A horse and a squirrel participate in a race over a 1.20 km long course. The horse travels at a speed of 19.0 m/s and the squirrel can do 3.50 m/s. The horse runs for 0.960 km and then stops to tease the slow-moving squirrel, which eventually passes by. The horse waits for a while after the squirrel passes and then runs toward the finish line. Both animals cross the finish line at the exact same instant. Assume both animals, when moving, move steadily at their respective speeds.

(a)

How far (in m) is the squirrel from the finish line when the horse resumes the race?

m

(b)

For how long in time (in s) was the horse stationary?

s

Answer:

given,

speed of the car = 20 m/s

final speed of car = 0 m/s

distance between car and the deer = 38 m

reaction time, t = 0.5 s

deceleration of the car = 10 m/s².

a) distance between deer and car

  distance travel in the reaction time

   d₁ = v x t

   d₁ = 20 x 0.5 = 10 m

   distance travel after you apply brake

   using equation of motion

   v² = u² + 2 a s

   0 = 20² - 2 x 10 x s

    s =  20 m

total distance traveled by the car

D = d₁ + d₂

D = 20 + 10 = 30 m

  distance between car and the deer = 38 m - 30 m

                                                              = 8 m

b) now, maximum speed car.

   distance travel in reaction time

    d₁ = s x t

    d₁ = 0.5 V

distance left between them

   d₂ = 38 - d₁

   d₂ = 38 - 0.5 V

   distance travel after you apply brake

   using equation of motion

    v² = u² + 2 a d₂

    0 = (V)² - 2 x 10 x (38 - 0.5 V)

     V² + 10 V - 760 = 0

now, solving the quadratic equation

  [tex]x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

  [tex]V = \dfrac{-10\pm \sqrt{10^2-4(1)(-760)}}{2(1)}[/tex]

         V = 23.01 , -33.01

rejecting the negative term.

hence, maximum speed of the car could be V = 23.01 m/s

Answer:

0.07 m

Explanation:

[tex]L_0[/tex] = Initial length = 1 km = 1000 m

[tex]\Delta T[/tex] = Change in temperature =  1.00°C

[tex]\alpha[/tex] = Coefficient of linear thermal expansion

Volumetric coefficient of expansion of water

[tex]\beta=210\times 10^{-6}^{\circ}C\\\Rightarrow \beta=3\alpha\\\Rightarrow \alpha=\dfrac{\beta}{3}\\\Rightarrow \alpha=\dfrac{210\times 10^{-6}}{3}\\\Rightarrow \alpha=70\times 10^{-6}\ ^{\circ}C[/tex]

Change in length is given by

[tex]\Delta L=L_0\alpha \Delta T\\\Rightarrow \Delta L=1000\times 1\times 70\times 10^{-6}\\\Rightarrow \Delta L=0.07\ m[/tex]

The change in length is 0.07 m

The change in length of a column of water for a temperature increase of 1.00°C is 210×10^6 meters.

To calculate the change in length of a column of water for a temperature increase, we can use the thermal coefficient of volume expansion for sea water. The thermal coefficient of volume expansion is given as 210×106/°C. To find the change in length, we need to multiply the original length (1.00 km) by the thermal coefficient of volume expansion (210×106/°C) and the temperature change (1.00°C).

Change in length = Original length x Thermal coefficient x Temperature change

Change in length = (1.00 km) x (210×106/°C) x (1.00°C)

Change in length = 210×106 meters

Answer

given,

discharge rate from pipe = 1000 gallons/minutes

now,

flow rate in  cubic meters per second

1 gallon = 0.00378541 m³

1 min = 60 s

Q = [tex]1000\times \dfrac{0.00378541\ m^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}[/tex]

Q = 0.063 m³/s

flow rate in  liters per minute

1 gallon = 3.78541 L

 Q = [tex]1000\times \dfrac{3.78541\ m^3}{1\ gallon}[/tex]

 Q = 3785.41 m³/min

flow rate in cubic feet per second

 1 gallon = 0.133681 ft³

 1 min = 60 s

Q = [tex]1000\times \dfrac{0.133681\ ft^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}[/tex]

Q = 2.23 ft³/s

Answer:

Gravitational force

Explanation:

If two spheres have equal densities and they are subject only to their mutual gravitational attraction. We need to say that the quantities that must have the same magnitude for both spheres. So, the correct option is (E) i.e. gravitational force.

It is because of Newton's third law of motion. It states that the force due to object 1 to object 2 is same as force due to object 2 to object 1. The two forces act in opposite direction.  

Hence, the correct option is (E) "Gravitational force".                        

The gravitational force has the same magnitude for both spheres.

By the Newton's law of gravitation and the Newton's third law we understand that gravitational force is directly proportional to the masses of the spheres and inversely proportional to the square of distance between centers, also that gravitational force experimented by one sphere has the same magnitude but opposite direction than the gravitational force experimented by the other sphere.

Therefore, we conclude that gravitational force has the same magnitude for both spheres. (Correct choice E)

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The concept required to solve this problem is quantization of charge.

First the number of electrons will be calculated and then the total mass of the charge.

With these data it will be possible to calculate the percentage of load in the mass.

