The ____________is a metal retainer that keeps the rolling elements evenly spaced. N A. seal N B. separator N C. bearing N D outer race

Answer:

a) After 26.67 seconds it returns back to the origin

b) Velocity when it returns to the origin = 60 m/s in the -x direction

Explanation:

a) Let the starting position be origin and time be t.

  After time t displacement, s = 0 m

  Initial velocity, u = 60 m/s

  Acceleration, a = -4.5 m/s²

 We have equation of motion s = ut + 0.5 at²

 Substituting

        s = ut + 0.5 at²

        0 = 60 x t + 0.5 x (-4.5) x t²        

        2.25t² - 60 t = 0

        t² - 26.67 t = 0

        t (t-26.67) = 0

      t = 0s or t = 26.67 s

So after 26.67 seconds it returns back to the origin

b) We have equation of motion v = u + at

  Initial velocity, u = 60 m/s

  Acceleration, a = -4.5 m/s²

  Time , t = 26.67

 Substituting

        v = 60 - 4.5 x 26.67 = -60 m/s

Velocity when it returns to the origin = 60 m/s in the -x direction

Answer:

T^4

Explanation:

According to the Stefan's law, the energy radiated per unit time and per unit area is directly proportional to the four power of teh absolute temperature of the black body.

E ∝ T^4

The intensity of radiation from an ideal radiator is not directly proportional to the absolute temperature T, but rather to T4, according to the Stefan-Boltzmann law of radiation.

The intensity of radiation from an ideal radiator, or blackbody, is not directly proportional to the absolute temperature T; instead, it follows the Stefan-Boltzmann law of radiation. This law states that the intensity of radiation, referred to as 'P' or power, is directly proportional to the fourth power of the absolute temperature (P = σAT4). Here, σ (sigma) represents Stefan-Boltzmann constant, A denotes surface area, and T signifies absolute temperature.

This means the answer to your question is none of the above, as the intensity of the radiation isn't directly proportional to T, but rather to T4.

As an example, considering two temperatures T₁ = 293 K and T₂ = 313 K, the rate of heat transfer with T₂ is about 30 percent higher than T₁. This showcases the considerable impact of temperature on radiation intensity.

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Answer:

Answer to the question:

Explanation:

Differences between ionic bond and covalent bond:

The ionic bond occurs between two different atoms (metallic and non-metallic), while the covalent bond occurs between two equal atoms (non-metallic).

In the covalent bond there is an electron compartment, while in the ionic bond there is an electron transfer.

Ionic bonds have a high melting and boiling point, while covalent bonds usually have a low point.

Final answer:

An ionic bond involves the transfer of electrons, resulting in a bond between a positive ion and a negative ion. A covalent bond involves the sharing of electrons between atoms, creating a stable electron arrangement for both atoms.

Explanation:

An ionic bond is formed by the transfer of electrons from one atom to another, resulting in a bond between a positive ion (cation) and a negative ion (anion). On the other hand, a covalent bond is formed by the sharing of electrons between atoms, creating a bond in which electrons are shared rather than transferred.

In an ionic bond, one atom gains electrons and becomes negatively charged, while the other atom loses electrons and becomes positively charged. This attraction between oppositely charged ions forms the bond. In a covalent bond, atoms share one or more pairs of electrons, resulting in a stable electron arrangement for both atoms.

Examples of compounds with ionic bonds include sodium chloride (NaCl) and magnesium oxide (MgO), while examples of compounds with covalent bonds include water (H2O) and methane (CH4).

Explanation:

First, convert 87 mi/h to ft/min.

87 mi/h × (5280 ft/mi) × (1 h / 60 min) = 7656 ft/min

The time to reach the home plate is:

t = 60 ft / 7656 ft/min

t = 0.00784 min

The number of revolutions made in that time is:

n = 1710 rev/min × 0.00784 min

n = 13.4 rev

Rounding to 2 significant figures, the ball makes 13 revolutions.

The speed of the object with mass 80 kg and moved under a constant force at the final location is 12 m/s.

The speed of a body is the rate at which it covers the total distance is in the time taken.

If the speed of the body has direction, then it is known as the velocity. The velocity is the vector quantity.

