In 1970, a rocket powered car called Blue Flame achieved a maximum speed of 1.00(10 km/h (278m/s).Suppose the magnitude of the car's constant acceleration is 5.56 m/s^2. If the car is initially at rest , what is the distance traveled. During its acceleration?

Answer:

As you learned, scientists standardized the short-hand way we represent elements, by their chemical symbol. One of the main reasons this was developed was because using letters was the easiest way to represent the elements. Another reason that we use chemical symbols is to allow us to write chemical formulas easily

Or

One set of chemical symbols means that every science student and scientist -- no matter what languages they speak -- can identify these chemicals.

I hope this helps

As you taught, scientists standardized the use of an element's chemical symbol as a shorthand for representing it. This was created in part because representing the elements with letters was the simplest method. Utilizing chemical symbols also makes it easier to formulate chemical formulas.

All identified chemical elements are arranged in rows (referred to as periods) and columns (referred to as groups) in the periodic table of chemical elements, also known as the periodic table, in ascending order of atomic number.

The periodic table is used by scientists to quickly refer to details about an element, such as its atomic mass and chemical symbol. Scientists can identify trends in element properties like electronegativity, ionization energy, and atomic radius thanks to the periodic table's arrangement.

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Answer:

36.25 N

Explanation:

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

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

where:

[tex]k=9\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant

[tex]q_1, q_2[/tex] are the magnitude of the two charges

r is the separation between the two charges

Moreover:

- The force is repulsive if the two  charges have same sign

- The force is attractive if the two charges have opposite sign

In this problem, we have 3 charges:

[tex]q_1=-5\mu C = -5\cdot 10^{-6}C[/tex] is the charge located at [tex]x=+10 cm = +0.10 m[/tex]

[tex]q_2=+6\mu C=+6\cdot 10^{-6}C[/tex] is the charge located at [tex]x=-8 cm =-0.08 m[/tex]

[tex]q_3=+2\mu C=+2\cdot 10^{-6}C[/tex] is the charge located at [tex]x=-2 cm=-0.02 m[/tex]

The force between charge 1 and charge 3 is:

[tex]F_{13}=\frac{kq_1 q_3}{(x_1-x_3)^2}=\frac{(9\cdot 10^9)(5\cdot 10^{-6})(2\cdot 10^{-6})}{(0.10-(-0.02))^2}=6.25 N[/tex]

And since the two charges have opposite sign, the force is attractive, so the force on charge 3 is to the right (towards charge 1).

The force between charge 2 and charge 3 is:

[tex]F_{23}=\frac{kq_2 q_3}{(x_2-x_3)^2}=\frac{(9\cdot 10^9)(6\cdot 10^{-6})(2\cdot 10^{-6})}{(-0.08-(-0.02))^2}=30.0 N[/tex]

And since the two charges have same sign, the force is repulsive, so the force on charge 3 is to the right (away from charge 2).

So the two forces on charge 3 have same direction (to the right), so the net force is the sum of the two forces:

[tex]F=F_{13}+F_{23}=6.25+30.0=36.25 N[/tex]

To solve this, use Coulomb's Law to calculate the separate forces each charge exerts on the -2 µC charge and then add these up using the principle of superposition to get total force.

This question involves the principle of superposition and Coulomb's Law in Physics. According to Coulomb's Law, the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In essence, it can be expressed as F = k|q1*q2|/r², where k is the Coulomb constant, q1 and q2 are the charges, and r is the distance between them.

You need to calculate the forces that the -2 µC charge experiences due to the -5 µC and the 6 µC charges separately, and then superpose (or add up) these forces to get the total force on the -2 µC charge.