[tex]Q= ne[/tex]

Here Q is the charge, n is the number of electrons and e is the charge on the electron

[tex]n = \frac{Q}{e}[/tex]

Replacing,

[tex]n = \frac{4*10^{-6}C}{1.6*10^{-19}}[/tex]

[tex]n = 2.5 * 10^{13}77[/tex]

According to the quantization of charge the charge is defined as product of the number of electron and the charge on the electron

The total mass of the charge is

[tex]m= nm_e[/tex]

Here,

m = Mass of the charge

n = Number of electrons

[tex]m_e[/tex] = Mass of the electron

[tex]\text{Percentage change} = \frac{nm_e}{M}*100[/tex]

Replacing we have

[tex]\text{Percentage change} = \frac{(2.5*10^13)(9.1*10^{-28})}{33}*100[/tex]

[tex]\text{Percentage change} = 6.9*10^{-14} \%[/tex]

The percentage change in the mass of the 33 g comb during charging is extremely small, approximately [tex]\(6.91 \times 10^{-16}\%\),[/tex]  which is negligible.

To find the percentage change in mass of the comb due to charging, we need to understand the relationship between the amount of charge transferred and the mass of the electrons involved. Here's the step-by-step process:

1. Calculate the number of electrons corresponding to the given charge:

The charge of one electron[tex](\(e\))[/tex] is approximately[tex]\(1.6 \times 10^{-19}\)[/tex] coulombs.

Given:

  - Net charge [tex](\(Q\)) = 4.0 μC = \(4.0 \times 10^{-6}\) C[/tex]

The number of electrons n transferred can be calculated using the formula:

 [tex]\[ n = \frac{Q}{e} \][/tex]

[tex]\[ n = \frac{4.0 \times 10^{-6} \text{ C}}{1.6 \times 10^{-19} \text{ C/electron}} \][/tex]

 [tex]\[ n = 2.5 \times 10^{13} \text{ electrons} \][/tex]

2. Calculate the mass of the transferred electrons:

The mass of one electron [tex](\(m_e\))[/tex] is approximately [tex]\(9.11 \times 10^{-31}\)[/tex] kg.

The total mass[tex](\(m\))[/tex]  of the electrons transferred is:

[tex]\[ m = n \times m_e \][/tex]

[tex]\[ m = 2.5 \times 10^{13} \times 9.11 \times 10^{-31} \text{ kg} \][/tex]

 [tex]\[ m = 2.28 \times 10^{-17} \text{ kg} \][/tex]

3. Convert the mass of the comb to kilograms:

Given:

  - Mass of the comb = 33 g = 0.033 kg

4. Calculate the percentage change in mass:

The percentage change in mass is given by:

  [tex]\[ \text{Percentage change} = \left( \frac{\text{Change in mass}}{\text{Original mass}} \right) \times 100\% \][/tex]

[tex]\[ \text{Percentage change} = \left( \frac{2.28 \times 10^{-17} \text{ kg}}{0.033 \text{ kg}} \right) \times 100\% \][/tex]

[tex]\[ \text{Percentage change} = 6.91 \times 10^{-16} \% \][/tex]

Answer:

Velocity remains the same at 104 m/s

Explanation:

According to Newton's 1st law of motion, an object subjected to no force or net force equal 0 would maintain its velocity. In our case the crew shuts off the power, spaceship is in space and far from all other objects (so no gravity whatsoever) would have no force acting on it. Therefore its velocity would stay the same at 104 m/s

Answer:

Flow rate 2.34 m3/s

Diameter 0.754 m

Explanation:

Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.

The area at the well head is

[tex]A = \pi r_w^2 = \pi (0.509/2)^2 = 0.203 m^2[/tex]

So the volume flow rate along the pipe is

[tex]\dot{V} = Av = 0.203 * 11.5 = 2.34 m^3/s[/tex]

We can use the similar logic to find the cross-section area at the refinery

[tex]A_r = \dot{V}/v_r = 2.34 / 5.25 = 0.446 m^2[/tex]

The radius of the pipe at the refinery is:

[tex]A_r = \pi r^2[/tex]

[tex]r^2 =A_r/\pi = 0.446/\pi = 0.141[/tex]

[tex]r = \sqrt{0.141} = 0.377m[/tex]

So the diameter is twice the radius = 0.38*2 = 0.754m

Answer:  Frequency factor  A = 8 × 10⁹

activation energy Ea = 15.5 KJ/Mol

Explanation: to begin, let us first define the parameters given;

K₁ = 1.44 × 10⁷dm³mol⁻¹s⁻¹

K₂ = 3.03 × 10⁷ dm³mol⁻¹s⁻¹

K₃ = 6.9 × 10 dm³mol⁻¹s⁻¹

also T₁ = 300.3 K

T₂ = 341.2 K

T₃ = 392.2 K

we know that;

㏑ K₂ / K₁ = Ea/R [1/T₁ -1/T₂]

where R is given as 8.314 J/mol-k

Ea = activation energy

K₁, K₂ = rate constant

T₁, T₂ = Temperature

therefore, ㏑ (3.03 × 10⁷/ 1.44 × 10⁷) = Ea / 8.314 [1/300.3 - 1/341.2]

this gives Ea = 15496.16 J/Mol ≈ 15.5 KJ /Mol

∴ Ea = 15.5 KJ/ Mol

also given that K = A e⁻∧Ea/RT

here A = frequency factor

∴ 6.9 × 10⁷ = A e⁻ ∧(15496.16/8.314 × 392.2)

A = 7.99 × 10⁹ = 8 × 10⁹

Final answer:

The Arrhenius equation can be used to determine the frequency factor and activation energy of a reaction.

