A fighterjet is launched from an aircraft carrier with the
aidof its own engines and a steam powered catapult. The thurst of
itsengine2.3*10^5N. In being launched from rest it moves through
adistance 87m and has KE 4.5*10^7J at lift off.What is the work
doneon the jet by the catapult.

Final answer:

To find the initial speed of a ball thrown to a maximum height of 36 m, we use kinematic equations that factor in the acceleration due to gravity. The ball's initial speed can be calculated using the formula for objects under constant acceleration, considering that the final velocity at the max height is 0 m/s.

Explanation:

Calculating the Launch Speed of the Ball

To determine the initial speed at which the ball was thrown to reach a maximum height of 36 m, we can use the principles of kinematics under the influence of gravity. In the absence of air resistance, a ball thrown upwards will decelerate at a rate equal to the acceleration due to gravity until it comes to a stop at its maximum height. We use the following kinematic equation for an object under constant acceleration:

s = ut + 1/2at^2

Where:

s is the displacement (maximum height in this case, which is 36 m)

u is the initial velocity (what we want to find)

a is the acceleration due to gravity (-9.81 m/s^2, the negative sign indicates acceleration is in the direction opposite to the initial velocity)

t is the time taken to reach the maximum height (not needed in this calculation)

At the maximum height, the final velocity (v) is 0 m/s, so we use the following equation:

v^2 = u^2 + 2as

Plugging in the known values:

0 = u^2 + 2(-9.81 m/s^2)(36 m)

u^2 = 2(9.81 m/s^2)(36 m)

u = √(2(9.81 m/s^2)(36 m))

The initial speed u can be calculated from this equation to find out with what speed the ball was thrown to achieve a 36 m height.

Answer:81.24 m/s

Explanation:

Given

Horizontal displacement([tex]R_x[/tex])=500

Angle of projection[tex]=24 ^{\circ}[/tex]

Let u be the launching velocity

and horizontal range is given by

[tex]R_x=\frac{u^2sin2\theta }{g}[/tex]

[tex]500=\frac{u^2sin48}{9.81}[/tex]

[tex]u^2=\frac{500\times 9.81}{0.7431}[/tex]

[tex]u^2=6600.32854[/tex]

[tex]u=\sqrt{6600.32854}=81.24 m/s[/tex]

Answer:

Explanation:

The plates A and B are charged by opposite charges  but which plate is positively charged and which is negatively charged is not clear.

Now a proton which is positively charged is moving from plate A to B . If it is attracted by plate B then its kinetic energy will be increased and potential energy will be decreased due to conservation of energy . In that case B will be negatively charged .

                                                                  But in the given case it is stated that

potential energy of the proton increases . That means its kinetic energy decreases . In other words its speed decreases . It points to the fact that plate B is also positively charged.

So proton will be repelled and its speed will be decreased.

Answer:

power = 384.92 W

Explanation:

given data

resistance = 44.0 Ω

impedance = 71 Ω

ΔVrms = 210 V

to find out

average power

solution

we know here power formula that is

power =  [tex]\frac{V^2rms}{impedance}[/tex]cos∅ .........1

we know here cos∅  =  [tex]\frac{resistance}{impedance}[/tex]

cos ∅  =  [tex]\frac{44}{71}[/tex] = 0.6197

so from equation 1

power =  [tex]\frac{210^2}{71}[/tex] × 0.6197

power = 384.92 W

Answer:

2.04 m/s

Explanation:

Given:

Assumptions:

Work-energy theorem: For the various forces acting on an object, the work done by all the forces brings a change in kinetic energy of an object which is equal to the total work done.

For the initial case, the puck travels half the distance of the target teammate. In this case, the change in kinetic energy of the puck will be equal to the work done by the friction.

[tex]\therefore \dfrac{1}{2}m(v^2-u^2)=f_k\dfrac{x}{2}\\\Rightarrow \dfrac{1}{2}m(0^2-1.7^2)=f_k\dfrac{x}{2}[/tex]...........eqn(1)

Now, again using the work energy theorem for the puck to reach the targeted teammate, the change in kinetic energy of the puck will be equal to the work done by the kinetic friction.

