A small insect is placed 3.75 cm from a +4.00 −cm -focal-length lens. Calculate the angular magnification.

Explanation:

When a body or object falls, basically two forces act on it:  

1. The force of air friction, also called "drag force" [tex]D[/tex]:  

[tex]D={C}_{d}\frac{\rho V^{2} }{2}A[/tex]  (1)

Where:  

[tex]C_ {d}[/tex] is the drag coefficient  

[tex]\rho[/tex] is the density  of the fluid (air for example)

[tex]V[/tex] is the velocity  

[tex]A[/tex] is the transversal area of the object

So, this force is proportional to the transversal area of ​​the falling element and to the square of the velocity.  

2. Its weight due to the gravity force [tex]W[/tex]:  

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

(2)

Where:  

[tex]m[/tex] is the mass of the object

[tex]g[/tex] is the acceleration due gravity  

So, at the moment when the drag force equals the gravity force, the object will have its terminal velocity:

[tex]D=W[/tex] (3)

[tex]{C}_{d}\frac{\rho V^{2} }{2}A=m.g[/tex]  (4)

[tex]V=\sqrt{\frac{2m.g}{\rho A{C}_{d}}}[/tex]  (5) This is the terminal velocity

As we can see, there is no "falling time" in this equation.

Therefore, the terminal velocity is not dependent on the falling time.

Answer:

The distance the block will slide before it stops is x= 3.4 m .

Explanation:

m1= 0,02 kg

V1= 400 m/s

m2= 2 kg

V2= ?

g= 9.8 m/s²

μ= 0.24

N = m2 * g = 19.6 N

Fr= μ * N

Fr= 4.704 N

due to the conservation of the amount of movement:

m1*V1 = m2*V2

V2= 4 m/s

Fr = m2*a

a= Fr/m2

a= -2.352 m/s²

matching momentum and amount of movement

Fr*t=m2*V2

t= 1.7 sec

x= V2*t - (a*t²)/2

x= 3.4 m

Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ =  400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction,  μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

[tex]v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f[/tex]

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

Finally, determine the distance the block will slide before it stops

From kinematic, distance is the product of speed and time

[tex]S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2[/tex]

Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

S = 6.6693 - 3.3350

S = 3.3343 m

Thus, the distance the block will slide before it stops is 3.3343 m

Answer:

23.5 J/Kg °C

Explanation:

The amount of heat required to raise the temperature of substance of mass 1 kg by 1 degree Celcius.

c = Q / m ΔT

Here, Q / m = 282 J/kg, ΔT = 12

So, the specific heat = 282 / 12 = 23.5 J/Kg °C

The resulting velocity of the tennis ball is calculated by first determining the impulse, which is the integral of the force over the contact time. The impulse is then equal to the change in momentum, allowing us to solve for the velocity. The calculated velocity of the ball after impact is 74.67 m/s.

To calculate the resulting velocity of the ball after being hit by the racket, we first need to determine the impulse imparted to the ball. The force applied by the racket is variable and given by F=at-bt², where a = 1290 N/ms, b = 330 N/ms², and t is the time in milliseconds. To calculate the impulse (J), we integrate the force over the contact time, from 0 to 2.80 ms.

Impulse, J, is the integral of F with respect to t, which gives us J = (1/2)at² - (1/3)bt³ evaluated from 0 to 2.80 ms. Plugging in the values:

J = (1/2)(1290 N/ms)(2.80 ms)² - (1/3)(330 N/ms²)(2.80 ms)³ = 1290(3.92) - 330(2.744) N = 5056.8 - 905.52 N = 4151.28 Nms

The impulse is equal to the change in momentum of the tennis ball. Considering the ball's mass m = 55.6 g = 0.0556 kg, and initial velocity, u = 0 m/s (since it's hit from rest), we apply the impulse-momentum theorem:

J = F - mv, therefore, v = J/m.

Substituting the values we have: v = 4151.28 Nms / 0.0556 kg = 74667.7 m/s = 74.67 m/s.

So, the resulting velocity of the tennis ball is 74.67 m/s after the racket's impact.

The resulting velocity of the ball after being hit by the racket is approximately [tex]\( 49466.325 \, \text{m/s} \)[/tex].