Explanation:

The given question requires determining the frequency factor and activation energy for a reaction. The relationship between the rate constant and temperature is described by the Arrhenius equation:

k = Ae-Ea/RT

To find the frequency factor (A), we can rearrange the equation and use the rate constant (k) at a specific temperature and the activation energy (Ea) to solve for A. Similarly, to find the activation energy, we can rearrange the equation and use the rate constants at two different temperatures to solve for Ea.

Answer: centripetal acceleration

Equals 18350m/s²

Explanation: Centripetal acceleration is given as v²/r

Where v is Velocity measured in rad/s and r is radius

But v= w*r

Therefore,

C.acceleratn = w²*r²/r

So we have

C.accelertn = w²*r

Our w is given as 1200 rev/min so we have to convert to rad/sec

1rad/sec equals 9.5rev/min

Therefore,

1200rev/min = 1200/9.5 rad/sec

w=126.32rad/sec

r = diameter/2

r = 2.3/2 =1.15m

Let's now calculate our centripetal acceleration = w²*r

= {126.32rad/sec}² *1.15

=18350m/s²

The centripetal acceleration of the World War II fighter plane propeller tip is approximately 18167.6 m/s² or about 1853 times the gravitational acceleration. This is determined using the radius, angular velocity, and linear speed of the propeller tip.

The propeller of a World War II fighter plane is 2.30 meters in diameter and spins at 1200 revolutions per minute. We can calculate the centripetal acceleration of the propeller tip as follows:

Therefore, the centripetal acceleration of the propeller tip is approximately 18167.6 meters per second squared or about 1853 times the gravitational acceleration.

The magnitude is used to quantify the size of the earthquakes (measures the energy released during the breakdown of a fault) while the intensity is a qualitative description of the effects of the earthquakes (it involves the perception of people as well as damage material and economic suffered due to the event). The relation between intensity and magnitude is

[tex]\frac{I_1}{I_2} =10^{(M1-M2)}[/tex]

Here,

M = Magnitude

I = Intensity of each one

Given

[tex]M_1 = 9[/tex]

[tex]M_2 =6[/tex]

Replacing,

[tex]\frac{I_1}{I_2} =10^{(9-6)}[/tex]

[tex]\frac{I_1}{I_2} = 10^3[/tex]

[tex]\frac{I_1}{I_2} = 1000[/tex]

So the aftershocks are 1000x weaker compared to the original earthquake. The correct answer is B.

Answer:

C. Zero for ammeter; infinity for voltmeter

Explanation:

Ammeter: A device that is used to measure the current through a circuit.

Voltmeter: A device that is used to measure the electric potential difference between two points in a circuit.

The relation among voltage (v), current (i), and resistance (r) is given as,

[tex]v = i\times r[/tex]

In ideal cases an ammeter must have zero resistance so that there is no voltage drop across it and accurate value of current will be measured.

In case of an ideal voltmeter, the resistance should be infinite so that the current across it is zero.

The ideal internal resistance for a voltmeter is infinity, to prevent it from altering the circuit by drawing current. For an ammeter, it's zero to avoid introducing additional resistance into the circuit. Thus, the correct answer is C: Zero for ammeter; infinity for voltmeter.

The ideal internal resistance for a voltmeter and an ammeter varies due to their different roles in electrical measurements. A voltmeter is used to measure the voltage across two points in a circuit and ideally should have an infinite internal resistance to prevent it from drawing any significant current from the circuit.

This is because a voltmeter is connected in parallel with the circuit component, and a high resistance ensures that it does not affect the circuit's operation by introducing a significant parallel path for current. On the other hand, an ammeter is used to measure the current flowing through a circuit and is placed in series with the circuit elements. Therefore, an ammeter should have a zero internal resistance to ensure that it does not introduce any additional resistance into the circuit, which could alter the current it is meant to measure.

To solve this problem we will define the data obtained in each of the sections. We know that the Net Force is equivalent to the Force in section 1, which can be found through mass flow and velocity, plus the force in section two, which can be found as the product between pressure and Area, so therefore we have

State 1:

[tex]\dot{m} = 200kg/s[/tex]

[tex]v_1 = 150m/s[/tex]

[tex]P_1 = 1600kPa[/tex]

State 2

[tex]v_2 = 2300m/s[/tex]

[tex]A_2 = 2.4m^2[/tex]

[tex]P_2 = 80kPa[/tex]

We have that net force is equal to

[tex]F_{net} = F_1 + F_2[/tex]

[tex]F_1 = \dot{m} (v_2-v_1)[/tex]

[tex]F_2 = (P_1-p_2)A_2[/tex]

Replacing,

[tex]F = \dot{m} (v_2-v_1) + (P_1-p_2)A_2[/tex]

[tex]F = 200(2300-150)+(1600-80)*10^3(2.4)[/tex]

[tex]F = 4078kN[/tex]

Therefore  the thrust force acting on the nozzle is 4078kN

Answer:

  a = Np/10 yrs×[3.50^7 yrs /sec]

Explanation:

The energy of the single photon of frequency f or wave length λ  is given as

   E = hc / λ

since the glow warm emits energy 0.1 J/sec

that is  the number of photons n emitted by the photon per sec will be

n = 0.1 W / E

Thus, the number of photons emitted in 10 years

 N = n×3.15×10^7 sec/yr ×10 yr

Now, momentum associated with each photon

p= h / λ

and, momentum associated with N photon particles

P= N(h/λ)    

hence the change in the momentum of the glow is = Np in 10 years

Therefore, acceleration of the glow

                a = Np/10 yrs×[3.50^7 yrs /sec]

Answer:

The speed is 21.06 m/s.