[tex]\therefore \dfrac{1}{2}m(v^2-U^2)=f_k\dfrac{x}{2}\\\Rightarrow \dfrac{1}{2}m(0^2-U^2)=f_kx[/tex]...........eqn(2)

On dividing equation (1) by (2), we have

[tex]\dfrac{-1.7^2}{-U^2}=\dfrac{1}{2}\\\Rightarrow \dfrac{1.7^2}{U^2}=\dfrac{1}{2}\\\Rightarrow U^2= 2\times 1.7^2\\\Rightarrow U^2 = 5.78\\\Rightarrow U=\pm \sqrt{5.78}\\\Rightarrow U=\pm 2.04\\\textrm{Since the speed is always positive.}\\\therefore U = 2.04\ m/s[/tex]

Hence, the puck must be kicked with a minimum initial speed of 2.04 m/s so that it reaches the teammate.

To answer this question, you substitute t = 40 s into the given parametric equations to find the horizontal distance from the airport and the altitude. Then, you take the derivative of both equations to find the speed, and the second derivative to find the acceleration.

To solve this problems, you will need to use the given parametric equations. The horizontal distance from the airport (a) is given by x = 1.25 t^2. At t = 40 s, you can simply substitute the value of t into the equation to find x. (b) The altitude of the jet is represented by y = 0.03 t^3.


Similarly, substitute t = 40 s into this equation to find y. (c) The speed of the jet can be found by calculating the derivative of both x and y with respect to t and then using these to find the magnitude of the velocity vector. (d) The acceleration of the jet can be found by taking the second derivative of both x and y with respect to t and again using these to find the magnitude of the acceleration vector.

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Answer:

A  = 3.55 m

Explanation:

given,

depth of water = 20 m                

Amplitude of wave at 20 m depth = 1.1 m        

Amplitude at the depth  of 6.2 m              

amplitude is inversely proportional to depth of water

so,                                      

[tex]\dfrac{A_1}{A_2} = \dfrac{d_2}{d_1}[/tex]

[tex]A_2= A_1\dfrac{d_1}{d_2}[/tex]

[tex]A_2= 1.1\dfrac{20}{6.2}[/tex]

A  = 3.55 m

hence, the amplitude of the wave at the depth of 6.2 m is 3.55 m.

Answer:

The electric force on q₁ is [tex]4.5\times10^{-3}\ N[/tex].

Explanation:

Given that,

Charge on [tex]q_{1}=-6.0\ \mu C[/tex]

Distance [tex]x= -3.0\ m[/tex]

Charge on [tex]q_{2}=1.0\ \mu C[/tex] at origin

Distance  [tex]x= 3.0\ m[/tex]

Charge on [tex]q_{3}=-1.0\ \mu C[/tex]

We need to calculate the electric force on q₁

Using formula of electric force

[tex]F_{12}=\dfrac{kq_{1}q_{2}}{r^2}[/tex]

Put the value into the formula

[tex]F_{12}=\dfrac{9\times10^{9}\times(-6.0\times10^{-6})\times1.0\times10^{-6}}{(3.0)^2}[/tex]

[tex]F_{12}=-0.006\ N[/tex]

Negative sign shows the attraction force.

We need to calculate the electric force F₁₃

[tex]F_{13}=\dfrac{9\times10^{9}\times(-6.0\times10^{-6})\times(-1.0\times10^{-6})}{(6.0)^2}[/tex]

[tex]F_{13}=0.0015\ N[/tex]

Positive sign shows the repulsive force.

We need to calculate the net electric force

[tex]F=F_{12}+F_{13}[/tex]

[tex]F=-0.006+0.0015[/tex]

[tex]F=0.0045\ N[/tex]

[tex]F=4.5\times10^{-3}\ N[/tex]

Hence, The electric force on q₁ is [tex]4.5\times10^{-3}\ N[/tex].

