correct option is (c). Hence, this is the required solution.

2.0×10^-3 of a meter is equivalent to 2.0 millimeters, hence option c is the correct answer.

The given length, 2.0×10^-3 of a meter, corresponds to a unit of length commonly used in the metric system. In this system, 1 meter is equal to 10^3 millimeters. Therefore, 2.0×10^-3 of a meter equates to 2.0 millimeters. This means the best answer from the provided options would be option c: 2.0 millimeters.

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Answer:[tex]\theta =41.409 ^{\circ}[/tex]

Explanation:

Given

Jack and Jill ran up the hill at 2.8 m/s

Horizontal component of Jill's velocity vector was 2.1 m/s

Let [tex]\theta [/tex]is the angle made by Jill's velocity with it's horizontal component

Therefore

[tex]2.8cos\theta =2.1[/tex]

[tex]cos\theta =\frac{2.1}{2.8}[/tex]

[tex]cos\theta =0.75[/tex]

[tex]\theta =41.409 ^{\circ}[/tex]

Vertical velocity is given by

[tex]V_y=2.8sin41.11=1.85 m/s[/tex]

The question can be solved using the Pythagorean theorem to calculate the vertical velocity component and apply trigonometry to find the angle of the hill. Remember the conversion from radians to degrees.

The question asks about the angle of the hill that Jack and Jill climbed, as well as the vertical component of Jill’s velocity. We know that Jill ran up the hill at a velocity of 2.8 m/s and the horizontal component of her velocity was 2.1 m/s. These two components form a right angle triangle where the hypotenuse is the total velocity (2.8 m/s), one side is the horizontal velocity (2.1 m/s), and the other side, which we're finding, is the vertical component of the velocity.

We can find the angle of the hill, θ, using the tangent function of trigonometry. Tan θ = opposite side / adjacent side. In this case, the 'opposite side' is the vertical velocity component we're after, and the 'adjacent side' is the horizontal component (2.1 m/s). To find the vertical velocity component, you can use the Pythagorean theorem, which states that: (Hypotenuse)² = (Adjacent Side)² + (Opposite Side)² or (2.8 m/s)² = (2.1 m/s)² + (vertical velocity)².

Once you solve for the vertical velocity through the Pythagorean theorem, you then insert it into the tangent equation to solve for the angle θ. Remember that when you use the arctangent function on a calculator to find the angle, the answer will likely be in radians, so you might need to convert them into degrees.

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