To find the resulting velocity of the ball after being hit by the racket, we can use the impulse-momentum theorem, which states that the change in momentum of an object is equal to the impulse applied to it:

[tex]\[ J = \Delta p \][/tex]

The impulse J is equal to the integral of the force F with respect to time t over the duration of the contact:

[tex]\[ J = \int_{0}^{2.80} F \, dt \][/tex]

Given that [tex]\( F = at - bt^2 \)[/tex], we can integrate [tex]\( F \)[/tex] with respect to [tex]\( t \)[/tex] to find [tex]\( J \)[/tex]:

[tex]\[ J = \int_{0}^{2.80} (at - bt^2) \, dt \][/tex]

[tex]\[ J = \left[ \frac{1}{2} at^2 - \frac{1}{3} bt^3 \right]_{0}^{2.80} \][/tex]

[tex]\[ J = \left( \frac{1}{2} a(2.80)^2 - \frac{1}{3} b(2.80)^3 \right) - \left( \frac{1}{2} a(0)^2 - \frac{1}{3} b(0)^3 \right) \][/tex]

[tex]\[ J = \left( \frac{1}{2} \cdot 1290 \cdot (2.80)^2 - \frac{1}{3} \cdot 330 \cdot (2.80)^3 \right) - 0 \][/tex]

[tex]\[ J = \left( \frac{1}{2} \cdot 1290 \cdot 7.84 - \frac{1}{3} \cdot 330 \cdot 21.952 \right) \][/tex]

[tex]\[ J = \left( 5056.64 - 2308.656 \right) \][/tex]

[tex]\[ J = 2747.984 \, \text{N} \cdot \text{ms} \][/tex]

Now, we know that impulse [tex]\( J \)[/tex] is also equal to the change in momentum [tex]\( \Delta p \)[/tex]  of the ball:

[tex]\[ J = \Delta p = mv - mu \][/tex]

Where:

- [tex]\( m \)[/tex] is the mass of the ball,

- v is the final velocity of the ball,

- u is the initial velocity of the ball (which we assume to be zero as the ball is initially at rest).

So, we can rearrange the equation to solve for v:

[tex]\[ v = \frac{J}{m} \][/tex]

Now, let's plug in the values to find v:

[tex]\[ v = \frac{2747.984 \, \text{N} \cdot \text{ms}}{0.0556 \, \text{kg}} \][/tex]

[tex]\[ v \approx 49466.325 \, \text{m/s} \][/tex]

So, the resulting velocity of the ball after being hit by the racket is approximately [tex]\( 49466.325 \, \text{m/s} \)[/tex].

Answer:

55536.6 J

Explanation:

Given:

Mass of the ice = 54g

Initial temperature = 0°C

Final Temperature = 100°C

Mass of the steam = 6.6g

Now the energy required for the transformation of the ice to vapor will involve the heat requirement in the following stages as:  

1) The energy required to melt ice = mass of ice × heat of fusion of water = 54g × 334 J/g  = 18036 J

  (because heat of fusion for water = 334 J/g)

2) The energy to heat water from 0 to 100 = mass of water × specific heat of water × change in temperature = 54g × 4.186 J/g°C × 100 °C = 22604.4 J  

lastly,

3) the energy required to vaporize 6.6g of water = mass of water × heat of vaporization of water = 6.6 × 2257 J/g = 14896.2 J  

Thus,

the total energy required to transform the ice cube to accomplish the transformation = 18036 + 22604.4 + 14896.2 = 55536.6 J.

Answer:

Speed of the airplane 10.0 s later = 12.2 m/s

Explanation:

Mass of Boeing 777 aircraft = 300,000 kg

Braking force = 445,000 N

Deceleration

            [tex]a=\frac{445000}{300000}=1.48m/s^2[/tex]

Initial velocity, u = 27 m/s

Time , t = 10 s

We have equation of motion, v =u +at

            v = 27 + (-1.48) x 10 = 27 - 14.8 = 12.2 m/s

Speed of the airplane 10.0 s later = 12.2 m/s

After applying the deceleration due to the braking force for 10 seconds, the speed of the Boeing 777 airplane reduces to 12.17 m/s.

Calculating the Final Speed of a Boeing 777 After Braking

To determine the speed of a Boeing 777 aircraft after a 10-second interval of braking, we will use the relationship between force, mass, initial velocity, and acceleration provided by Newton’s second law of motion. The braking force applied on the aircraft is 445,000 N acting opposite to the direction of motion. We can calculate the deceleration using the formula:

F = ma

Where F is the force, m is the mass, and a is the acceleration (deceleration in this case since the force is applied opposite to the direction of motion). The mass (m) of the airplane is 300,000 kg, so we can solve for 'a' as follows:

a = F / m = 445,000 N / 300,000 kg = 1.483 m/s² (deceleration)

Next, we use the kinematic equation to find the final velocity (vf) after 10 seconds:

vf = vi + at

Where vi is the initial velocity, and t is the time. We know that the initial velocity (vi) is 27.0 m/s and the time (t) is 10.0 s.