The correct answer is that the electric force on q1 is 2.55 x 10^5 N, directed towards the origin.

To find the electric force on q1, we need to calculate the net electric force due to q2 and q3. According to Coulomb's law, the force between two point charges is given by:

[tex]\[ F = k \frac{|q_1 q_2|}{r^2} \][/tex]

[tex]where \( F \) is the magnitude of the force, \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \) N m^2/C^2), \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between the charges.[/tex]

First, we calculate the force between q1 and q2:

[tex]\[ F_{12} = k \frac{|q_1 q_2|}{r_{12}^2} \][/tex]

[tex]\[ F_{12} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{|(-6.0 \times 10^{-6} \text{ C})(1.0 \times 10^{-6} \text{ C})|}{(-3.0 \text{ m})^2} \][/tex]

[tex]\[ F_{12} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{(6.0 \times 10^{-6} \text{ C})(1.0 \times 10^{-6} \text{ C})}{(9.0 \text{ m}^2)} \][/tex]

[tex]\[ F_{12} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{6.0 \times 10^{-12} \text{ C}^2}{9.0 \text{ m}^2} \][/tex]

[tex]\[ F_{12} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) (6.67 \times 10^{-13} \text{ C}^2/\text{m}^2) \][/tex]

[tex]\[ F_{12} = 5.99 \times 10^{-3} \text{ N} \][/tex]

Since q1 and q2 have opposite signs, the force [tex]\( F_{12} \)[/tex] is attractive, meaning it is directed towards q2, which is towards the origin.

Next, we calculate the force between q1 and q3:

[tex]\[ F_{13} = k \frac{|q_1 q_3|}{r_{13}^2} \][/tex]

[tex]\[ F_{13} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{|(-6.0 \times 10^{-6} \text{ C})(-1.0 \times 10^{-6} \text{ C})|}{(-3.0 \text{ m} - 3.0 \text{ m})^2} \][/tex]

[tex]\[ F_{13} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{(6.0 \times 10^{-6} \text{ C})(1.0 \times 10^{-6} \text{ C})}{(6.0 \text{ m})^2} \][/tex]

[tex]\[ F_{13} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{6.0 \times 10^{-12} \text{ C}^2}{36.0 \text{ m}^2} \][/tex]

[tex]\[ F_{13} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) (1.67 \times 10^{-12} \text{ C}^2/\text{m}^2) \][/tex]

[tex]\[ F_{13} = 1.50 \times 10^{-2} \text{ N} \][/tex]

Since q1 and q3 have the same sign, the force [tex]\( F_{13} \)[/tex] is repulsive, meaning it is directed away from q3, which is also towards the origin in this case.

Now, we add the forces vectorially. Since both forces are directed towards the origin, we can simply add their magnitudes:

[tex]\[ F_{net} = F_{12} + F_{13} \][/tex]

[tex]\[ F_{net} = 5.99 \times 10^{-3} \text{ N} + 1.50 \times 10^{-2} \text{ N} \][/tex]

[tex]\[ F_{net} = 2.099 \times 10^{-2} \text{ N} \][/tex]

Converting this to a more standard scientific notation:

[tex]\[ F_{net} = 2.55 \times 10^5 \text{ N} \][/tex]

Therefore, the electric force on q1 is [tex]\( 2.55 \times 10^5 \)[/tex] N, directed towards the origin.

Answer:

1.61 second

Explanation:

Angle of projection, θ = 53°

maximum height, H = 7.8 m

Let T be the time taken by the ball to travel into air. It is called time of flight.

Let u be the velocity of projection.

The formula for maximum height is given by

[tex]H = \frac{u^{2}Sin^{2}\theta }{2g}[/tex]

By substituting the values, we get

[tex]7.8= \frac{u^{2}Sin^{2}53 }{2\times 9.8}[/tex]

u = 9.88 m/s

Use the formula for time of flight

[tex]T = \frac{2uSin\theta }{g}[/tex]

[tex]T = \frac{2\times 9.88\times Sin53 }{9.8}[/tex]

T = 1.61 second

Final answer:

The boulder will land approximately 26.34 meters from the base of the building.

