The efficiency of a Stirling cycle depends on the temperatures of the hot and cold isothermal parts of the cycle.If you increase the upper temperature, keeping the lower temperature the same, does the efficiency increase,decrease, or remain the same?A. increaseB. decreaseC. remain the same

Answer:

.09 mm

Explanation:

In case of diffraction by single slit with width a the width of central maxima is given below

width = 2 λD/a

where λ is wavelength of light , D is distance of screen , a is width o slit

substituting the given values

14 x 10⁻³ = [tex]\frac{2\times530\times10^{-9}\times1.2}{a}[/tex]

a = .09 mm

b) If a is greater , width of central maxima will be less wide.

Answer:

66.4767 m

Explanation:

[tex]L_0[/tex] = Original length of power line = 145 km

[tex]\alpha[/tex] = Coefficient of linear expansion = [tex]1.62\times 10^{-5}\ /^{\circ}C[/tex]

Initial temperature = 13.7°C

Final temperature = 42°C

Change in length of a material is given by

[tex]\Delta L=\alpha L_0\Delta T\\\Rightarrow \Delta L=1.62\times 10^{-5}\times 145000\times (42-13.7)\\\Rightarrow \Delta=66.4767\ m[/tex]

The change in length will be 66.4767 m

To solve this problem it is necessary to apply the concepts related to Newton's second law and the kinematic equations of movement description.

Newton's second law is defined as

[tex]F = ma[/tex]

Where,

m = mass

a = acceleration

From this equation we can figure the acceleration out, then

[tex]a = \frac{F}{m}[/tex]

[tex]a = \frac{11*10^3}{80}[/tex]

[tex]a = 137.5m/s[/tex]

From the cinematic equations of motion we know that

[tex]v_f^2-v_i^2 = 2ax[/tex]

Where,

[tex]v_f =[/tex]Final velocity

[tex]v_i =[/tex]Initial velocity

a = acceleration

x = displacement

There is not Final velocity and the acceleration is equal to the gravity, then

[tex]v_f^2-v_i^2 = 2ax[/tex]

[tex]0-v_i^2 = 2(-g)x[/tex]

[tex]v_i =\sqrt{2gx}[/tex]

[tex]v_i = \sqrt{2*9.8*4.8}[/tex]

[tex]v_i = 9.69m/s[/tex]

From the equation of motion where acceleration is equal to the velocity in function of time we have

[tex]a = \frac{v_i}{t}[/tex]

[tex]t = \frac{v_i}{a}[/tex]

[tex]t =\frac{9.69}{137.5}[/tex]

[tex]t = 0.0705s[/tex]

Therefore the time required is 0.0705s

The minimum time for the mass to come to rest at the end of the fall is about 0.071 s

[tex]\texttt{ }[/tex]

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

[tex]\large {\boxed {F = ma }[/tex]

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

[tex]\texttt{ }[/tex]

[tex]\large {\boxed {F = \Delta (mv) \div t }[/tex]

F = Force ( Newton )

m = Object's Mass ( kg )

v = Velocity of Object ( m/s )

t = Time Taken ( s )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

mass of climber = m = 80 kg

height of fall = h = 4.8 m

net force = ∑F = 11 k N = 11000 N

Asked:

minimum time = t = ?

Solution:

FIrstly , we could find the initial velocity of climber as he caught by the rope:

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

[tex]v^2 = 0^2 + 2(9.8)(4.8)[/tex]

[tex]v^2 = 94.08[/tex]

[tex]v = \frac{28}{3}\sqrt{5} \texttt{ m/s}[/tex]

[tex]\texttt{ }[/tex]

Next , we will use Newton's Law of Motion to calculate the minimum time:

[tex]\Sigma F = \Delta p \div t[/tex]

[tex]\Sigma F = m \Delta v \div t[/tex]

[tex]11000 = 80 (\frac{28}{3}\sqrt{5}) \div t[/tex]

[tex]t = 80 (\frac{28}{3}\sqrt{5}) \div 11000[/tex]

[tex]t \approx 0.071 \texttt{ s}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Dynamics

Answer:[tex]5.67\times 10^{3} m/s[/tex]

Explanation:

Given

mass of satellite [tex]m=155 kg[/tex]

Satellite is orbiting 5995 km above Earth surface

mass of Earth [tex]M=5.97\times 10^{24} kg[/tex]

Radius of Earth [tex]R=6370 km[/tex]

here [tex]r=R+5995=6370+5995=12,365 km[/tex]

Gravitational Force will Provide the centripetal Force

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

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

[tex]v=\sqrt{\frac{6.67\times 10^{-11}\times 5.97\times 10^{24}}{12365\times 10^{3}}}[/tex]

[tex]v=\sqrt{32.203\times 10^6}[/tex]

[tex]v=5.67\times 10^3m/s=5674.7 km/s[/tex]            

Answer:

Length of the pendulum will be 3.987 m

Explanation:

We have given time period of the pendulum T = 8 sec

Acceleration due to gravity [tex]g=9.81m/sec^2[/tex]

We have to find the length of the simple pendulum

We know that time period of the simple pendulum is given by

[tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]

[tex]8=2\times 3.14 \sqrt{\frac{l}{9.81}}[/tex]

[tex]l=3.987m[/tex]

So length of the pendulum will be 3.987 m

To solve this problem it is necessary to apply the concepts related to wavelength depending on the frequency and speed. Mathematically, the wavelength can be expressed as

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

Where,

v = Velocity

f = Frequency,

Our values are given as

L = 3.6m

v= 192m/s

f= 320Hz

Replacing we have that

[tex]\lambda = \frac{192}{320}[/tex]

[tex]\lambda = 0.6m[/tex]

The total number of 'wavelengths' that will be in the string will be subject to the total length over the size of each of these undulations, that is,

[tex]N = \frac{L}{\lambda}[/tex]

[tex]N = \frac{3.6}{0.6}[/tex]

[tex]N = 6[/tex]

Therefore the number of wavelengths of the wave fit on the string is 6.