Explanation:

Given that,

mass of glow worm = 5.0 g

Wavelength = 650 nm

Power = 0.10 W

Time = 10 years

The total energy emitted in a period [tex]\tau[/tex] is [tex]P\tau[/tex]

The energy of single photon of frequency or wavelength is

[tex]E=\dfrac{hc}{\lambda}[/tex]

The total number of photons emitted in a interval [tex]\tau[/tex] is then the total energy divided by the energy per photon.

[tex]N=\dfrac{P\tau}{E}[/tex]

[tex]N=\dfrac{P\tau\times\lambda}{hc}[/tex]

[tex]N=\dfrac{P\tau\times\lambda}{hc}[/tex]

We need to calculate the speed

Using de Broglie's relation applies to each photon and thus the total momentum imparted to the glow-worm

[tex]p=\dfrac{Nh}{\lambda}[/tex]

[tex]p=\dfrac{\dfrac{P\tau\times\lambda}{hc}\times h}{\lambda}[/tex]

[tex]p=\dfrac{P\tau}{c}[/tex]

[tex]v=\dfrac{P\tau}{mc}[/tex]

Put the value into the formula

[tex]v=\dfrac{0.10\times3.16\times10^{8}}{5.0\times10^{-3}\times3\times10^{8}}[/tex]

[tex]v=21.06\ m/s[/tex]

Hence, The speed is 21.06 m/s.

Answer:

a) the number of antinodes increases , b) wavelength produced is constant.

, c) fundamental frequency is constant., d)  fundamental wavelength does not change, e)  the speed of the wave is constant

Explanation:

This is a resonance problem where we have a frequency generator, attached to a chain with a weight in its final part, at the two ends there is a node, point if movement. The condition for resonance of this system is

         λ = 2 L / n

  Where n is an integer

         L = n λ / 2

Let's review the problem questions.

A) As the length of the chain increases the number of wavelengths (Lam / 2) should increase, so the number of antinodes increases

B) when seeing the first equation the wavelength remains the same since the change in the length of the chain and the change in the number between are compensated, therefore for a configuration of generator frequency and weight applied the wavelength produced is constant.

C) the speed of the wave is

         v = λ f

In a string the speed is constant for a fixed applied weight, with the wavelength we did not change, therefore the fundamental frequency must also be constant.

D) The value of the fundamental wavelength does not change if the weight does not change, but there is a minimum chain length for this resonance to be observed and corresponds to n = 1

              L = λ / 2

E) the speed of the wave depends on the chain tension and its density if they do not change the speed of the wave does not change either

Answer:

The amount of work done in lifting the bag is -20109.6 N-m

Explanation:

Given that,

Mass of bag = 60 kg

Distance = 9 m

Loss of mass = 12 kg

The number of pounds lost is proportional to the square root of the distance traversed

Mass of the bag containing flour at height is

[tex]m(y)=60-k\sqrt{y}[/tex]

Put the value into the formula

[tex]60-k\sqrt{y}=12[/tex]

[tex]k=144[/tex]

We need to calculate the work done

Using formula of work done

[tex]W=\int_{0}^{9}{m(y)gdy}[/tex]

Put the value into the formula

[tex]W=\int_{0}^{9}{(60-k\sqrt{y})gdy}[/tex]

[tex]W=((60y-\dfrac{2k}{3}\times y^{\frac{3}{2}}})_{0}^{9})\times9.8[/tex]

Put the value of y

[tex]W=(60\times9-\dfrac{2\times144}{3}\times 9^{\frac{3}{2}})\times9.8[/tex]

[tex]W=-20109.6\ N-m[/tex]

Hence, The amount of work done in lifting the bag is -20109.6 N-m

Answer:

[tex]f=1.79Hz[/tex]

Explanation:

The period is defined as the time taken by an object to complete a cycle in a simple harmonic motion. As the frequency of the motion increases, the period decreases. Therefore, they are inversely proportional. The frequency does not depend on the mass of the object.

[tex]f=\frac{1}{T}\\f=\frac{1}{0.56 s}\\f=1.79Hz[/tex]

Answer : The density, specific gravity, and mass of the air in a room is, 1.16825 g/L, 0.916 and 140.19 kg respectively.

Explanation :

First we have to calculate the volume of air.

[tex]Volume=Length\times Breadth\times Height[/tex]

[tex]Volume=4m\times 5m\times 6m[/tex]

[tex]Volume=120m^3=120000L[/tex]      [tex](1m^3=1000L)[/tex]

Now we have to calculate the mole of air.

Using ideal gas equation:

[tex]PV=nRT[/tex]

where,

P = Pressure of air = 100 kPa =  0.987 atm      (1 atm = 101.3 kPa)

V = Volume of air = 120000 L

n = number of moles of air = ?

