The cockroach Periplaneta americana can detect a static electric field of magnitude 8.50 kN/C using their long antennae. If the excess static charge on a cockroach is modeled as point charges located at the end of each antenna, what magnitude of charge q would each antenna possess in order for each antennae to experience a force of magnitude 2.00 μN from the external electric field? Calculate q in units of nanocoulombs (nC) .

The equations of motion are used to find the motion characteristics of an object with time

A) The time it takes the motorcycle to reach the car is approximately 5.39 seconds

B) The distance the motorcycle travel from the moment it starts to accelerate to the moment it catches up with the car is approximately 155 meters

The reason the above values are correct is as follows:

The given parameters are;

The initial speed of the car and the motorcycle, v₁ = 18.0 m/s

The initial distance between the motorcycle following the car = 58.0 m

The time after which the motorcycle starts to accelerate, t₁ = 5.00 secs

The acceleration of the motorcycle = 4.00 m/s²

The time at which the motorcycle catches up with the car = t₂

Required:

A) The time it takes the motorcycle to accelerate before it catches up with the car, which is to find t₂ - t₁

Method:

The distance moved by motorcycle to reach the car = The distance moved by the car in the same time + 58.0 meters

Solution;

The appropriate equation of motion to use are [tex]s = u \cdot t + \dfrac{1}{2} \cdot a\cdot t^2[/tex] and [tex]d = u\cdot t[/tex]

Where;

s = The distance between the motorcycle and the car = 58.0 m

u = The initial velocity of the motorcycle = 18.0 m/s

t = Δt = t₂ - t₁ = The time the motorcycle accelerates

a = The acceleration of the motorcycle = 4.00 m/s²

d = The distance moved by the car

By the method, we have;

[tex]s = d + 58.0[/tex]

[tex]u \cdot t + \dfrac{1}{2} \cdot a\cdot t^2 = u \cdot t[/tex]

Plugging in the values into the equation gives;

[tex](18.0 \times t) + \dfrac{1}{2} \times 4 \times t^2 = (18.0 \times t) + 58.0[/tex]

Cancelling the like term, (18.0 × t), on both sides of the equation gives;

[tex]\dfrac{1}{2} \times 4 \times t^2 = 58.0[/tex]

[tex]2\cdot t^2 = 58.0[/tex]

[tex]t = \Delta t = \sqrt{\dfrac{58.0}{2} } = \sqrt{29} \approx 5.39[/tex]

The time it takes the motorcycle to reach the car, t ≈ 5.39 seconds

The time it takes the motorcycle to reach the car is approximately 5.39 seconds

B) Required:

The distance the motorcycle travels from the moment it starts to accelerate, t₁ to the time it catches up with the car, t₂

Method:

The distance travelled during the time interval t₂ - t₁ should be calculated

Solution:

The distance, s, travelled during the time interval [tex]t_2 - t_1 = \Delta t = t[/tex] is given as follows;

[tex]s = u \cdot t + \dfrac{1}{2} \cdot a \cdot t^2[/tex]

From part (A), u = 18.0 m/s t = [tex]\sqrt{29}[/tex] s, and a = 4.00 m/s², therefore;

[tex]s = 18.0 \times \sqrt{29} + \dfrac{1}{2} \times 4.00 \times (\sqrt{29} )^2 \approx 154.933[/tex]

Rounded to three significant figures, the distance the motorcycle travel, s = 155 meters

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To solve the question, we use the equations of motion to equate the distances traveled by the car and motorcycle. After solving the quadratic equation, we find the time the motorcycle catches up with the car and the distance it traveled.

The question involves a situation where a motorcycle is following a car traveling at a constant speed on a highway. Initially, both the car and the motorcycle are traveling at the same speed of 18.0 m/s, and there is a distance of 58.0 m between them. After 5.00 seconds, the motorcycle begins to accelerate at 4.00 m/s² until it catches up with the car. To find the time t2-t1, which is the duration the motorcycle catches up to the car, and the distance the motorcycle covers from the moment it starts to accelerate until it catches up with the car, we use the equations of motion.

We know that the car travels with a constant velocity, so the distance it covers in time t after the motorcycle starts accelerating can be given by:

d_car = v_car × t

For the motorcycle, which starts to accelerate at time t1, the distance it covers can be given by:

d_moto = d_initial + v_initial × (t - t1) + ½ × a × (t - t1)²

Given that the initial distance d_initial is 58.0 m, v_initial is 18.0 m/s, and a is 4.00 m/s², we can equate the distances for both the car and motorcycle when the motorcycle catches up:

18.0 × t = 58.0 + 18.0 × (t - 5.00) + ½ × 4.00 × (t - 5.00)²

By solving this quadratic equation, we can find the value of t, and then t2-t1 would simply be t - 5.00 seconds. Finally, to find out the distance the motorcycle traveled, we substitute the value of t into the distance equation for the motorcycle.

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Answer:

θ = 3.38 degrees

r = 0.48mi

Explanation:

a) Start by converting 1.7 mi into feet.

1mile = 5280 feet,

so 1.7 mi = 8976 ft

[tex]y=rsin\theta,[/tex]

so [tex]\theta = sin^{1] {530}{8976}[/tex]

and θ = 3.38 degrees

b )Now to gain another 150 ft of elevation, again use the

equation y=rsin(θ)

150 = rsin(3.38)

r = 2544.187 ft.

