An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an angle of θ = 30◦ to the ground. The coefficient of kinetic friction μk = 0.2. At the bottom of the plane, the mass slides along a rough surface with a coefficient of kinetic friction μk = 0.3 until it comes to rest. The goal of this problem is to find out how far the object slides along the rough surface. What is the work done by the friction force while the mass is sliding down the in- clined plane? (Is it positive or negative?) (b) What is the work done by the gravitational force while the mass is sliding down the inclined plane? (Is it positive or negative?)

Answer:

the velocity of car when it passes the truck is u = 16.33 m/s

Explanation:

given,

constant speed of truck  = 28 m/s

acceleration of car = 1.2 m/s²

passes the truck in 545 m

speed of the car when it just pass the truck = ?

time taken by the truck to travel 545 m

              time =[tex]\dfrac{distance}{speed}[/tex]

              time =[tex]\dfrac{545}{28}[/tex]

              time =19.46 s

velocity of the car when it crosses the truck

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

[tex]545= u\times 19.46 + \dfrac{1}{2} \times 1.2 \times 19.46^2[/tex]

u = 16.33 m/s

the velocity of car when it passes the truck is u = 16.33 m/s

Answer:

- 1.65 × 10⁻⁷ C

Explanation:

Given:

Sides of the cube = 30 cm long

Electric flux, φ = - 3.1 kN.m²/C = - 3.1 × 10³ N.m²/C

Now,

For the cube,

The charge using the Gauss law is given as:

q = ε₀ × (nφ)

Here,

q is the charge

ε₀ = 8.85 × 10⁻¹² C²/N.m²

n is the number of sides for the cube = 6

Thus,

q = 8.85 × 10⁻¹² × 6 × (- 3.1 × 10³ )

or

q = - 1.65 × 10⁻⁷ C

Answer:

a) 32.58 m/s²

b) 161.84 m/s

Explanation:

Initial velocity = u = 0

Final velocity = v = 145 m/s

Time taken = t = 4.45 s

s = Displacement of dragster = 402 m

a = Acceleration

[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{145-0}{4.45}\\\Rightarrow a=32.58\ m/s^2[/tex]

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as-u^2}\\\Rightarrow v=\sqrt{2\times 32.58\times 402-0^2}\\\Rightarrow v=161.84\ m/s[/tex]

The final velocity is greater than the velocity used to find the average acceleration due to the gear changes. The first gear in a dragster has the most amount of toque which means the acceleration will be maximum. The final gears have less torque which means the acceleration is lower here. The final gears have less acceleration but can spin faster which makes the dragster able to reach higher speeds but slowly.

Answer:

time required to mow lawn is 1274.06 minutes

Explanation:

given data

average mow = 1400 square meters each hour

area = 320000 square feet

to find out

How many minutes will take to mow lawn

solution

we know that here 1 square feet is equal to 0.092903 square meter

so 320000 square feet will be = 0.09203 × 320000 = 29728.9728 square meter

so time required is express as

time required = [tex]\frac{distance}{speed}[/tex]

time required = [tex]\frac{29728.9728}{1400}[/tex]

time required = 21.23 hours

so time required = 21.23 × 60 min = 1274.06

time required to mow lawn is 1274.06 minutes

Answer:

[tex]t = 3.49 s[/tex]

[tex]a = -5.02 m/s^2[/tex]

Explanation:

As we know that initial speed of the cheetah is given as

[tex]v_i = 17.5 m/s[/tex]

finally it comes to rest so final speed is given as

[tex]v_f = 0[/tex]

now we know that distance covered by cheetah while it stop is given as

[tex]d = 30.5 m[/tex]

now by equation of kinematics we know that

[tex]d = (\frac{v_f + v_i}{2})t[/tex]

here we have

[tex]30.5 m = (\frac{0 + 17.5}{2}) t[/tex]

[tex]30.5 = 8.75 t[/tex]

[tex]t = 3.49 s[/tex]

Now in order to find the acceleration we know that

[tex]v_f - v_i = at[/tex]

[tex]0 - 17.5 = a(3.49)[/tex]

[tex]a = -5.02 m/s^2[/tex]

To determine the time it takes for a cheetah running at 17.5 m/s to stop over a distance of 30.5 m, kinematic equations are used. The acceleration is calculated to be approximately -4.98 m/s², and the time to come to rest is roughly 3.49 seconds.

To find out how long it takes for a cheetah running at 17.5 m/s to come to rest and what is the cheetah's acceleration while doing so, we can use the following kinematic equation:

[tex]v^2 = u^2 + 2as[/tex]

Where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance covered. Given that the cheetah comes to rest, v = 0 m/s, u = 17.5 m/s, and s = 30.5 m.