Answer:

The phase angle is 0.0180 rad.

(c) is correct option.

Explanation:

Given that,

Voltage = 12 V

Angular velocity = 50 Hz

Capacitance [tex]C= 20\times10^{-2}\ F[/tex]

Inductance [tex]L=20\times10^{-3}\ H[/tex]

Resistance [tex]R=  50\ Omega[/tex]

We need to calculate the impedance

Using formula of impedance

[tex]z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}[/tex]

[tex]z=\sqrt{50^2+(50\times20\times10^{-3}-\dfrac{1}{50\times20\times10^{-2}})^2}[/tex]

[tex]z=50.00[/tex]

We need to calculate the phase angle

Using formula of phase angle

[tex]\theta=\cos^{-1}(\dfrac{R}{z})[/tex]

[tex]\theta=\cos^{-1}(\dfrac{50}{50.00})[/tex]

[tex]\theta=0.0180\ rad[/tex]

Hence, The phase angle is 0.0180 rad.

The kinetic energy of the spaceship increases by approximately 0.379 × 10^9 Joules as it moves from R1 = 7300 km to R2 = 6700 km in the gravitational field of Earth.

To find the change in kinetic energy of the spaceship as it moves from position R1 = 7300 km to position R1 = 6700 km in the gravitational field of Earth, we can use the gravitational potential energy and the fact that the mechanical energy is conserved. The change in kinetic energy is equal to the negative of the change in potential energy.

The gravitational potential energy (U) is given by the formula:

U = -G * (m1 * m2) / R

Where:

- G is the universal gravitational constant (approximately 6.674 × 10^(-11) N·(m/kg)^2).

- m1 is the mass of the Earth (approximately 5.972 × 10^24 kg).

- m2 is the mass of the spaceship (8600 kg in this case).

- R is the distance from the centre of the Earth.

Now, let's calculate the potential energy at R1 = 7300 km and R2 = 6700 km:

For R1 = 7300 km:

R1 = 7300 km = 7,300,000 meters

U1 = -G * (m1 * m2) / R1

U1 = -6.674 × 10^(-11) N·(m/kg)^2 * (5.972 × 10^24 kg * 8600 kg) / 7,300,000 m

U1 ≈ -5.125 × 10^9 J

For R2 = 6700 km:

R2 = 6700 km = 6,700,000 meters

U2 = -G * (m1 * m2) / R2

U2 = -6.674 × 10^(-11) N·(m/kg)^2 * (5.972 × 10^24 kg * 8600 kg) / 6,700,000 m

U2 ≈ -5.504 × 10^9 J

Now, let's find the change in potential energy (ΔU) as the spaceship moves from R1 to R2:

ΔU = U2 - U1

ΔU ≈ (-5.504 × 10^9 J) - (-5.125 × 10^9 J)

ΔU ≈ -0.379 × 10^9 J

So, the change in potential energy is approximately -0.379 × 10^9 Joules.

The change in kinetic energy is equal to the negative of the change in potential energy, so:

Change in kinetic energy = -ΔU

Change in kinetic energy ≈ 0.379 × 10^9 J

Therefore, the kinetic energy of the spaceship increases by approximately 0.379 × 10^9 Joules as it moves from R1 = 7300 km to R2 = 6700 km in the gravitational field of Earth.

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Answer:

[tex]h=6.377\times10^{-34}kgm^2/s[/tex]

Explanation:

The maximum kinetic energy of the photoelectrons is given by the formula [tex]K_M=hf-\phi[/tex].

We have two situations where for [tex]f_1=547.5\times10^{12}Hz[/tex] we get [tex]K_{M1}=1.26\times10^{-19}J[/tex] and for [tex]f_2=738.8\times10^{12}Hz[/tex] we get [tex]K_{M2}=2.48\times10^{-19}J[/tex], so we have:

[tex]K_{M1}=hf_1-\phi[/tex]

[tex]K_{M2}=hf_2-\phi[/tex]

We can eliminate [tex]\phi[/tex] by substracting the first equation to the second:

[tex]K_{M2}-K_{M1}=hf_2-\phi-(hf_1-\phi)=h(f_2-f_1)[/tex]

Which means:

[tex]h=\frac{K_{M2}-K_{M1}}{f_2-f_1}=\frac{2.48\times10^{-19}J-1.26\times10^{-19}J}{738.8\times10^{12}Hz-547.5\times10^{12}Hz}=6.377\times10^{-34}kgm^2/s[/tex]

Answer:

C) the magnitude of the acceleration is a minimum.

Explanation:

As we know that ,the general equation of the simple harmonic motion given as

The displacement x given as

x=X sinω t

Then the velocity v will become

v= X ω cosωt

The acceleration a

a= - X ω² sinω t

The speed of the particle will be maximum when cosωt will become 1 unit.