Plugging the numbers in:

vf = 27.0 m/s - (1.483 m/s² × 10.0 s) = 27.0 m/s - 14.83 m/s = 12.17 m/s

Therefore, the speed of the Boeing 777 airplane 10 seconds later is 12.17 m/s.

Explanation:

The Coulomb's law states that the force acting on two charges is directly proportional to the product of charges and inversely proportional to the square of distance between them . Mathematically, it is given by

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

Where

k is the electrostatic constant

q₁ and q₂ are charges

r is the distance between them.

The SI unit of electric force is Newton. It can be attractive or repulsive. The attraction or repulsion depend on charges. If both charges are positive, the force is repulsive and if both are opposite charges, the force is attractive.

Answer:

EMF = 2.4 V

Explanation:

As we know by Lenz law that induced EMF is given by rate of change in magnetic flux in the coil

As we know that magnetic flux is given by

[tex]\phi = NBA[/tex]

here we know that

[tex]N = 300[/tex]

[tex]Area = 0.20 \times 0.20 = 0.04 m^2[/tex]

now we for induced EMF we have

[tex]EMF = NA\frac{dB}{dt}[/tex]

here we have

[tex]EMF = (300)(0.04)(\frac{0.90 - 0.50}{2})[/tex]

[tex]EMF = 2.4 V[/tex]

The magnitude of the induced emf in a coil, while the magnetic field is changing, can be calculated using Faraday's law. The law states that the induced emf is equal to the negative of the rate of change of the magnetic flux linked through the circuit. We use this formula to calculate the change in magnetic field, take into accounts of coil's area and number of turns then divide by the change in time.

The magnitude of the induced emf in a coil, while the magnetic field is changing, can be calculated using Faraday's law of electromagnetic induction, which states that the induced emf is equal to the negative rate of change of the magnetic fluxlinked through the circuit.

In this case, the magnetic flux, Φ, through the coil can be calculated using the formula Φ = BAN, where B is the magnetic field, A is the area of each turn (which can be calculated using the given side length) and N is the number of turns (300 in this case).

To calculate the rate of change of the magnetic flux use the formula ΔΦ/Δt, where ΔΦ is the change in magnetic flux and Δt is the change in time. In this case, the magnetic field changes from 0.50 T to 0.90 T in 2.0 s, so the change in magnetic flux is (B_final - B_initial)*A*N, and Δt is 2.0 s.

The magnitude of the induced emf (which is also the negative rate of change of the magnetic flux) is given by Faraday's law, |ε| = |ΔΦ/Δt|.?

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Answer:

[tex]U = 1.83 \times 10^5 J[/tex]

Explanation:

Total internal energy of the gas is given by the formula

[tex]U = \frac{f}{2}nRT[/tex]

here as we know that neon gas is monoatomic gas

so we will have

[tex]f = 3[/tex]

n = 50 moles

[tex]T = 20 ^0C = 20 + 273 = 293 K[/tex]

now from above equation we will have

[tex]U = \frac{3}{2}(50)(8.31)(293)[/tex]

[tex]U = 1.83 \times 10^5 J[/tex]

Answer:

Part a)

[tex]T = 425.6 K[/tex]

Part b)

[tex]KE_{avg} = 8.81\times 10^{-21} J[/tex]

Part c)

in order to find the average speed we need to know about the the gas molar mass or we need to know which gas it is.

Explanation:

Part a)

As per ideal gas equation we know that

[tex]PV = nRT[/tex]

here we know that

[tex]P = 1.60 \times 10^6 Pa[/tex]

n = 3.80 moles

[tex]V = 8.40 L = 8.40 \times 10^{-3} m^3[/tex]

now from above equation we have

[tex]T = \frac{PV}{nR}[/tex]

[tex]T = \frac{(1.60 \times 10^6)(8.40 \times 10^{-3})}{(8.31)(3.80)}[/tex]

[tex]T = 425.6 K[/tex]

Part b)

Average kinetic energy of the gas is given as

[tex]KE_{avg} = \frac{3}{2}KT[/tex]

here we know that

[tex]K = 1.38 \times 10^{-23}[/tex]

T = 425.6 K

now we have

[tex]KE_{avg} = \frac{3}{2}(1.38 \times 10^{-23})(425.6)[/tex]

[tex]KE_{avg} = 8.81\times 10^{-21} J[/tex]

Part c)

in order to find the average speed we need to know about the the gas molar mass or we need to know which gas it is.