We can rearrange the formula to solve for a:

[tex]0 = (17.5 m/s)^2 + 2a(30.5 m)[/tex]

[tex]a = - (17.5 m/s)^2 / (2 * 30.5 m)[/tex]

a = - 4.98 m/s2 (The acceleration is negative because it's a deceleration)

Now, using the formula v = u + at to find the time, t, it takes to come to rest:

[tex]0 = 17.5 m/s - (4.98 m/s^2 * t)\\t = 17.5 m/s / 4.98 m/s^2[/tex]

t = 3.51 s (approximately 3.49 s)

Answer:

W_planet=(W_earth*a)/g

Explanation:

The mass of the object does not change. It is the same at Earth and at the other planet.

W_earth=Mg       weight at Earth

W_planet=Ma       weight at other planet

If we divide the last equations:

W_planet/W_earth=a/g

W_planet=(W_earth*a)/g

Answer:

The length of the rod is 0.6 m

Solution:

Mass of the copper rod, m = 100 g = 0.1 kg

Angular speed at lowest position, [tex]\omega_{L} = 7 rad/s[/tex]

Now,

By using the law of conservation of energy, the overall mechanical energy of the system taken about the center of mass remain conserve.

Thus at the initial position of the rod, i.e., horizontal:

[tex]\frac{1}{2}mv_{c}^{2} + \frac{1}{2}I\omega^{2} + mg\frac{L}{2} =0 + mg\frac{L}{2}[/tex]

[tex]\frac{1}{2}mv_{c}^{2} + \frac{1}{2}I\omega^{2} = 0[/tex]

(Since, [tex]v_{c} = 0[/tex] and [tex]\omega = 0[/tex] at horizontal position).

where

Tranlational Kinetic energy about center of mass = [tex]\frac{1}{2}mv_{c}^{2}[/tex]

Rotational K.E about the center of mass, [tex]\frac{1}{2}I\omega^{2}[/tex]

Potential energy about Center of Mass = [tex]mg\frac{L}{2}[/tex]

Now, applying the law of conservation at the lowest point of the rod:

[tex]\frac{1}{2}mv'_{c}^{2} + \frac{1}{2}I\omega'^{2} + 0 =\frac{1}{2}m(\frac{\omega' L}{2})^{2} + \frac{1}{2}I\omega'^{2}[/tex]

[tex]\frac{1}{2}mv'_{c}^{2} + \frac{1}{2}I\omega'^{2}  =\frac{1}{2}m(\frac{\omega' L}{2})^{2} + \frac{1}{2}\frac{mL^{12}}{12}\omega'^{2} = \frac{mL^{2}\omega'^{2}}{6}[/tex]

where

Moment of inertia about the center of mass at the lowest position is [tex]\frac{mL^{2}}{12}[/tex]

[tex]v'_{c} = \frac{\omega L}{2}[/tex]

Thus

From these, potential energy about center of mass = [tex]\frac{mL^{2}\omega'^{2}}{6}[/tex]

[tex]mg\frac{L}{2} = \frac{mL^{2}\omega'^{2}}{6}[/tex]

At the lowest point, [tex]\omega' = 7 rad/s[/tex]

Thus

[tex]mg\frac{L}{2} = \frac{mL^{2}\times 7^{2}}{6}[/tex]

[tex]g = \frac{49L}{3}[/tex]

g = 9.8[tex]m/s^{2}[/tex]

[tex]9.8 = \frac{49L}{3}[/tex]

L = 0.6 m

Answer:[tex]\theta =27.26^{\circ}[/tex]

Explanation:

Given

Vector A has magnitude of 5 units

Vector B has magnitude of 9 units

Dot product of A and B is 40

i.e.

[tex]A\cdot B=|A||B|cos\theta [/tex]

[tex]40=5\times 9\times cos\theta [/tex]

[tex]cos\theta =\frac{8}{9}=0.889[/tex]

[tex]\theta =27.26^{\circ}[/tex]

Answer:

boy average velocity to catch the ball just before it hits the ground is 8.53 m/s

Explanation:

given data

vertical distance = 75.1 m

horizontal distance = 33.4 m

to find out

Find the average velocity to catch the ball just before it hits the ground

solution

we know here ball is coming downward and boy is running to catch it

so first we calculate the time of ball to reach at ground that is express by equation of motion

s = ut + 0.5 × at²    .....................1

here u is initial speed that is zero and a = 9.8 and s is distance

put here all value to get time t  in equation 1

75.1 = 0 + 0.5 × (9.8)t²  

t = 3.914 s

so

boy speed is = [tex]\frac{distance}{time}[/tex]

speed = [tex]\frac{33.4}{3.914}[/tex]

speed = 8.53 m/s

so average velocity to catch the ball just before it hits the ground is 8.53 m/s

Answer:

ΔTmin = 3.72 °C

Explanation:

With a 16-bit ADC, you get a resolution of [tex]2^{16}=65536[/tex] steps. This means that the ADC will divide the maximum 10V input into 65536 steps:

ΔVmin = 10V / 65536 = 152.59μV

Using the thermocouple sensitiviy we can calculate the smallest temperature change that 152.59μV represents on the ADC:

[tex]\Delta Tmin = \frac{\Delta Vmin}{41 \mu V/C}= 3.72 C[/tex]

Answer:

(a) [tex]L_{D} = 1.052\times 10^{- 4} m[/tex]

N = [tex]4.87\times 10^{4}[/tex]

(b) [tex]L'_{D} = 5.531\times 10^{- 6} m[/tex]

N' = [tex]7.087\times 10^{- 4}[/tex]

(c) [tex]L''_{D} = 4.43\times 10^{- 13} m[/tex]

N'' = [tex]3.63\times 10^{- 14}[/tex]

Solution:

As per the question, we have to calculate the Debye, [tex]L_{D}[/tex] length and N for the given cases.

Also, we utilize the two relations:

1. [tex]L_{D} = \sqrt{\frac{KT\epsilon_{o}}{ne^{2}}}[/tex]

2. N = [tex]\frac{4}{3}n\pi(L_{D})^{3}[/tex]

Now,

(a) n = [tex]10^{10} cm^{- 3}\times (10^{- 2})^{- 3} = 10^{16} m^{- 3}[/tex]

KT = 2 eV

Then

[tex]L_{D} = \sqrt{\frac{2\times 1.6\times 10^{- 19}\times 8.85\times 10^{- 12}}{10^{16}(1.6\times 10^{- 19})^{2}}}[/tex]

(Since,

e = [tex]1.6\times 10^{- 19} C[/tex]

[tex]\epsilon_{o} = 8.85\times 10^{- 12} F/m[/tex])

Thus

[tex]L_{D} = 1.052\times 10^{- 4} m[/tex]

Now,

N = [tex]\frac{4}{3}\times 10^{16}\pi(1.052\times 10^{- 4})^{3} = 4.87\times 10^{4}[/tex]

(b) n = [tex]10^{6} cm^{- 3}\times (10^{- 2})^{- 3} = 10^{12} m^{- 3}[/tex]

KT = 0.1 eV

Then

[tex]L'_{D} = \sqrt{\frac{0.1\times 1.6\times 10^{- 19}\times 8.85\times 10^{- 12}}{10^{12}(1.6\times 10^{- 19})^{2}}}[/tex]

[tex]L'_{D} = 5.531\times 10^{- 6} m[/tex]

N' = [tex]\frac{4}{3}\times 10^{12}\pi(5.531\times 10^{- 6})^{3} = 7.087\times 10^{- 4}[/tex]

(c) n = [tex]10^{17} cm^{- 3}\times (10^{- 2})^{- 3} = 10^{23} m^{- 3}[/tex]

KT = 800 eV

[tex]L''_{D} = \sqrt{\frac{800\times 1.6\times 10^{- 19}\times 8.85\times 10^{- 12}}{10^{23}(1.6\times 10^{- 19})^{2}}}[/tex]

[tex]L''_{D} = 4.43\times 10^{- 13} m[/tex]

N'' = [tex]\frac{4}{3}\times 10^{23}\pi(4.43\times 10^{- 13})^{3} = 3.63\times 10^{- 14}[/tex]

Answer with Explanation:

The position of the particle as a function of time is given by [tex]s(t)=t^2-26t+133[/tex]

Part 1) The position as a function of time is shown in the below attached figure.

Part 2) By the definition of velocity we have

[tex]v=\frac{ds}{dt}\\\\\therefore v(t)=\frac{d}{dt}\cdot (t^2-26t+133)\\\\v(t)=2t-26[/tex]

The velocity as a function of time is shown in the below attached figure.

Part 3) The displacement of the particle in the first 19 seconds is given by [tex]\Delta x=s(19)-s(0)\\\\\Delta x=(19^2-26\times 19+133)-(0-0+133)=-133millimeters[/tex]

Part 4) The distance covered in the first 19 seconds can be found by evaluating the integral

[tex]s=\int _{0}^{19}\sqrt{1+(\frac{ds}{dt})^2}\\\\s=\int _{0}^{19}\sqrt{1+(2t-26)^2}\\\\\therefore s=207.03meters[/tex]

Part 4) As we can see that the position-time graph is parabolic in shape hence we conclude that the motion is uniformly accelerated motion.