Explanation:

To determine the horizontal displacement of the boulder, we need to analyze the horizontal and vertical components of its motion separately.

First, we can determine the time it takes for the boulder to hit the ground using the vertical component of its motion. The initial vertical velocity can be found by multiplying the initial velocity (25 m/s) by the sin of the launch angle (45°). We can then use the equation h = v0yt + 0.5gt2 to solve for the time, where h is the vertical displacement (30 m - 0 m = 30 m), v0y is the initial vertical velocity, g is the acceleration due to gravity (-9.8 m/s2), and t is the time. Solving for t, we get t ≈ 1.87 s.

Next, we can determine the horizontal displacement by multiplying the horizontal component of the initial velocity (25 m/s cos 45°) by the time of flight (1.87 s). Multiplying the values, we get approximately 26.34 m. Therefore, the boulder lands approximately 26.34 m from the base of the building.

Answer:

The force required by the brakes is [tex]3.67\times 10^{3} N[/tex]

Solution:

As per the question:

Mass of the truck, M = 2200 kg

Height, h = 50 m

distance moved by the truck before stopping, x = 300 m

[tex]g = 10 m/s^{2}[/tex]

Force required by the brakes to stop the truck, [tex]F_{b}[/tex] can be calculated by using the law of conservation of energy.

Now,

Kinetic Energy(K.E) downhil, K.E = reduction in potential energy, [tex]\Delta PE[/tex]

[tex]\Delta PE = Mgh = 2200\times 10\times 50 = 1100 kJ[/tex]

Work done is provided by the decrease in K.E,i.e.,  change in potential energy.

W = [tex]F\times x = 300 F = 1100\times 10^{3}[/tex]

F = [tex]3.67\times 10^{3} N[/tex]

Answer:48.52 kJ

Explanation:

Given

Resistance[tex]=100 \Omega [/tex]

temperature increases from [tex]10^{\circ}C to 20^{\circ}C[/tex]

Voltage=50 V

Heat given(H)[tex]=\frac{V^2t}{R}[/tex]

Where V=voltage applied

t=time

R=Resistance

[tex]H=\frac{50^2\times 60\times 60}{100}=90 kJ[/tex]

Heat absorbed by water is

[tex]Q=mc(\Delta T)[/tex]

where

m=mass of water

c=specific heat of water

[tex]\Delta T[/tex]=change in temperature

[tex]Q=1000\times 4.184\times (20-10)=41.48 kJ[/tex]

Therefore  90-41.48=48.52 kJ is not absorbed by water and leaves the system into the surroundings.

Answer:

A) t = 7.0 s    

B) x = 25 m  

C) v = 10 m/s

Explanation:

The equations for the position and velocity of an object traveling in a straight line is given by the following expressions:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

A)When both friends meet, their position is the same:

x bicyclist = x friend

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

If we place the center of the frame of reference at the point when the bicyclist starts following his friend, the initial position of the bicyclist will be 0, and the initial position of the friend will be his position after 2 s:

Position of the friend after 2 s:

x = v · t

x = 3.6 m/s · 2 s = 7.2 m

Then:

1/2 · a · t² = x0 + v · t       v0 of the bicyclist is 0 because he starts from rest.

1/2 · 2.0 m/s² · t² = 7.2 m + 3.6 m/s · t

1  m/s² · t² - 3.6 m/s · t - 7.2 m = 0

Solving the quadratic equation:

t = 5.0 s

It takes the bicyclist (5.0 s + 2.0 s) 7.0 s to catch his friend after he passes him.

B) Using the equation for the position, we can calculate the traveled distance. We can use the equation for the position of the friend, who traveled over 7.0 s.

x = v · t

x = 3.6 m/s · 7.0 s = 25 m

(we would have obtained the same result if we would have used the equation for the position of the bicyclist)

C) Using the equation of velocity:

v = a · t

v = 2.0 m/s² · 5.0 s = 10 m/s

The heat transferred to the cold end of the square steel bar as a function of time can be calculated using a derivation of the heat conduction formula, factoring in the steel bar's thermal conductivity, cross-sectional area, temperature differential, time, and length.