It means that sinωt will become zero.So acceleration and displacement will be minimum.

Therefore when speed is maximum then acceleration will be minimum.

At the mean position the speed of the particle is maximum that is why kinetic energy also will be maximum and the potential energy will be minimum.

Therefore option C is correct.

In simple harmonic motion, the speed is greatest when the displacement is a maximum.

In simple harmonic motion, the speed is greatest at the point in the cycle when the displacement is a maximum. When an object is at its maximum displacement, it has the maximum potential energy, and as it moves towards the equilibrium position, the potential energy is converted into kinetic energy, leading to an increase in speed.

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Answer:

a. the absolute value of the change in the momentum of the small car is 0.078  

b. the velocity of the larger car after the collision is 0.1513 m/s

Explanation:

The linear momentum P is calculated as:

P = MV

Where M is the mass and V the velocity

Therefore, for calculated the change of the linear momentum of the small cart, we get:

[tex]P_{fc}-P_{ic} =[/tex]ΔP

where [tex]P_{ic}[/tex] in the inicial momentum and [tex]P_{fc}[/tex] is the final momentum of the small cart. Replacing the values, we get:

0.10 kg (0.76) -0.10(1.54) = -0.078 kg m/s

The absolute value: 0.078 kg m/s

On the other hand, using the law of the conservation of linear momentum, we get:

[tex]P_i = P_f[/tex]

Where [tex]P_i[/tex] is the linear momentum of the sistem before the collision and [tex]P_f[/tex] is the linear momentum after the collision.

[tex]P_i=P_{ic}\\P_f=P_{fc} + P_{ft}[/tex]

Where [tex]P_{fc}[/tex] is the linear momentum of the small cart after the collision and [tex]P_{ft}[/tex] is the linear momentum of the larger cart after the collision

so:

(0.10 kg)(1.54 m/s) = (0.10 kg)(-0.76 m/s) + (1.52 kg)(V)

Note: we choose the first direction of the small car as positive.

Solving for V:

[tex]\frac{0.10(1.54)+0.10(0.76)}{1.52} = V[/tex]

V = 0.1513 m/s

Final answer:

The magnitude of the change in momentum for the small cart is 0.078 kg·m/s. The speed of the larger cart after the collision is 0 m/s.

Explanation:

(a) What is the magnitude (absolute value) of the change in momentum for the small cart?

To find the change in momentum for the small cart, we can use the formula:

Change in momentum = mass × change in velocity

The mass of the small cart is 0.10 kg and its initial velocity is 1.54 m/s. After the collision, its velocity changes to -0.76 m/s (since it recoils in the opposite direction). Therefore, the change in velocity is: (0.76 m/s - 1.54 m/s) = -0.78 m/s.

Substituting the values into the formula, we get:

Change in momentum = 0.10 kg × (-0.78 m/s) = -0.078 kg·m/s

Since the question asks for the magnitude (absolute value) of the change in momentum, the answer is 0.078 kg·m/s.

(b) What is the speed (absolute value of the velocity) of the larger cart after the collision?

In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. Since the larger cart is initially at rest, its momentum before the collision is zero. Therefore, its momentum after the collision must also be zero.

Since momentum is equal to mass × velocity, the mass of the larger cart is 1.52 kg and its final velocity after the collision is denoted as Vf. We can set up the equation:

1.52 kg × Vf = 0 kg·m/s

Solving for Vf, we find that the velocity of the larger cart after the collision is also zero. Therefore, the answer is 0 m/s.

Answer:

It's held together by the nuclear force.

Explanation:

There are more elemental forces than just the electromagnetic one. In this case, it is the nuclear force (called also strong force) the one that holds the nucleus together because it is stronger than the electromagnetic force over such short distances as the one inside the atomic nucleus.

Answer:

The strong nuclear force, which is attractive over short distances like the nucleus, and stronger than electricity, holds the nucleus together.

Explanation:

Answer:

3. F = m g

Explanation:

The centripetal force the person is exerting on the wall is;

                                         [tex]f = \frac{mv^{2}}{r}[/tex]

The maximum force due to  static friction is f = µ R ,

where R is  the normal force exerted by  the wall on the person. For the person to be held up against the wall, the maximal  friction force must be larger than the force of  gravity mg.

The actual force is now equal to or less than the maximum µR

                                                 f ≤ fmax = µR

When the angular velocity increase beyond the minimum angular velocity ωmin that is required  to hold the person against the wall,  the frictional force does to increase any more. Therefore, f still equals mg regardless of  the maximum frictional force.  

The magnitude of the frictional force between the person and the wall can be determined by considering Newton's laws of motion. According to Newton's second law, the net force on an object is equal to its mass multiplied by its acceleration. In this case, the person is being held against the wall with an angular velocity, which implies a centripetal acceleration. The frictional force between the person and the wall acts as the centripetal force, so we can equate the two.

The magnitude of the frictional force between the person and the wall can be determined by considering Newton's laws of motion. According to Newton's second law, the net force on an object is equal to its mass multiplied by its acceleration. In this case, the person is being held against the wall with an angular velocity, which implies a centripetal acceleration. The frictional force between the person and the wall acts as the centripetal force, so we can equate the two:

Ffriction = mω²r

Where Ffriction is the frictional force, m is the mass of the person, ω is the angular velocity, and r is the distance between the person and the axis of rotation (which is the radius of the cylinder in this problem). Therefore, the correct answer is option 9, F = 3mg.