Answer:

speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s

initial kinetic energy = final kinetic energy

[tex]KE = 6.66 J[/tex]

Explanation:

Since there is no external force on the system of two balls so here total momentum of two balls initially must be equal to the total momentum of two balls after collision

So we will have

momentum conservation along x direction

[tex]m_1v_{1i} + m_2v_{2i} = m_1v_{1x} + m_2v_{2x}[/tex]

now plug in all values in it

[tex]1.47 \times 3.01 + 0 = 1.47 v_1cos68 + 1.47 v_2cos22[/tex]

so we have

[tex]3.01 = 0.375v_1 + 0.927v_2[/tex]

similarly in Y direction we have

[tex]m_1v_{1i} + m_2v_{2i} = m_1v_{1y} + m_2v_{2y}[/tex]

now plug in all values in it

[tex]0 + 0 = 1.47 v_1sin68 - 1.47 v_2sin22[/tex]

so we have

[tex]0 = 0.927v_1 - 0.375v_2[/tex]

[tex]v_2 = 2.47 v_1[/tex]

now from 1st equation we have

[tex]3.01 = 0.375 v_1 + 0.927(2.47 v_1)[/tex]

[tex]v_1 = 1.13 m/s[/tex]

[tex]v_2 = 2.78 m/s[/tex]

so speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s

Also we know that since this is an elastic collision so here kinetic energy is always conserved to

initial kinetic energy = final kinetic energy

[tex]KE = \frac{1}{2}(1.47)(3.01^2)[/tex]

[tex]KE = 6.66 J[/tex]

The distance between the two ships after two hours can be found using vector addition and the law of cosines, and the relative speed of ship A as seen by ship B is determined by subtracting the velocity vector of ship B from that of ship A.

The question asks to determine the distance between two ships two hours after their departure from the same port and the relative speed of ship A as seen from ship B. To solve this, we need to apply the concepts of vector addition and use trigonometry.

First, we find the vectors of the individual ships' movements. Ship A travels at 40 mph in a direction 35° west of north, and Ship B travels at 20 mph 80° east of north.

To find the distance between the two ships after two hours, we'll calculate each ship's displacement (speed multiplied by time) for two hours and then use trigonometry to determine the resultant displacement vector between the two ships. Using the law of cosines, we can find the distance between the ships. To find the relative speed of ship A from B, we will subtract the velocity vector of B from A.

These steps will yield the answers to the question after proper calculations.

Answer:

[tex]3.33\times 10^{-4}[/tex] C

Explanation:

[tex]E[/tex] = Maximum electric field strength = [tex]3\times 10^{6}[/tex] N/C

[tex]r[/tex] = Radius of the sphere  = [tex]1 [/tex] m

[tex]Q[/tex] = maximum charge stored by the sphere  = ?

Considering that the total charge is stored at the center of the sphere, the electric field at the surface of sphere can be given as

[tex]E=\frac{kQ}{r^{2}}[/tex]

Inserting the values for the variables in the above equation

[tex]3\times 10^{6}=\frac{(9\times 10^{9})Q}{1^{2}}[/tex]

[tex]3\times 10^{6}=(9\times 10^{9})Q[/tex]

Dividing both side by [tex](9\times 10^{9})[/tex]

[tex]\frac{3\times 10^{6}}{9\times 10^{9}}= \frac{9\times 10^{9}}{9\times 10^{9}}Q[/tex]

[tex]Q = \frac{3\times 10^{6}}{9\times 10^{9}}[/tex]

[tex]Q = 3.33\times 10^{-4}[/tex] C

To calculate the electric field at the center of the square, we use the Coulomb's law formula for each charge located at the corners of the square. We then compute the vector sum of these electric fields, taking into account the direction of each field.

First, we use Coulomb's law formula to calculate the electric field created by a single charge, which is E = kQ/r², where E is the electric field, k is Coulomb's constant (8.99 x 10⁹ N.m²/C²), Q is the charge and r is the distance from the charge.

We calculate the electric field at the center of the square due to each charge separately. The electric field due to the -38.6 μC charge is E1 = kQ/r² = 8.99 x 10⁹ N.m²/C² * -38.6 x 10⁻⁶C/(0.425m/√2)², and similarly, the electric field due to each -27.0μC charge is E2 = kQ/r² = 8.99 x 10⁹ N.m²/C² * -27.0 x 10⁻⁶C/(0.425m/√2)².