Convert r = 2544.187 ft into miles by dividing by 5280,

and r = 0.48mi

Answer:

a) 0.1 lb/ft³

b) 20 lb

c) 88.788 N

Explanation:

Given:

Volume of the object = 200 ft³

Specific volume = 10 ft³/ lb

Now,

Density = [tex]\frac{\textup{1}}{\textup{Specific volume}}[/tex]

or

Density = [tex]\frac{1}{10}[/tex] = 0.1 lb/ft³

also,

Specific volume = [tex]\frac{\textup{Volume}}{\textup{Mass}}[/tex]

or

10 = [tex]\frac{200}{\textup{Mass}}[/tex]

or

Mass = 20 lb

And,

Weight =  Mass in kg × Acceleration due to gravity

Mass = 20 lb

and 1 lb = 0.453 kg

thus,

20 lb = 20 × 0.453 = 9.06 kg

Therefore,

weight = 9.06 × 9.8 = 88.788 N

For the object,

Specific volume of the object is inverse to the density of the object.

Given information-

The volume of the object is 200 ft cubed.

The specific volume of the object is 10 ft cubed.

The density of a object is the inverse of its specific volume, thus density of the given object is,

[tex]\rho=\dfrac{1}{\rm specific volume} \\\rho=\dfrac{1}{10} \\\rho=0.1 \rm ib/ft^3[/tex]

The mass of a object is the ratio of the volume to the specific volume, thus mass of the given object is,

[tex]m=\dfrac{\rm volume}{\rm specific volume} \\m=\dfrac{200}{10} \\m=20 \rm ib[/tex]

As 1 Ib is equal to the 0.453 kg.

The weight of a object is the product of the mass of the object to the acceleration due to gravity, thus weight of the given object is,

[tex]W=0.453\times20\times9.81\\W=88.788\rm N[/tex]

Thus for the object,

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Answer:

(a) 1.73 s

(b) 14.75 m

(c) 3.36 s

(d) double

(e) 63.32 m

Explanation:

Vertical component of initial velocity, uy = 17 m/s

Horizontal component of initial velocity, ux = 18.3 m/s

(A) At highest point of trajectory, the vertical component of velocity is zero. Let the time taken is t.

Use first equation of motion in vertical direction

vy = uy - gt

0 = 17 - 9.8 t

t = 1.73 seconds

(B) Let the highest point is at height h.

Use III equation of motion in vertical direction

[tex]v^{2}=u^{2}-2gh[/tex]

0 = 17 x 17 - 2 x 9.8 x h

h = 14.75 m

(C) The time taken by the ball to return to original level is T.

Use second equation of motion i vertical direction.

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

h = 0 , u = 17 m/s

0 = 17 t - 0.5 x 9.8 t^2

t = 3.46 second

(D) It is the double of time calculated in part A

(E) Horizontal distance = horizontal velocity x total time

d = 18.3 x 3.46 = 63.32 m

Answer:

Acceleration, a = [tex]97.52 m/s^{2}[/tex]

t = 0.08 s

Solution:

As per the question:

The velocity of the player, v = 7.9 m/s

distance, d = 0.32 m

Now, consider the direction of the initial velocity of the player as positive and the acceleration to be constant:

Also, the final velocity of the player, v' = 0 m/s as he finally stops

Using the third eqn of motion:

[tex]v'^{2} = v^{2} - 2ad[/tex]

[tex]0 = 7.9^{2} - 2a\times 0.32[/tex]

a = [tex]97.52 m/s^{2}[/tex]

Also, From eqn (1) of motion:

v' = v - at

0 = 7.9 - 97.52t

t = 0.08 s

Answer:

the radii of curvature is 30 cm.

Explanation:

given,

object is place at = 45 cm

image appears at = 90 cm

focal length = ?

refractive index = 1.5

radii of curvature = ?

[tex]\dfrac{1}{f} = \dfrac{1}{u} +\dfrac{1}{v}[/tex]

[tex]\dfrac{1}{f} = \dfrac{1}{45} +\dfrac{1}{90}[/tex]

f = 30 cm

using lens formula

[tex]\dfrac{1}{f} = (n-1)(\dfrac{1}{R_1} -\dfrac{1}{R_2})[/tex]

[tex]R_1 = R\ and\ R_2 = -R[/tex]

[tex]\dfrac{1}{f} = (n-1)(\dfrac{1}{R} +\dfrac{1}{R})[/tex]

[tex]R = (n -1)\ f [/tex]

[tex]R = 2(1.5 -1)\ 30[/tex]

R = 30 cm

hence, the radii of curvature is 30 cm.

Answer:

The possible value of power fluctuation is 3.9811 W

Solution:

As per the question:

Power output, P = 10 W

Power fluctuation is from +3 dB to -3 dB

Therefore,

Power fluctuation in dB = 6 dB

Now,

P (in dB) = [tex]10log_{10}P'[/tex]

6 dB = [tex]10log_{10}P'[/tex]

[tex]0.6 = log_{10}P'[/tex]

[tex]P' = 10^{0.6} W = 3.9811 W[/tex]

Answer:

The ball will be at 700 m above the ground.