The constant force applied on the object is  <200, 460, -150> N. It can be written in the vector form as,

[tex]\vec F=200\hat i+460\hat j-150\hat k[/tex]

The initial position vector of the object for the location <7, -34, -7> can be given as,

[tex]d_i=7\hat i-34\hat j-7\hat k[/tex]

The final position vector of the object is for the location <12, -42, -11> m can be given as,

[tex]d_f=12\hat i-42\hat j-11\hat k[/tex]

Thus, the total distance traveled by the constant force is,

[tex]d=(12\hat i-42\hat j-11\hat k)-(7\hat i-34\hat j-7\hat k)\\d=5\hat i-8\hat j-4\hat k[/tex]

The work done by a object is the product of force and displacement. Thus, work done on the object is,

[tex]W=\vec F.\vec d \\W=(200\hat i+460\hat j-150\hat k).(5\hat i-8\hat j-4\hat k)\\W=1000-3680+600\\W=-2080 \rm J[/tex]

Now, the work done on a object is equal to the kinetic energy of the body. Thus, work done on the object is,

[tex]W=\dfrac{1}{2}m(v^2_f-v^2_i)[/tex]

Here, v(f) is the final speed and v(i) is the initial speed of the object. As the mass of the object is 80 kg and initial speed is 14 m/s. Thus, put the values as,

[tex]-2080=\dfrac{1}{2}(80)(v^2_f-14^2)\\v_f=12\rm m/s[/tex]

Thus, the speed of the object with mass 80 kg and moved under a constant force at the final location is 12 m/s.

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To find the speed of the object at the final location, we need to calculate the net force acting on it using the equation F = ma. By plugging in the values, we can calculate the final speed of the object.

To determine the speed of the object at the final location, we need to calculate the net force acting on the object using the equation F = ma, where F is the force, m is the mass, and a is the acceleration.

Given that the force acting on the object is <200, 460, -150> N, and the mass of the object is 80 kg, we can calculate the acceleration using the formula a = F/m. After calculating the acceleration, we can use it to find the final speed of the object using the equation v^2 = u^2 + 2as, where v is the final speed, u is the initial speed, a is the acceleration, and s is the displacement.

By plugging in the given values, we can calculate the final speed of the object at the location <12, -42, -11> m.

Explanation:

It is given that,

A particle starts from rest and has an acceleration function as :

[tex]a(t)=(5-10t)\ m/s^2[/tex]

(a) Since, [tex]a=\dfrac{dv}{dt}[/tex]

v = velocity

[tex]dv=a.dt[/tex]

[tex]v=\int(a.dt)[/tex]

[tex]v=\int(5-10t)(dt)[/tex]

[tex]v=5t-\dfrac{10t^2}{2}=5t-5t^2[/tex]

(b) [tex]v=\dfrac{dx}{dt}[/tex]

x = position

[tex]x=\int v.dt[/tex]

[tex]x=\int (5t-5t^2)dt[/tex]

[tex]x=\dfrac{5}{2}t^2-\dfrac{5}{3}t^3[/tex]

(c) Velocity function is given by :

[tex]v=5t-5t^2[/tex]

[tex]5t-5t^2=0[/tex]

t = 1 seconds

So, at t = 1 second the velocity of the particle is zero.

To find the velocity function of the particle, integrate the acceleration function, which results in 5t - 5t^2. The position function is found by integrating the velocity function, yielding 1/2 * 5t^2 - 1/3 5t^3. The velocity is zero at t = 0 and t = 1 second.

We can solve for the velocity function and the position function by integrating the acceleration function.

Velocity Function

The velocity function is the integral of the acceleration function. Since the acceleration function given is a(t) = 5 - 10t m/s2, the velocity function v(t) is obtained by integrating a(t) with respect to time:

v(t) = ∫ (5 - 10t) dt = 5t - 5t2 + C

Since the particle starts from rest, the initial velocity is 0, which means C = 0. Therefore, the velocity function is v(t) = 5t - 5t2.

Position Function

The position function s(t) is the integral of the velocity function. We integrate v(t) to obtain the position function:

s(t) = ∫ (5t - 5t2) dt = ½ 5t2 - ⅓ 5t3 + K

Since the particle starts from position zero, K = 0 and the position function is s(t) = ½ 5t2 - ⅓ 5t3.

Zero Velocity

To find when the velocity is zero, we set the velocity function equal to zero and solve for t:
5t - 5t2 = 0t = 0 or t = 1 s.

Answer:

Temperature, T = 1542.10 K

Explanation:

It is given that,

The black body radiation emitted from a furnace peaks at a wavelength of, [tex]\lambda=1.9\times 10^{-6}\ m[/tex]

We need to find the temperature inside the furnace. The relationship between the temperature and the wavelength is given by Wein's law i.e.