The force between the -5 µC charge and the 2 µC charge at position -2 cm, F1 = k|-5*2|/12² = k*10/144

The force between the 6 µC charge and the 2 µC charge, F2 = k*6*2/6² = k*12/36

Add up these forces to get the total force as follows: F = F1 + F2

Using the known value for k (the Coulomb constant) F = (8.9875 × 10^9 N·m²/C²) ( (10/144) + (12/36) ) N

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testing the model and analyzing the results of the tests

Explanation:

Answer: The stratosphere and the thermosphere are responsible for the majority of the solar radiation absorption.

Explanation:

Answer:The stratosphere and the thermosphere are responsible for the majority of the solar radiation absorption.

Answer:

Mixing a milkshake

Explanation:

Becuse it’s physics becuse your using muscle and moving it and changing it by force.

Train has the most momentum if they were all traveling 50 mph

Explanation:

Answer: The temperature at which the intrinsic concentration exceeds the impurity density by factor of 10 is 636 K.

Explanation:

The given data is as follows.

         [tex]N_{d} = 10^{19} per cm^{-3}[/tex]

         [tex]N_{a} = 10^{15} per cm^{-3}[/tex]

As we are given that [tex]n_{i}[/tex]exceeds impurity density by a factor of 10.

Therefore,   [tex]n_{i} = 10N_{d}[/tex]

   [tex]10^{20} = 3.87 \times 10^{6} \times T^{\frac{3}{2}}e^({\frac{-7014}{T}})[/tex]

[tex]T^{\frac{3}{2}}e^({\frac{-7014}{T}}) = \frac{10^{20}}{3.87 \times 10^{6}}[/tex]

          T = 1985 K

Also,  [tex]n_{i} = 10N_{d}[/tex]

       [tex]10^{6} = 3.87 \times 10^{16} \times T^{\frac{3}{2}}e^({\frac{-7014}{T}})[/tex]

                T = 636 K

Thus, we can conclude that the temperature at which the intrinsic concentration exceeds the impurity density by factor of 10 is 636 K.

The actual numerical solution for the N-type sample yields a temperature of approximately 277 K, and for the P-type sample, the temperature is approximately 160 K.

To find the temperature at which the intrinsic concentration [tex]\( n_i \)[/tex]exceeds the impurity density by a factor of 10, we use the relation:

[tex]\[ n_i^2 = N_d N_a \times 10^2 \][/tex]

where [tex]\( n_i \)[/tex] is the intrinsic carrier concentration,[tex]\( N_d \)[/tex] is the donor density (for N-type), and [tex]\( N_a \)[/tex] is the acceptor density (for P-type). The intrinsic carrier concentration [tex]\( n_i \)[/tex] is temperature-dependent and can be approximated by:

[tex]\[ n_i(T) = n_i(T_{ref}) \left( \frac{T}{T_{ref}} \right)^{3/2} \exp \left( -\frac{E_g}{2k} \left( \frac{1}{T} - \frac{1}{T_{ref}} \right) \right) \][/tex]

where [tex]\( T \)[/tex] is the absolute temperature, [tex]\( T_{ref} \)[/tex]is a reference temperature (usually 300 K), [tex]\( E_g \)[/tex] is the energy bandgap of silicon at the reference temperature (approximately 1.12 eV for silicon), and [tex]\( k \)[/tex] is Boltzmann's constant [tex](\( 8.617 \times 10^{-5} \) eV/K)[/tex].

For the N-type silicon sample, we have [tex]\( N_d = 10^{19} \) cm\( ^{-3} \)[/tex], and we want to find[tex]\( T \)[/tex] such that[tex]\( n_i = 10 \times N_d \)[/tex]. We can rewrite the equation as:

[tex]\[ n_i^2(T) = (10 \times N_d)^2 \][/tex]

[tex]\[ n_i(T_{ref})^2 \left( \frac{T}{T_{ref}} \right)^3 \exp \left( -\frac{E_g}{k} \left( \frac{1}{T} - \frac{1}{T_{ref}} \right) \right) = 100 \times N_d^2 \][/tex]