Answer:

The resistivity of the material used to make the rod is ρ= 7.5 * 10⁻⁷ Ω.m

Explanation:

R= 0.2 Ω

L= 0.8 m

S= 1.5mm*2mm= 3 mm² = 3 * 10⁻⁶ m²

ρ = (R*S)/L

ρ= 7.5 * 10⁻⁷ Ω.m

Answer:

23.99N

Explanation:

Given:

Pressure created by the heart = 120mm of hg

converting the pressure into the standard unit of N/m²

1mm of hg = [tex]\frac{1}{760}atm=\frac{1}{760}\times 1.013\times10^5N/m^2[/tex]

now, 120 mm of hg in N/m² will be

120mm of hg = [tex]\frac{12}{760}\times 1.013\times10^5N/m^2[/tex]

also

given effective area = 15.0 cm² = 15 × 10⁻⁴m²

Now,

Force = Pressure × Area

thus,

Force exerted will be =  [tex]\frac{120}{760}\times 1.013\times10^5N/m^2[/tex] ×  15 × 10⁻⁴m²

or

Force exerted will be = 23.99N

The force that it exerts is about 24.0 N

[tex]\texttt{ }[/tex]

Let's recall Hydrostatic Pressure formula as follows:

[tex]\boxed{ P = \rho g h}[/tex]

where:

P = hydrosatic pressure ( Pa )

ρ = density of  fluid ( kg/m³ )

g = gravitational acceleration ( m/s² )

h = height of a column of liquid ( m )

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

blood pressure = P = 120 mmHg = 0.12 mHg

effective area = A = 15.0 cm² = 15.0 × 10⁻⁴ m²

density of mercury = ρ = 13600 kg/m³

gravitational acceleration = g = 9.8 m/s²

Asked:

force = ?

Solution:

We will use this following formula to solve this problem:

[tex]P = F \div A[/tex]

[tex]F = P A[/tex]

[tex]F = \rho g h A[/tex]

[tex]F = 13600 \times 9.8 \times 0.12 \times ( 15.0 \times 10^{-4} )[/tex]

[tex]F \approx 24.0 \texttt{ N}[/tex]

[tex]\texttt{ }[/tex]

The force that it exerts is about 24.0 N

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Pressure

Answer:

Part a)

[tex]\lambda = 300 m[/tex]

Part b)

[tex]E = 2.7 N/C[/tex]

Part c)

[tex]I = 9.68 \times 10^{-3} W/m^2[/tex]

[tex]P = 3.22 \times 10^{-11} N/m^2[/tex]

Explanation:

Part a)

As we know that frequency = 1 MHz

speed of electromagnetic wave is same as speed of light

So the wavelength is given as

[tex]\lambda = \frac{c}{f}[/tex]

[tex]\lambda = \frac{3\times 10^8}{1\times 10^6}[/tex]

[tex]\lambda = 300 m[/tex]

Part b)

As we know the relation between electric field and magnetic field

[tex]E = Bc[/tex]

[tex]E = (9 \times 10^{-9})(3\times 10^8)[/tex]

[tex]E = 2.7 N/C[/tex]

Part c)

Intensity of wave is given as

[tex]I = \frac{1}{2}\epsilon_0E^2c[/tex]

[tex]I = \frac{1}{2}(8.85 \times 10^{-12})(2.7)^2(3\times 10^8)[/tex]

[tex]I = 9.68 \times 10^{-3} W/m^2[/tex]

Pressure is defined as ratio of intensity and speed

[tex]P = \frac{I}{c} = \frac{9.68\times 10^{-3}}{3\times 10^8}[/tex]

[tex]P = 3.22 \times 10^{-11} N/m^2[/tex]

Answer:

Its potential energy decreases and itselectricpotential increases.

Explanation:

The electric potential does not depend on the charge but only on the magnitude of the electric field. In particular:

- Electric potential decreases when moving in the same direction of the electric field lines

- Electric potential increases when moving in the opposite direction to the field lines

So in this case, since the electron is moving in a direction opposite to the field, the electric potential increases.

However, the electric potential energy of a charge is given by

[tex]U=qV[/tex]

where

q is the charge

V is the electric potential

Here we said that the electric potential is increasing: however, the charge q of an electron is negative. This means that the product (qV) is increasing in magnitude but it is negative, so the potential energy of the electron is decreasing.

For an electron moving in a direction opposite to the electric field its potential energy decreases and its electric potential increases.

Electric potential is the amount of work needed to move a unit charge from a point to a specific point against an electric field.it is denoted by U.

However, the electric potential energy of a charge is given by

[tex]\rm{U=qv}[/tex]

q is denoted for the charge.

V is denoted for the electric potential.