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The spring's force constant is [tex]\( 15.30 \, \text{N/m} \).[/tex]

To find the spring's force constant, we can use the formula for the period of oscillation of a mass-spring system:

[tex]\[ T = 2\pi \sqrt{\frac{m}{k}} \][/tex]

where [tex]\( T \)[/tex] is the period of oscillation, m is the mass attached to the spring, and [tex]\( k \)[/tex] is the spring constant.

Given:

[tex]\[ m = 0.200 \, \text{kg} \]\[ \text{Time for one complete oscillation } = 2.60 \, \text{s} \][/tex]

First, let's find the period [tex]\( T \)[/tex] using the given time for one complete oscillation:

[tex]\[ T = 2.60 \, \text{s} \][/tex]

Now, we can rearrange the formula for the period to solve for the spring constant [tex]\( k \):[/tex]

[tex]\[ k = \frac{4\pi^2 m}{T^2} \][/tex]

Substitute the known values:

[tex]\[ k = \frac{4\pi^2 \cdot 0.200 \, \text{kg}}{(2.60 \, \text{s})^2} \][/tex]

Calculate:

[tex]\[ k = \frac{4\pi^2 \cdot 0.200 \, \text{kg}}{6.76 \, \text{s}^2} \]\[ k = \frac{7.854 \, \text{kg} \cdot \text{m}^{-1}}{6.76 \, \text{s}^2} \]\[ k \approx 15.30 \, \text{N/m} \][/tex]

So, the spring's force constant is [tex]\( 15.30 \, \text{N/m} \).[/tex]

Complete Question:
In a physics lab, you attach a 0.200-kg air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is 2.60 s. Find the spring’s force constant

Answer:

B) AC current can be used in transformers while DC current cannot.

Explanation:

AC current is used for long distance transportation of electrical energy with amplifying the voltage and reducing the current in the conductors so that there is minimum loss of energy in the form of heat.

The relation between heat energy and electrical current is given by Joule's law as:

[tex]Q=i^2.R.t[/tex]

where:

The above mentioned variation  of current and voltage in the transmission lines of AC is achieved by a device called transformer. It consist of a rectangular shaped steel core common to two of the insulated wire winding of which one side acts as input (primary coil) and the other one acts as secondary (output).

AC i.e. alternating current can only be used in the transformers because they have the continuously varying amplitude which in turn provides the change in flux with respect to the time inducing an emf in the  nearby secondary coil of the transformer without any physically moving parts.

The alternating nature of AC current aloows changes in magnet fields to be used to transform electricity and make it more efficient to transport over long distances. AC current can be used in transfromers while DC current cannot.

Answer:

1000 Nm

2000 Nm

1.00007 seconds

Explanation:

I = Moment of inertia = 5 kgm²

[tex]\alpha[/tex] = Angular acceleration

[tex]\omega_f[/tex] = Final angular velocity

[tex]\omega_i[/tex] = Initial angular velocity

t = Time taken

Torque is given by

[tex]\tau=I\alpha\\\Rightarrow \tau=5\times 200\\\Rightarrow \tau=1000\ Nm[/tex]

The torque of the disc would be 1000 Nm

If [tex]\alpha=400\ rad/s^2[/tex]

[tex]\tau=I\alpha\\\Rightarrow \tau=5\times 400\\\Rightarrow \tau=2000\ Nm[/tex]

The torque of the disc would be 2000 Nm

From equation of rotational motion

[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{3820\times \frac{2\pi}{60}-0}{400}\\\Rightarrow t=1.00007\ s[/tex]

It would take 1.00007 seconds to reach 3820 rpm

To develop this problem it is necessary to apply the concepts related to kinetic energy and gravitational potential energy.

By conserving energy we know that

[tex]PE = KE[/tex]

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

Where,

m = mass

v = Velocity

G = Gravitational universal constant

M = Mass of Spherical asteroid

m = mass of object

r = Radius

Mass of a Sphere can be expressed as,

[tex]M= \rho* (\frac{4}{3}\pi r^3 )[/tex]

Replacing we have that,

[tex]\frac{1}{2}*2.8^2 - 6.67*10^{-11}*\frac{4\pi*2000*r^3}{3r} = 0[/tex]

[tex]6.67*10^{-11}*(\frac{4\pi}3{}2000*r^2)=\frac{1}{2}*2.8^2[/tex]

[tex]r = \sqrt{\frac{1}{2}*\frac{2.8^2}{6.67*10^{-11}*2000*4/3*\pi}}[/tex]

[tex]r=2648 m[/tex]

Therefore the diameter is 5296 m.

b) Applying the concept of gravitational force and centripetal force we have to

[tex]F_g = F_c[/tex]

[tex]\frac{G M m}{r^2} = \frac{m v^2}{r}[/tex]

[tex]\frac{G M}{r} = v^2[/tex]

[tex]6.67*10^{-11}*\frac{(\frac{4\pi}{3}2000*r^3)}{r} = 2.8^2[/tex]

[tex]r=3746 m[/tex]

Therefore the diameter would be 7492 m

The diameter of the largest asteroid from which a froghopper could jump free can be calculated using the escape velocity formula. Similarly, for the froghopper to enter a circular orbit, its horizontal speed needs to equal the required orbital speed, calculated using the gravitational field strength at the surface of the asteroid.