We know that[tex]\( n_i(T_{ref}) \)[/tex] is approximately [tex]\( 1.5 \times 10^{10} \) cm\( ^{-3} \) at \( T_{ref} = 300 \)[/tex] K. Plugging in the values, we get:

[tex]\[ \left( \frac{T}{300 \text{ K}} \right)^3 \exp \left( -\frac{1.12 \text{ eV}}{8.617 \times 10^{-5} \text{ eV/K}} \left( \frac{1}{T} - \frac{1}{300 \text{ K}} \right) \right) = \frac{100 \times (10^{19} \text{ cm}^{-3})^2}{(1.5 \times 10^{10} \text{ cm}^{-3})^2} \][/tex]

Solving this equation numerically for [tex]\( T \)[/tex] gives us the temperature for the N-type silicon sample.

For the P-type silicon sample, we follow the same steps but with [tex]\( N_a = 10^{15} \) cm\( ^{-3} \)[/tex], and we find the temperature at which [tex]\( n_i = 10 \times N_a \)[/tex]. The equation to solve is:

[tex]\[ \left( \frac{T}{300 \text{ K}} \right)^3 \exp \left( -\frac{1.12 \text{ eV}}{8.617 \times 10^{-5} \text{ eV/K}} \left( \frac{1}{T} - \frac{1}{300 \text{ K}} \right) \right) = \frac{100 \times (10^{15} \text{ cm}^{-3})^2}{(1.5 \times 10^{10} \text{ cm}^{-3})^2} \][/tex]

Solving this equation numerically for [tex]\( T \)[/tex] gives us the temperature for the P-type silicon sample.

The actual numerical solution for the N-type sample yields a temperature of approximately 277 K, and for the P-type sample, the temperature is approximately 160 K. These are the temperatures at which the intrinsic carrier concentration exceeds the impurity density by a factor of 10 for each sample.

Explanation:

The fluctuating sound heard when two objects vibrate with different frequencies is called beats. It is given that guitar string produces 3 beats/s when sounded with a 352 Hz tuning fork and 8 beats/s when sounded with a 357 Hz tuning fork.

It is assumed to find the vibrational frequency of the string.

For 3 beats/s, beat frequency can be :

352 - 3 or 352 + 3 = 349 Hz or 355 Hz

For 8 beats/s, beat frequency can be :

357 - 8 or 357 + 8 = 349 Hz or 365 Hz

It means that the vibrational frequency is 349 Hz.

Answer:

[tex]F_Q=\frac{Mg}{3}[/tex]

Explanation:

We are given that

Length of beam=L

Mass of object=M

We have to find the FQ , the  vertical force that pier exerts on the right end of the bridge .

Torque about P

[tex]F_Q(3L)-MgL=0[/tex]

[tex]F_Q(3L)=MgL[/tex]

[tex]F_Q=\frac{MgL}{3L}[/tex]

[tex]F_Q=\frac{Mg}{3}[/tex]

Hence, the force that exerted pier Q exerts on the right end  of the bridge is given by

[tex]F_Q=\frac{Mg}{3}[/tex]

the minimum speed the asteroid should have to just graze the surface of the Earth is approximately 9426 m/s.

To calculate the minimum speed the asteroid should have so it will just graze the surface of the Earth, we can use the concept of gravitational potential energy and kinetic energy.

At the point where the asteroid just grazes the surface of the Earth, its gravitational potential energy will be converted entirely into kinetic energy. The minimum speed required will ensure that this kinetic energy is just enough to overcome the gravitational attraction and reach the Earth's surface.

The gravitational potential energy (U) of an object at a distance (r) from the center of the Earth is given by the formula:

[tex]\[ U = -\frac{GMm}{r} \][/tex]

Where:

- ( U ) is the gravitational potential energy,

- ( G ) is the gravitational constant [tex](\( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)),[/tex]

- ( M ) is the mass of the Earth[tex](\( 5.972 \times 10^{24} \, \text{kg} \)),[/tex]

- ( m ) is the mass of the asteroid (which cancels out in this context), and

- ( r ) is the distance from the center of the Earth.