The total electric field is simply the vector sum of the individual fields which would involve figuring out the geometric relationship between the individual electric fields. An important note is, electric fields from negative charges are directed towards the charges, so the directions of the fields must be taken into account while adding.

https://brainly.com/question/14423246

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Answer:

[tex]E=2.0*10^{11}N/m^{2}[/tex]

Explanation:

Relation between stress and Force:

[tex]\sigma=\frac{F}{A}=\frac{F}{\pi*d^{2}/4}[/tex]

Relation between stress and strain:

Young's modulus is defined by the ratio of longitudinal stress σ , to the longitudinal strain ε:

[tex]E=\frac{\sigma}{\epsilon}[/tex]

[tex]\epsilon=\frac{\Delta l}{l}[/tex]

So:

[tex]E=\frac{F*l}{\pi*d^{2}/4*\Delta l}=\frac{50*10^{6}*3}{\pi*(0.4^{2}/4)*5.967*10^{-3}}=2*10^{11}N/m^2[/tex]

Answer:

[tex]\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.[/tex]

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force between two static point charges  [tex]\rm q_1[/tex] and [tex]\rm q_1[/tex], separated by a distance [tex]\rm r[/tex], is given by

[tex]\rm F = \dfrac{kq_1q_2}{r^2}.[/tex]

where k is the Coulomb's constant.

Initially,

[tex]\rm r = 0.160\ m\\F_i = -1.30\ N.\\\\and \ \ |q_2|>|q_1|.[/tex]

The negative sign is taken with force F because the force is attractive.

Therefore, the initial electrostatic force between the charges is given by

[tex]\rm F_i = \dfrac{kq_1q_2}{r^2}.\\-1.30=\dfrac{kq_1q_2}{0.160^2}\\\rm\Rightarrow q_2 = \dfrac{-1.30\times 0.160^2}{q_1k}\ \ \ ..............\ (1).[/tex]

Now, the objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial positions.

The force is now repulsive, therefore, [tex]\rm F_f = +1.30\ N.[/tex]

The new charges on the two objects are

[tex]\rm q_1'=q_2' = \dfrac{q_1+q_2}{2}.[/tex]

The new force is given by

[tex]\rm F_f = \dfrac{kq_1'q_2'}{r^2}\\+1.30=\dfrac{k\left (\dfrac{q_1+q_2}{2}\right )\left (\dfrac{q_1+q_2}{2}\right )}{0.160^2}\\\Rightarrow \left (\dfrac{q_1+q_2}{2}\right )^2=\dfrac{+1.30\times 0.160^2}{k}\\(q_1+q_2)^2=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+q_2^2+2q_1q_2=\dfrac{4\times 1.30\times 0.160^2}{k}\\\\[/tex]

Using (1),

[tex]\rm q_1^2+\left ( \dfrac{-1.30\times 0.160^2}{q_1k}\right )^2+2\left (\dfrac{-1.30\times 0.160^2}{k} \right )=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+\dfrac 1{q_1^2}\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0[/tex]

[tex]\rm q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )=0\\q_1^4-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2=0[/tex]

[tex]\rm q_1^4-q_1^2\left (2.22\times 10^{-11} \right )+\left ( 1.37\times 10^{-23}\right ) =0\\\Rightarrow q_1^2 = \dfrac{-(-2.22\times 10^{-11})\pm \sqrt{(-2.22\times 10^{-11})^2-4\cdot (1)\cdot (1.37\times 10^{-23})}}{2}\\=1.11\times 10^{-11}\pm 1.046\times 10^{-11}.\\=6.4\times 10^{-13}\ \ \ or\ \ \ 2.156\times 10^{-11}\\\Rightarrow q_1 = \pm 8.00\times 10^{-7}\ C\ \ \ or\ \ \ \pm 4.64\times 10^{-6}\ C.[/tex]

Using (1),

When [tex]\rm q_1 = \pm 8.00\times 10^{-7}\ C[/tex],

[tex]\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 8.00\times 10^{-7}\times 9\times 10^9}=\mp4.6\times 10^{-6}\ C.[/tex]

When [tex]\rm q_1=\pm 4.6\times 10^{-6}\ C[/tex],

[tex]\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 4.64\times 10^{-6}\times 9\times 10^9}=\mp7.97\times 10^{-7}\ C\approx 8.0\times 10^{-7}\ C.[/tex]

Since, [tex]\rm |q_2|>|q_1|[/tex]

Therefore, [tex]\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.[/tex]

Answer:

(a):  [tex]\rm meter/ second^2.[/tex]