To find the diameter of the largest spherical asteroid from which a froghopper could jump free, we need to understand the escape speed, the speed necessary to overcome an object's gravitational field. The formula for escape velocity is √(2*G*M/R), where G is the gravitational constant, M is the mass of the asteroid, and R is its radius (half of the diameter).

In this case, we can solve the formula for the radius, R = 2*G*M/v², where v is the takeoff speed of a froghopper (2.8 m/s). The mass of the spherical asteroid can be calculated from the given density (2.0 g/cm³ or 2000 kg/m³) and the volume of a sphere (4/3*π*R³). Substituting the mass into the formula will provide the radius, and hence diameter, of the asteroid.

For a froghopper to enter a circular orbit just above the surface of the asteroid, its horizontal speed must equal the circular orbital speed, which depends on the gravitational field strength at the surface of the asteroid. It's determined by the formula v = √(G*M/R), which can be rearranged to solve for the required diameter of the asteroid.

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Answer:

option A

Explanation:

given,

drive 4 km at 30 km/h and

another 4 km at 50 km/h

average speed for the whole 8 Km trip = ?

time for the trip of first 4 km

 distance = speed x time

[tex]t_1 = \dfrac{d}{s}[/tex]

[tex]t_1 = \dfrac{4}{30}[/tex]

     t₁= 0.1333 hr

time from another 4 km trip

[tex]t_2 = \dfrac{d}{s}[/tex]

[tex]t_2= \dfrac{4}{50}[/tex]

     t₂ = 0.08 hr

average speed

     [tex]s= \dfrac{d}{t}[/tex]  

     [tex]s= \dfrac{8}{0.1333 + 0.08}[/tex]  

     [tex]s= \dfrac{8}{0.2133}[/tex]  

            s = 37.50 Km/hr

Less than 40 Km/h

the correct answer is option A

Answer:

3.16 × [tex]10^{-7}[/tex] W/[tex]m^{2}[/tex]

Explanation:

β(dB)=10 × [tex]log_{10}(\frac{I}{I_{0} })[/tex]

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

β=55 dB

Therefore plugging into the equation the values,

55=10 [tex]log_{10}(\frac{I}{ [tex]10^{-12}[/tex]})[/tex]

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

[tex]10^{5.5}[/tex]= [tex]\frac{I}{10^{-12} }[/tex]

316227.76×[tex]10^{-12}[/tex]= I

I= 3.16 × [tex]10^{-7}[/tex] W/[tex]m^{2}[/tex]

Answer:[tex]a=1.75\times 10^{21}\left ( 2\hat{i}5.08\hat{j}\right )[/tex]

Explanation:

Given

Electric Field [tex]\vec{E}=2\hat{i}+5.4\hat{j}[/tex]

[tex]\vec{B}=0.4\hat{k}\ T[/tex]

velocity [tex]\vec{v}=8\hat{i} m/s[/tex]

mass of electron [tex]m=9.1\times 10^{-31} kg[/tex]

Force on a charge Particle moving in Magnetic Field

[tex]F=e\left [ \vec{E}+\left ( \vec{v}\times \vec{B}\right )\right ][/tex]

[tex]a=\frac{e\left [ \vec{E}+\left ( \vec{v}\times \vec{B}\right )\right ]}{m}[/tex]

[tex]a=\frac{1.6\times 10^{-19}\left [ 2\hat{i}+5.4\hat{j}+\left ( 8\hat{i}\times 0.4\hat{k}\right )\right ]}{9.1\times 10^{-31}}[/tex]  

[tex]a=1.75\times 10^{21}\left ( 2\hat{i}+5.08\hat{j}\right )\ m/s^2[/tex]

The Lorentz force law is used to calculate the force exerted on an electron in an electromagnetic field. This force is then divided by the mass of the electron to calculate the acceleration.

The acceleration of an electron moving in both electric and magnetic fields can be found using the Lorentz force law, which states that the force exerted on a charged particle in an electromagnetic field is the vector sum of the electric and magnetic forces. The formula is F = q(E + v × B), where q is the charge of the particle, E is the electric field, v is the velocity of the particle, and B is the magnetic field. The acceleration a is then calculated as the force divided by the mass m of the electron, which is known to be 9.11 x 10^-31 kg.

First, calculate the force F due to the electric field E (since F = qE) and the magnetic field B (since F = qvBsinθ). Using the given values, E = (2.00î + 5.40ĵ) V/m, v = 8.0î m/s, and B = 0.400k T, we find that Fe = qE = -(1.60 x 10^-19 C)(2.00î + 5.40ĵ) N (the charge of an electron is -1.60 x 10^-19 C), and Fb = qvBsinθ = - (1.60 x 10^-19 C)(8.0î m/s)(0.400 T), since the angle θ between v and B is 90 degrees. Then, the total force F is the vector sum of Fe and Fb.

Finally, calculate the acceleration a by dividing the total force F by the mass m of the electron to get a = F/m.

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Answer:

A) To true. he pressure at the bottom of the pool decreases by exactly the same amount as the atmospheric pressure decreases

Explanation:

Let us propose the solution of this problem before seeing the final statements. The pressure increases with the depth of raposin due to the weight of water that is above the person and also the pressure exerted by the atmosphere on the entire pool, the equation describing this process is

    P =[tex]P_{atm}[/tex] + ρ g y

Where [tex]P_{atm}[/tex] is the atmospheric pressure, ρ  the water density, and 'y' the depth measured from the surface.