The above relation shows that the electric potential is inversely proportional to the electric potential energy.

The electric potential is independent of the charge. but it depends on the direction in which the charge is moving. If the charge moves in the direction of the electric field electric potential decreases and vice versa.

Here the electron is moving in a direction opposite to the field, the electric potential increases while electric potential energy decreases.

To learn more about the electric potential energy refers to the link.

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Answer:L=33.93

R.E.=152.30

Explanation:

Given

mass of disk[tex]\left ( m\right )[/tex]=21kg

thickness[tex]\left ( t\right )[/tex]=0.5m

radius[tex]\left ( r\right )[/tex]=0.6m

t=0.7 sec for every complete rotation

therefore angular velocity[tex]\left ( \omega \right ) =\frac{\Delta \theta }{\Delta t}[/tex]

[tex]\left ( \omega \right )=\frac{ 2\pi }{0.7}=8.977[/tex] rad/s

Rotational angular momentum is given by

[tex]L=I\omega [/tex]

[tex]I[/tex]=[tex]\frac{mr^2}{2}[/tex]

I=3.78 [tex]kg-m^2[/tex]

[tex]L=3.78\times 8.977[/tex]

L=33.93 [tex]\hat{j}[/tex]

considering disk is rotating in z-x plane

Rotational kinetic energy=[tex]\frac{I\omega ^{2}}{2}[/tex]=152.30 J

The rotational angular momentum of the disk is 33.89 kg.m²/s directed into the page, and the rotational kinetic energy of the disk is 151.5 J.

To calculate the rotational angular momentum of the disk, we will first need to compute the moment of inertia and angular velocity. The moment of inertia is given by the formula for a solid disk, which is I = 1/2mR². Here, m is the mass of the disk (in this case, 21 kg), and R is the radius of the disk (which is 0.6 m). So, I = 0.5*21*(0.6)² = 3.78 kg.m².

Furthermore, the angular velocity (ω) can be calculated using the formula 2π/T, where T is the period, which is the time it takes for one complete rotation (0.7 s in this situation). Therefore, ω = 2π/0.7 = 8.96 rad/s.

Finally, the rotational angular momentum (L), is given by the product of the moment of inertia and angular velocity, L = I*ω. Therefore, L = 3.78 kg.m² * 8.96 rad/s = 33.89 kg.m²/s.

The direction of the rotational angular momentum is along the axis of rotation, which depends on the right-hand rule. Since it is rotating clockwise when viewed from above, the vector is directed into the page (negative z-direction).

To compute the rotational kinetic energy (KE), we use the formula KE = 1/2 I ω². Substituting the values, we get KE = 0.5*3.78 kg.m²*(8.96 rad/s)² = 151.5 J.

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Answer:

100 Degrees is boiling point.

Explanation:

Answer:

Loss in kinetic energy is 86.4 J.

Explanation:

It is given that,

Mass of first car, m₁ = 3 kg

Velocity of first car, v₁ = 9 m/s

Mass of second car, m₂ = 2 kg

Velocity of second car, v₂ = -3 m/s (opposite direction)

If the cars are locked together after the collision with a speed of 4.20 m/s, V = 4.2 m/s

It is a case of inelastic collision. Some of the kinetic energy will be lost in the form of heat energy, sound energy etc.

Initial kinetic energy, [tex]K_i=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2[/tex]

[tex]K_i=\dfrac{1}{2}\times 3\ kg\times (9\ m/s)^2+\dfrac{1}{2}\times 2\ kg\times (-3\ m/s)^2[/tex]

[tex]K_i=130.5\ J[/tex]

Final kinetic energy, [tex]K_f=\dfrac{1}{2}(m_1+m_2)V^2[/tex]

[tex]K_f=\dfrac{1}{2}\times (3\ kg+2\ kg)\times (4.2\ m/s)^2[/tex]

[tex]K_f=44.1\ J[/tex]

Kinetic energy lost, [tex]\Delta K=K_f-K_i[/tex]

[tex]\Delta K=44.1-130.5[/tex]

[tex]\Delta K=-86.4\ J[/tex]

So, 86.4 J of kinetic energy is lost. Hence, this is the required solution.

The loss in kinetic energy during a collision can be calculated by computing the initial and final kinetic energies of the objects and then finding the difference.

In order to compute the loss in kinetic energy, we first need to compute the initial and final kinetic energies of both toy cars. Initially, the kinetic energy of the 3 kg toy car is (1/2) * 3 kg * (9 m/s)^2 and that of the 2 kg toy car is (1/2) * 2 kg * (3 m/s)^2. Summing these, we obtain the initial kinetic energy (call this Ki). Following the collision, the toy cars stick together and move with a common velocity of 4.20 m/s. Therefore, the final kinetic energy (call this Kf) of them together is (1/2) *  (3 kg + 2 kg) * (4.20 m/s)^2. The loss in kinetic energy is then Ki - Kf.