The kinetic energy (K) of an object moving at a speed (v) is given by the formula:

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

At the point of grazing, the gravitational potential energy is equal to the kinetic energy:

[tex]\[ -\frac{GMm}{r} = \frac{1}{2} mv^2 \][/tex]

The mass of the asteroid (m) cancels out from both sides of the equation. We can rearrange this equation to solve for the minimum speed (v):

[tex]\[ v = \sqrt{\frac{2GM}{r}} \][/tex]

Substitute the known values:

[tex]- \( G = 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \),\\- \( M = 5.972 \times 10^{24} \, \text{kg} \), and\\- \( r = 24000 \, \text{km} = 24000 \times 10^3 \, \text{m} \).[/tex]

[tex]\[ v = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{24000 \times 10^3}} \][/tex]

Calculate the minimum speed (v):

[tex]\[ v \approx \sqrt{8.89 \times 10^8} \]\[ v \approx 9426 \, \text{m/s} \][/tex]

So, the minimum speed the asteroid should have to just graze the surface of the Earth is approximately 9426 m/s.

Final answer:

The incorrect statement about polarized materials is that two polarized filters with their axes parallel will block all light, while in reality, they pass all light that is polarized along the axis.

Explanation:

Answer:

0.0176m

Explanation:

Given that,

railroad track has a length of 40 m

temperature is T₁ −5 ◦C

temperature is T₂ 35 ◦C

linear expansion coefficient of steel is 11 × 10−6 ( ◦C)−1

Lo = 40 m

T₁ = -5° C

T₂ = 35° C

dT = T₂ - T₁

    = 35 - (-5)

     = 40°C

L = Lo*(1+alpha*dT)

dL = Lo*alpha*dT

dT =  40°C

alpah = 11 x 10⁻⁶

Lo = 40 m

dL = 40 × 11 x 10⁻⁶  × 40

 = 0.0176m

Answer:

ΔL = 0.0176m

Explanation:

We are given;

Length of railroad track; L = 40 m

First Temperature; T1 = −5 ◦C

Second temperature; T2 = 35 ◦C

linear expansion coefficient of steel; α = 11 × 10^(−6) (◦C)^(-1)

The increase in length is given by the equation;

ΔL = α•L•ΔT

Where,

α is linear coefficient

L is length

ΔT is change in temperature.

ΔT = first temperature - Second Temperature

Thus, ΔT = 35 - (-5) = 35 + 5 = 40°C

Thus,plugging in relevant values,

ΔL = α•L•ΔT = 11 × 10^(−6)•40•40

ΔL = 0.0176m

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of [tex]m_2[/tex] would be four times that of [tex]m_1[/tex].

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

[tex]\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2[/tex].

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is [tex]T[/tex], then the angular velocity of the SHM would be

[tex]\displaystyle \omega = \frac{2\pi}{T}[/tex].

Assume that the mass starts with a zero displacement and a positive velocity. If [tex]A[/tex] represent the amplitude of the SHM, then the displacement of the mass at time [tex]t[/tex] would be:

[tex]\mathbf{x}(t) = A\sin(\omega\cdot t)[/tex].

The velocity of the mass at time [tex]t[/tex] would be:

[tex]\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)[/tex].

The acceleration of the mass at time [tex]t[/tex] would be:

[tex]\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)[/tex].

Let [tex]m[/tex] represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time [tex]t[/tex] would be:

[tex]\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)[/tex],

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant [tex]k[/tex] will be equal to:

[tex]\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}[/tex].

Since [tex]\displaystyle \omega = \frac{2\pi}{T}[/tex], it can be concluded that:

[tex]\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}[/tex].

For the first mass [tex]m_1[/tex], if the time period is [tex]T_1[/tex], then the spring constant would be:

[tex]\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2[/tex].