(b):  [tex]\rm meter/ second^3.[/tex]

(c):  [tex]\rm 2ct-3bt^2.[/tex]

(d):  [tex]\rm 2c-6bt.[/tex]

(e):  [tex]\rm t=\dfrac{2c}{3b}.[/tex]

Explanation:

Given, the position of the particle along the x axis is

[tex]\rm x=ct^2-bt^3.[/tex]

The units of terms [tex]\rm ct^2[/tex] and [tex]\rm bt^3[/tex] should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of [tex]\rm ct^2=meter[/tex]

Therefore, unit of [tex]\rm c= meter/ second^2.[/tex]

(b):

Unit of [tex]\rm bt^3=meter[/tex]

Therefore, unit of [tex]\rm b= meter/ second^3.[/tex]

(c):

The velocity v and the position x of a particle are related as

[tex]\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.[/tex]

(d):

The acceleration a and the velocity v of the particle is related as

[tex]\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.[/tex]

(e):

The particle attains maximum x at, let's say, [tex]\rm t_o[/tex], when the following two conditions are fulfilled:

Applying both these conditions,

[tex]\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.[/tex]

For [tex]\rm t_o = 0[/tex],

[tex]\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c[/tex]

Since, c is a positive constant therefore, for [tex]\rm t_o = 0[/tex],

[tex]\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0[/tex]

Thus, particle does not reach its maximum value at [tex]\rm t = 0\ s.[/tex]

For [tex]\rm t_o = \dfrac{2c}{3b}[/tex],

[tex]\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.[/tex]

Here,

[tex]\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}<0.[/tex]

Thus, the particle reach its maximum x value at time [tex]\rm t_o = \dfrac{2c}{3b}.[/tex]

The units of constant c are in meters per second squared per second squared (m/s^2). The units of constant b are in meters per second cubed (m/s^3). The velocity can be found by taking the derivative of the position function, and the acceleration can be found by taking the derivative of the velocity function. To find the time when the particle reaches its maximum x value, set the velocity equal to zero and solve for t.

(a) The units of constant c can be determined by analyzing the equation x = ct2 - bt3. Since x is in meters and t is in seconds, the units of x/t2 will be in meters per second squared. Therefore, the units of constant c will be in meters per second squared per second squared (m/s2).

(b) Similar to part (a), the units of constant b can be determined by analyzing the equation x = ct2 - bt3. Since x is in meters and t is in seconds, the units of x/t3 will be in meters per second cubed. Therefore, the units of constant b will be in meters per second cubed (m/s3).

(c) The velocity v can be found by taking the derivative of the position function x with respect to time t. In this case, v = d/dt(ct2 - bt3) = 2ct - 3bt2.

(d) The acceleration a can be found by taking the derivative of the velocity function v with respect to time t. In this case, a = d/dt(2ct - 3bt2) = 2c - 6bt.

(e) To find the time when the particle reaches its maximum x value, we can set the velocity v equal to zero and solve for t. 2ct - 3bt2 = 0. Solving this equation will give us the time t when the particle reaches its maximum x value.

Answer:

4000 J

Explanation:

The energy possessed by virtue of motion of the body is called its kinetic energy.

When a body is in motion , it has kinetic energy.

Mass of the object, m = 80 kg

velocity of the object, v = 10 m/s

The formula for the kinetic energy is given by

[tex]K=\frac{1}{2}mv^{2}[/tex]

K = 0.5 x 80 x 10 x 10

K = 4000 J

Thus, the kinetic energy of the object is 4000 J.

To convert 1g of ice at absolute zero to 1g of boiling water, we need to calculate the heat required in three stages: heating the ice to melting point, melting the ice to water, and heating the water to boiling point. Adding the heat required for these three stages, we find the total amount of heat required is approximately 316.38 cal.

To find the total heat needed to turn 1 g of ice at absolute zero to boiling water, we calculate it in three parts. First, we calculate the heat required to raise the temperature of ice from absolute zero (-273.15 degrees Celsius) to 0 degrees Celsius using the formula: Q = mcΔT where m = 1g, c = 0.5 cal/(g*C), and ΔT = 273.15 °C. This gives us 136.58 cal.

Second, we need to factor in the heat needed to convert the ice at 0 degree Celsius to water at 0 degree Celsius. The heat of fusion for ice is known as 79.8 cal/g, so for 1 gram, the heat would be 79.8 cal.

Third, we calculate the heat needed from water at 0 degrees to water boiling point (100 degrees C). Here the formula remains the same but specific heat capacity changes to 1 cal/(g*C), which results in 100 cal.