This question relates to the transfer of heat through a square steel bar using conduction. The heat transfer can be calculated using the formula Q = k·A·(T1 - T2)·t / L, where Q is the heat transferred, k is the thermal conductivity of the material, A is the cross-sectional area, T1 and T2 are the temperatures at both ends of the material respectively, t is the time of heat transfer, and L is the length of the material.

In this case, the steel bar is square, so its cross-sectional area A is calculated by squaring the side length w, so A = w². So, the exact formulation to calculate the heat transferred to the cold end of the bar as a function of time will be Q= 14.6 J/(s⋅m⋅°C)⋅(0.18 m)²⋅(88°C - T2)·t / 1.7 m.

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The heat transferred to the cold end of the steel bar is calculated by finding the rate of heat transfer and multiplying it by the time. The rate of heat transfer is approximately 23.5729 W.

To find the heat transferred to the cold end of the bar as a function of time, we use the formula for the rate of heat transfer through a material.

Given data:

The rate of heat transfer Q/t can be calculated using the formula:

[tex]Q/t = k * A * (T_1 - T_2) / L[/tex]

Plugging in the values:

[tex]Q/t = 14.6 J/(s.m.^oC) * 0.0324 m^2 * (88^oC - 0^oC) / 1.7 m[/tex]

[tex]Q/t = 14.6 * 0.0324 * 88 / 1.7[/tex]

[tex]Q/t = 23.5729 W[/tex]

Therefore, the heat transferred to the cold end of the bar as a function of time is:

[tex]Q = (23.5729 W) * t[/tex]

The heat transferred to the cold end of a square steel bar over time can be calculated by finding the rate of heat transfer and multiplying it by the time. The rate of heat transfer for the given data is approximately 23.5729 W.

Answer:

The strength of the electric field is [tex]1.35\times10^{4}\ N/C[/tex].

Explanation:

Given that,

Speed [tex]v= 5.05\times10^{5}\ m/s[/tex]

Time [tex]t= 3.90\times10^{-7}\ s[/tex]

Angle = 45°

We need to calculate the acceleration

Using equation of motion

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

[tex]5.05\times10^{5}=0+a\times3.90\times10^{-7}[/tex]

[tex]a =\dfrac{5.05\times10^{5}}{3.90\times10^{-7}}[/tex]

[tex]a=1.29\times10^{12}\ m/s^2[/tex]

We need to calculate the strength of the electric field

Using relation of newton's second law and electric force

[tex]F= ma=qE[/tex]

[tex]ma = qE[/tex]

[tex]E=\dfrac{ma}{q}[/tex]

Put the value into the formula

[tex]E=\dfrac{1.67\times10^{-27}\times1.29\times10^{12}}{1.6\times10^{-19}}[/tex]

[tex]E=1.35\times10^{4}\ N/C[/tex]

Hence, The strength of the electric field is [tex]1.35\times10^{4}\ N/C[/tex].

Answer:

[tex]OF=\rho _wVg[/tex]

Explanation:

The bouyancy force an object feels (OF) when submerged ina fluid is always the weight of the liquid the object displaces. This weight will be the mass of fluid displaced ([tex]m_d[/tex]) multiplied by the acceleration of gravity g. The mass displaced will be the density of the fluid [tex]\rho_w[/tex] multiplied by the volume of fluid displaced [tex]V_d[/tex]. If the object is fully submerged, then this volume will be the same as the volume of the object V. We write all these steps in equations:

[tex]OF=m_dg=\rho _wV_dg=\rho _wVg[/tex]

Answer:

22.93 seconds

Explanation:

s = Displacement = 3000 - 420 = 2580 m = The distance she will free fall

a = Acceleration due to gravity = 9.81 m/s²

u = Initial velocity = 0

v = Final velocity

t = Time taken

Equation of motion

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 2580=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2580\times 2}{9.81}}\\\Rightarrow t=22.93\ s[/tex]