Similarly, for the second mass [tex]m_2[/tex], if the time period is [tex]T_2[/tex], then the spring constant would be:

[tex]\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2[/tex].

Since the two springs are the same, the two spring constants should be equal to each other. That is:

[tex]\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2[/tex].

Simplify to obtain:

[tex]\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2[/tex].

1. 3 significant figures

2. 6 significant figures

3. 4 significant figures

Answer:

Question incomplete

This is the complete question

You have a collection of 1.0 kΩ resistors.

How can you connect four of them to produce an equivalent resistance of 0.25 kΩ?

Explanation:

Given that,

We have collection of resistor

R = 1.0 kΩ.

And we want to design an equivalent resistance of

R = 0.25kΩ

We know that,

Series connection of add up resistance and this increases the value of resistance, I.e

Req = R1 + R2 + R3 +.....

While

Parallel resistance add up the reciprocal of resistance and this reduces resistance

So, 1/Req = 1/R1 + 1/R2 + 1/R3 +......

Now, we want to design 0.25kΩ

Connect four of the resistor in parallel will give a equivalent resistance of 0.25 kΩ

1/Req = 1/R1 + 1/R2 + 1/R3 + 1/R4

R1 = R2 = R3 = R4 = 1.0 kΩ

1/Req = 1/1 + 1/1 + 1/1 + 1/1

1/Req = 1+1+1+1

1/Req = 4

Taking reciprocal of both sides..

Req = ¼ kΩ

This is the required equivalent resistance

So, connect the four of the resistor in parallel will give the required 0.25 kΩ resistance

Answer:

The answer is D.  The electrons will flow from an area of high potential energy to an area of low potential energy.

X== low potential energy

Y = high potential energy

z = flow of electrons

Explanation:

The electrons will flow from an area of high potential energy to an area of low potential energy.

Potential energy is the energy that an item retains as a result of its location in relation to other objects, internal tensions, electric charge, or even other elements. Although it has connections towards the Greek philosopher Aristotle's notion of potentiality.

The gravitational potential energy of such an item, the elastic potential energy of a stretched spring, as well as the electric potential energy of such an electric charge inside an electric field are examples of common forms of potential energy. The electrons will flow from an area of high potential energy to an area of low potential energy.

X== low potential energy

Y = high potential energy

z = flow of electrons

Therefore, the electrons will flow from an area of high potential energy to an area of low potential energy.

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Answer:

The current is flowing To The Left. The magnetic field is flowing out of the Bottom of the screen and into the Top of the screen.

Answer:

C. 1500.

Explanation:

750 / .5 = 1500.

Hope this helps & best of luck!

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Answer:

water, air, vegetation, and light. also a good temperature thats not too hot or too cold

Explanation:

it would need oxygen, trees, places for animals to build habitats, and water. The temperature also cant be too high.

Answer:

X

Explanation:

X represents the location of the image produced by the lens.

A converging lens produced a virtual image when the object is placed in front of the focal point. For such a position, the image is magnified and upright, thus allowing for easier viewing.

Virtual images are always located behind the mirror. Virtual images can be either upright or inverted. Virtual images can be magnified in size, reduced in size, or the same size as the object. Virtual images can be formed by concave, convex, and plane mirrors.

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Answer:

For a plane mirror, the image distance equals the object distance, so the image distance will increase as the object distance increases

The height of the image stays the same and the image distance increases.)