Adding the calculated heat amounts for each phase gives the total heat required. Hence, the total amount of heat needed is 136.58 cal + 79.8 cal + 100 cal = 316.38 cal.

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Answer and Explanation:

The clothing after spinning in the dryer cling together. This is because in the dryer they are rubbed against each other and due to this rubbing, electrons are transferred from one to the other clothes and acquire charge as a result of charging by friction thus producing static electricity.

As the material of the clothes in the dryer is different, clinging will be more.

The sticking of these clothes together is known as Static cling.

In case, the clothing are of same material, the static electricity produced as a result of frictional charging would be less and hence less static cling would occur.

Clothing clings together in the dryer due to static electricity. Clinging would be expected to be more common if the clothing were made of the same material. Static electricity occurs when different materials rub against each other and build up opposite charges.

The clothing tends to cling together after going through the dryer due to static electricity. When fabrics rub against each other in the dryer, they become charged with static electricity. The charged clothes then attract and stick to one another, causing them to cling together.

If all the clothing were made of the same material like cotton, you would expect more clinging compared to drying different kinds of clothing together. This is because different materials have different tendencies to build up static electricity. When different fabrics with different tendencies to build up static electricity are dried together, they can counteract each other's charges, reducing the overall amount of clinging.

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Answer:

The magnitude of the resultant vector is 13.656 units.

Explanation:

The vector A can be represented vectorially as

 [tex]\overrightarrow{r}_{a}=6.00\widehat{i}-4.40\widehat{j}[/tex]

Similarly vector B can be represented vectorially as

[tex]\overrightarrow{r}_{b}=3.30\widehat{i}-5.60\widehat{j}[/tex]

Thus upon adding the 2 vectors we get

[tex]\overrightarrow{r}_{a}+\overrightarrow{r}_{b}=6.00\widehat{i}-4.40\widehat{j}+3.30\widehat{i}-5.60\widehat{j}\\\\=(6.00+3.30)\widehat{i}-(4.40+5.60)\widehat{j}\\\\=9.30\widehat{i}-10.00\widehat{j}[/tex]

Now the magnitude of the vector is given by:

|r|=[tex]\sqrt{x^{2}+y^{2}}\\\\|r|=\sqrt{9.30^{2}+(-10)^{2}}\\\\\therefore |r|=13.65units[/tex]

Answer:

Explanation:

Initial velocity of train u = 3.4 m⁻¹ , Acceleration a = .065 ms⁻² ,

time t = 9.75 x 60 = 585 s.

v = u + at

= 3.4 + .065 x 585

= 41.425 m / s

distance travelled during the acceleration ( s )

s = ut + 1/2 at²

= 3.4 x 585 + .5 x .065 x 585²

= 1989 + 11122.31

= 13111.31 m .

again for slowing process

u = 41.425 m / s

v = u -at

0 = 41.425 - 0.625 t

t = 66.28 s

v² = u² - 2as

0 = 41.425² - 2 x .625 s

s = 1372.82 m

The final velocity of the freight train is approximately 21.2 m/s. The train takes approximately 33.92 seconds to come to a stop. The train travels approximately 15,087 meters during the acceleration and 3,655 meters during the deceleration.

To find the final velocity of the freight train, we can use the equation:

v = u + at

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

Plugging in the given values, we have:

v = 3.4 m/s + (0.065 m/s2)(9.75 min)(60 s/min)

Solving this equation, we get the final velocity of the freight train to be approximately 21.2 m/s.

To find the time it takes for the train to stop, we can use the same equation:

v = u + at

But this time, the initial velocity is the final velocity from part (a), and the acceleration is the deceleration rate of 0.625 m/s2.

Plugging in the given values, we have:

0 = 21.2 m/s + (0.625 m/s2)t

Solving this equation, we find that it takes approximately 33.92 seconds for the train to come to a stop.

To find the distance traveled during the acceleration, we can use the equation:

s = ut + (1/2)at2

where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.

Plugging in the given values, we have:

s = 3.4 m/s(9.75 min)(60 s/min) + (1/2)(0.065 m/s2)(9.75 min)(60 s/min)2

Solving this equation, we find that the train traveled approximately 15,087 meters during the acceleration.

To find the distance traveled during the deceleration, we can use the same equation:

s = ut + (1/2)at2

But this time, the initial velocity is the final velocity from part (a), and the acceleration is the deceleration rate of 0.625 m/s2.

Plugging in the given values, we have:

s = 21.2 m/s(33.92 s) + (1/2)(-0.625 m/s2)(33.92 s)2

Solving this equation, we find that the train traveled approximately 3,655 meters during the deceleration.