Explanation:

We can use the following kinematic equation

[tex]y(t) = \ y_0 \ + \ v_0 \ t \ + \frac{1}{2} \ a \ t^2[/tex].

where y(t) represent the height from the ground. For our problem, the initial height will be:

[tex]y_0 \ = \ 1000 m[/tex].

The initial velocity:

[tex]v_0 = - 20 \frac{m}{s}[/tex],

take into consideration the minus sign, that appears cause the ball its thrown down.  The same minus appears for the acceleration:

[tex]a=-10\frac{m}{s}[/tex]

So, the equation for our problem its:

[tex]y(t) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ t \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ t^2[/tex].

Taking t=6 s:

[tex]y(6 \ s) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ * \ 6 \ s \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ (6 s)^2[/tex].

[tex]y(6 \ s) = \ 1000 m \ - 120 m - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ 36 s^2[/tex].

[tex]y(6 \ s) = \ 1000 m \ - 120 m - 180 m[/tex].

[tex]y(6 \ s) = \ 1000 m \ - 300 m[/tex].

[tex]y(6 \ s) = \ 700 m [/tex].

So this its the height of the ball 6 seconds after being thrown.

Given a ball thrown straight downward from a 1000 m high cliff with an initial speed of 20 m/s and considering an acceleration due to gravity of -10 m/s², it can be calculated using a kinematics equation that the ball will be 760 m above the ground after 6 seconds.

The subject of this question is in the field of Physics, specifically kinematics. The problem is set up to have an object that is thrown downward with an initial speed and is then under the influence of gravity, causing it to accelerate downward. To calculate how far above from the ground the ball will be at a certain time, we use the equation for position in uniformly accelerated motion which is:

y = y₀ + v₀×t + 0.5×a×t²

Here, y₀ represents the initial height, v₀ is the initial speed, a is the acceleration due to gravity, and t is the time. Since the boy is standing at the edge of a 1000 m cliff, y₀ = 1000 m. The ball is thrown downwards with an initial speed of 20 m/s, so v₀=-20 m/s (we take down as the negative direction). On earth, the acceleration due to gravity is approximately -10 m/s² and a = -10 m/s². The time t = 6 seconds is given in the problem.

Substituting these values into our equation gives y = 1000 + (-20×6) + 0.5×(-10) (6²), which simplifies to y = 760 m. Therefore, the ball is 760 meters above the ground 6 seconds after it was thrown.

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Answer:

7.1 m

Explanation:

Given:

Distance traveled by the student in the first attempt = [tex](2.9 \pm 0.1)\ m[/tex]

Distance traveled by the student in the second attempt = [tex](3.9 \pm 0.2)\ m[/tex]

So, the maximum distance that the student could travel in this attempt = [tex](2.9+0.1)\ m = 3.0\ m[/tex]

So, the maximum distance that the student could travel in this attempt = [tex](3.9+0.2)\ m = 4.1\ m[/tex]

Since the student first moves straight in a particular direction, rests for a while and then moves some distance in the same direction.

So, the largest distance that the student could possibly be from the starting point would be the largest distance of the final position of the student from the starting point.

And this distance is equal to the sum of the maximum distance possible in the first attempt and the second attempt of walking which is 7.1 m.

Hence, the largest distance that the student could possibly be from the starting point is 7.1 m.

Answer:

The value of resistance will be 42.08 ohm    

Explanation:

We have given capacitance [tex]C=10\mu F=10\times 10^{-6}F[/tex]

Frequency f = 1 kHz = 1000 Hz

Impedance Z = 45 ohm

Capacitive reactance [tex]X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}=\frac{1}{2\times 3.14\times 1000\times 10^{-5}}=15.92ohm[/tex]

We know that impedance is given by [tex]Z=\sqrt{R^2+X_C^2}[/tex]

[tex]45=\sqrt{R^2+15.92^2}[/tex]

Squaring both side [tex]2025=R^2+253.4464[/tex]

R = 42.08 ohm

Final answer:

The pressure ratio between different diameter pipes can be calculated based on the square of their diameters.

Explanation:

To find the volume rate of flow (Q), we can use the principle of continuity that states the volume flow rate must be constant throughout the pipe. Q can be calculated using the area of cross-section (A) and velocity (v) with the formula Q = A1 x v1 = A2 x v2.

The ratio in water pressure between the larger and smaller water pipes can be determined by considering the cross-sectional areas of the pipes. As pressure is inversely proportional to the cross-sectional area of the pipe, the ratio in pressure is equal to the ratio of the squares of the pipe diameters. In this case, the ratio of pressures would be 36:1.