[tex]\lambda\propto \dfrac{1}{T}[/tex]

or

[tex]\lambda=\dfrac{b}{T}[/tex]

b = Wein's displacement constant

[tex]\lambda=\dfrac{2.93\times 10^{-3}}{T}[/tex]

[tex]T=\dfrac{2.93\times 10^{-3}}{\lambda}[/tex]

[tex]T=\dfrac{2.93\times 10^{-3}}{1.9\times 10^{-6}\ m}[/tex]

T = 1542.10 K

So, the temperature inside the furnace is 1542.10 K. Hence, this is the required solution.

Using Wien's Displacement Law, the temperature inside a furnace emitting radiation that peaks at a wavelength of 1.9 x 10^-6 m is approximately 1525 Kelvin.

The temperature of the furnace can be determined using Wien's Displacement Law, which states that the product of the temperature of a black body and the wavelength at which the radiation it emits is most intense is approximately equal to 2.898 x 10^-3 m.K. In this specific case, the peak wavelength of the energy emitted from the furnace is given as 1.9 x 10^-6 m. Therefore, the temperature inside the furnace can be calculated using the formula: T = Amax / λ, where Amax = 2.898 x 10^-3 m.K and λ = 1.9 x 10^-6 m. Calculating from the given formula, T ≈ 1525 K. Therefore, the temperature inside the furnace is approximately 1525 Kelvin.

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Answer:

980 dyne

Explanation:

Volume = 1 cm^3, d = 0.92 g / cm^3, D = 1 g/cm^3

In the equilibrium condition, the buoyant force is equal to the weight of the block.

Buoyant force = Volume of block x density of water x g

Buoyant force = 1 x 1 x 980 = 980 dyne

Answer:

0.198 m/s

Explanation:

m1 = 0.7 kg, u1 = 0.15 m/s,

v1 = 0.08 m/s

m2 = 0.5 kg, u2 = 0.1 m/s

Let the speed of 0.5 kg is v2 after the collision.

By using the conservation of momentum

Momentum before collision = momentum after collision

m1 u1 + m2 u2 = m1 v1 + m2 v2

0.7 x 0.15 + 0.5 x 0.1

= 0.7 × 0.08 + 0.5 × v2

0.099 = 0.5 v2

v2 = 0.198 m/s

Answer:

a) Maximum height reached above ground = 2.8 m

b) When he reaches maximum height he is 2 m far from end of the ramp.

Explanation:

a) We have equation of motion v²=u²+2as

   Considering vertical motion of skateboarder.  

   When he reaches maximum height,

          u = 6.6sin58 = 5.6 m/s

          a = -9.81 m/s²

          v = 0 m/s

  Substituting

         0²=5.6² + 2 x -9.81 x s

          s = 1.60 m

  Height above ground = 1.2 + 1.6 = 2.8 m

b) We have equation of motion v= u+at

   Considering vertical motion of skateboarder.  

   When he reaches maximum height,

          u = 6.6sin58 = 5.6 m/s

          a = -9.81 m/s²

          v = 0 m/s

  Substituting

         0= 5.6 - 9.81 x t

          t = 0.57s

  Now considering horizontal motion of skateboarder.  

  We have equation of motion s =ut + 0.5 at²

          u = 6.6cos58 = 3.50 m/s

          a = 0 m/s²

          t = 0.57  

  Substituting

         s =3.5 x 0.57 + 0.5 x 0 x 0.57²

         s = 2 m

  When he reaches maximum height he is 2 m far from end of the ramp.

The highest point reached by the skateboarder is 1.612 meters above the ground. When the skateboarder reaches the highest point, the horizontal distance from this point to the end of the ramp is  3.568 meters.

Given:

Initial velocity (v₀) = 6.6 m/s

Launch angle (θ) = 58°

Height of the ramp (h) = 1.2 m

Acceleration due to gravity (g) = 9.8 m/s²

(a) To find the maximum height reached by the skateboarder:

Δy = v₀y² / (2g)

v₀y = v₀ × sin(θ)

v₀y = 6.6 × sin(58°)

v₀y = 5.643 m/s

Δy = (5.643 )² / (2 × 9.8)

Δy ≈ 1.612 m

Therefore, the highest point reached by the skateboarder is 1.612 meters above the ground.

