A long wave travels in 20m depth of water. If the amplitude of this wave in 20m depth water is 1.1 m , what is the amplitude of the same wave in 6.2 m depth? Assume the wave is traveling perpendicular to the coast.

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:

[tex]\frac{W}{F} = 8.8\times10^7[/tex]

Explanation:

According to newton's law of gravitation

[tex]F=G\frac{m_1\times m_2}{r^2}[/tex]

here m_1=m_2=15000 kg

r= 3 meters and G= 6.67[tex]6.67\times10^{-11}[/tex]

putting values we get

[tex]F= 6.67\times10^{-11}\frac{15000^2}{3^2}[/tex]

solving the above equation we get

Force of gravitation F= [tex]1.6675\times 10^{-3}[/tex] newton

weight of the one of the truck W = mg= 15000×9.81 N

=147000 N

therefore [tex]\frac{W}{F} = \frac{147000}{1.66\times10^{-3}}[/tex]

=8.8×10^{7)

Answer:

The refractive index of the sphere is 2.66

Solution:

The refractive index, [tex]n_{w} = 1.33[/tex] and since the sphere is surrounded by water.

Therefore, according to the question, the parallel rays that affect one of the faces of the sphere converges on the second vortex:

Thus the image distance from the pole  of surface 1, v' = 2R

where

R = Radius of the sphere

Now, using the eqn:

[tex]\frac{n_{w}}{v} + \frac{n}{v'} = \frac{n - n_{w}}{R}[/tex]

[tex]0 + \frac{n}{2R} = \frac{n - 1.33}{R}[/tex]

Since, v is taken as infinite

n = 2.66

Answer:

(A) Torque required is 21.205 N-m

(b) Wok done will be equal to 1199.1286 j

Explanation:

We have given moment of inertia [tex]I=12kgm^2[/tex]

Wheel deaccelerate from 135 rpm to 0 rpm

135 rpm = [tex]135\times \frac{2\pi }{60}=14.1371rad/sec[/tex]

Time t = 8 sec

So angular speed [tex]\omega _i=135rpm[/tex] and [tex]\omega _f=0rpm[/tex]

Angular acceleration is given by [tex]\alpha =\frac{\omega _f-\omega _i}{t}=\frac{0-14.1371}{8}=--1.7671rad/sec^2[/tex]

Torque is given by torque [tex]\tau =I\alpha[/tex]

[tex]=12\times 1.7671=21.205N-m[/tex]

Work done to accelerate the vehicle is

[tex]\Delta w=K_I-K_F[/tex]

[tex]\Delta W=\frac{1}{2}\times 12\times 14.137^2-\frac{1}{2}\times 12\times0^2=1199.1286J[/tex]

Answer:

The time is 5.71 sec.

Explanation:

Given that,

Acceleration [tex]a= -4.20 m/s^2[/tex]

Initial velocity = 24.0 m/s

We need to calculate the time

Using equation of motion

v = u+at[/tex]

Where, v = final velocity

u = inital velocity

t = time

a = acceleration

Put the value into the formula

[tex]0 =24.0 +(-4.20)\times t[/tex]

[tex]t = \dfrac{-24.0}{-4.20}[/tex]

[tex]t=5.71\ sec[/tex]

Hence, The time is 5.71 sec.

Final answer:

The acceleration due to gravity on this planet is approximately [tex]g = 9.82 m/s^2.[/tex]

Explanation:

To determine the acceleration due to gravity on an unknown planet using a pendulum, you can utilize the formula for the period of a simple pendulum: T = 2π[tex]\sqrt{(L/g}[/tex], where T is the period, L is the length, and g is the acceleration due to gravity.

With the given length of the pendulum being 0.81 m and the period being 1.138 s, you can rearrange the formula to solve for  [tex]g = (4*22/7)^2[/tex][tex](L/T^2)[/tex]. Plugging in the known values, [tex]g = (4*22/7)^2[/tex][tex](0.81 m / 1.138^2 s^2)[/tex].

Computing the value, we find that the acceleration due to gravity on this planet is approximately [tex]g = 9.82 m/s^2.[/tex]

Answer:

Answered

Explanation:

The girl whirling the ball should let go off ball when the ball is at a position such that tangent to the circle is in the direction of the target.

the tangent at any point in a circular path indicates the direction of velocity at that point. And the moment when the centripetal force is removed the ball will follow the tangential path at that moment.

Final answer:

The girl should release the ball when it aligns with the target in a straight line from her, following principles of inertia and circular motion. In a scenario where a ball winds around a post, it would speed up due to the conservation of angular momentum, aligning with Michelle's prediction.