Answer:

The value of acceleration equals [tex]0.0556m/s^{2}[/tex]

Explanation:

The  distances covered in each of the three phases are calculated as under

1) Distance covered in accelerating phase for a period of 5 minutes or 300 seconds ( 1 minute = 60 seconds)

Using second equation of kinematics

[tex]s=ut+\frac{1}{2}at^{2}\\\\s_{1}=0\times 300+\frac{1}{2}\times a\times (300)^{2}\\\\s_{1}=0.5\times a\times (300)^{2}[/tex]

2) Distance covered in 5 minutes while travelling at a constant speed which is The speed after 5 minutes of travel is obtained by first equation of kinematics as

[tex]v=u+at\\\\v=0+a\times 300\\\\v=300a[/tex]

Thus distance traveled equals

[tex]s_{2}=300a\times 300\\\\s_{2}=(300)^{2}a[/tex]

3)

The phase when the car stops the distance it covers during this phase can be obtained using third equation of kinematics as

[tex]v^{2}=u^{2}+2as\\\\\therefore s=\frac{v^{2}-u^{2}}{2a}\\\\s_{3}=\frac{0-(300a)^{2}}{2\times -a}\\\\s_{3}=\frac{300^{2}a}{2}[/tex]

Now the sum of [tex]s_{1}+s_{2}+s_{3}[/tex] equals 10 kilometers or 10000 meters.

Thus we get

[tex]0.5\times a\times 300^{2}+300^{2}a+300^{2}\times \frac{a}{2}=10000\\\\a(0.5\times 300^{2}+300^{2}+0.5\times 300^{2})=10000\\\\\therefore a=\frac{10000}{(0.5\times 300^{2}+300^{2}+0.5\times 300^{2})}=0.0556m/s^{2}[/tex]

To solve for the acceleration, we can use the constant acceleration equation and find the distance traveled during each phase.

The constant acceleration equation can be used to solve this problem. The total distance traveled by the train is equal to the sum of the distances traveled during each phase.

In the first phase, the train starts at rest and accelerates for 5 minutes. The distance traveled during this phase can be found using the equation s = 0.5at^2, where s is the distance, a is the acceleration, and t is the time. Since the initial velocity is 0, we can simplify the equation to s = 0.5at^2.

In the second phase, the train travels at a constant speed for 5 minutes. The distance traveled during this phase is equal to the product of the speed and the time.

In the third phase, the train decelerates with a negative acceleration. The distance traveled during this phase can be found using the equation s = ut + 0.5at^2, where u is the initial velocity, a is the acceleration, and t is the time. Since the final velocity is 0, we can simplify the equation to s = ut.

The total distance traveled is given as 10 km. By plugging in the values for the distances and using the given times for each phase, we can solve for the acceleration, a, and find that a = 0.014 km/s^2.

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Answer:

(a) 5.6g

(b) 15.1g

Explanation:

Let us assume:

Part (a):

While Dr. John Paul Stapp accelerates the rocket to its top speed, we have

[tex]u = 0 m/s\\v = 282\ m/s\\t = 5.1\ s\\\therefore a = \dfrac{v-u}{t}\\\Rightarrow  a = \dfrac{282-0}{5.1}\\\Rightarrow  a =55.29\\\textrm{Dividing both sides by }g\\\Rightarrow \dfrac{a}{g}=\dfrac{55.29}{9.80}\\\Rightarrow \dfrac{a}{g}=5.6\\\Rightarrow a=5.6g[/tex]

Hence, the acceleration of the rocket in the direction of motion is 5.6g.

Part (b):

While Dr. John Paul Stapp accelerates the rocket from its top speed to rest, we have

[tex]u = 282 m/s\\v = 0\ m/s\\t = 1.9\ s\\\therefore a = \dfrac{v-u}{t}\\\Rightarrow  a = \dfrac{0-282}{1.9}\\\Rightarrow  a =-148.42\\\textrm{Dividing both sides by }g\\\Rightarrow \dfrac{a}{g}=\dfrac{-148.42}{9.80}\\\Rightarrow \dfrac{a}{g}=-15.1\\\Rightarrow a=-15.1g[/tex]

Here, the negative sign represents that the motion of acceleration of the rocket is opposite to the direction of motion.

Hence, the acceleration of the rocket in the direction opposite to that of motion is 15.1g.

Final answer:

The response provides the calculation for Dr. John Paul Stapp's acceleration in his direction of motion and the acceleration opposite to his motion in terms of multiples of g.

Explanation:

Acceleration:

(a) Acceleration in his direction of motion: To find the acceleration in Stapp's direction of motion, we can use the formula a = Δv / Δt. Plugging in the values, we get a = (282 m/s) / (5.1 s) = 55.3 m/s². Converting this to multiples of g, we have 55.3 m/s² / 9.81 m/s² ≈ 5.6 g.

(b) Acceleration opposite to his direction of motion: The deceleration can be calculated using the same formula a = Δv / Δt. Substituting the values, we have a = (282 m/s) / (1.9 s) = 148.4 m/s². Expressing this in multiples of g, we get 148.4 m/s² / 9.81 m/s² ≈ 15.1 g.

Answer:

The person should use a convex lens of power -0.625 D.

Explanation:

Given that, the far point of the person, with myopic eye, is 160 cm.

It means that the image of an object, which is at infinity, is formed at this point 160 cm distance away from the person's eye.

If the object is infinity, then the object distance, u = -[tex]\infty[/tex].

Image distance, v = -160 cm.

u and v are taken to be negative because the object and the image of the object are on the side of the eye from where the light is coming.

Let the focal length of the lens which is to be used for the correction of this myopic eye be f, then using lens equation,

[tex]\rm \dfrac 1f = \dfrac 1v-\dfrac 1u=\dfrac 1{-160 }-\dfrac1{-\infty}=\dfrac 1{-160}-0=\dfrac 1{-160}.\\\Rightarrow f=-160\ cm.[/tex]

The negative focal length indicates that the lens should be convex.