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Answer:

The statue compress the stand by 3.68 x 10⁻⁵ m.

Explanation:

Change in length

            [tex]\Delta L=\frac{PL}{AY}[/tex]

Load, P = 3500 x 9.81 = 34335 N

Young's modulus, Y = 2.3 x 10¹⁰ N/m²

Area, A = 7.3 x 10⁻² m²

Length, L = 1.8 m

Substituting

         [tex]\Delta L=\frac{PL}{AE}=\frac{34335\times 1.8}{7.3\times 10^{-2}\times 2.3\times 10^{10}}=3.68\times 10^{-5}m[/tex]

The statue compress the stand by 3.68 x 10⁻⁵ m.

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

[tex]Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}[/tex]

Where,

[tex]C_d[/tex] is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = [tex]\frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2[/tex]

A₂ = Area at the throat

A₂ = [tex]\frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2[/tex]

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

[tex]Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}[/tex]

or

[tex]Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}[/tex]

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Answer:

it is at height of y = 0.533 m from ground

Explanation:

As per law of reflection we know that angle of incidence = angle of reflection

so here we have

[tex]tan\theta_i = tan\theta_r[/tex]

here we know that

[tex]tan\theta_i = \frac{y}{d}[/tex]

also we have

[tex]tan\theta_r = \frac{H - y}{D}[/tex]

now we have

[tex]\frac{H - y}{D} = \frac{y}{d}[/tex]

here we have

[tex]\frac{1.6 - y}{5.8} = \frac{y}{2.9}[/tex]

[tex]3y = 1.6[/tex]

[tex]y = 0.533 m[/tex]

To see the bottom of the object in the mirror, the bottom of the mirror should be at a height of half the height of the person.

To see the bottom of the object in the mirror, the person should be able to see an image of the bottom of the object reflected in the mirror.

Using the law of reflection, we can determine the height h at which the bottom of the mirror should be. The angle of incidence for the light from the bottom of the object is equal to the angle of reflection, so the height of the mirror should be equal to the height of the person's eyes, H, plus the height of the bottom of the object, h. We can calculate h using similar triangles:

h/H = d/D

where d is the distance from the object to the mirror and D is the distance from the person to the mirror. Substituting the given values, we have:

h = (d/D) * H

Substituting d = D/2, we get:

h = (D/2D) * H = H/2

Therefore, the bottom of the mirror should be at a height of half the height of the person.

Answer:

55000000 years

Explanation:

Rate of North American Plate moving (Velocity) = 20 mm/yr

Degrees New York has to move = 10° west

New York's longitude = 110 km/°

Distance of New York = 110×10 = 1100 km

= 1100×10⁶ mm

[tex]\text {Time taken by the continental plate}=\frac {\text {Distance of New York}}{\text {Rate of North American Plate moving (Velocity)}}\\\Rightarrow \text {Time taken by the continental plate}=\frac{1100\times 10^6}{20}\\\Rightarrow \text {Time taken by the continental plate}=55\times 10^6\ years[/tex]

∴ Time taken by New York to move 10° west is 55000000 years

Answer:

(a) 9.36 kHz

(b) 3.12 kHz

Explanation:

(a)

V = speed of sound

[tex]v[/tex] = speed of airplane = (0.5) V

f = actual frequency of sound emitted by airplane = 4.68 kHz = 4680 Hz

f' = Frequency heard by the stationary listener

Using Doppler's effect

[tex]f' = \frac{Vf}{V-v}[/tex]

[tex]f' = \frac{V(4680)}{V-(0.5)V)}[/tex]

f' = 9360 Hz

f' = 9.36 kHz

(b)

V = speed of sound

[tex]v[/tex] = speed of airplane = (0.5) V

f = actual frequency of sound emitted by airplane = 4.68 kHz = 4680 Hz

f' = Frequency heard by the stationary listener

Using Doppler's effect

[tex]f' = \frac{Vf}{V+v}[/tex]

[tex]f' = \frac{V(4680)}{V+(0.5)V)}[/tex]

f' = 3120 Hz

f' = 3.12 kHz

The frequencies heard by the stationary listener when the airplane is approaching and when it is moving away can be calculated using the formula for the Doppler effect. The formula differs slightly depending on whether the source of the sound (in this case, the airplane) is moving towards or away from the observer (the listener).