(b) The time of flight can be calculated using the equation:

t = 2 × v₀y / g

t = 2 × 5.643  / 9.8

t = 1.153 s

Δx = v₀x × t

First, we need to find the initial horizontal velocity (v₀x):

v₀x = v₀ × cos(θ)

v₀x = 6.6 × cos(58°)

v₀x = 3.099 m/s

Δx = 3.099 * 1.153

Δx = 3.568 m

Therefore, when the skateboarder reaches the highest point, the horizontal distance from this point to the end of the ramp is  3.568 meters.

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Answer:

Part a)

[tex]\tau = 23.1 Nm[/tex]

Part b)

[tex]\tau = 17.05 Foot pound force[/tex]

Explanation:

As we know that torque is defined as the product of force and its perpendicular distance from reference point

so here we have

[tex]\tau = \vec r \times \vec F[/tex]

now we have

[tex]\tau = (0.140)(165)[/tex]

[tex]\tau = 23.1 Nm[/tex]

Part b)

Now we know the conversion as

1 meter = 3.28 foot

1 N = 0.225 Lb force

now we have

[tex]\tau = 23.1 Nm[/tex]

[tex]\tau = 23.1 (0.225 Lb)(3.28 foot)[/tex]

[tex]\tau = 17.05 Foot pound force[/tex]

The torque exerted when tightening a bolt with a force of 165 N at a distance of 0.140 m from the center of the bolt is 23.1 N·m. This torque can be converted to approximately 17.0 ft·lb.

To find the torque exerted when tightening a bolt, you can use the equation T = RF sin(θ). In this case, the force is 165 N and the distance from the center of the bolt is 0.140 m. Since you are pushing perpendicularly, the angle θ is 90 degrees. Thus, the torque exerted is 165 N x 0.140 m x sin(90°) = 23.1 N·m.

To convert this torque to foot-pounds, you can use the conversion factor 1 N·m = 0.73756 ft·lb. Therefore, the torque is 23.1 N·m x 0.73756 ft·lb/N·m ≈ 17.0 ft·lb.

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Answer:

B) 20N.s is the correct answer

Explanation:

The formula for the impulse is given as:

Impulse = change in momentum

Impulse = mass × change in speed

Impulse = m × ΔV

Given:

initial speed  = 40m/s

Final speed = -60 m/s (Since the the ball will now move in the opposite direction after hitting the bat, the speed is negative)

mass = 0.20 kg

Thus, we have

Impulse = 0.20 × (40m/s - (-60)m/s)

Impulse = 0.20 × 100 = 20 kg-m/s or 20 N.s

The impulse experienced by the ball is,

[tex]\rm Impulse = 20\; Nsec[/tex]

Given :

Mass = 0.20 Kg

Initial Speed = 40 m/sec

Final Speed = -60 m/sec (minus sign shows that ball move in opposite direction)

Solution :

We know that the formula of Impulse is,

[tex]\rm Impulse = m\times \Delta V[/tex]

where, m is the mass and [tex]\rm \Delta V[/tex] is change in velocity.

[tex]\rm Impulse = 0.20\times(40-(-60))[/tex]

[tex]\rm Impulse = 20\; Nsec[/tex]

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Answer:

3.95 m

Explanation:

m = 1 kg, h = 100 m, k = 125 N/m

Let the spring is compressed by y.

Use the conservation of energy

potential energy of the mass is equal to the energy stored in the spring

m x g x h = 1/2 x ky^2

1 x 9.8 x 100 = 0.5 x 125 x y^2

y^2 = 15.68

y = 3.95 m

Answer:

Part a)

[tex]\Delta V_{max} = 14 \times 10^3 Volts[/tex]

Part b)

[tex]C = 4.96 \times 10^{-11} farad[/tex]

Part c)

[tex]Q = 0.69 \mu C[/tex]

Part d)

[tex]E = 4.86 \times 10^{-3} J[/tex]

Explanation:

Part a)

As we know that dielectric constant of pyrex glass is 5.6 and its dielectric breakdown strength is given as

[tex]E = 14 \times 10^6 V/m[/tex]

now we have

[tex]E . d = \Delta V[/tex]

[tex](14 \times 10^6)(0.001) = \Delta V[/tex]

so we have

[tex]\Delta V_{max} = 14 \times 10^3 Volts[/tex]

Part b)

Capacitance is given as

[tex]C = \frac{k\epsilon_0 A}{d}[/tex]

[tex]C = \frac{5.6(8.85 \times 10^{-12})(10 \times 10^{-4}}{0.001}[/tex]