The power of the lens is given by

[tex]\rm P=\dfrac{1}{f}=\dfrac{1}{-160\ cm }=\dfrac{1}{-1.6\ m}=-0.625\ D.[/tex]

So, the person should use a convex lens of power -0.625 D.

Answer:

- 1.428 D

Explanation:

A myopic person is able to see the nearby objects clearly but cannot see the far off objects very clearly. It can be cured by using concave lens of suitable power.

far point = 70 cm

v = - 70 cm (position of image from the lens)

u = ∞ (position of object from lens)

Let f be the focal length of the lens

Use lens equation

[tex]\frac{1}{f}=\frac{1}{v}-\frac{1}{u}[/tex]

[tex]\frac{1}{f}=\frac{1}{-70}-\frac{1}{∞}[/tex]

f = - 70 cm

[tex]P=\frac{1}{f}[/tex]

Where, P is the power of lens

P = - 1.428 Dioptre

Thus, the power of lens used is - 1.428 D.

Answer: a) 1.6 * 10 ^-7 N/C (upward) ; b) -2.5*10^-6N (downward)

Explanation: In order to solve this proble we have to use the Coulomb law given by:

F=q*E from this expression we have

E=F/q=4.8*10^-6/30 C= 1.6 * 10 ^-7 N/C

The force on the particle charge by -1.6 X10 C place intead of the initial charge we have

F=q*E= -16 C* 1.6 * 10 ^-7 N/C= -2.5*10^-6N

Answer:

a) v=2.743m/s

b) [tex]a_c = 8.363m/s^2[/tex]

c) T=2.543N

Explanation:

First, calculate the height of the ball at the starting point:

[tex]y' = 0.9cos(55)[/tex]

[tex]y' = 0.516[/tex]

At this point, just in the moment the ball is released, all the energy of the system is potencial gravitational energy. When it is at the bottom all the potencial energy is transformed into kinetic energy:

[tex]E_p=E_k\\mgh=\frac{mv^2}{2}[/tex]

Solving for v:

[tex]v=\sqrt{2gh}[/tex]

if h is the height loss: (l-y')

v=2.743m/s

The centripetal acceleration is the acceleration caused by the tension force exercised by the string, and is pointing outside of the trayectory path (at the lowest point, directly dawn):

[tex]a_c=\frac{v^2}{r}[/tex]

[tex]a_c = 8.363m/s^2[/tex]

To calculate tension, just make the free body diagram of forces in the ball, noticing the existence of the centripetal acceleration:

[tex]\sum{F_y}=ma_c=T-W\\T=ma_c+W\\T=m(a_c+g)\\T=0.14(8.363+9.8)\\T=2.543N[/tex]

Answer:

a)  Y component of the vector =15.54 m

b) Vector magnitude = 21.6 m

Explanation:

The given vector makes 44 degree angle with Y axis, as given. This is same as 90 -44 = 46 degrees with the horizontal or X axis.

b) X component of the given vector = [tex]A_{x}[/tex] = A cos 46 =15

⇒ A = 15/cos 16 = 21.6 m = Total vector magnitude.

a) Y component of the vector = 21.6 sin 46 = 15.54 m

b) A = 21.6 m

The y-component of vector A (Ay) is approximately 10.3 m, assuming a positive y-direction due to the angle being in the second quadrant. The magnitude of vector A, found using the Pythagorean theorem with its components, is approximately 18.0 m.

To find the y-component of vector A (Ay), we use the trigonometric relationship Ay = A sin(θ), where A is the magnitude of the vector and θ is the angle it makes relative to the x-axis. However, since the angle given is clockwise from the y-axis, we must first convert it to the counterclockwise angle from the x-axis, which would be θ = 90° + 44.0° = 134.0°.

Given that Ax = -15.0 m, and assuming θ is measured counterclockwise, we can find the magnitude of A using the Pythagorean theorem: A = √(Ax² + Ay²). To find Ay, we rearrange the equation for the y-component to Ay = A sin(θ) and solve for Ay using the magnitude we just found. Our direction angle θ is in the second quadrant, meaning Ay should be positive.

Since we only have Ax, we first find A as follows: A = √((-15.0 m)² + Ay²). To determine Ay, we need to look at signs and quadrants. Ax is negative, and because the angle is measured clockwise from the y-axis, Ay must be positive. Therefore, we can infer that Ay is approximately 10.3 m by implication of the Pythagorean theorem, knowing that the other possible value is negative, which does not fit the quadrant.

To find the magnitude of A, we use the derived components: A = √((-15.0 m)² + (10.3 m)²), yielding approximately 18.0m.

Answer:

a)X=70.688m

b)v=18.8m/S

Explanation:

the truck has constant speed while the car moves in uniformly accelerated motion

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t  (1)\\\\

\frac{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\

X=Xo+ VoT+0.5at^{2}    (3)\\

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve .

For the automobile

we use the ecuation number 3

Xo=0

Vo=0

a=2.5

X=0.5(2.5)t^2

X=1.25t^2

for the truck

x=Vt

X=9.4t

the distance is the same we can match the previous equations

1.25t^2=9.4t

t=9.4/1.25=7.52s

using the ecuation for the truck

x=9.4*7.52=70.688m

b) for this point we can use the ecuation number 1 for automobile

Vf=Vo+at

Vf=0+2.5(7.52)

Vf=18.8m/S

Answer:

1010 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-68^2}{2\times 250}\\\Rightarrow a=-9.248\ m/s^2[/tex]

[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-68}{-9.248}\\\Rightarrow t=7.35\ s[/tex]

Time taken by the thunderbird to stop is 7.35 seconds

Time the thunderbird was at the pit is 5 seconds

[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{68^2-0^2}{2\times 420}\\\Rightarrow a=5.5\ m/s^2[/tex]

[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{68-0}{5.5}\\\Rightarrow t=12.36\ s[/tex]

Time taken to accelerate back to 68 m/s is 12.36 seconds

Total time to this point is 7.35+5+12.36 = 24.71 seconds

The Mercedes Benz is moving at a constant velocity hence it has no acceleration and we use the formula

Distance = Speed × Time

⇒Distance = 68 × 24.71 = 1680 m

The thunderbird has covered 250+420 = 670 m

So, the distance between them is 1680-670 = 1010 m

Final answer:

To find out how far the Thunderbird has fallen behind the Mercedes Benz, we calculate the time taken for its deceleration, pit stop, and acceleration, then determine how far the Mercedes traveled in that time. The Thunderbird falls behind by approximately 1346.4 meters.

Explanation:

The Thunderbird's driver decelerates to a stop, pauses for 5 seconds, then accelerates back to speed. The Mercedes Benz continues at a constant speed of 68.0 m/s throughout this time. We will calculate the time taken for the Thunderbird to decelerate and accelerate, and use this to determine how far the Mercedes has traveled in the same period, thus finding out how far the Thunderbird has fallen behind.

Deceleration Phase

First, we need to find the Thunderbird's deceleration. Using the formula v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity (68.0 m/s), a is the acceleration, and s is the distance (250 m), we solve for a to find the deceleration rate. Rearranging the formula gives us a = (v^2 - u^2) / 2s. Plugging in the numbers, we find that the deceleration rate is approximately -9.184 m/s^2.

To find out how long this deceleration took, we use the formula v = u + at, solving for t gives us t = (v - u) / a. This calculation reveals that the deceleration phase lasted approximately 7.4 seconds.

Acceleration Phase

Assuming the Thunderbird accelerates at the same rate but in the positive direction over 420 m, we'll use the same method to find out the acceleration time. The acceleration time is also approximately 7.4 seconds.

Total time for stop and acceleration equals deceleration time + pit stop time + acceleration time, which totals 19.8 seconds.

The Mercedes Benz, traveling at 68.0 m/s, would have traveled 1346.4 meters in this time.

Therefore, the Thunderbird has fallen behind by 1346.4 meters, considering both the stopping and accelerating phases and the pit stop time.

Answer:

displacement is 481.3 m  with 40.88° north east

Explanation:

given data

walk 1 = 300 m north = 300 j

walk 2 = 400 m northwest =400 cos(45) i + sin(45) j

walk 3 = 700 m east southeast  = 700 cos(22.5) i- sin(22.5) j

to find out

displacement and direction

solution

we consider here direction x as east and direction y as north

so

displacement = distance 1 + distance 2 + distance 3

displacement = 300 j + 400 cos(45) i + sin(45) j + 700 cos(22.5) i- sin(22.5) j  

we get here

displacement = 363.9 i + 315 j

so magnitude

displacement = [tex]\sqrt{363.9^{2}+315^{2}  }[/tex]

displacement = 481.3 m

and angle will be = arctan(315/369.9)

angle = 0.713494059 rad

angle is 40.88 degree

so displacement is 481.3 m  with 40.88° north east

Answer:

Magnitude of displacement = 481.24 m

Direction = 40.88 degrees north east.

Explanation:

Let east is along real axis and north is along imaginary axis. So,

First walk = d1 = J300 m

Second walk = d2 = -400Cos(45) + J400Sin(45) = (-282.84 + J282.84) m  

Third walk = d3 = 700Cos(22.5) – J700Sin(22.5) = (646.71 – J267.88) m

Total displacement = d = d1 + d2 + d2 = (363.86 + J314.96)m

Magnitude = √((363.86)^2+(314.96)^2) = 481.24 m

Direction = arctan(314.96/363.86) = 40.88 degrees north east.

That statement is false.

Angular velocity is the rate at which that angle changes. As long as the reference line doesn't move, it doesn't matter where it is. The rate at which the angle changes is still the same, constant number.

The angular velocity of a rigid body that undergoes plane motion depends on the reference line from which its angle of rotation is measured is False.

When anything is moving, its velocity tells us how quickly that something's location is changing from a certain vantage point and as measured by a particular unit of time.

When a point travels down a path and completes a certain distance in a predetermined amount of time, its average speed over that time is equal to the distance covered divided by the travel time. A train traveling 100 kilometers in two hours, for instance, is doing it at an average speed of 50 km/h.

The pace at which that angle change is known as the angular velocity. It doesn't matter where the reference line is, as long as it stays put. The rate of change of the angle remains constant and the same.

Therefore, the angular velocity of a rigid body that undergoes plane motion depends on the reference line from which its angle of rotation is measured is False.