This question involves the Doppler Effect, which describes how the frequency of a wave changes for an observer moving relative to the source of the wave. We are given that the airplane is moving at half the speed of sound and emits a sound of frequency 4.68 kHz.

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Answer:

increase the charge

Explanation:

The force acting on charge particle when moving perpendicular to the magnetic field

F = q v B

The centripetal force is given by

F =  mv^2 / r

Comparing both, we get

r = m v / B q

That means the radius of the circular path depends on mass of the charge particle, velocity of the charge particle, magnetic field strength and charge of the particle.

To decrease the radius:

1. increase the charge

2. increase the magnetic field strength

3. decrease the speed

4. decrease the mass

The factor that will decrease the radius of the circular path is increase the charge on the particle.

The magnetic force on the charged particle is calculated as follows;

F = qvB

Where;

The force on the particle moving in circular path is given as;

F = mv²/r

qvB = mv²/r

qB = mv/r

r = mv/qB

Thus, the factor that will decrease the radius of the circular path is increase the charge on the particle.

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Answer:

(a) the constant of the spring is 27095.375 J/m2

(b) 0.052 m/s

(c) F=17341.04 N

Explanation:

Hello

The law of conservation of energy states that the total amount of energy in any isolated system (without interaction with any other system) remains unchanged over time.

if we assume that there is no friction, then, the kinetic energy of the train (due to movement) will be equal to the energy accumulated in the spring

Step 1

energy of the train (kinetic)

[tex]E_{k}=\frac{m*v^{2} }{2}\\  E_{k}=\frac{2.15*10^{5}kg*(0.71\frac{m}{s}) ^{2} }{2}\\E_{k}=54190.75 J\\ \\[/tex]

step 2

energy of the spring

[tex]E_{s}=\frac{Kx^{2} }{2}\\[/tex]

where K is the constant of the spring and x the length compressed.

[tex]E_{s}=\frac{Kx^{2} }{2}=54190.75 J\\\frac{Kx^{2} }{2}=54190.75 J\\\\k=\frac{2*54190.75 J}{x^{2}}\\k=\frac{2*54190.75 J}{(2.0 m)^{2} }\\ k=27095.375\frac{J}{m^{2} } \\\\[/tex]

(a) the constant of the spring is 27095.375 J/m2

(b)

[tex]x=0.640 m\\\\E_{s}=\frac{27095.375 (\frac{j}{m^{2} })*(0.640m)^{2}  }{2}\\ E_{s}=5549.1328 J\\[/tex]

equal to train energy

[tex]E_{k}=\frac{2.15*10^{5}kg*(v) ^{2} }{2}\\\frac{2.15*10^{5}kg*(v) ^{2} }{2}=5549.1328 J\\v^{2}=\frac{2*5549.1328 J}{2.15*10^{5}kg}\\v=0.052 \frac{m}{s} \\\\[/tex]

(b) 0.052 m/s

(c)

[tex]F= kx\\\\F=27095.375 \frac{J}{m^{2} }*(0.640 m)\\ F=17341.04 N\\F=17341.04[/tex]

(c) F=17341.04 N

I hope it helps

The force constant of the spring is 38.03 N/m. If the train compressed the spring 0.640 m, the train would be going at a speed of 0.418 m/s. The force the spring exerts when compressed 0.640 m is 24.34 N.

To solve these questions we use conservation of energy principles and the spring equation, F = k*x. The initial kinetic energy of the train is converted into potential energy in the spring at the point of maximum compression.

(a) What is the force constant of the spring?
Here we equate kinetic energy with potential energy by using the equations KE = 0.5 * m * v^2 and PE = 0.5 * k * x^2. We can solve for k to get k = m * v^2 / x^2, so it's (2.15 x 10^5 kg * (0.710 m/s)^2)/(2.00 m)^2 = 38.03 N/m.

(b) What speed would the train be going if it only compressed the spring 0.640 m?
Here we rearrange the previous equation for velocity, v = sqrt(k * x^2 / m) which we then place values into to calculate v = sqrt(38.03 N/m * (0.640 m)^2)/(2.15 x 10^5 kg) = 0.418 m/s.

(c) What force does the spring exert when compressed 0.640 m?
Here we refer back to Hooke's Law (F = k*x), which states that the force required to compress or extend a spring by some distance x is proportional to that distance. The spring constant (k) is the proportionality constant in this relationship. So, the force is F = (38.03 N/m * 0.640 m) = 24.34 N.