[tex]C = 4.96 \times 10^{-11} farad[/tex]

Part c)

Now we have

[tex]Q = C\Delta V[/tex]

[tex]Q = (4.96 \times 10^{-11})(14 \times 10^3)[/tex]

[tex]Q = 0.69 \mu C[/tex]

Part d)

[tex]Energy = \frac{1}{2}CV^2[/tex]

[tex]E = \frac{1}{2}(4.96 \times 10^{-11})(14 \times 10^3)^2[/tex]

[tex]E = 4.86 \times 10^{-3} J[/tex]

Answer:

0.68 kg-m²

Explanation:

F = Force applied by the muscle = 2615 N

r = effective perpendicular lever arm = 2.85 cm = 0.0285 m

α = Angular acceleration of the forearm = 110.0 rad/s²

I  = moment of inertia of the boxer's forearm = ?

Torque is given as

τ = I α                                   eq-1

Torque is also given as

τ = r F                                   eq-2

using eq-1 and eq-2

r F =  I α

(0.0285)(2615) = (110.0) I

I = 0.68 kg-m²

Answer:

Has been removed 1.458 moles.

Explanation:

n1= 1.8 mol

p1= 27.3 atm

p2= 5.2 atm

n2= ?

n2= n1 * p2/p1

n2= 0.342 moles

Δn= n1-n2

Δn= 1.458 moles

Answer:46.03 KJ

Explanation:

no of moles(n)=5

temperature of air(T)=[tex]20^{\circ}[/tex]

pressure(p)=1 atm

final volume is [tex]\frac{V}{10}[/tex]

We know work done in adaibatic process is given by

W=[tex]\frac{P_iV_i-P_fV_f}{\gamma -1}[/tex]

[tex]\gamma[/tex] for air is 1.4

we know [tex]P_iV_i^{\gamma }= P_fV_f^{\gamma }[/tex]

[tex]1\left ( V\right )^{\gamma }[/tex]=[tex]P_f\left (\frac{v}{10}\right )^{\gamma }[/tex]

[tex]10^{\gamma }=P_f[/tex]

[tex]P_f=25.118 atm[/tex]

W=[tex]\frac{1\times 0.1218-25.118\times 0.01218}{1.4 -1}[/tex]

W=-46.03 KJ

it means work is done on the system

Answer:

[tex]T = 1.26 \times 10^8 K[/tex]

Explanation:

As we know that rms speed of ideal gas is given by the formula

[tex]v_{rms} = \sqrt{\frac{3RT}{M}}[/tex]

here we know that

[tex]v_{rms} = 2.67 \times 10^5 m/s[/tex]

molecular mass of gas is given as

[tex]M = 44 g/mol = 0.044 kg/mol[/tex]

now from above formula we have

[tex]2.67\times 10^5 = \sqrt{\frac{3(8.31)T}{0.044}}[/tex]

now we have

[tex]T = 1.26 \times 10^8 K[/tex]

The root mean square speed of a molecule in an ideal gas can be used to determine the temperature of the gas when vrms is known. This involves solving the formula for temperature using the given vrms and mass of a molecule. We need to ensure that the molar mass is properly converted to the mass of a molecule and that we use SI units.

The subject question is related to the concept of root mean square speed (vrms) in gas physics and involves calculations using the ideal gas model. According to this concept, we know that vrms is given by the equation vrms = sqrt(3kB T / m), where:

Given that vrms = 2.67 x 105 m/s and molar mass (M) = 44.0 g/mol (which must be converted to kg as SI units are used in this context), we can solve the formula for T to find the temperature T = vrms^2×m / (3×kB) after substituting the known values into this equation. Note that one must convert the molar mass to mass of a molecule by dividing it by Avogadro's number (6.022 x 10²³ mol⁻¹).

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Answer:

40N

Explanation:

frictional force = mgsin©

frictional force = 8*10*sin30

frictional force = 8*10*0.5= 40N//

Answer:

250 N

Explanation:

m = Mass of cart = 50 kg

F = Pushing Force = 250 N

g = Acceleration due to gravity = 9.81 m/s²

μ = Coefficient of friction

[tex]F_s =\text{Static frictional force between the floor and the crate}= 275\ N[/tex]

N = Weight of cart = mg

N = 50×9.81

N = 490.5 N

[tex]F_s =\mu N\\\Rightarrow \mu=\frac{F_s}{N}\\\Rightarrow \mu=\frac{275}{490.5}\\\Rightarrow \mu=0.56[/tex]

Here it can be seen that the pushing force is less than the static frictional force so the crate will not move

∴ The frictional force on the crate while the person pushes on it is 250 N

Answer:

Linear Perspective.