To know more about velocity:

https://brainly.com/question/19979064

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A) -0.15 s

First of all, we need to velocity of the swimmer in absence of current. This is given by:

[tex]v_0 = \frac{d}{t}[/tex]

where

d = 50.0 m

t = 25.0 s

Substituting,

[tex]v_0 = \frac{50.0}{25.0}=2.0 m/s[/tex]

In lane 1, the velocity of the current is

[tex]v_c = +1.2 cm/s = +0.012 m/s[/tex]

where the + sign means it is in the same direction as the swimmer. Therefore, the net velocity of the swimmer in lane 1 will be

[tex]v=v_0+v_c = 2.0 + 0.012 = = 2.012 m/s[/tex]

And so, the time the swimmer will take to cover the 50.0 m will be:

[tex]t=\frac{d}{v}=\frac{50.0}{2.012}=24.85 s[/tex]

So, the time would change by

[tex]\Delta t = 24.85 - 25.0 = -0.15 s[/tex]

which means that the swimmer will be 0.15 s faster.

B) +0.15 s

To solve this part, we just need to consider that the current goes in the opposite direction, so its velocity actually is:

[tex]v_c = -12 cm/s = -0.012 m/s[/tex]

where the negative sign indicates the opposite direction.

So, the net velocity of the swimmer in lane 8 is

[tex]v=v_0+v_c = 2.0 - 0.012 = = 1.988 m/s[/tex]

And so, the time the swimmer will take to cover the 50.0 m will be:

[tex]t=\frac{d}{v}=\frac{50.0}{1.988}=25.15 s[/tex]

So, the time would change by

[tex]\Delta t = 25.15 - 25.0 = +0.15 s[/tex]

which means that the swimmer will be 0.15 s slower.

Answer:

The rise from A to B is 0.887

Solution:

As per the question:

The following reading of an inverted staff is given as:

A = 2.915

B = -2.028

Here, for inverted staff, the greater reading shows greater elevation and lesser reading shows lower elevation.

Thus

The rise from A to B is given as:

A - B = 2.915 - 2.028 = 0.887

Answer:

398 m/s

Explanation:

The acceleration is given by:

a = K/(L - x)^2

K = 100 m^3/s^2

L = 0.169 m

This acceleration will result in a force:

F = m * a

F = m * K/(L - x)^2

This force will perform a work:

W = F * L

The ball will advance only until x = L - D/2

[tex]W = m * K \int\limits^{L - D/2}_0 {\frac{1}{(L - x)^2}} \, dx[/tex]

This work will be converted to kinetic energy

W = Ek

Ek = 1/2 * m * v^2

[tex]1/2 * m * v^2 = m * K \int\limits^{L - D/2}_0 {\frac{1}{(L - x)^2}} \, dx[/tex]

[tex]1/2 * v^2 = K \int\limits^{L - D/2}_0 {\frac{1}{(L - x)^2}} \, dx[/tex]

First we solve thr integral:

[tex]K \int\limits^{L - D/2}_0 {\frac{1}{(L - x)^2}} \, dx[/tex]

We use the replacement

u = L - x

du = -dx

And the limits

When x = L - D/2, u = D2/2, and when x = 0, u = L

[tex]-K \int\limits^{D/2}_L {u^-2}} \, du[/tex]

K / u evaluated between L and D/2

2*K / D - K / L

Then

1/2 * v^2 = 2*K / D - K / L

1/2 * v^2 = K * (2/D - 1/L)

v^2 = 2*K*(2/D - 1/L)

[tex]v = \sqrt{2*K*(2/D - 1/L)}[/tex]

[tex]v = \sqrt{2*100*(2/0.0025 - 1/0.169)} = 398 m/s[/tex]

Answer:

[tex]acceleration=0.8 \frac{m}{s^{2} } \\distance=112.5 m[/tex]

Explanation:

We can find these answers following the equations of motion.

[tex]a=[/tex]Δ[tex]V/t[/tex]

Where Δ[tex]V[/tex] is the difference between the final speed and the initial speed. And [tex]t[/tex] is the time spent

We replace the terms:

[tex]a=\frac{18\frac{m}{s} -12\frac{m}{s} }{7.5s}[/tex]

We solve the difference:

[tex]a=\frac{6\frac{m}{s}}{7.5s}[/tex]

We divide the terms, so we can have the answer:

[tex]a=0.8 \frac{m}{s^{2}}[/tex]

2. To find the distance traveled by the truck, we use the equation:

[tex]x=V_0t+\frac{1}{2}at^{2}[/tex]

Where [tex]x[/tex] is the distance traveled, [tex]V_0[/tex] is the initial speed, [tex]a[/tex] is the acceleration and [tex]t[/tex] is the time.

We replace the terms:

[tex]x=(12\frac{m}{s}*7.5s)+\frac{1}{2}[0.8\frac{m}{s^{2} }*(7.5)^{2} ][/tex]

We multiply and solve the exponential:

[tex]x=90m+\frac{1}{2}(0.8\frac{m}{s^{2} }*56.25s^{2} )[/tex]

Then, we multiply the terms left:

[tex]x=90m+22.5m[/tex]

And add, so we can have the answer:

[tex]x=112.5m[/tex]

Answer:

[tex]Q=-8.85*10^{-12}*0.36 C =-3.186*10^{-12} C[/tex]

(negative charge)

Explanation:

Hi!