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Answer:

[tex]Distance_{vaccum}=5.19cm[/tex]

Explanation:

The speed of light in these mediums shall be lower than that in vacuum thus the total time light needs to cross both the media are calculated as under

Total time = Time taken through ice + Time taken through quartz

Time taken through ice = Thickness of ice / (speed of light in ice)

[tex]T_{ice}=\frac{2.20\times 10^{-2} \times \mu _{ice}}{V_{vaccum}}[/tex]

[tex]T_{quartz}=\frac{1.50\times 10^{-2} \times \mu _{quartz}}{V_{vaccum}}[/tex]

Thus in the same time the it would had covered a distance of

[tex]Distance_{vaccum}=Totaltime\times V_{vaccum}\\\\Distance_{vaccum}=10^{-2}[2.20\mu _{ice+1.50\mu _{quartz}}][/tex]

we have

[tex]\mu _{ice}=1.309\\\\\mu _{quartz}=1.542[/tex]

Applying values we have

[tex]Distance_{vaccum}=10^{-2}[2.20\times 1.309+1.50\times 1.542][/tex]

[tex]Distance_{vaccum}=5.19cm[/tex]

The light, after going through ice and quartz, would have traveled approximately 5.198 cm in vacuum. This is calculated considering indices of refraction for each material and their thicknesses.

The subject of this question is in the domain of Physics, specifically optics. The problem wants us to compare the distances light travels in different media to its travel in vacuum. The key to solve this matter is knowing the speed of light changes depending on the medium it passes through: it's slower in media like ice or quartz than in vacuum, due to these materials' indices of refraction.

The index of refraction for ice is approximately 1.31 and for crystalline quartz is approximately 1.544. The index of refraction (n) is the ratio of the speed of light in vacuum (c), to the speed of light in the medium (v), or n = c/v. To find out the distances light travels in vacuum for the time it takes to travel through the ice and quartz, we need to multiply the thickness of each material by their respective indices of refraction.

For the ice, it's 2.20 cm * 1.31 = 2.882 cm and for the quartz, it's 1.50 cm * 1.544 = 2.316 cm. So the total distance light would have traveled in vacuum in the same time it passes through both sheets is approximately 2.882 cm + 2.316 cm = 5.198 cm.

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Answer:

There is a dependency relationship between the refractive index of each substance and the radiation wavelength.

The refractive index in a given medium is inversely proportional to the wavelength of a color.

For example:

The rays of the red color have a wavelength greater than the rays of the blue color, therefore they have a lower refractive index and consequently a light scattering less than the blue.

Snell's law :

n₂/n₁ = v₁/v₂ =  λ₁ /λ₂

*n: (refractive index)

v: (speed of light propagation)

λ: (wavelength)

Answer:

E

Explanation:

F = G * m1 * m2 / r^2     Increase the distance by 3

F1 = G * m1 * m2 / (3r)^2

F1 = G * m1 * m2 / (9*r^2) What this means is the the force decreases by a factor of 9, but we are not done.

F2 = G * 3m1 * 3m2 / (9 r^2)

F2 = G * 9 m1 * m2 / (9 r^2)

In F2 the 9s cancel out and we are left with

F2 = G * m1 * m2/r^2 which is the same thing.

F2 equals F

Answer:

E. The gravitational force remains unchanged

Explanation:

Answer:

105.6857 nm

Explanation:

Wave length of laser pointer = λ = 685 nm = 685×10⁻⁹ m

Distance between screen and slit = L = 5.4 m

Width of bright band = 2x = 7 cm

⇒x = 3.5 cm = 3.5×10⁻² m

The first minimum occurs at

sin θ = λ/d

θ = λ/d                 (since the angle is very small, sin θ ≈ θ)

Width of slit

d = λL/x

⇒d = 685×10⁻⁹×5.4/3.5×10⁻²

⇒d = 1056.857×10⁻⁷ m

∴ The width of the slit is 105.6857 nm

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

                                               = 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

                                               = 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2                

                                                = 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J

Answer:

0.24 kgm²

Explanation:

[tex]L[/tex] = length of the bat = 81.3 cm = 0.813 m

[tex]m[/tex]  = mass of the bat = 0.96 kg

[tex]d[/tex]  = distance of the center of mass of bat from the axis of rotation = 55.9 cm = 0.559 m

[tex]T[/tex]  = Period of oscillation = 1.35 sec

[tex]I[/tex] = moment of inertia of the bat

Period of oscillation is given as

[tex]T = 2\pi \sqrt{\frac{I}{mgd}}[/tex]

[tex]1.35 = 2(3.14) \sqrt{\frac{I}{(0.96)(9.8)(0.559)}}[/tex]

[tex]I[/tex] = 0.24 kgm²