Explanation:

Linear perspective cue to depth and it is in respect to both texture gradient or the next depth cue and relative size. In linear perspective, parallel lines that go along seems to have converged at distance very far, eventually reaching a vanishing point at the horizon. For example the railway track appears to be converging at a point very distant from the observer but it didn't.

Answer:

d. linear perspective

Explanation:

According to a different source, these are the options that come with this question:

a. visual acuity

b. texture gradient

c. retinal disparity

d. linear perspective

Linear perspective refers to a way in which depth is created. This term is particularly employed by artists who create an illusion of depth on a flat surface. This perspective leads all parallel lines to be lost in a single vanishing point on the horizon line. In this example, Gina is using the linear perspective cue to understand the depth of the scene she is observing.

(a) The velocity of the car-truck system after collision is 18.08 m/s.

(b) The increase in internal energy of the car and truck is 2,835,031.2 J.

The velocity of the system after collision is determined by applying the principle of conservation of linear momentum as shown below;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2800(36) + 4000(-15) = vx(2800 + 4000)

40,800 = 6800vx

vx = 6 m/s

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2800(0) + 4000(29) = vz(2800 + 4000)

116,000 = 6800vz

vz = 17.06 m/s

K.E(car) = ¹/₂mv²

K.E(car) = ¹/₂ x (2800) x (36)²

K.E(car) = 1,814,400 J

v(truck) = √(15² + 29²) = 32.65

K.E(truck) = ¹/₂ x (4000) x (32.65)²

K.E(truck) = 2,132,045 J

K.E(total) =  1,814,400 J + 2,132,045 J = 3,946,445 J

K.E =  ¹/₂(m₁ + m₂)v²

K.E = ¹/₂ x (2800 + 4000) x (18.08)²

K.E = 1,111,413.8 J

U = ΔK.E

U = 3,946,445 J - 1,111,413.8 J

U = 2,835,031.2 J

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Final answer:

To determine the velocity of the stuck-together car and truck after the collision, apply the principle of conservation of momentum. Consider the change in kinetic energy and the production of thermal energy and deformation.

Explanation:

To determine the velocity of the stuck-together car and truck just after the collision, we can apply the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision. We can find the final velocity by summing the momenta of the car and truck and dividing by their combined mass.

In this case, the car's momentum is the product of its mass and velocity, and the truck's momentum is the product of its mass and velocity. Adding these momenta together and dividing by the combined mass gives us the final velocity of the stuck-together car and truck just after the collision.

For the increase in internal energy of the car and truck, we need to consider the change in kinetic energy and the production of thermal energy and deformation. The change in kinetic energy can be calculated by finding the difference between the initial and final kinetic energies of the car and truck. The thermal energy and deformation depend on factors such as the materials involved and the severity of the collision.

Answer:

The answer is 3Q

Explanation:

The metal temperature increases in a linear way, we could get a difference between final and initial temperature

[tex]DT=FinalTemperature-InitialTemperature[/tex]

We get a temperature difference of 2 degrees per each heat addition.  If we add the same heat 3 times more, it will increase to 12 degrees

Answer:

The quantity of heat required to increase the temperature of the object from  6°C to 12°C is 3Q

Explanation:

Heat capacity is the quantity of heat required to increase the temperature of an object.

Q = mcΔθ

where;

Q is the quantity of heat

m is the mass of the object

c is specific heat capacity of the object

Δθ is change in temperature = T₂ - T₁

For the first sentence of this question;

Q = mc(6-4)

Q = mc(2)

Q = 2mc

For the second sentence of this question;

Let Q₂ be the quantity of heat required to increase the temperature of the object from  6°C to 12°C

Q₂ = mcΔθ

Q₂ = mc(12-6)

Q₂ = mc(6)

Q₂ = 6mc

Q₂ = 3(2mc)

Recall, Q = 2mc

Thus, Q₂ = 3Q

Answer:

Height, h = 13.02 meters

Explanation:

It is given that,

Spring constant of the spring, k = 750 N/m

It is compressed by 35 cm relative to its unstained length, x = 35 cm = 0.35 m

Mass of the object, m = 0.36 kg

When the spring is released, the mass is launched vertically in the air. We need to find the height attained by the mass at this position. On applying the conservation of energy as :

Energy stored in the spring = change on potential energy

[tex]\dfrac{1}{2}kx^2=mgh[/tex]

[tex]h=\dfrac{kx^2}{2mg}[/tex]

[tex]h=\dfrac{750\ N/m\times (0.35\ m)^2}{2\times 0.36\ kg\times 9.8\ m/s^2}[/tex]

h = 13.02 meters

So, the mass will reach a height of 13.02 meters. Hence, this is the required solution.