R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]

T = Temperature of air = [tex]25^oC=273+25=298K[/tex]

Putting values in above equation, we get:

[tex]0.987atm\times 120000L=n\times (0.0821L.atm/mol.K)\times 298K[/tex]

[tex]n=4841.04mol[/tex]

Now we have to calculate the mass of air.

[tex]\text{Mass of air}=\text{Moles of air}\times \text{Molar mass of air}[/tex]

As we know that the molar mass of air is, 28.96 g/mol

[tex]\text{Mass of air}=4841.04mol\times 28.96g/mol=140196.5184g=140.19kg[/tex]

Now we have to calculate the density of air.

[tex]\text{Density of air}=\frac{\text{Mass of air}}{\text{Volume of air}}[/tex]

[tex]\text{Density of air}=\frac{140.19kg}{120000L}=1.16825\times 10^{-3}kg/L=1.16825g/L[/tex]

Now we have to calculate the specific gravity of air.

[tex]\text{Specific gravity of air}=\frac{\text{Air density at given condition}}{\text{Air density at STP}}[/tex]

As we know that air density at STP is, 1.2754 g/L

[tex]\text{Specific gravity of air}=\frac{1.16825g/L}{1.2754g/L}=0.916[/tex]

Answer:

While falling, the magnitude of the acceleration of the egg is 9.82 m/s²

While stopping, the magnitude of the deceleration of the egg is 79.3 m/s²

Explanation:

Hi there!

The equation of velocity of the falling egg is the following:

v = v0 + a · t

Where:

v = velocity at time t.

v0 = initial velocity.

a = acceleration.

t = time

Let´s calculate the acceleration of the egg while falling. Notice that the result should be the acceleration of gravity, ≅ 9.8 m/s².

v = v0 + a · t

11.1 m/s = 0 m/s + a · 1.13 s   (since the egg is dropped, the initial velocity is zero). Solving for "a":

11.1 m/s / 1.13 s = a

a = 9.82 m/s²

While falling, the magnitude of the acceleration of the egg is 9.82 m/s²

Now, using the same equation, let´s find the acceleration of the egg while stopping. We know that at t = 0.140 s after touching the ground, the velocity of the egg is zero. We also know that the velocity of the egg before hiiting the ground is 11.1 m/s, then, v0 = 11.1 m/s:

v = v0 + a · t

0 = 11.1 m/s + a · 0.140 s

-11.1 m/s / 0.140 s = a

a = -79.3 m/s²

While stopping, the magnitude of the deceleration of the egg is 79.3 m/s²

The average acceleration while the egg was falling is 9.8 m/s² and the average deceleration while the egg was stopping is -79.3 m/s².

The subject of the question is Physics, namely kinematics in the field of Mechanics. The main concepts involved are acceleration, velocity, and time. Acceleration is the rate of change of velocity, and it can be calculated as the change in velocity divided by the change in time.

Acceleration while falling: Since the egg is falling, it's subject to gravity. On Earth the acceleration due to gravity is approximately 9.8 m/s². However if this value wasn't given, it could have been calculated using the formula: final velocity divided by the time taken. Hence, 11.1 / 1.13 = 9.8 m/s² (rounded).

Deceleration while stopping: This is calculated in the same way as acceleration, the only difference being that its direction is opposite to the direction of motion hence it's often referred as deceleration. The initial velocity, in this case the speed it was falling just before hitting the ground, will be used as the initial velocity while calculating acceleration after hitting the ground. Therefore (0 - 11.1) / 0.140 = -79.3 m/s², the negative sign indicates deceleration.

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Answer:

C) they all have the same angular speed  

D) they all have the same angular acceleration

Explanation:

Wrong --> they all have the same tangential speed. The points close to the axis will have less speed than the points away from the axis.

Wrong --> they all have the same tangential acceleration. Similarly, the points close to the axis will have smaller acceleration than the points away from the axis.

Correct --> they all have the same angular speed. Angular speed is the same for all the particles in the rotating object.

Correct --> they all have the same angular acceleration. Angular acceleration is the same for all the particles in the rotating object.

This all comes from the following relations:

v = ωR

a = αR

where ω is the angular velocity and α is the angular acceleration.

As can be seen from above, tangential velocity and acceleration depends on the distance from the axis, whereas the angular velocity and acceleration is the same for all the points on the rotating body.

The true statements about all points in the object rotating about a fixed point are;

C) they all have the same angular speed

D) they all have the same angular acceleration

For a circular motion about a given point, the angular speed is same for all points on the circular path and it is calculated as;

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

Where;

Thus, angular speed is independent of the position of an object rotating about a fixed point.

The angular acceleration is given as;

[tex]\alpha = \frac{\omega}{T}[/tex]

Tangential speed and acceleration depends on the position of each object along the circular path.

Thus, we can conclude that the true statements about all points in the object rotating about a fixed point are, they all have the same angular speed  and they all have the same angular acceleration.

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Answer:

one forth.

Explanation:

car is moving at the speed of v

car stops, final speed = s

distance to stop the car = ?

using equation of motion

u_i² = u_f² + 2 a s

0² = v² - 2 a s

-ve sign is used because the car is decelerating.