Let's examine this equation in we see that the total pressure is directly proportional to the atmospheric pressure and depth

Now we can examine the claims

A) To true. State agreement or with the equation above

B) False. Pressure changes with atmospheric pressure

C) False. It's the opposite

D) False. They are directly proportional

A The pressure at the bottom of the pool decreases by exactly the same amount as the atmospheric pressure decreases

Hydrostatic pressure is pressure caused by the weight of a liquid.

The weight of a liquid is affected by the force of gravity.

If a liquid is placed in a container, the higher the liquid content in the container, the heavier the liquid content is and the greater the liquid pressure at the bottom of the container.

The hydrostatic pressure of a liquid can be formulated:

[tex]\large{\boxed{\bold {P_h ~ = ~ \rho.g.h}}[/tex]

Ph = hydrostatic pressure (N / m², Pa)

ρ = density of liquid (kg / m³)

g = acceleration due to gravity (m / s²)

h = height / depth of liquid surface (m)

If the container is open, then the atmospheric pressure (P₀) can be entered into the equation.

[tex]\large{\boxed {\bold {P_h ~ = ~ P_o ~ + ~ \rho.g.h}}[/tex]

The magnitude of P₀ is usually equal to = 1.01.10⁵ Pascal (Pa = N / m²) = 1 atm

For example submarines  :

The deeper a submarine is, the greater the hydrostatic pressure it experiences, so that the hull / wall of the submarine is made thick to withstand that pressure.

As a storm approaches, the atmospheric pressure drops

Because the hydrostatic pressure is proportional to the atmospheric pressure, the pressure at the bottom of the pool will also be reduced

Isaac Newton's investigations of gravity

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Gravitational force

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increase the gravitational force between two objects

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Answer:

1. [tex]X_{cm}=-8.57m[/tex]

2. [tex]Y_{cm}=-14.29m[/tex]

3. [tex]V_{cm} = 16.66m/s[/tex]

4. α = 59.05°

5. [tex]V_{cm} = 16.66m/s[/tex]

6. α = 59.05°

Explanation:

The position of the center of mass 1s before the collision is:

[tex]X_{cm}=\frac{X_c*mc+X_t*m_t}{m_c+m_t}[/tex]

where

[tex]X_c=-20m[/tex];  [tex]m_c=1500kg[/tex];  

[tex]X_t=0m[/tex];   [tex]m_t=2000kg[/tex];

Replacing these values:

[tex]X_{cm}=-8.57m[/tex]

[tex]Y_{cm}=\frac{Y_c*mc+Y_t*m_t}{m_c+m_t}[/tex]

where

[tex]Y_c=0m[/tex];  [tex]m_c=1500kg[/tex];  

[tex]Y_t=-25m[/tex];   [tex]m_t=2000kg[/tex];

Replacing these values:

[tex]Y_{cm}=-14.29m[/tex]

The velocity of their center of mass is:

[tex]V_{cm-x}=\frac{V_{c-x}*mc+V_{t-x}*m_t}{m_c+m_t}[/tex]

where

[tex]V_{c-x}=20m/s[/tex];  [tex]m_c=1500kg[/tex];  

[tex]V_{t-x}=0m/s[/tex];   [tex]m_t=2000kg[/tex];

Replacing these values:

[tex]V_{cm-x}=8.57m/s[/tex]

[tex]V_{cm-y}=\frac{V_{c-y}*mc+V_{t-y}*m_t}{m_c+m_t}[/tex]

where

[tex]V_{c-y}=0m[/tex];  [tex]m_c=1500kg[/tex];  

[tex]V_{t-y}=-25m[/tex];   [tex]m_t=2000kg[/tex];

Replacing these values:

[tex]V_{cm-y}=-14.29m[/tex]

So, the magnitude of the velocity is:

[tex]V_{cm}=\sqrt{V_{cm-x}^2+V_{cm-y}^2}[/tex]

[tex]V_{cm}=16.66m/s[/tex]

The angle of the velocity is:

[tex]\alpha =atan(V_{cm-y}/V_{cm-x})[/tex]

[tex]\alpha=59.05\°[/tex]

Since on any collision, the velocity of the center of mass is preserved, then the velocity after the collision is the same as the previously calculated value of 16.66m/s at 59.0° due north of east

Calculations using the physics concept of center of mass determine the positions and velocities of the system of two vehicles before and after collision. The x and y coordinates of the center of mass are found to be -12m and -15m respectively, with a combined vCM direction of 180 degrees relative to the positive x-axis.