Explanation:

For plane mirrors, the object distance (is equal to the image distance. That is the image is the same distance behind the mirror as the object is in front of the mirror. If you stand a distance of 2 meters from a plane mirror, you must look at a location 2 meters behind the mirror in order to view your image

The final rotational kinetic energy of the rotating disk is 9.95 J

Explanation:

Given data,

mass 0.7 kg radius 10 cm rotates at the speed of 15 rev/sec

We have the formula,

Kf= 1/2 I ω f²

I=1/2 Mr²2

ωf= (0.8) ωo

substitute in the formula we get

Kf= 1/2 (1/2 MR²2) (0.8) ² ( ωo)²

=(0.16) M R²2 ωo²2

=(0.16)(0.7)(0.10)²(15[2π)] ²

Kf=9.95 J

The final rotational kinetic energy of the rotating disk is 9.95 J

Answer:

(a) 2.42 km

(b) 119.74°

Explanation:

(a)

The shortest distance from starting to end point is the hypotenuse

[tex]C=\sqrt{a^{2}+b^{2}}[/tex]

Where a is is base and b is height. Substituting 1.2 km for a and 2.1km for b then

[tex]C=\sqrt{1.2^{2}+2.1^{2}}=2.41867732448956 km\approx 2.42 km[/tex]

(b)

The angle is given by the [tex]\theta[/tex] as indicated in the sketch

[tex]\theta=\frac {2.1}{1.2}=60.2551187030578\approx 60.26^{\circ}[/tex]

Towards East it will be 180-60.26=119.74°

Answer:

[tex]T = 12910.5\,N[/tex] for a craft with a mass of 1500 kg.

Explanation:

Let consider that craft has a mass of 1500 kg. The submersible craft is modelled after the Newton's Laws, whose equation of equilibrium is:

[tex]\Sigma F = T - W +F_{D} = 0[/tex]

The tension experimented by the cable while the craft is lowering to the seafloor is:

[tex]T = W - F_{D}[/tex]

[tex]T = (1500\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)-1800\,N[/tex]

[tex]T = 12910.5\,N[/tex]

Answer:

(a) Tension on both side of wire

[tex]T_1=46.706 N[/tex]

[tex]T_2=51.710 N[/tex]

(b) acceleration of the Box

[tex]a = 3.336 \frac{m}{sec^2}[/tex]

(c) The horizontal and vertical components

Horizontal component [tex]T_1=46.706 N[/tex]

Vertical Component  =130.19 N

Explanation:

Refer attached figure for details.

[tex]T_1\ \&\ T_2\ are\ tensions\ in\ the\ string\ and\ a\ is\ the\ acceleration\ of\ the\ masses.[/tex]

Applying Newton's 2 law of motion for 14 kg block in horizontal direction

[tex]T_1 = 14\ a[/tex]-----------(i)

Similarly, applying Newton's 2 law of motion for 8 kg block in vertical direction

[tex]8 g - T_2 =8 a[/tex]-----(ii)

Consider the case of pulley,

[tex]\tau_e_x_t= I\alpha--------(iii)\\\\Where,\\\tau_e_x_t =Torque\ acting\ on\ the\ pulley\\I=moment\ of\ inertia\ of\ pulley\\\alpha= angular\ acceleration[/tex]

where,

[tex]I= \frac{MR^2}{2} (for\ pulley\ disk)[/tex]

[tex]I=\frac{3\cdot0.5^2}{2} =0.375\ kgm^2[/tex] (since mass of the pulley = 3 kg & Radius = 0.5 m)

&[tex]\tau_e_x_t= Net\ force \cdot Distance\ from\ application\ point[/tex]

Hence [tex]\tau_e_x_t = (T_2-T_1) \cdot \frac{1}{2} = 0.375\cdot\alpha[/tex]

[tex]T_2-T_1=0.75\cdot\alpha[/tex]--------(iv)

Relation between linear acceleration (a) and angular acceleration (α) is as follows,

[tex]a = R\alpha=0.5\cdot\alpha \ (R\ is \ radius\ of \ pulley)[/tex]

[tex]\alpha=2a[/tex]--------------------(v)

Putting the value of (v) in to (iv)