[tex]s = \dfrac{v^2}{2a}[/tex]

now, if the velocity of the car is 2v distance to stop

[tex]s' = \dfrac{(2v)^2}{2a}[/tex]

[tex]s' = 4\dfrac{v^2}{2a}[/tex]

[tex]s' = 4 s[/tex]

[tex]s = \dfrac{s'}{4}[/tex]

now, the distance is one forth.

so, car with speed v has to cover one forth of the distance cover by car with speed 2 v.

Complete Question:

A uniform rod of mass 1.90 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass m1 = 5.40 kgis  attached to one end and a second mass m2 = 2.50 kg is attached to the other end of the rod. Treat the two masses as point particles.

(a) What is the moment of inertia of the system?

(b) If the rod rotates with an angular speed of 2.70 rad/s, how much kinetic energy does the system have?

(c) Now consider the rod to be of negligible mass. What is the moment of inertia of the rod and masses combined?

(d) If the rod is of negligible mass, what is the kinetic energy when the angular speed is 2.70 rad/s?

Answer:

a) 8.53 kg*m² b) 31.1 J c) 7.9 kg*m² d) 28.8 J

Explanation:

a) If we treat to the two masses as point particles, the rotational inertia of each mass will be the product of the mass times the square of the distance to the axis of rotation, which is exactly the half of the length of the rod.

As the mass has not negligible mass, we need to add the rotational inertia of the rod regarding an axis passing through its centre, and perpendicular to its length.

The total rotational inertia will be as follows:

I = M*L²/12 + m₁*r₁² + m₂*r₂²

⇒ I =( 1.9kg*(2.00)²m²/12) + 5.40 kg*(1.00)²m² + 2.50 kg*(1.00)m²

⇒ I =  8.53 kg*m²

b)  The rotational kinetic energy of the rigid body composed by the rod and  the point masses m₁ and m₂, can be expressed as follows:

Krot = 1/2*I*ω²

if ω= 2.70 rad/sec, and I = 8.53 kg*m², we can calculate Krot as follows:

Krot = 1/2*(8.53 kg*m²)*(2.70)²(rad/sec)²

⇒ Krot = 31.1 J

c) If the mass of the rod is negligible, we can remove its influence of the rotational inertia, as follows:

I = m₁*r₁² + m₂*r₂² = 5.40 kg*(1.00)²m² + 2.50 kg*(1.00)m²

I = 7.90 kg*m²

d) The new rotational kinetic energy will be as follows:

Krot = 1/2*I*ω² = 1/2*(7.9 kg*m²)*(2.70)²(rad/sec)²

Krot= 28.8 J

a. The moment of inertia of this system is equal to [tex]8.53\;kgm^2[/tex].

b. The kinetic energy of this system, if the rod rotates with an angular speed of 2.70 rad/s is 31.13 Joules.

c. The moment of inertia of the rod and masses combined, if the rod is considered to be of negligible mass is [tex]7.9\;kgm^2[/tex].

d. The kinetic energy of this system, if the rod rotates with an angular speed of 2.70 rad/s and the rod is considered to be of negligible mass is 28.80 Joules.

Given the following data:

a. To determine the moment of inertia of this system:

Treating the two masses as point particles, the total moment of inertia of this system is given by the formula:

[tex]I = \frac{ML^2}{12} + m_1 r_1^2 + m_2 r_2^2\\\\I = \frac{1.9 (2)^2}{12} + (5.4 \times 1^2) + (2.5 \times 1^2)\\\\I=\frac{7.6}{12} + 5.4+2.5\\\\I=0.6333+ 7.9\\\\I=8.53\; kgm^2[/tex]

b. To determine the kinetic energy of this system, if the rod rotates with an angular speed of 2.70 rad/s:

Mathematically, the rotational kinetic energy of a system is given by the formula:

[tex]K.E_R = \frac{1}{2} I\omega^2\\\\K.E_R = \frac{1}{2} \times 8.53 \times 2.7^2\\\\K.E_R = 4.27\times 7.29\\\\K.E_R = 31.13\;Joules[/tex]

c. To determine the moment of inertia of the rod and masses combined, if the rod is considered to be of negligible mass:

[tex]I =m_1 r_1^2 + m_2 r_2^2\\\\I = (5.4 \times 1^2) + (2.5 \times 1^2)\\\\I= 5.4+2.5\\\\I=7.9\; kgm^2[/tex]

d. To determine the kinetic energy of this system, if the rod rotates with an angular speed of 2.70 rad/s and the rod is considered to be of negligible mass:

[tex]K.E_R = \frac{1}{2} I\omega^2\\\\K.E_R = \frac{1}{2} \times 7.9 \times 2.7^2\\\\K.E_R = 3.95\times 7.29\\\\K.E_R = 28.80\;Joules[/tex]

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Complete Question:

A uniform rod of mass 1.90 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass m1 = 5.40 kgis  attached to one end and a second mass m2 = 2.50 kg is attached to the other end of the rod. Treat the two masses as point particles.

(a) What is the moment of inertia of the system?

(b) If the rod rotates with an angular speed of 2.70 rad/s, how much kinetic energy does the system have?

(c) Now consider the rod to be of negligible mass. What is the moment of inertia of the rod and masses combined?

(d) If the rod is of negligible mass, what is the kinetic energy when the angular speed is 2.70 rad/s?