Time taken by the skydiver to cover the distance between 3000 m to an altitude of 420 m neglecting air resistance is 22.93 seconds

Answer:[tex]v_0=v\sqrt{\frac{g-vc}{g+vc}}[/tex]

Explanation:

Given

v=initial velocity

resisting acceleration =cv

also gravity is opposing the upward motion

Therefore distance traveled during upward motion

[tex]v^2_f-v^2=2as[/tex]

Where a=cv+g

[tex]0-v^2=2(cv+g)s[/tex]

[tex]s=\frac{v^2}{2(cv+g)}[/tex]

Now let v_0 be the velocity at the ground

[tex]v^2_0-0=2(g-vc)s[/tex]

substituting s value

[tex]v^2_0=v^2\frac{g-vc}{(cv+g)}[/tex]

[tex]v_0=v\sqrt{\frac{g-vc}{g+vc}}[/tex]

The magnitudes of vectors A and B can be determined by resolving each vector into its x and y components, setting up equations for their sum, and solving the equations simultaneously.

To find the magnitudes of vectors A and B when their sum with vector C results in a zero vector (A + B + C = 0), we can break down each vector into its x and y components and then solve the component equations separately.

Since vector C points in the negative y-direction and has a magnitude of 21.0 m, its components are Cx = 0 m and Cy = -21.0 m.

Vector A has a positive y-component and makes an angle of α = 43.4° with the positive x-axis. Thus, the components of A are Ax = A cos(α) and Ay = A sin(α).

Vector B also has a positive y-component and makes an angle of β = 27.7° with the negative x-axis, which is the same as 180° - β with the positive x-axis. So, B's components are Bx = -B cos(180° - β) = -B cos(β) and By = B sin(180° - β) = B sin(β).

The equations for the x and y components of the vector sum being zero are:

Substituting the component expressions in, we get:

These are two equations with two unknowns (the magnitudes A and B), which can be solved simultaneously to find the values of A and B. The solution involves elementary algebra and the use of trigonometric identities.

Answer:

v = 1.15*10^{7} m/s

Explanation:

given data:

charge/ unit area[tex] = \sigma = 1.99*10^{-7} C/m^2[/tex]

plate seperation = 1.69*10^{-2} m

we know that

electric field btwn the plates is[tex] E = \frac{\sigma}{\epsilon}[/tex]

force acting on charge is F = q E

Work done by charge q id[tex] \Delta X =\frac{ q\sigma \Delta x}{\epsilon}[/tex]

this work done is converted into kinectic enerrgy

[tex]\frac{1}{2}mv^2 =\frac{ q\sigma \Delta x}{\epsilon}[/tex]

solving for v

[tex]v = \sqrt{\frac{2q\Delta x}{\epsilon m}[/tex]

[tex]\epsilon = 8.85*10^{-12} Nm2/C2[/tex]

[tex]v = \sqrt{\frac{2 1.6*10^{-19}1.99*10^{-7}*1.69*10^{-2}}{8.85*10^{-12} *9.1*10^{-31}}[/tex]

v = 1.15*10^{7} m/s

Answer:

The rock will rise 2.3 m above the top of the window

Explanation:

The equations for the position and velocity of the rock are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the rock at time t

v0 = initial velocity

y0 = initial height

g = acceleration due to gravity

t = time

v = velocity at time t

If we place the center of the frame of reference at the bottom of the window, then, y0 = 0 and at t = 0.22 s, y = 1.7 m. With this data, we can calculate v0:

1.7 m = 0.22 s · v0 - 1/2 · 9.8 m/s² · (0.22 s)²

Solving for v0:

v0 = 8.8 m/s

Now that we have the initial velocity, we can calculate the time at which the rock reaches its maximum height, knowing that at that point its velocity is 0.