[tex]T_2 -T_1= 1.5 a[/tex]---------(vi)

adding equation (i),(ii) & (vi) gives

8g =22 a + 1.5 a

[tex]a = 3.336 \frac{m}{sec^2}[/tex]

now putting the value of a in equation (i) & (ii) we get

[tex]T_1=46.706 N[/tex]

[tex]T_2=46.706 +1.5 \cdot 3.336[/tex] = 51.710 N

(a) Hence Tension on both side of wire

[tex]T_1=46.706 N[/tex]

[tex]T_2=51.710 N[/tex]

(b) acceleration of the Box

[tex]a = 3.336 \frac{m}{sec^2}[/tex]

(c) The horizontal and vertical components

Horizontal component [tex]T_1=46.706 N[/tex]

Vertical Component = [tex]T_2+8\cdot g[/tex] =51.710 + 8 x 9.81 =130.19 N

Answer:

Tension, T1 = 46.2 N and T2 = 52 N, where as acceleration = 3.3 ms^-2.

Forces on the pulley are 46.2 N , 81.4 N horizontal and vertical

respectively.

Explanation:

Given:

Mass of the box on rest,[tex]m_1[/tex] = 14 kg

Mass of the attached box,[tex]m_2[/tex] = 8 kg

Mass of the pulley, [tex]m_3[/tex] = 3 kg

Diameter of the pulley, [tex]d[/tex] = 1 m

Radius of the pulley, [tex]r[/tex] = 0.5 m

Here we will be using the concept of net force ([tex]F_n_e_t[/tex]),net torque ([tex]\tau_n_e_t[/tex]) and acceleration of the pulley .

A FBD is attached with.

Lets find the tension on the wire using Fnet.

⇒ [tex]T_1=m_1(a)[/tex]                                                                                       ...for m1

⇒ [tex]m_2g-T_2=m_2(a)[/tex] can be written as [tex]T_2=m_2g-m_2(a)[/tex]                ...for m2

Considering clockwise torque as negative and anticlockwise torque as positive.

Moment of inertia (I) of the disk/pulley = [tex]\frac{m_3r^2}{2}[/tex] and [tex]\alpha=\frac{a}{r}[/tex] .

Now using net torque on the pulley we can say that.

⇒ [tex](T_2-T_1)r=I\alpha[/tex]

⇒ [tex](T_2-T_1)r=\frac{m_3r^2}{2}\times \frac{a}{r}[/tex]

⇒ [tex](T_2-T_1)=\frac{m_3a}{2}[/tex]

⇒ Plugging T1 and T2 .

⇒ [tex]m_2g-m_2a-m_1a=\frac{m_3a}{2}[/tex]

⇒ Isolating a from the rest.

⇒ [tex]m_2g=\frac{m_3a}{2}+m_2a+m_1a[/tex]

⇒ [tex]m_2g=a\ [\frac{m_3}{2}+m_2+m_1][/tex]

⇒ [tex]\frac{m_2g}{\frac{m_3}{2} +m_2+m_1} =a[/tex]

⇒ Plugging the numeric value

⇒ [tex]\frac{(8\times 9.8)}{(\frac{3}{2} +8+14)} =a[/tex]

⇒ [tex]3.3 =a[/tex]

⇒ Acceleration = 3.3 [tex]ms^-^2[/tex]

So,

(a).

Tension in the wire

⇒ [tex]T_1=m_1(a)=14\times 3.3 =46.2\ N[/tex]

⇒ [tex]T_2=m_2g-m_2(a)=8(9.8-3.3)=52\ N[/tex]

(b).

The acceleration of the box is 3.3 ms^-2.

(c).

Forces on the pulley.

Horizontal force, [tex]P_H[/tex] = [tex]T_1[/tex] = [tex]46.2\ N[/tex]

Vertical force,[tex]P_V[/tex] = [tex]T_2+m_3g[/tex] = [tex]52+3(9.8)[/tex] = [tex]81.4\ N[/tex]

The values are as follows:

Tension as T1 = 46.2 N and T2 = 52 N ,where as acceleration =3.3 ms^-2.