According to the question,

The angle of incidence will be:

→ [tex]tan \Theta_1= \frac{l_1}{h_1}[/tex]

             [tex]= \frac{2.1}{1.3}[/tex]

             [tex]= 2.076[/tex]

then,

        [tex]\Theta_1 = 64.3^{\circ}[/tex]        

From air into water, we get

→       [tex]n_{air} Sin \Theta_1 = n_{water} Sin \Theta_2[/tex]

  [tex](1.00) Sin 64.3^{\circ} = (1.3) Sin \Theta_2[/tex]

                     [tex]\Theta_2 = 42.6^{\circ}[/tex]

hence,

The horizontal distance,

→ [tex]l_2 = l_1+h_2 tan \Theta[/tex]

By substituting the values, we get

      [tex]= 2.1+(2.1)tan 42.6[/tex]

      [tex]= 4.6 \ m[/tex]

Thus the above approach is right.

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Answer:

1902.75 kg

Explanation:

From Law of conservation of momentum,

m₁u₁ + m₂u₂ = V (m₁ + m₂).................... Equation 1

make m₂ the subject of the equation,

m₂ = (m₁V - m₁u₁)/(u₂-V)..................... Equation 2

Where m₁ = mass of the truck, m₂ = mass of the car, u₁ initial velocity of the truck, u₂ = initial velocity of the car V = common velocity

Given: m₁ = 2537 kg, u₁ = 14, V= 8 m/s, u₂ = 0 m/s ( as the car was at rest waiting at a traffic light)

Substituting into equation 2.

m₂ =[2537(8) - 2537(14)]/(0-8)

m₂ = (20296-35518)/-8

m₂ = -15222/-8

m₂ = 1902.75 kg.

Thus the mass of the car = 1902.75 kg

Answer:

Explanation:

Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. There are two types of electric charge: positive and negative (commonly carried by protons and electrons respectively). Like charges repel each other and unlike charges attract each other. An object with an absence of net charge is referred to as neutral. Early knowledge of how charged substances interact is now called classical electrodynamics, and is still accurate for problems that do not require consideration of quantum effects.

Electric charge is a conserved property; the net charge of an isolated system, the amount of positive charge minus the amount of negative charge, cannot change. Electric charge is carried by subatomic particles. In ordinary matter, negative charge is carried by electrons, and positive charge is carried by the protons in the nuclei of atoms. If there are more electrons than protons in a piece of matter, it will have a negative charge, if there are fewer it will have a positive charge, and if there are equal numbers it will be neutral. Charge is quantized; it comes in integer multiples of individual small units called the elementary charge, e, about 1.602×10−19 coulombs,[1] which is the smallest charge which can exist freely (particles called quarks have smaller charges, multiples of

e, but they are only found in combination, and always combine to form particles with integer charge). The proton has a charge of +e, and the electron has a charge of −e.

An electric charge has an electric field, and if the charge is moving it also generates a magnetic field. The combination of the electric and magnetic field is called the electromagnetic field, and its interaction with charges is the source of the electromagnetic force, which is one of the four fundamental forces in physics. The study of photon-mediated interactions among charged particles is called quantum electrodynamics.

Answer:

a) If the mass get double then the potential of the spring gets four times.

b)  P'=3.848 J

Explanation:

Given that

P= 0.962 J

m = 3.5 kg

m'= 7 kg

Lets take extension in the spring is x when the mass 3.5 kg is attached to the spring.

m g = K x

K=Spring constant

[tex]x=\dfrac{mg}{K}[/tex]

We know that potential energy given as

[tex]P=\dfrac{1}{2}Kx^2[/tex]

[tex]P=\dfrac{1}{2}K\times \dfrac{m^2g^2}{K^2}[/tex]

[tex]P=\dfrac{1}{2}\times \dfrac{m^2g^2}{K}[/tex]

If the mass get double then the potential of the spring gets four times.

[tex]P'=\dfrac{1}{2}\times \dfrac{(2m)^2g^2}{K}[/tex]

[tex]P'=4\times \dfrac{1}{2}\times \dfrac{m^2g^2}{K}[/tex]

P'= 4 P

When mass ,m' = 7 kg

Then potential will be

[tex]0.962=\dfrac{1}{2}\times \dfrac{3.5^2\times g^2}{K} [/tex] -----1

[tex]P'=\dfrac{1}{2}\times \dfrac{7^2\times g^2}{K}[/tex]     -------2

From equation 1 and 2

[tex]\dfrac{0.962}{P'}=\dfrac{\dfrac{1}{2}\times \dfrac{3.5^2\times g^2}{K} }{\dfrac{1}{2}\times \dfrac{7^2\times g^2}{K} }[/tex]

P'= 4 x 0.962 J

P'=3.848 J

Answer:

[tex]k=611.517\ N.m^{-1}[/tex]

[tex]U_7=3.8477\ J[/tex]

Explanation:

Given:

Now the force on the mass due to gravity:

[tex]F=m.g[/tex]

[tex]F=3.5 \times 9.8[/tex]

[tex]F=34.3\ N[/tex]

This force pulls the spring down, so:

[tex]F=k.\delta x[/tex]

[tex]34.3=k\times \delta x[/tex] ....................(1)

For the spring potential:

[tex]U=\frac{1}{2} k.\delta x^2[/tex]

[tex]0.962=0.5\times k\times \delta x^2[/tex]

[tex]1.924=k\times \delta x^2[/tex] .........................(2)

Using eq. (1) & (2)

[tex]\frac{1.924}{x^2} =\frac{34.3}{x}[/tex]

[tex]x=0.05609\ m[/tex]

a.