Then:

v = v0 + g · t

0 = 8.8 m/s - 9.8 m/s² · t

-8.8 m/s / -9.8 m/s² = t

t = 0.90 s

Now, we can calculate the max height of the rock:

y = y0 + v0 · t + 1/2 · g · t²

y = 8.8 m/s · 0.90 s - 1/2 · 9.8 m/s² · (0.90 s)²

y = 4.0 m

Then the rock will rise (4.0 m - 1.7 m) 2.3 m above the top of the window

Answer:

Explanation:

Given

maximum height=0.380 m

initial velocity=[tex]v_0[/tex]

[tex]H_{max}=\frac{v_0^2}{2g}[/tex]

[tex]0.380=\frac{v_0^2}{2\times 9.81}[/tex]

[tex]v_0=2.73 m/s[/tex]

The time of flight will be [tex]t=\frac{2v_0}{g}[/tex]

time to reach top +time to reach bottom will be same

[tex]t=\frac{2\times 2.73}{9.81}[/tex]

t=0.556 s

The initial velocity of the flea as it leaves the ground is approximately 1.95 m/s. The flea is in the air for approximately 0.40 seconds.

To find the initial velocity of the flea as it leaves the ground, we can use the equation for vertical motion: v^2 = v0^2 + 2aΔy. In this equation, v is the final velocity (which is 0 at the highest point), v0 is the initial velocity, a is the acceleration due to gravity (-9.80 m/s^2), and Δy is the change in height (0.380 m). Rearranging the equation gives us: v0^2 = 2aΔy. Plugging in the values for a and Δy and solving for v0, we get v0 = sqrt(2aΔy).

For the second part of the question, we can use the equation for the time of flight: Δt = 2v0/a. Plugging in the values for v0 and a, we get: Δt = 2 * v0 / a. Calculating the value of Δt will give us the time the flea is in the air.

Using the given values for a and Δy in the equation for v0 and evaluating the expression gives us the answer to the first part of the question: v0 = sqrt(2 * 9.80 * 0.380). To find the time of flight, we can plug in the values for v0 and a into the equation for Δt: Δt = 2 * 1.95 / 9.80. Evaluating this expression gives us the answer to the second part of the question: Δt = 0.40 s. Therefore, the initial velocity of the flea as it leaves the ground is approximately 1.95 m/s and the flea is in the air for approximately 0.40 seconds.

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Answer:

Density of 127 I = [tex]\rm 1.79\times 10^{14}\ g/cm^3.[/tex]

Also, [tex]\rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.[/tex]

Explanation:

Given, the radius of a nucleus is given as

[tex]\rm r=kA^{1/3}[/tex].

where,

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

[tex]\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.[/tex]

For the nucleus 127 I,

Mass, M = [tex]\rm 2.1\times 10^{-22}\ g.[/tex]

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

[tex]\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.[/tex]

On comparing with the density of the solid iodine,

[tex]\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.[/tex]

Answer:

Time, t = 3.04 hours

Explanation:

Given that,

Average speed of the marathon, v = 8.6 mi/h

Distance covered by the marathon runner, d = 26.22 mi

We need to find the time taken by the marathon runner to run that distance. Time taken is given by :

[tex]t=\dfrac{d}{v}[/tex]

[tex]t=\dfrac{26.22\ mi}{8.6\ mi/h}[/tex]

t = 3.04 hours

So, the time taken by the marathon runner is 3.04 hours. Hence, this is the required solution.

Answer:

yes driver exceed the car speed

Explanation:

given data

speed of car = 38 m/s

speed limit = 75.0 mi/h

to find out

Is the driver exceeding the speed limit

solution

we know car speed is 38 m/s and limit is 75 mi/hr

so for compare the speed limit we convert limit and make them same

as we know

1 m/s = 2.236 mi/hr

so

car speed 38 m/s = 38 × 2.236 = 85.003 mi/hr

as this car speed is exceed the speed limit that is 75 mi/hr

yes driver exceed the car speed

The direction of vector R is [tex]138.47^o[/tex]counterclockwise from the +x-direction and the angle is measured in counterclockwise direction.

For the direction of vector R, which is the angle measured counterclockwise from the +x-direction, we utilize trigonometry.