The maximum height reached by the mass when the spring is released is 13.02 m.

The given parameters;

Apply the principle of conservation energy to determine the maximum height reached by the mass when the spring is released.

mgh = ¹/₂kx²

where;

[tex]h = \frac{kx^2}{2mg} \\\\h = \frac{(750) \times (0.35)^2}{2(0.36\times 9.8)} \\\\h = 13.02 \ m[/tex]

Thus, the maximum height reached by the mass when the spring is released is 13.02 m.

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The speed of the car on Curve B is obtained by solving for the radius in Curve A's equation, then substituting that into the formula for Curve B. The formula involved is based on principles of circular motion and forces.

The subject of this question is Physics, specifically the principles of circular motion and the forces at play within. Given the information from Curve A, we can ascertain that the speed of the car on Curve B can be found using principles of physics. The formula to find the speed at which the car can travel around a banked curve without relying on friction is: v = sqrt[rgtan(Θ)]. Where v is the speed, r is the radius of the curve, g is the acceleration due to gravity, and Θ is the angle of the banked curve.

Since the radius is the same for both curves, and the speed is known for Curve A, we can set it up so: 19.1 = sqrt[r * 9.8 * tan(12.7)]. We then solve for r (the radius), and apply it to Curve B: v = sqrt[r * 9.8 * tan(15.1)]. By substituting in the value of r obtained from the first equation, we can calculate the speed for Curve B.

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To find the speed at which the car can travel around curve B without relying on friction, we can use the same formula but with a banked angle of 15.1°.

The speed at which a car can travel around a banked curve without relying on friction can be calculated using the ideal banking angle formula. The formula is given by v = √(g * r * tan(θ)), where v is the speed, g is the acceleration due to gravity, r is the radius of the curve, and θ is the banked angle. In this case, curve A is banked at 12.7° and the car can travel at a speed of 19.1 m/s. To find the speed at which the car can travel around curve B without relying on friction, we can use the same formula but with a banked angle of 15.1°.

Plugging in the values into the formula, v = √(9.8 * r * tan(15.1)). Since the radius is the same for both curves, we can solve for v: 19.1 = √(9.8 * r * tan(12.7))

Solving the equation for v, we find that the car can travel around curve B without relying on friction at a speed of approximately 21.2 m/s.

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Answer:

605 km

Explanation:

Hello

the same units of measure should be used, then

Step 1

convert  42 m/s ⇒   km/h

1 km =1000 m

1 h = 36000 sec

[tex]42 \frac{m}{s}*\frac{1\ km}{1000\ m}=0.042\ \frac{km}{s}\\ 0.042\ \frac{km}{s}\\[/tex]

[tex]0.042\ \frac{km}{s}*\frac{3600\ s}{1\ h} =151.2 \frac{km}{h}\\ \\Velocity =151.2\ \frac{km}{h}[/tex]

Step 2

find kilometers traveled after 4  hours

[tex]V=\frac{s}{t}\\ \\[/tex]

V,velocity

s, distance traveled

t. time

now, isolating s

[tex]V=\frac{s}{t} \\s=V * t\\[/tex]

and replacing

[tex]s=V * t\\s=151.2\frac{km}{h}*4 hours\\ s=604.8 km\\[/tex]

S=604.8 Km

Have a great day

Answer:

The intensity level of sound 2 is 93.3\ W/m².

Explanation:

Given that,

Intensity of sound 1 = 45.0 W/m²

Intensity level of sound 2 = 3.2 dB

[tex]I_{0}=10^{-12}\ w/m^2[/tex]

We need to calculate the intensity

Using equation of the sound level intensity

[tex]I_{dB}=10 log\dfrac{I}{I_{0}}[/tex]

[tex]I_{dB}=10 log(\dfrac{45.0}{10^{-12}})[/tex]

[tex]I_{dB}=136.5 dB[/tex]

The intensity of sound 2 is greater than 3.2 dB.