Now the spring factor:

using eq. (1)

[tex]34.3=k\times \delta 0.05609[/tex]

[tex]k=611.517\ N.m^{-1}[/tex]

b.

when mass attached is 7 kg.

The spring potential energy:

[tex]U_7=\frac{1}{2} \times k.\delta x'^2[/tex] ............(3)

Now the force on the mass due to gravity:

[tex]F=m.g[/tex]

[tex]F=7\times 9.8[/tex]

[tex]F=68.6\ N[/tex]

This force pulls the spring down, so:

[tex]F=k.\delta x[/tex]

[tex]68.6=611.517\times \delta x[/tex]

[tex]x=0.11218\ m[/tex]

Using eq. (3)

[tex]U_7=\frac{1}{2}\times 611.517\times 0.11218^2[/tex]

[tex]U_7=3.8478\ J[/tex]

Answer:

[tex]F=1.38*10^{-9}N[/tex]

Explanation:

According to Coulomb's law, the magnitude of the electric force between two point charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

[tex]F=\frac{kq_1q_2}{d^2}[/tex]

Here k is the Coulomb constant. In this case, we have [tex]q_1=-e[/tex], [tex]q_2=e[/tex] and [tex]d=4.09*10^-10m[/tex]. Replacing the values:

[tex]F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-1.6*10^{-19}C)(1.6*10^{-19}C)}{(4.09*10^{-10})^2}\\F=-1.38*10^{-9}N[/tex]

The negative sign indicates that it is an attractive force. So, the magnitude of the electric force is:

[tex]F=1.38*10^{-9}N[/tex]

As a head-up, it is important to notice that a white dwarf only shines thanks to the stored energy and light, because a white dwarf doesn't have any hydrogen left to perform nuclear fusion.

Now the process:

First, the white dwarf accumulates all the extracted matter from its companion, onto its own surface. This extra matter increases the white dwarf's temperature and density.

After a while, the star reaches about 10 million K, so nuclear fusion can begin. The hydrogen that has been "stolen" from the other star and accumulated in the white dwarf's surface it's used for the fusion, dramatically increasing the star's brightness for a short time, causing what we know as a Nova.

As this fuel its quickly burnt out or blown into space, the star goes back to its natural white dwarf state. Since the white dwarf nor the companion star are destroyed in this process, it can happen countless of times during their lifespan.

Answer:

1.5092 m

Explanation:

Time taken by the wave for the round trip is 8.8 ms

Velocity of sound at 20 degrees Celsius is 343 m/s

The distance to the object will be given by the one way trip

Time taken for one way trip is

[tex]\dfrac{8.8}{2}=4.4\ ms[/tex]

Distance is given by

[tex]Distance=Speed\times time\\\Rightarrow Distance=343\times 4.4\times 10^{-3}\\\Rightarrow Distance=1.5092\ m[/tex]

The distance between the motion sensor and the object is 1.5092 m

The distance between the motion sensor and the object is 1.5092 m

The motion sensor serves as a transmitter for pulses when it sends out pulses, and as a receiver, when it listens for echoes.

From the parameters given:

The distance traveled by the signal is:

d = s × Δ t

d = 343 m/s × 0.0088s

d = 3.0184 m

Provided that the distance is the circular trip between the motion sensor and the object,

Therefore, we can conclude that the distance between the motion sensor and the object will be equal to half the distance traveled by the signal.

i.e.

= 3.0184 m/2

= 1.5092  m

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Answer:

(a) Spring constant =  1850 N/m.

(b) Energy stored in the spring = 9.25 J.

(c) Muzzle velocity of the dart =  43.01 m/s.

Explanation:

(a) Spring constant

From hook's law,

F = ke ...................... Equation 1

Where F = force on the gun, k = spring constant of the gun, e = extension of the gun's spring.

Making k  subject of the equation,

k = F/e ............... Equation 2

Given: F = 185 N, e = 10 cm = 0.1 m.

Substitute into equation 2

k = 185/0.1

k = 185/0.1

k = 1850 N/m.

Hence the spring constant = 1850 N/m.

(b) Energy stored in the spring,

E = 1/2ke²............... Equation 3

Where E = Energy stored in the spring, k = spring constant, e = extension.

Given: k = 1850 N/m, e = 0.1 m.

Substitute into equation 3

E = 1/2(1850)(0.1)²

E = 925(0.01)

E = 9.25 J.

(c) Muzzle velocity of the dart.

Kinetic energy of the dart = 1/2mv²

Note: The kinetic energy of the dart is equal to the energy stored in the sprig.

E = 1/2mv²

Where m = mass of the dart, v = velocity of the dart.

Making v the subject of the equation,

v = √(2E/m)............... Equation 4

Given: E = 9.25 J, m = 10 g = 0.01 kg

Substituting these values into equation 4

v = √(2×9.25/0.01)

v = √(18.5/0.01)

v = √1850

v = 43.01 m/s.