Given:

x-component of vector R [tex](R_x) = -23.2\ units[/tex]

y-component of vector R [tex](R_y) = 21.4\ units[/tex]

The angle is determined using the arctangent function:

[tex]\theta = tan^{-1}(R_y / R_x)[/tex]

Substituting the given values in the equation:

[tex]\theta = tan^{-1}(21.4 / -23.2).[/tex]

Calculating using a calculator:

[tex]\theta=-41.53^o[/tex]

Since angles are measured counterclockwise from the positive x-direction, adding 180° to the calculated angle gives the direction in that context. The final angle is calculated as:

[tex]\theta= -41.53^o + 180^o \\\theta= 138.47^o\\[/tex]

Therefore, the direction of vector R is [tex]138.47^o[/tex]counterclockwise from the +x-direction.

To know more about the vector:

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The angle of vector R relative to the positive x-axis can be calculated using the arctangent function with the given x and y components, modified to reflect the vector's presence in quadrant II of the Cartesian plane.

To determine the direction of vector R given its x and y components, we can use the arctangent function to find the angle of the vector relative to the positive x-axis. Since the x-component (Rx) is -23.2 units and the y-component (Ry) is 21.4 units, we calculate the angle θ using the formula θ = tan⁻¹(Δy/Δx). Plugging in our values, we get θ = tan⁻¹(21.4 / -23.2). Remember to adjust the angle based on the signs of Rx and Ry since tan⁻¹ only provides results for quadrants I and IV, and our vector lies in quadrant II. This angle will be measured counterclockwise from the positive x-direction.

Answer:

chapter book

Explanation:

Answer:

v = √rg.

Explanation:

The Minimum speed of the stone that can have to the stone when it is rotated in a vertical circle is √rg.

Mathematical Proof ⇒

at the top point on the circle we have

T + mg = m v²/r

We know that minimum speed will be at the place when its tension will be zero.

∴ v² = rg

⇒ v = √rg.

So, the minimum speed or the critical speed is given as  v = √rg.

Answer:

[tex]v=\sqrt{rg}[/tex]

Explanation:

radius of circle = R

Let T be the tension in the string.

At highest point A, the tension is equal to or more than zero, so that it completes the vertical circle. tension and weight is balanced by the centripetal force.

According to diagram,

[tex]T + mg = \frac{mv^{2}}{R}[/tex]

T ≥ 0

So, [tex]mg = \frac{mv^{2}}{R}[/tex]

Where, v be the speed at the highest point, which is called the critical speed.

[tex]v=\sqrt{rg}[/tex]

Thus, the critical speed at the highest point to complete the vertical circle is  [tex]v=\sqrt{rg}[/tex].

Answer:

The continental drift speed is   6.18 cm/yr

Explanation:

In order to calculate the drift speed of the continents, we can assume that it is constant for what the relationship meets

       v = d / t

Where v is the speed, t the time and d the distance

To find the distance we use the closest points that are in the South Atlantic, after reviewing a world map, these points are Brazil and Namibia that has an approximate distance of 7415 km

To start the calculation let's reduce the magnitude

    d = 7415 km (1000m/1km) (100cm/1m)

    d = 7.415 10⁸ 8 cm

    t = 120000000 years  

    t = 1.2 10⁸  year

With these values ​​we calculate the average speed

   v = 7.415 10 3 / 1.2 108 [km / yr.]

   v = 6.18 10-5 Km / yr

   v = 7.415 10 8 / 1.2 10 8 [cm / yr]

   v = 6.18 cm / yr.

The continental drift speed is   6.18 cm/yr

Answer:

b) 58 kg

Explanation:

Gravitational potential energy is the energy that an object has due to its state or position in which it rests.

Gravitational potential energy = U = mass x gravity x height = m g h = 474 J

Height of the stool = h = 0.84 m

rearranging m g h  and solving for the mass m gives m =

474 / (9.8)(0.84)  = 57.6 kg = 58 kg (rounded to 2 significant digits).