Therefore,

[tex]I_{2}=136.5+3.2[/tex]

[tex]I_{2}=139.7\ dB[/tex]

Calculate the intensity of sound 2

[tex]I_{2}=10log(\dfrac{I'}{10^{-12}})[/tex]

[tex]I'=10^{-12}(10^{\frac{139.7}{10}})[/tex]

[tex]I'=93.3\ W/m^2[/tex]

Hence, The intensity level of sound 2 is 93.3\ W/m².

Decibels, measured in dB, quantifies the intensity level of sound. They relate to the intensity of sound in a logarithmic manner. More information and advanced calculations are needed to find the actual intensity of Sound 2 based on the given decibel difference

Let's begin by understanding that decibels (dB) measure the intensity level of sound. The decibel level (dB) is a logarithmic unit used to describe a ratio of two values of a physical quantity, here, the sound intensity. One key fact is that each time the intensity increases by a factor of 10; the decibel level rises by 10 dB. Thus, the difference in decibels corresponds to the logarithmic ratio of the two intensities involved.

In the given problem, Sound 2 is 3.2 dB greater than Sound 1. It means the ratio of their intensities has a logarithmic relation with a value of 3.2. However, to convert this decibel difference into an actual value for Sound 2's intensity, additional calculations involving logarithms and the initial intensity value for Sound 1 are needed. As these calculations involve advanced math, it'd be more appropriate to take this problem to a higher level course or consult a logarithm table or a scientific calculator.

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Hey there!:

Take the speed of sound to be 343m/s.

Direct frequency perceived by observer:

(343 + 17) / 343) * 90Hz = 94.460Hz

Change in frequency = ( 4.460 - 90 ) = 4.460Hz.  

90 - (4.460 x 2) = 81.08 Hz. indirect frequency heard by observer.  

Therefore :

Beat frequency = (94.460 - 81.08) = 13.38Hz.  

Hope this helps!

Answer:

Velocity of airplane is 500 km/h

Velocity of wind is 40 km/h

Explanation:

[tex]V_a[/tex]= Velocity of airplane in still air

[tex]V_w[/tex]= Velocity of wind

Time taken by plane to travel 1150 km against the wind is 2.5 hours

[tex]V_a-V_w=\frac {1150}{2.5}\\\Rightarrow V_a-V_w=460\quad (1)[/tex]

Time taken by plane to travel 450 km against the wind is 50 minutes = 50/60 hours

[tex]V_a+V_w=\frac {450}{50}\times 60\\\Rightarrow V_a-V_w=540\quad (2)[/tex]

Subtracting the two equations we get

[tex]V_a-V_w-V_a-V_w=460-540\\\Rightarrow -2V_w=-80\\\Rightarrow V_w=40\ km/h[/tex]

Applying the value of velocity of wind to the first equation

[tex]V_a-40=460\\\Rightarrow V_a =500\ km/h[/tex]

∴ Velocity of airplane in still air is 500 km/h and Velocity of wind is 40 km/h

The given mathematical problem involves calculating speeds. The speed of the airplane in still air was found to be 501 km/h and the wind speed was calculated to be 41 km/h, by using the concept of speed equals distance divided by time.

The subject is related to measurements of speed, which is essentially a mathematical problem, specifically involving algebraic calculations. In order to solve this problem, we will use the concept that speed equals distance divided by time.

Firstly, convert all time measurements to the same unit. Since the speed of an airplane is typically measured in km/h, it will be convenient to convert the 50 minutes to hours (50/60 = 0.83 hours).

When moving against the wind, the combined speed (airplane speed minus wind speed) is 1150 km / 2.5 h = 460 km/h. When moving with the wind, the combined speed (airplane speed plus wind speed) becomes 450 km / 0.83 h = 542 km/h.

Now you can find the speed of the airplane in still air by averaging the two combined speeds: (460 km/h + 542 km/h) / 2 = 501 km/h. The speed of the wind can be found by subtracting the airplane's speed from the higher combined speed: 542 km/h - 501 km/h = 41 km/h. Therefore, the speed of the airplane in still air is 501 km/h and the wind speed is 41 km/h.

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Answer:

option (C)

Explanation:

EMF stand for electro motive force. the emf of a battery is the potential between the two electrodes when it is not use in the circuit.

The terminal potential difference is the potential difference between the electrodes of a cell when it is in use.

EMF is only when the current is very large in the battery.

The emf of a battery is equal to its terminal potential difference only when a large cwrrent is in the battery.

C) only when a large cwrrent is in the battery.