Answer:

A) The time it takes for the T-shirt to reach the fan is 2.3582s

B) The height the fan is from the ground is 10.2438m

Explanation:

First, we need to find the velocity vectors for each axe:

[tex]v_{x}=30*cos(32\textdegree)=25.44\frac{m}{s}\\ v_{x}=30*sin(32\textdegree)=15.90\frac{m}{s}[/tex]

Next, using the movement equation for the x axis, we get the time:

[tex]D=v_{x}*t\\t=\frac{D}{v_{x}}=2.3585s[/tex]

Finally, using the movement equation for the y axis, we find the height (assuming the gravity as 9,8[tex]\frac{m}{s^{2}}[/tex])

[tex]h=v_{y}*t-\frac{g*t^{2}}{2}=10.2438\frac{m}{s^{2}}[/tex]

Answer:12.206 cm,[tex]\theta =54.99^{\circ}[/tex]

Explanation:

Given

Insect walks 15 cm to the right

so its position vector is[tex]r_1=15i[/tex]

Now it moves 10 cm up so its new position vector

[tex]r_2=15i+10j[/tex]

Now it moves 8 cm left so its final position vector is

[tex]r_3=15\hat{i}+10\hat{j}-8\hat{i}=7\hat{i}+10\hat{j}[/tex]

so its displacement is given by

[tex]|r_3|=\sqrt{7^2+10^2}=\sqrt{149}=12.206 cm[/tex]

For direction, let \theta is the angle made by its position vector with x axis

[tex]tan\theta =\frac{10}{7}=1.428[/tex]

[tex]\theta =54.99^{\circ}[/tex]

Answer:

d) Sofia.

Explanation:

If they walked down a straight road, their displacement will be 4 miles. However, their position depends on the origin of a coordinate system which could be located anywhere. If the origin of the coordinate system was at the starting point of their walk, then and only then, you can say that their position is also 4 miles.

Answer:

256 x 10³ Nm

Explanation:

Force on each of the charges of 8 C due to uniform electric field in x -y plane  will be equal and opposite because one is positive charge and the other is negative charge. They will make a couple

Each of the forces of the couple will act in x - y plane and will be equal to

F = Q E = 8 X 8 X 10³

= 64 X 10³ N

Arm of the couple = 4 m

Moment of couple about z axis.

= 64 x 10³ x 4

= 256 x 10³ Nm.

The magnitude of the net torque applied to the rod in a perpendicular electric field is calculated using the product of the force on one charge by the electric field, the distance from the origin, and considering that there are two opposite forces creating the torque.

In this Physics problem, we are asked to calculate the magnitude of the net torque applied to a thin rod which has point charges at its ends, placed in a perpendicular electric field. The charges are +8.0C and -8.0C, and the electric field has a magnitude of 8.0 × 10³ N/C. Torque (τ) is given by the cross product of the position vector (r) and the force vector (F), or τ = r × F. Given that the electric field E is perpendicular to the rod and applying a force (F = E × q) on each charge, the torque for each charge will be τ = rF sin(θ), where θ is the angle between r and the direction of the force, which in this case is 90 degrees.

The charge at the positive end of the rod experiences a downward force, while the charge at the negative end experiences an upward force. Both forces will be equal in magnitude but opposite in direction, which leads to a torque that tries to rotate the rod around the z-axis. Therefore, the magnitude of net torque will be twice that of one charge, since the rod has a charge on each end and the forces act at a distance (2.0 m from the origin for both charges). Thus, the magnitude of the net torque can be found with τ = 2 × (2.0 m × 8.0 × 10³ N/C × 8.0C).

The area of a room with length 9.72 m and width 5.3 m is 52 m² when rounded to two significant figures.

The area of a room can be calculated using the formula: Area = Length × Width. For this room, with a length of 9.72 m and a width of 5.3 m, plug the values into the formula: Area = 9.72 m × 5.3 m. Calculate to get an area of 51.516 m². However, since the values given for length and width only have two significant figures, we should round our answer to have the same. So, the area of the room, rounded to two significant figures, is 52 m².

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Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = [tex]\frac{\textup{V}}{\textup{R}}[/tex]

or

I₀ = [tex]\frac{\textup{6.53}}{\textup{594}}[/tex]

or

I₀ = 0.0109 A

also,

I = [tex]I_0[1-e^{-\frac{t}{\tau}}][/tex]

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = [tex]0.0109[1-e^{-\frac{\tau}{\tau}}][/tex]

or

I = [tex]0.0109[1-2.717^{-1}][/tex]

or

I = 0.00688 A

or

I = 6.88 mA