To answer the questions presented, it's important to recall the concept of center of mass in physics. The center of mass of a system of particles (like two cars in this case) is the point at which its mass may be considered to be concentrated, when calculating moments and linear momentum. The positions of center of mass are given by:

xCm = (m1*x1 + m2*x2) / (m1 + m2)

yCm = (m1*y1 + m2*y2) / (m1 + m2)

And the center of mass velocity can be found using:

vCm = (m1*v1 + m2*v2) / (m1 + m2)

Assuming east as positive x-direction and north as positive y-direction, substituting the given values into the equations above, we find that:

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To solve this problem it is necessary to apply the continuity equations for which it is defined that the proportion of Area in the initial section is equal to the final section. In other words,

[tex]A_1 v_1 = A_2 v_2[/tex]

Where,

[tex]A_i =[/tex] Cross sectional area at each section

[tex]v_i =[/tex]Velocities of fluid at each section

The total area of the branch is eighteen times of area of smaller artery. The average cross-sectional area of each artery is [tex]0.7cm^2.[/tex]

Therefore the Cross-sectional area at the end is

[tex]A_2= 18*0.7cm^2[/tex]

[tex]A_2 = 12.6cm^2[/tex]

Applying the previous equation we have then

[tex]A_1 v_1 = A_2 v_2[/tex]

[tex](1.7cm^2) v_1 = (12.6cm^2)v_2[/tex]

The ratio of the velocities then is

[tex]\frac{v_1}{v_2} = \frac{1.7}{12.6}[/tex]

[tex]\frac{v_1}{v_2} = 0.135[/tex]

Therefore the factor by which the velocity of blood will reduce when it enters the smaller arteries is 0.1349

The incandescent lightbulb emits approximately [tex]\(1.51 \times 10^{19}\)[/tex] visible-light photons per second.

To find the number of photons emitted per second, you can use the following steps:

1. Calculate the energy of one photon using the formula:

  [tex]\[ E = hf \][/tex]

  where:

  - [tex]\( E \)[/tex] is the energy of one photon,

  - [tex]\( h \)[/tex] is Planck's constant [tex](\( 6.626 \times 10^{-34} \ \text{Joule}\cdot\text{s} \))[/tex],

  - [tex]\( f \)[/tex]is the frequency of the light.

  The frequency [tex](\( f \))[/tex] can be determined using the speed of light [tex](\( c \))[/tex] and the wavelength [tex](\( \lambda \))[/tex]:

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

  where:

  - [tex]\( c \)[/tex] is the speed of light [tex](\( 3.00 \times 10^8 \ \text{m/s} \))[/tex],

  - [tex]\( \lambda \)[/tex] is the wavelength of the light.

2. Find the number of photons emitted per second using the power of the visible light [tex](\( P_{\text{visible}} \))[/tex] and the energy of one photon:

 [tex]\[ \text{Number of photons per second} = \frac{P_{\text{visible}}}{E} \][/tex]

  where:

  - [tex]\( P_{\text{visible}} \)[/tex] is the power of the visible light (5 W),

  - [tex]\( E \)[/tex] is the energy of one photon.

Let's perform the calculations:

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

[tex]\[ f = \frac{3.00 \times 10^8 \ \text{m/s}}{600 \times 10^{-9} \ \text{m}} \][/tex]

[tex]\[ f = 5.00 \times 10^{14} \ \text{Hz} \][/tex]

Now, calculate the energy of one photon:

[tex]\[ E = hf \][/tex]

[tex]\[ E = (6.626 \times 10^{-34} \ \text{Joule}\cdot\text{s}) \times (5.00 \times 10^{14} \ \text{Hz}) \][/tex]

[tex]\[ E = 3.313 \times 10^{-19} \ \text{Joule} \][/tex]

Finally, calculate the number of photons emitted per second:

[tex]\[ \text{Number of photons per second} = \frac{5 \ \text{W}}{3.313 \times 10^{-19} \ \text{Joule/photon}} \][/tex]

[tex]\[ \text{Number of photons per second} \approx 1.51 \times 10^{19} \ \text{photons/s} \][/tex]

So, the incandescent lightbulb emits approximately [tex]\(1.51 \times 10^{19}\)[/tex] visible-light photons per second.

Answer:

T= 1.71×10^{-3} sec= 1.71 mili sec

t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec

Explanation:

First of all we write equation for current oscillation in LC circuits. Note, the maximum current (I_0)=  5.5 mA is the amplitude of this function. Then, we continue to solve for the angular frequency(ω). Afterwards, we calculate the time period T. qo = maximum charge on capacitor. = 1.5× 10 ^− 6 C

a) I(t) = -ωqosin(ωt+φ)

⇒Io= ωqo

⇒ω= Io/qo

also we know that T= 2π/ω

⇒T= [tex]2\pi\frac{q_0}{I_0}[/tex]

now putting the values we get

= [tex]2\pi\frac{1.5\times10^{-6}}{5.5\times10^{-3}}[/tex]

= 1.71×10^{-3} sec

b) note that the time [tex]t_{fc}[/tex] it takes the capacitor to from uncharge to fully charged is one fourth of the period . That is

[tex]t_{fc}= \frac{T}{4}[/tex]

[tex]t_{fc}= \frac{ 1.71\times10^{-3} }{4}[/tex]

t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec

Answer:

18.2 cm

Explanation:

The kinetic energy of the block

= .5 x 3.49 x 10⁻² x 10.3²

= 1.85 J

If x be the compression created in the spring due to this kinetic energy

1.85 = 1/2 k x²

= .5 x 111 x²

x² = 3.3 x 10⁻²

x = 0.182 m .

= 18.2 cm

This will be the amplitude of oscillation under SHM.

To solve the problem it is necessary to apply the concepts related to wavelength and error calculation.

The calculation of the error is usually defined by

[tex]z = \frac{\lambda_o-\lambda_r}{\lambda_r}[/tex]

Where

[tex]\lambda_o =[/tex] Wavelength from the observer

[tex]\lambda_r =[/tex] Wavelength from the rest

The previous formula is exactly equal to,

[tex]z=  \frac{\lambda_o}{\lambda_r}-1[/tex]

[tex](z+1) = \frac{\lambda_o}{\lambda_r}[/tex]

[tex]\frac{\lambda_o}{\lambda_r}=(10+1)[/tex]

[tex]\lambda_o=11 \lambda_r[/tex]

Therefore the observed wavelength will be 11 times longer that the emitted wavelength.

If we make the comparison for the ultraviolet region (from 10nm to 400nm) we get that,

[tex]\lambda_o=11 \lambda_r[/tex]

[tex]\lambda_o=11 (\frac{10nm*400nm}}{2})[/tex]

[tex]\lambda_o = 2255nm[/tex]

That is the infrared region of electromagnetic spectra.

The wavelength of light we observe from these galaxies compares to its original wavelength when it was emitted, a function of the relative velocity between the emitting source and the observing receiver.

The redshift (z) corresponds to a change in the way in which the frequency of light photon waves is observed in the spectroscope as a function of the relative velocity between the emitting source and the observing receiver (a measure of distance, of according to cosmological expansion, based on recessional velocity).

A decrease in frequency occurs as the distance from the source galaxy increases, causing the light ray to be captured as a standard color. The deviation or distance measurement is represented by the letter z.

With this information, we can conclude that the wavelength of light we observe from these galaxies compares to its original wavelength when it was emitted, a function of the relative velocity between the emitting source and the observing receiver.

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Answer

given,

constant speed of cart on right side = 2 m/s

diameter of nozzle = 50 mm = 0.05 m

discharge flow through nozzle = 0.04 m³

One-fourth of the discharge flows down the incline

three-fourths flows up the incline

Power = ?

Normal force i.e. Fn acting on the cart

[tex]F_n = \rho A (v - u)^2 sin \theta[/tex]

v is the velocity of jet

Q = A V

[tex]v = \dfrac{0.04}{\dfrac{\pi}{4}d^2}[/tex]

[tex]v= \dfrac{0.04}{\dfrac{\pi}{4}\times 0.05^2}[/tex]

v = 20.37 m/s

u be the speed of cart assuming it to be u = 2 m/s

angle angle of inclination be 60°

now,

[tex]F_n = 1000 \times \dfrac{\pi}{4}\times 0.05^2\times (20.37 - 2)^2 sin 60^0[/tex]

F n = 2295 N

now force along x direction

[tex]F_x = F_n sin 60^0[/tex]

[tex]F_x = 2295 \times sin 60^0[/tex]

[tex]F_x = 1987.52\ N[/tex]

Power of the cart

P = F x v

P = 1987.52 x 20.37

P = 40485 watt

P = 40.5 kW

The power produced by the stream is; 40.5 kW

We are given;

Constant speed of cart on right side; u = 2 m/s

Diameter of nozzle; d = 50 mm = 0.05 m

Discharge flow through nozzle; Q = 0.04 m³/s

Formula for the normal force acting on the cart is;

Fₙ = ρA(v - u)²sin θ

Let us find v the speed of the jet from;

Q = Av

v = Q/A

v = 0.04/(π * 0.05²/4)

v = 20.37 m/s

From online research about this question, the angle of inclination is 60°. Thus;

Fₙ = ρA(v - u)²sin θ

Fₙ = 1000 * (π * 0.05²/4) * (20.37 - 2)sin 60

Fₙ = 2295 N

Force along the x-direction will be;

Fₓ = Fₙ * sin 60

Fₓ = 2295 * sin 60

Fₓ = 1987.52 N

Power of the Cart is;

P = Fₓ * v

P = 1987.52 * 20.37

P = 40.5 kW

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To solve this problem it is necessary to apply the concepts related to Pressure, Strength and Area.

We know by definition that Pressure is the amount of Force expressed per unit area, that is,

[tex]P = \frac{F}{A}[/tex]

Where,

F = Force

A = Cross-sectional Area

The net pressure on the bottle would be given by the difference between the internal pressure and the atmospheric pressure, therefore

[tex]P_{net} = P_{in}-P_{atm}[/tex]

[tex]P_{net} = 2.5atm-1atm[/tex]

[tex]P_{net} = 1.5atm (\frac{101325Pa}{1atm})[/tex]

[tex]P_{net} = 151987.5Pa[/tex]

The given radio is,

[tex]r = 3cm^2[/tex]

Hence the Cross-sectional Area would be

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

[tex]A = \pi(3*10^{-2})^2[/tex]

[tex]A = 2.827*10^{-3}m^2[/tex]

Applying the equation for Pressure we have that

[tex]P = \frac{F}{A}[/tex]

[tex]151987.5= \frac{F}{2.827*10^{-3}}[/tex]

[tex]F = 429.66N[/tex]

Therefore the frictional force on the cork due to the neck of the bottle is 429.66N.

Answer:

The gauge pressure is 6670.8 Pascal

Explanation:

Given the height is 0.68 m

The Gauge pressure is determine by the formula,

Gauge pressure , P = hdg

Where h = 0.68 m

d = 1000

g = 9.81

Substituting the values in the formula of Gauge pressure, we get

Gauge Pressure = hdg =[tex]0.68 \times 1000 \times 9.81[/tex]

Gauge Pressure = 6670.8

Therefore, the gauge pressure is 6670.8 Pascal