To solve this problem we use Gauss Law for electric fields, which relates the flux of electric field E through a closed surface S with the charge Q enclosed by that surface:

[tex]\int_{S} \vec{E}\cdot \vec{n} \;dA = \frac{Q}{\epsilon_0}[/tex]

[tex]\vec{n} = \text{outwards surface normal}\\\epsilon_0 = 8.85*10^{-12}\frac{C^2}{Nm^2}[/tex]

In this problem S is a cube of side length 2cm. The integral is easy because the electric field is uniform in each face, and normal to the face. The total integral is the sum of the integrals in each of the six faces.

[tex]\int_{S} \vec{E}\cdot \vec{n} \;dA = \frac{Q}{\epsilon_0} = 3(-500\frac{N}{C} (2cm)^2) + 3(200\frac{N}{C} (2cm)^2)[/tex]

[tex](-1500*4*10^{-4}\frac{Nm^2}{C} + 600*4*10^{-4}\frac{Nm^2}{C} )=-0.36\frac{Nm^2}{C}[/tex]

[tex]Q=-8.85*10^{-12}*0.36 C =-3.186*10^{-12} C[/tex]

Answer:

E the electric field remains unchanged.

Explanation:

potential difference between the plates of a capacitor

V = Q / C  Where Q is charge on the capacitor and C is capacity of capacitor

Here Q is unchanged.

C the capacity has value equal to ε A / d

Here d is distance between plates. when  it is halved, capacitance C becomes double.

V = Q /C , When C becomes double V becomes half

E = V/d , E is electric field between plates having separation of d.

When V becomes half and d also becomes half , there is no change in the value of E.  

Hence E the electric field remains unchanged.

The magnitude of the electric field between two charged parallel plates remains constant when the plate separation is reduced from d to d/2, provided the charge density on the plates stays the same. Thus, the magnitude of the electric field halfway between the plates would still be E.

The student asked what the magnitude of the electric field would be if the separation between two large, flat, parallel, charged plates is reduced from d to d/2, assuming the charge on the plates remains constant.

According to Essential Knowledge 2.C.5, the electric field (E) between two oppositely charged parallel plates with uniformly distributed electric charge is constant in magnitude and direction at points far from the edges of the plates.

The magnitude of the electric field E between parallel plates is given by the equation E = σ/ε₀, where σ is the surface charge density and ε₀ is the permittivity of free space. Since the surface charge density (σ = Q/A, charge per area) does not change when the separation between the plates is altered and the permittivity of free space (ε₀) is a constant, the magnitude of the electric field between the plates remains the same even if the separation is reduced to d/2.

Therefore, the magnitude of the electric field halfway between the plates when the separation is reduced to d/2 would still be E.

Answer:

75 m /s

Explanation:

Acceleration of ball a = change in velocity / time

=( 90 - 60)/ 3 = 10 m s⁻²

Initial velocity u = 60 m/s

a = 10 m s⁻²

time t = 3

distance traveled s = ut + 1/2 at²

= 60 x 3 + .5 x 10 x 3 x 3

= 225 m

Average velocity = total distance traveled /time

= 225 / 3 = 75 m/s

Final answer:

To estimate the surface temperature of a star modeled as an ideal blackbody, we can use the Stefan-Boltzmann law. The law states that the power radiated by a blackbody is proportional to its surface area and temperature to the fourth power. We can rearrange the equation to solve for temperature: T = (I/(σ*A))^1/4.

Explanation:

To estimate the surface temperature of a star modeled as an ideal blackbody, we can use the Stefan-Boltzmann law. The law states that the power radiated by a blackbody is proportional to its surface area and temperature to the fourth power. We can rearrange the equation to solve for temperature: T = (I/(σ*A))^1/4, where T is the temperature, I is the intensity of radiation, σ is the Stefan-Boltzmann constant, and A is the surface area.

Plugging in the given values, we have T = (0.055W/m² / (5.670 x 10^-8 W/(m²K^4) * 4π(5.0 x 10^8m)^2))^1/4.

Answer:

net electrostatic force = 46.76 N

and direction is vertical downward

Explanation:

given data

charge =  +5μC

corners identical charges = -6 μC

length of side = 0.10 m

to find out

What is the net electrostatic force

solution

here apex is the vertex where the two sides of equal length meet

so upper point A have charge +5μC and lower both point B and C have charge -6 μC

so

force between +5μC and -6 μC is express as

force  = [tex]k \frac{q1q2}{r^2}[/tex]       ..............1

put here electrostatic constant  k = 9 × [tex]10^{9}[/tex] Nm²/C² and q1 q2 is charge given and r is distance 0.10 m

so

force  = [tex]9*10^{9} \frac{5*6*10^(-12)}{0.10^2}[/tex]

force = 27 N

so net force is vector addition of both force

force = [tex]\sqrt{x^{2}+x^{2}+2x^{2}cos60}[/tex]

here x is force 27 N

force = [tex]\sqrt{27^{2}+27^{2}+2(27)^{2}cos60}[/tex]

net electrostatic force = 46.76 N

and direction is vertical downward