Explanation:

The question about when a girl should release a ball while whirling it around her head in a horizontal plane to hit a target involves understanding circular motion and projectile motion principles in physics. According to the laws of physics, the ball will continue to move tangentially to the circle at the point of release because of inertia. Thus, she should let go of the string when the ball is directly in line with the target, assuming no air resistance and that the path follows a straight line in the horizontal direction.

The conservation of angular momentum and the conservation of energy are relevant in situations where objects are in circular motion, like a ball tied to a string being whirled around. For an object in circular motion, when the radius of the motion decreases (as in the ball winding around a post), it will speed up because angular momentum is conserved. This implies Michelle’s view on the ball having to speed up as it approaches the post is correct, contrasting Astrid’s expectation of constant speed due to energy conservation.

Answer:

285 seconds

Explanation:

Jenny speed is 3.8 m/s

Alyssa speed in 4.0 m/s

Alyssa starts after 15 seconds

Find the distance covered by Jenny, when Alyssa starts

Distance=Speed*time

Distance covered by Jenny in 15 seconds= 3.8×15=57m

Relative speed of the two members heading same direction will be;

4.0m/s-3.8m/s=0.2m/s

To find the time Alyssa catch up with Jenny you divide the distance to be covered by Alyssa by the relative speed of the two

Distance=57m, relative speed=0.2m/s  t=57/0.2 =285 seconds

=4.75 minutes

Answer:

Explanation:

[tex]v_{Jenny} =3.8m/s[/tex]

[tex]t_{Jenny}=t[/tex]

[tex]d_{Jenny}=d[/tex]

[tex]v_{Alyssa}=4.0m/s[/tex]

[tex]t_{Alyssa}=t-15[/tex], because she has a difference of 15 seconds.

[tex]d_{Alyssa}=d[/tex]

Both move at a constant speed, that means there's no acceleration, their speed is always the same.

Now, the equation of each movement is

[tex]d=3.8t[/tex] and [tex]d=4(t-15)[/tex], then we solve this two.

We replace the first equation into the second one

[tex]3.8t=4(t-15)\\3.8t=4t-20\\20=4t-3.8t\\0.2t=20\\t=\frac{20}{0.2}\\ t=100[/tex]

That means after 100 seconds Alyssa catch up with Jenny.

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:

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

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: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:

The charge q₃ must be placed at X = +2.5 cm

Explanation:

Conceptual analysis

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

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

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

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

Equivalences

1µC= 10⁻6 C

1cm= 10⁻² m

Data

k = 8.99*10⁹ N×m²/C²

q₁ =+3 µC =3*10⁻⁶ C

q₂ = -10 µC =-10*10⁻⁶ C

q₃= +6µC =+6*10⁻⁶ C

d₁ = 3cm =3×10⁻² m

d₂ = 4cm = 4×10⁻² m

Graphic attached

The attached graph shows the field due to the charges:

E₁:Field at point P due to charge q₁. As the charge is positive ,the field leaves the charge. The direction of E1 is (+ x).

E₂: Field at point P due to charge q₂. As the charge is positive ,the field leaves the charge. The direction of E1 is (+ x).

Problem development

E₃: Field at point P due to charge q₃. As the charge q₃ is positive, the field leaves the charge.

The direction of E₃ must be (- x) so that the electric field can be equal to zero at point P since E₁ and E₂ are positive, then, q₃must be located to the right of point P.

We make the algebraic sum of fields at point P due to the charges q1, q2, and q3:

E₁+E₂-E₃=0

[tex]\frac{k*q_{1} }{d_{1}^{2}  } +\frac{k*q_{2} }{d_{2}^{2}  } -\frac{k*q_{3} }{d_{3}^{2}  } =0[/tex]

We eliminate k

[tex]\frac{q_{1} }{d_{1} ^{2} } +\frac{q_{2} }{d_{2} ^{2} }+\frac{q_{3} }{d_{3} ^{2} }=0[/tex]

We replace data

[tex]\frac{3*10^{-6} }{(3*10^{-2})^{2} } +\frac{10*10^{-6} }{(4*10^{-2})^{2} } +\frac{6*10^{-6} }{d_{3} ^{2} } =0[/tex]

we eliminate 10⁻⁶

[tex]\frac{3}{9*10^{-4} } +\frac{10}{16*10^{-4} } =\frac{6}{d_{3}^{2}  }[/tex]

[tex](\frac{1}{10^{-4} }) *(\frac{1}{3} +\frac{5}{8}) =\frac{6}{d_{3}^{2}  }[/tex]

[tex]\frac{23*10^{4} }{24} =\frac{6}{d_{3} ^{2} }[/tex]

[tex]d_{3} =\sqrt{\frac{6*24}{23*10^{4} } }[/tex]

[tex]d_{3} =2.5*10^{-2} m\\d_{3} =2.5 cm[/tex]

The charge q₃ must be placed at X = +2.5 cm

Answer:

43.3 m

Explanation:

d1 = 25.1 m in 15.4° west of north

d2 = 38.8 m in 23.1° south of west

Write the displacements in vector form

[tex]\overrightarrow{d_{1}}=25.1\left ( -Sin15.4\widehat{i}+Cos15.4\widehat{j} \right )=-6.67\widehat{i}+24.2\widehat{j}[/tex]

[tex]\overrightarrow{d_{2}}=38.8\left ( -Cos23.1\widehat{i}-Sin23.1\widehat{j} \right )=-35.69\widehat{i}-15.22\widehat{j}[/tex]

The resultant displacement is given by

[tex]\overrightarrow{d}=\overrightarrow{d_{1}}+\overrightarrow{d_{2}}[/tex]

[tex]\overrightarrow{d}}=\left ( -6.67-35.69 \right )\widehat{i}+\left ( 24.2-15.22 \right )\widehat{j}[/tex]

[tex]\overrightarrow{d}}=\left ( -42.36 \right )\widehat{i}+\left ( 8.98 \right )\widehat{j}[/tex]

The magnitude of the resultant displacement is given by

[tex]d=\sqrt{8.98^{2}+\left ( -42.36 \right )^{2}}=43.3 m[/tex]

Thus, you are 43.3 m far from your starting point.

Answer:

(A) only I

Explanation:

According to Newton's first law of motion, a particle in motion will continue moving in a straight line at constant speed or will stay at rest, if at rest, as long as there's no external force acting on it.

ALL of the statements except #2 COULD be true. (Choice-D)

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:

Option c

Explanation:

Magnetic field lines form loops starting from north pole to south pole outside the magnet and from south pole to north pole inside the magnet.

Thus the field is such that it is directed outwards from the North pole and directed inwards to the South pole of the magnet.

A compass in a magnetic field will will comply with the magnet's North pole directing towards the magnetic field.

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:

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.

Final answer:

The bead rests at a height of 9.03 x 10^7 meters above the fixed charge.

Explanation:

To find the height above the fixed charge where the bead rests in equilibrium, we need to consider the electric forces acting on the bead. The electric force is given by the equation:

F = k * (q1 * q2) / r^2

Where F is the force between the two charges, q1 and q2 are the charges, r is the distance between them, and k is the electrostatic constant. In this case, the two charges are the fixed charge at the base of the rod and the charge on the bead. Setting the gravitational force equal to the electric force, we can solve for the height.

First, we need to convert the given charge of 20 nC to coulombs by dividing it by 10^9:

q2 = 20 nC / 10^9 = 20 * 10^-9 C

Next, we calculate the gravitational force and the electric force:

F_gravity = m * g

F_electric = k * q1 * q2 / r^2

Since the bead is in equilibrium, the two forces must be equal:

m * g = k * q1 * q2 / r^2

Now, we can solve for the height:

h = sqrt(k * q1 * q2 / (m * g))

Plugging in the given values:

h = sqrt((9 * 10^9 N * m^2 / C^2) * (20 * 10^-9 C) / (0.050 x 10^-3 kg * 9.8 m/s^2))

Simplifying:

h = sqrt(4 * 10^11 / (0.049 x 10^-3))

h = sqrt(8.16 x 10^14)

h = 9.03 x 10^7 m

Therefore, the bead rests at a height of 9.03 x 10^7 meters above the fixed charge.

Answer:

chapter book

Explanation:

Answer:

[tex]a_{c}=v^{2}/R[/tex]

The radius of curvature changes so that centripetal acceleration is similar along the entire roller coaster.

Explanation:

We know that the centripetal acceleration is directly proportional to the tangential velocity and inversely proportional to the radius of curvature:

[tex]a_{c}=v^{2}/R[/tex]

By energy conservation (and common sense), we know that the speed at the top of the roller coaster is smaller. Therefore if the roller coaster has similar accelerations (therefore also similar normal forces) at the top and at the bottom, it is necessary that the difference in speed be compensated with the radius of curvature, i.e. smaller radius at the top than at the bottom.

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 impulse is (10.88 i^ + 7.04 j^) N s

maximum force on the ball is  (4.53 10 2 i^ + 2.93 102 j ^) N  

Explanation:

In a problem of impulse and shocks we must use the impulse equation

       I = dp = pf-p₀         (1)

       p = m V

With we have vector quantities, let's decompose the velocities on the x and y axes

      V₀ = -19 m / s

      θ₀ = 40.0º  

      Vf = 46.0 m / s

      θf = 30.0º

Note that since the positive direction of the x-axis is from the batter to the pitcher, the initial velocity is negative and the angle of 40º is measured from the axis so it is in the third quadrant

      Vcx = Vo cos θ

      Voy = Vo sin θ

      Vox= -19 cos (40) = -14.6 m/s

      Voy = -19 sin (40) =  -12.2 m/s

      Vfx = 46 cos 30 = 39.8 m/s

      Vfy = 46 sin 30 =  23.0 m/s

   a) We already have all the data, substitute and calculate the impulse for each axis

      Ix = pfx -pfy

      Ix = m ( vfx -Vox)

      Ix = 0.200 ( 39.8 – (-14.6))

      Ix = 10.88 N s

      Iy = m (Vfy -Voy)

      Iy = 0.200 ( 23.0- (-12.2))

      Iy=  7.04 N s

In vector form it remains

       I =  (10.88 i^ + 7.04 j^) N s

   b) As we have the value of the impulse in each axis we can use the expression that relates the impulse to the average force and your application time, so we must calculate the average force in each interval.

         I = Fpro Δt

In the first interval

        Fpro = (Fm + Fo) / 2

With the Fpro the average value of the force, Fm the maximum value and Fo the minimum value, which in this case is zero

         Fpro = (Fm +0) / 2

In the second interval the force is constant

          Fpro = Fm

In the third interval

         Fpro = (0 + Fm) / 2

Let's replace and calculate

         I =  Fpro1 t1 +Fpro2 t2  +Fpro3 t3

         I = Fm/2 4 10⁻³ + Fm 20 10⁻³+ Fm/2 4 10⁻³  

         I = Fm  24 10⁻³ N s

         Fm = I / 24 10⁻³

         Fm = (10.88 i^ + 7.04 j^) / 24 10⁻³

         Fm = (4.53 10² i^ + 2.93 10² j ^) N

maximum force on the ball is  (4.53 10 2 i^ + 2.93 102 j ^) N  

The impulse delivered to the ball is (-10.87 î + 7.04 ĵ) N·s. The maximum force on the ball is -452.92 N, based on the given force-time relationship.

Solution:

(a) To find the impulse delivered to the ball, we can use vector components of velocity and the formula for impulse.

Initial velocity vector of the ball:

vix = 19.0 m/s × cos(40.0°) = 14.55 m/s

viy = 19.0 m/s × sin(-40.0°) = -12.22 m/s

Final velocity vector of the ball:

vfx = -46.0 m/s × cos(30.0°) = -39.78 m/s

vfy = 46.0 m/s × sin(30.0°) = 23.00 m/s

Change in velocity vector:

Δvx = vfx - vix = -39.78 m/s - 14.55 m/s = -54.33 m/s

Δvy = vfy - viy = 23.00 m/s - (-12.22 m/s) = 35.22 m/s

Impulse vector (I = m × Δv):

Ix = 0.200 kg × (-54.33 m/s) = -10.87 N·s

Iy = 0.200 kg × (35.22 m/s) = 7.04 N·s

Total impulse vector:

I = (-10.87 î + 7.04 ĵ) N·s

(b) To determine the maximum force (Fmax) on the ball, consider the force-time relationship given:

Force increases linearly for 4.00 ms

Force holds constant for 20.0 ms

Force decreases linearly to zero in another 4.00 ms

The total duration of the force application is 28 ms (4 + 20 + 4 = 28 ms). The impulse is the area under the force-time graph, which is:

Impulse (I) = (1/2 × Fmax × 4.00 ms) + (Fmax × 20.0 ms) + (1/2 × Fmax × 4.00 ms)

-10.87 N·s = (0.002 × Fmax + 0.020 × Fmax + 0.002 × Fmax)

-10.87 N·s = 0.024 × Fmax

Fmax = -10.87 N·s / 0.024 s = -452.92 N

The negative sign indicates the direction opposite to the assumed positive direction.

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

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:

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.

Answer:

84.44N

Explanation:

Hi!

The force F between two charges q₁ and q₂ at a distance r from each other is given by Coulomb's law:

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

The force on the negative charge q₁ is the sum of the forces from the other two charges. This forces have equal magnitude as both distances are 48cm. The magnitud is:

[tex]F_{1,2} =F_{1,3} = -k_c\frac{(26\mu C)^2}{(48cm)^2}=-9*10^9Nm^2C^{-2}*0.54*\frac{10^{-12}C^2}{10^{-4}m^2}=-48.75N\\[/tex]

(negative means attractive)

The sum of the forces, because of symmetry reasons actos along line L (see the figure), and its magnitud is:

[tex]F = 2*48.75*\cos(30\º)N = 84.44N[/tex]

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