Forces on the pulley are 46.2 N , 81.4 N horizontal and vertical

respectively.

Answer:

A heavyweight suspended within a moving box needs to overcome inertia which leads to a slight delay in the motion of the weight as the box moves.

Explanation:

A seismograph is an instrument used to record earthquake waves

A heavyweight suspended within a moving box needs to overcome inertia which leads to a slight delay in the motion of the weight as the box moves.

The first earthquake waves arrive at a seismograph station, a short time after the earthquake occurs.

Answer:

Weights, vibrating rod, pendulum : sensitive to vibrations.

Explanation:

Seismograph is an instrument used to measure earthquakes by recording seismic waves. It provides us all details about earthquake - centre, time, depth, energy.

The device is sensitive to vibrations. It consists of a vibrating rod connected to a pendulum, that vibrates due to earthquake shaking. The weight is also complementary attached with rotating drum & pen, to record ground motion. The seismograph output is then recorded & processed on paper.

The answer for the following question is explained below.

Explanation:

Work:

A force causing the movement or displacement of an object.

Given:

mass of the person (m) = 65 kg

height of the cliff (h) = 2000 m

To calculate:

work done (W)

We know;

According to the formula:

  W = m × g × h

Where;

m represents mass of the person

g represents the acceleration due to gravity

where the value of g is;

  g = 10 m/ s²

h represents the height of the cliff

From the above formula;

  W = 65 × 10 × 2000

 W = 130,000 J

  W = 130 Kilo Joules

Therefore the work done is 130 kilo Joules.

The hydrostatic force on one end of the trough can be found using the formula F = pghA, where p is the density of the liquid, g is the acceleration due to gravity, h is the height of the liquid, and A is the area of the end of the trough. In this case, the ends of the trough are equilateral triangles with sides 4m long.

To find the hydrostatic force on one end of the trough, we can use the formula F = pghA, where F is the force, p is the density of the liquid, g is the acceleration due to gravity, h is the height of the liquid, and A is the area of the end of the trough. Since the ends of the trough are equilateral triangles, each side has a length of 4m. The area of an equilateral triangle is given by A = sqrt(3)/4 * s^2, where s is the length of the side. In this case, A = sqrt(3)/4 * (4m)^2 = 4sqrt(3)m^2. Plugging in the values, we have F = (835kg/m^3)*(9.8m/s^2)*h*(4sqrt(3)m^2).

Answer:

The two word term is "ICE SHOVES"

Explanation:

An ice shove also known as ice surge, ice heave, ivu, or shoreline ice pileup is simply a surge of ice which occurs from an ocean or large lake and drops on the shore. These ice shoves are usually caused by ocean currents, strong winds, or temperature differences which tend to push the ice to the shore. The ice shoves can create ice piles up to 12 metres high and more.

Ice shoves has been recorded in at least three lakes in the past years. They are Lake Dauphin (Manitoba, Canada), Mille Lacs Lake (Minnesota), Lake Winnebago (Wisconsin), etc.

Answer:

ICE SHOVE

Explanation:

An Ice Shove is a situation or phenomenon that occurs when the wind carries or pushes melting ice from frozen lakes down to the lake shore.

Agents involved in the pushing of melting ice down to the lake shore includes: currents in the ocean, very strong winds as well as variations and fluctuations in the temperature of the lake.

Ice shoves can result in a pile up and accumulation of ice on the lake shore to be at about 30-40 feet high.

Ice shoves can result in negative effects in the lives of living organisms. Such negative effects includes:

a. Damages to the shoreline of the lakes

b. Damages to the habitats of any living organisms by the lake.

c. Damages to structures (buildings) that can be found at the lake shore.

Answer:

65 years

Explanation:

The arc that ends at 4000 Hz intersects the 20 db hearing intensity at 65 years.

Answer:

65

Explanation: