Light with a wavelength of 587.5 nm illuminates a single slit 0.750 mm in width.
(a.) At what distance from the slit should a screen be located if the first minimum in the diffraction pattern is to be 0.850 mm from the center of the screen?
b.) What is the width of the central maximum?
(c.) Sketch the intensity distribution for the diffraction pattern observed on the viewing screen.

Answer:

0.674 s = t

Explanation:

Assuming that the door is completely open, exena need to rotate the door 90°.

Now, using the next equation:

T = I∝

Where T is the torque, I is the moment of inertia and ∝ is the angular aceleration.

Also, the torque could be calculated by:

T = Fd

where F is the force and d is the lever arm.

so:

T = 220N*1.25m

T = 275 N*m

Addittionaly, the moment of inertia of the door is calculated as:

I = [tex]\frac{1}{3}Ma^2[/tex]

where M is the mass of the door and a is the wide.

I  =[tex]\frac{1}{3}(750/9.8)(1.25)^2[/tex]

I = 39.85 kg*m^2

Replacing in the first equation and solving for ∝, we get::

T = I∝

275 = 39.85∝

∝ = 6.9 rad/s

Now, the next equation give as a relation between θ (the angle that exena need to rotate) ∝ (the angular aceleration) and t (the time):

θ = [tex]\frac{1}{2}[/tex]∝[tex]t^2[/tex]

Replacing the values of θ and ∝ and solving for t, we get:

[tex]\sqrt{\frac{2(\pi/2)}{6.9 rad/s}}[/tex] = t

0.674 s = t

Answer:

Explained.

Explanation:

Only the first question has been answered

In a period from left to right the nuclear charge increases and hence nucleus size is compressed. Thus,  atomic radius decreases.

In transition elements, electrons in ns^2 orbital remain same which is the outer most orbital having 2 electrons and the electrons are added to (n-1) d orbital. So, outer orbital electron experience almost same nuclear attraction and thus size remains constant.

The atomic radius of main-group elements decreases as you move to the right across a period due to increased positive charge, while the atomic radius of transition elements remains relatively constant.

The atomic radius of main-group elements decreases as you move to the right across a period because the number of protons in the nucleus increases. This increased positive charge pulls the electrons closer to the nucleus, reducing the size of the atom. In contrast, the atomic radius of transition elements remains relatively constant as you move across a period because their outermost electrons are in different energy levels or subshells. The addition of protons does not significantly affect the size of the atom.

Moving across a period, the number of protons in the nucleus increases, leading to a greater positive charge. This increased positive charge exerts a stronger pull on the electrons, pulling them closer to the nucleus and resulting in a smaller atomic radius.

However, electron shielding, or the repulsion between electrons in different energy levels, also plays a role. As you move across a period, the number of electrons in the same energy level (shell) remains constant, providing consistent shielding effects. This partial counteraction to the increased positive charge contributes to the overall trend of decreasing atomic radius.

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Answer:21.3 m/s

Explanation:

Given

speed [tex]u=18.6 m/s[/tex]

mass of fish [tex]m_f=2.30 kg[/tex]

Altitude [tex]h=5.50 m[/tex]

Time taken to cover h

[tex]h=ut+\frac{at^2}{2}[/tex]

[tex]5.5=\frac{9.8\times t^2}{2}[/tex]

[tex]t^2=1.122[/tex]

[tex]t=1.05 s[/tex]

Vertical velocity after [tex]t=1.05 s[/tex]

[tex]v_y=0+gt[/tex]

[tex]v_y=9.8\times 1.05=10.38 m/s[/tex]

Horizontal velocity will remain same [tex]u=18.6 m/s[/tex]

Net velocity [tex]v_{net}=\sqrt{u^2+v_y^2}[/tex]

[tex]v_{net}=\sqrt{18.6^2+10.38^2}[/tex]

[tex]v_{net}=\sqrt{453.76}=21.30 m/s[/tex]

Answer:

Exophthalmos

Explanation:

Exophthalmos is a disorder which can be either bilateral or unilateral. Sometimes it is also known by other names like Exophthalmus, Excophthamia, Exobitism.

It is basically the bulging of eye anterior out of orbit which if left unattended may result in eye openings even while sleeping consequently resulting in comeal dryness and damage which ultimately may lead to blindness.

It is commonly caused by trauma or swelling of eye surrounding tissues resulting from trauma.

Answer:

Van der waals forces.

Explanation:

When we spread glue to stick any two substances as A and B with adhesives C. then there are adhesive force between substance A and C and adhesive force between substance B and C and cohesive force between C itself will act. In all adhesive and cohesive forces van der waals forces will apply at molecular level because there is no chemical bonding between adhesive and surface but lots of small attractive forces.

Answer:

Van der Waals force

Explanation:

The technical name given to the force that holds glue to its bonding materials is called Van der Waals force.

The forces of Van der Waals is defined by  attraction and repulsion between atoms, molecules, and surfaces and other intermolecular forces. They differ from covalent and ionic bond in that they are caused by correlations in the varying polarizations of the nearby particles (as a result of quantum dynamics).

Answer:

0.02268 m

Explanation:

[tex]m_1[/tex] = Mass of turkey slices = 0.1 kg

[tex]m_2[/tex] = Mass of plate = 0.4 kg

[tex]u_1[/tex] = Initial Velocity of turkey slices = 0 m/s

[tex]u_2[/tex] = Initial Velocity of plate = 0 m/s

[tex]v_1[/tex] = Final Velocity of turkey slices

[tex]v_2[/tex] = Final Velocity of plate

k = Spring constant = 200 N/m

x = Compression of spring

g = Acceleration due to gravity = 9.81 m/s²

Equation of motion

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.25+0^2}\\\Rightarrow v=2.21472\ m/s[/tex]

The final velocity of the turkey slice is 2.21472 m/s = v₁

For the spring

[tex]x=\frac{m_1g}{k}\\\Rightarrow x=\frac{0.1\times 9.81}{200}\\\Rightarrow x=0.004905\ m[/tex]

As the linear momentum is conserved

[tex]m_1v_1=(m_1+m_2)v_2\\\Rightarrow v_2=\frac{m_1v_1}{m_1+m_2}\\\Rightarrow v_2=\frac{0.1\times 2.21472}{0.1+0.4}\\\Rightarrow v_2=0.442944\ m/s[/tex]

Here the kinetic and potential energy of the system is conserved

[tex]\frac{1}{2}(m_1+m_2)v_2^2+\frac{1}{2}kx^2=\frac{1}{2}kA^2\\\Rightarrow A=\sqrt{\frac{(m_1+m_2)v_2^2+kx^2}{k}}\\\Rightarrow A=\sqrt{\frac{(0.1+0.4)0.442944^2+200\times 0.004905^2}{200}}\\\Rightarrow A=0.02268\ m[/tex]

The amplitude of oscillations is 0.02268 m

Answer:

Speed by walking is 33 miles per hour

And speed of by car is 88 miles per hour

Explanation:

We have given that it takes same time to walk 1212 miles as 3232 miles by car

Now let the speed by walk is x

As speed by car is 55 miles per hour faster than by walk = x+55

As time is same and we know that time is given as [tex]time=\frac{distance}{speed}[/tex]

So [tex]\frac{1212}{x}=\frac{3232}{x+55}[/tex]

[tex]1212(x+55)=3232x[/tex]

[tex]1212x+66660=3232x[/tex]

x = 33 miles per hour

So speed by walking is 33 miles per hour

And so speed of car = 33+55 =88 miles per hour

Answer:

Phase Difference

Explanation:

When the sound waves have same wavelength, frequency and amplitude we just need the phase difference between them at a particular location to determine whether the waves are in constructive interference or destructive interference.

Interference is a phenomenon in which there is superposition of two coherent waves at a particular location in the medium of propagation.

When the waves are in constructive interference then we get a resultant wave of maximum amplitude and vice-versa in case of destructive interference.

Answer:

Doug speed will be 65 km/hr

Explanation:

Let the Thor's speed is x km/hr

So Doug's speed = x+1 km/hr

We have given that Doug and Thor take same time to cover 390 km and 384 km respectively

We know that time is given by

[tex]time=\frac{distance}{speed}[/tex]

So time taken by Doug to cover the distance

[tex]time=\frac{390}{x+1}[/tex]

And time taken by Thor to cover the distance

[tex]time=\frac{384}{x}[/tex]

As both times are equal

So [tex]\frac{390}{x+1}=\frac{384}{x}[/tex]

[tex]6x=384[/tex]

[tex]x=64km/hr[/tex]

So Doug speed will be 64+1 = 65 km/hr

When the pulling force points at an angle θ above the horizontal, the frictional force acting on the box is -20N. By using the equation for frictional force and the weight of the box, we can determine that the coefficient of friction is 0.133. To find the angle θ, we use trigonometric ratios and find that it is 150°.

Given that the weight of the box is 150N and the pulling force has a magnitude of 110N, we can determine the angle θ at which the pulling force is directed. Let's assume the angle θ is above the horizontal. The weight of the box, 150N, is equal to the normal force acting on the box. The frictional force between the box and the floor can be calculated as the difference between the force of the pulling and the weight of the box, which is 110N - 150N = -40N. Since the kinetic frictional force acting on the box is twice as large when the pulling force points horizontally, the frictional force when the pulling force points at an angle θ is -20N.

We can use the equation for frictional force, which is F_friction = μN, where F_friction is the frictional force, μ is the coefficient of friction, and N is the normal force. As the frictional force is -20N, we can substitute this value into the equation and solve for the coefficient of friction. Therefore, -20N = μ(150N), which gives us μ = -20N/150N = -0.133. Since the coefficient of friction is always positive, the actual value of μ is 0.133.

Now, let's use trigonometric ratios to find the angle θ. Since the weight of the box acts vertically downward and the pulling force has a horizontal component of 110N and a vertical component of -150N × sin(θ), the vertical components of the weight and the pulling force must cancel each other. Therefore, -150N × sin(θ) = 150N, which simplifies to sin(θ) = -1/2. Taking the inverse sine of -1/2, we get θ = -30° or 150°. However, since the pulling force is directed above the horizontal, the angle must be 150°.

Answer:

Sensory transduction

Explanation:

The term sensory transduction refers to the conversion process where the sensory energy is converted in order to change the potential of a membrane.

In other words, it can defined as the process of energy conversion such that stimulus can be transmitted or received by the sensory receptors and the nervous system may initiate with the sensory receptors.

Transduction takes in all of the five receptors of the body. Thus skin is also one of the receptors and hence conversion of heat energy into impulses takes place with the help of thermo-sensory neuron.

Answer:

a)v= 1.6573 m/s

Explanation:

a) Considering center of the disc as our reference point. The potential energy as well as the kinetic energy are both zero.

let initially the block is at a distance h from the reference point.So its potential energy is -mgh as its initial KE is zero.

let the block descends from h to h'

During this descend

PE of the block = -mgh'            {- sign indicates that the block  is descending}                

KE= 1/2 mv^2

rotation KE of the disc=  1/2Iω^2

Now applying the law of conservation of energy we have

[tex]-mgh = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2-mgh'[/tex]

[tex]mg(h'-h) = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2[/tex] ................i

Rotational inertia of the disc = [tex]\frac{1}{2}MR^2[/tex]

Angular speed ω =[tex]\frac{v}{R}[/tex]

by putting vales of  ω and I we get

so, [tex]\frac{1}{2}I\omega^2= \frac{1}{4}Mv^2[/tex]

Now, put this value of rotational KE in the equation i

[tex]mg(h'-h) = \frac{1}{4}(2m+M)v^2[/tex]

⇒[tex]v= \sqrt{\frac{4mg(h'-h)}{2m+M} }[/tex]

Given that (h'-h)= 0.5 m M= 360 g m= 70 g

[tex]v= \sqrt{\frac{4\times70\times 9.81\times 0.5}{140+360} }[/tex]

v= 1.6573 m/s\

b) The rotational Kinetic energy of the disc is independent of its radius hence on changing the radius there is no change in speed of the block.

Answer:

The speed of the block is 1.65 m/s.

Explanation:

Given that,

Radius = 12 cm

Mass  of pulley= 360 g

Mass of block = 70 g

Distance = 50 cm

(a). We need to calculate the speed

Using energy conservation

[tex]P.E=K.E[/tex]

[tex]P.E=mgh[/tex]

[tex]K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2[/tex]

[tex]K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\times(\dfrac{v}{r})^2[/tex]

[tex]K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times0.5Mr^2\times(\dfrac{v}{r})^2[/tex]

[tex]K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times0.5M\times v^2[/tex]

[tex] K.E=\dfrac{1}{2}v^2(m+0.5M)[/tex]

Put the value into the formula

[tex]mgh=\dfrac{1}{2}v^2(m+0.5M)[/tex]

[tex]v^2=\dfrac{2mgh}{m+0.5M}[/tex]

[tex]v=\sqrt{\dfrac{2mgh}{m+0.5M}}[/tex]

[tex]v=\sqrt{\dfrac{2\times70\times10^{-3}\times9.8\times50\times10^{-2}}{70\times10^{-3}+0.5\times360\times10^{-3}}}[/tex]

[tex]v=1.65\ m/s[/tex]

(b), We need to calculate the  speed of the block

When r = 5.0 cm

Here, The speed of the block is independent of radius of pulley.

Hence, The speed of the block is 1.65 m/s.

Answer:

A) 0.5cm  B) 1/30

Explanation:

The weight of the man = mass * acceleration due to gravity where the mass is 78kg and acceleration due to gravity is 9.81m/s^2

W = m * g = 78 * 9.81= 686.7 N

The force acting on the tendon is 8 times of the weight

Force = 8 * weight of the body = 8 * 686.7 = 5493.6 N

Young modulus of the tendon(e) = (F/A)/ (DL/L) where A is the cross sectional area in square meters, DL is the change in length of the tendon in meters and L is the original length of the tendon

e = (FL)/(ADL) cross multiply and make DL subject of the formula

DL = (FL) / (AL)

Convert the cross sectional area A into square meters and the length also

A = 110 / 1000000 since 1/1000 m = 1mm, 1/1000000 m^2 = 1 mm^2 and 1/100m = 1 cm

A = 0.00011 m ^2 and L = 0.15m

Substitute the values in the derived equation

DL = (5493.6 * 0.15)/ (1.5 * 10^ 9 * 1.1* 10^-4)

DL = 824.04 / 1.65 * 10^ 5

DL = 499.42 * 10^-5 = 499.42 *10^ -5 / 100 to convert it to meters

DL = 0.49942cm approx 0.5cm

B) fraction of the DL to L  = 0.5 / 15 = 1/30

Answer:

0.429 L of water

Explanation:

First to all, you are not putting the value of the energy given to vaporize water, so, to explain better this problem, I will assume a value of energy that I took in a similar exercise before, which is 970 kJ.

Now, assuming that the water density is 1 g/mL, this is the same as saying that 1 g of water = 1 mL of water

If this is true, then, we can assume that 1 kg of water = 1 L of water.

Knowing this, we have to use the expression to get energy which is:

Q = m * ΔH

Solving for m:

m = Q / ΔH

Now "m" is the mass, but in this case, the mass of water is the same as the volume, so it's not neccesary to do a unit conversion.

Before we begin with the calculation, we need to put the enthalpy of vaporization in the correct units, which would be in grams. To do that, we need the molar mass of water:

MM = 18 g/mol

The enthalpy in mass:

ΔH = 40.7 kJ/mol / 18 g/mol = 2.261 kJ/g

Finally, solving for m:

m = 970 / 2.261 = 429 g

Converting this into volume:

429 g = 429 mL

429 / 1000 = 0.429 L of water

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The correct answers are as follows:

1) The magnitude of the net force on the block at the moment it is released is given by Hooke's Law for springs in parallel, which states that the net force is the sum of the forces exerted by each spring. Since the block is pulled to the left, the force exerted by the left spring is to the right, and the force exerted by the right spring is to the left. Thus, the net force[tex]\( F \)[/tex] is:

[tex]\[ F = k_{\text{left}} \cdot x + k_{\text{right}} \cdot x \] \[ F = (31 \, \text{N/m} + 53 \, \text{N/m}) \cdot 0.27 \, \text{m} \] \[ F = 84 \, \text{N/m} \cdot 0.27 \, \text{m} \] \[ F = 22.68 \, \text{N} \][/tex]2) The effective spring constant [tex]\( k_{\text{eff}} \)[/tex] of the two springs in parallel is the sum of the individual spring constants:

[tex]\[ k_{\text{eff}} = k_{\text{left}} + k_{\text{right}} \] \[ k_{\text{eff}} = 31 \, \text{N/m} + 53 \, \text{N/m} \] \[ k_{\text{eff}} = 84 \, \text{N/m} \][/tex]

3) The period of oscillation [tex]\( T \)[/tex] for a mass-spring system is given by:

[tex]\[ T = 2\pi \sqrt{\frac{m}{k_{\text{eff}}}} \] \[ T = 2\pi \sqrt{\frac{7.4 \, \text{kg}}{84 \, \text{N/m}}} \] \[ T = 2\pi \sqrt{\frac{7.4}{84}} \] \[ T = 2\pi \sqrt{0.0881} \] \[ T = 2\pi \cdot 0.2968 \] \[ T \approx 1.86 \, \text{s} \][/tex]

4) The time it takes for the block to return to equilibrium for the first time is half of the period of oscillation:

[tex]\[ t = \frac{T}{2} \] \[ t = \frac{1.86 \, \text{s}}{2} \] \[ t \approx 0.93 \, \text{s} \][/tex]

5) The speed of the block as it passes through the equilibrium position can be found using the conservation of energy. The total mechanical energy is constant, so the potential energy at the release point is converted into kinetic energy at the equilibrium position:

[tex]\[ \frac{1}{2} k_{\text{eff}} x^2 = \frac{1}{2} m v^2 \] \[ k_{\text{eff}} x^2 = m v^2 \] \[ v^2 = \frac{k_{\text{eff}} x^2}{m} \] \[ v = \sqrt{\frac{k_{\text{eff}} x^2}{m}} \] \[ v = \sqrt{\frac{84 \, \text{N/m} \cdot (0.27 \, \text{m})^2}{7.4 \, \text{kg}}} \] \[ v = \sqrt{\frac{84 \cdot 0.0729}{7.4}} \] \[ v = \sqrt{\frac{6.1296}{7.4}} \] \[ v \approx \sqrt{0.8284} \] \[ v \approx 0.909 \, \text{m/s} \][/tex]

6) The magnitude of the acceleration[tex]\( a \)[/tex]of the block as it passes through equilibrium is given by Newton's second law:

[tex]\[ F = m \cdot a \] \[ a = \frac{F}{m} \] \[ a = \frac{22.68 \, \text{N}}{7.4 \, \text{kg}} \] \[ a \approx 3.065 \, \text{m/s}^2 \][/tex]

7) The position[tex]\( x(t) \) of the block at a time \( t \)[/tex] after it is released can be found using the equation for simple harmonic motion: [tex]\[ x(t) = A \cos(2\pi f t) \] \[ x(t) = 0.27 \cos\left(\frac{2\pi}{1.86} \cdot 1.06\right) \] \[ x(t) = 0.27 \cos(3.61\pi) \] \[ x(t) \approx 0.27 \cdot (-1) \] \[ x(t) \approx -0.27 \, \text{m} \][/tex]

8) The net force on the block at time [tex]\( t \)[/tex] is given by Hooke's Law, taking into account the position of the block:

[tex]\[ F(t) = k_{\text{eff}} \cdot x(t) \] \[ F(t) = 84 \, \text{N/m} \cdot (-0.27 \, \text{m}) \] \[ F(t) \approx -22.68 \, \text{N} \][/tex]

9) The total energy stored in the system[tex]\( E \)[/tex] is the potential energy at the maximum displacement, which is equal to the kinetic energy at the equilibrium position:

[tex]\[ E = \frac{1}{2} k_{\text{eff}} x^2 \] \[ E = \frac{1}{2} \cdot 84 \, \text{N/m} \cdot (0.27 \, \text{m})^2 \] \[ E = 42 \, \text{N/m} \cdot 0.0729 \, \text{m}^2 \] \[ E \approx 3.065 \, \text{J} \][/tex]

10) The period of oscillation is independent of the amplitude of the motion and depends only on the mass and the spring constant. Therefore, if the block had been given an initial push, the period of oscillation would not change. The correct answer is: the period would not change.

Answer:

v1 = 2.76 m/s and v2 = - 0.32 m/s

Explanation:

m1 = 0.140 kg

m2 = 0.299 kg

u1 = 0.80 m/s

u2 = - 2.28 m/s

Let the speed after collision is v1 and v2.

Use conservation of momentum

m1 x u1 + m2 x u2 = m1 x v1 + m2 x v2

0.140 x 0.80 - 0.299 x 2.28 = 0.140 x v1 + 0.299 x v2

0.112 - 0.68 = 0.14 v1 + 0.299 v2

0.14 v1 + 0.299 v2 = - 0.568 ..... (1)

By the use of coefficient of restitution, the value of e = 1 for elastic collision

[tex]e=\frac{v_{1}-v_{2}}{u_{2}-u_{1}}[/tex]

u2 - u1 = v1 - v2

- 2.28 - 0.8 = v1 - v2

v1 - v2 = 3.08

v1 = 3.08 + v2

Put in equation (1)

0.14 (3.08 + v2) + 0.299 v2 = - 0.568

0.43 + 0.44 v2 = - 0.568

v2 = - 0.32 m/s

and

v1 = 3.08 - 0.32 = 2.76 m/s

Thus, v1 = 2.76 m/s and v2 = - 0.32 m/s

Answer:

a. C will decrease

b. Q will remain the same

c. E will decrease

d. Delta-V will increase

Explanation:

Justification for C:

As we know that for parallel plate capacitors, capacitance is calculated using:

C = (ϵ_r *  ϵ_o * A) / d   - Say it Equation 1

Where:

ϵ_r - is the permittivity of the dielectric material between two plates

ϵ_o - Electric Constant

A - Area of capacitor's plates

d - distance between capacitor plates

From equation 1 it is clear that capacitance will decrease if distance between the plates will be increased.

Justification of Q

As charge will not be able to travel across the plates, therefore it will remain the same

Justification of E

As we know that E = Delta-V / Delta-d, thus considering Delta-V is increasing on increasing Delta-d (As justified below) as both of these are directly proportional to each other, therefore Electric field (E) will remain constant as capacitors' plates are being separated.

Moreover, as the E depends on charge density which remains same while plates of capacitor are being separated therefore E will remain the same.

Justification of Delta-V

As we know that Q = C * V, therefore considering charge remains the same on increasing distance between plates, voltage must increase to satisfy the equation.

Answer:

There is a constant emf induced in the loop.

Explanation:

In the uniform magnetic field suppose the rectangular wire loop of length L and width b is moved out with a uniform velocity v. suppose any instance x length of the loop is out of the magnetic field and L-x length is inside the loop.

Area of loop outside the field = b(L-x)

we know that flux φ= BA

B= magnitude of magnetic field , A=  area

and emf [tex]\epsilon= \frac{d\phi}{dt}[/tex]

[tex]\epsilon=B\frac{dA}{dt}[/tex]

[tex]\epsilon=B\frac{db(L-x)}{dt}[/tex]

[tex]\epsilon=Bb\frac{d(L-x)}{dt}[/tex]

B,b and L are constant and dx/dt = v

⇒ε = -Bbv

which is a constant hence There is a constant emf induced in the loop.

The rock will be at 90.4 m from the top of the cliff.

Explanation:

The rock is thrown with the “initial velocity” 3 m/s. We need to find how much distance does the rock traveled in 4 seconds (t).

From the “kinematic equations” take

[tex]s=u t+\frac{1}{2} a t^{2}[/tex]

Where, “s” is distance traveled, “u” initial velocity of the object, “t” time the object traveled and “a” acceleration due to gravity is [tex]9.8 \mathrm{m} / \mathrm{s}^{2}.[/tex]

Substitute the given values in the above formula,

[tex]s=3 \times 4+\frac{1}{2} \times 9.8 \times 4^{2}[/tex]

[tex]s=12+\frac{1}{2} \times 9.8 \times 16[/tex]

[tex]s=12+\frac{1}{2} \times 156.8[/tex]

[tex]s=12+78.4[/tex]

[tex]s=90.4[/tex]

The rock is at height of 90.4 m from the top of the cliff.

Answer:

Distance in mm will be 0.3718 mm

Explanation:

We have given charge surface charge density [tex]\rho _s=5\mu c/m^2=5\times 10^{-6}\mu c/m^2[/tex]

We know that electric field due to surface charge density is given by

[tex]E=\frac{\rho _S}{2\epsilon _0}=\frac{5\times 10^{-6}}{2\times 8.85\times 10^{-12}}=2.824\times 10^5Volt/m[/tex]

We have given potential difference V = 105 volt

We know that potential difference is given by [tex]V=Ed[/tex]

So [tex]105=2.824\times 10^5\times d[/tex]

[tex]d=37.181\times 10^{-5}m=0.3718mm[/tex]

Answer:

(a)20.65g

(b)0.19m

Explanation:

(a) The total mass would be it's mass per length multiplied by the total lenght

0.355(50 + 23*0.355) = 20.65 g

(b) The center of mass would be at point c where the mass on the left and on the right of c is the same

Hence the mass on the left side would be half of its total mass which is 20.65/2 = 10.32 g

[tex]c(50 + 23c) = 10.32[/tex]

[tex]23c^2 + 50c - 10.32 = 0 [/tex]

[tex]c \approx 0.19m[/tex]

Answer:

ΔU= *-26 KJ

Explanation:

Given that

Work done by motor W= 30 KJ

Heat gains by motor Q= 4 KJ

Sign convention:

 If heat is added to the system then it is taken as positive and if heat is rejected from the system then it is taken as negative.

If work done by the system then it is taken as positive and if work is done on the system then it is taken as negative.

From first law of thermodynamics

Q = W + ΔU

ΔU=Change in internal energy

Q=Heat transfer

W=Work

Now by putting the values

4 = 30 + ΔU

ΔU= -26 KJ

Answer:

Internal energy ∆U=-26KJ

Explanation:

Given that:

Work done by the motor=+30KJ

Heat gained by the motor=+4KJ

In solving thermodynamical questions it is reasonable to use the sign convention this

Heat is positive if it is added to a system,but becomes negative if the system rejects heats.

Work is positive if the system does work,but becomes negative if work is done on the system.

Using the thermodynamics first law

∆U=Q-W

∆U= 4-30=-26KJ

Answer:

L = 3.8 m

Explanation:

As we know that the frequency of sound is given as

[tex]f = 686 Hz[/tex]

speed of the sound is given as

[tex]v = 332 + 0.6 t[/tex]

[tex]v = 332 + (0.6 \times 20)[/tex]

[tex]v = 344 m/s[/tex]

now we have wavelength of sound is given as

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

[tex]\lambda = \frac{344}{686}[/tex]

[tex]\lambda = 0.50 m[/tex]

now we have path difference at initial position given as

[tex]\Delta L = \sqrt{L^2 + d^2} - L[/tex]

[tex]\Delta L = \sqrt{3^2 + 2.5^2} - 2.5[/tex]

[tex]\Delta L = 3.9 - 2.5 = 1.4 m[/tex]

now we know that for minimum sound intensity we have

[tex]\Delta L = \frac{2N + 1}{2}\lambda[/tex]

[tex]\Delta L = \frac{2N + 1}{2}(0.50)[/tex]

so we have

N = 2

[tex]\Delta L = 1.25 m[/tex]

so we have

[tex]\sqrt{2.5^2 + L^2} - L = 1.25[/tex]

[tex]2.5^2 + L^2 = L^2 + 1.25^2 + 2.5L[/tex]

[tex]L = 1.875 m[/tex]

Now for N = 1

[tex]\Delta L = 0.75 m[/tex]

so we have

[tex]\sqrt{2.5^2 + L^2} - L = 0.75[/tex]

[tex]2.5^2 + L^2 = L^2 + 0.75^2 + 1.5L[/tex]

[tex]L = 3.8 m[/tex]

so the next minimum intensity will be at L = 3.8 m

Answer:

1.832 kgm^2

Explanation:

mass of potter's wheel, M = 7 kg

radius of wheel, R = 0.65 m

mass of clay, m = 2.1 kg

distance of clay from centre, r = 0.41 m

Moment of inertia = Moment of inertia of disc + moment f inertia of the clay

I = 1/2 MR^2 + mr^2

I = 0.5 x 7 x 0.65 x 0.65 + 2.1 x 0.41 x 0.41

I = 1.47875 + 0.353

I = 1.832 kgm^2

Thus, the moment of inertia is 1.832 kgm^2.

The moment of inertia of the system about the axis of spin is mathematically given as

I = 1.832 kgm^2

Question Parameter(s):

A potter's wheel has the shape of a solid uniform disk of mass of 7 kg
and a radius of 0.65 m
A small 2.1 kg lump of very dense clay

the wheel at a distance of 0.41 m from the axis.

Generally, the equation for the moment of inertia   is mathematically given as

I = 1/2 MR^2 + mr^2

I = 0.5 x 7 (0.65)^2 + 2.1 (0.41)^2

I = 1.47875 + 0.353

I = 1.832 kgm^2

In conclusion moment of inertia is

I = 1.832 kgm^2

Read more about Inertia

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Answer:

C. both forces have the same magnitude

Explanation:

Here the action force is equal to the reaction force in accordance with the Newton's third law of motion.

Also when we apply the conservation of momentum so that the momentum bullet and the momentum of the gun are equal and according to the second law of motion by Newton, we have force equal to the rate of change in momentum.

We have the equation for momentum as:

[tex]p=m.v[/tex]

Newton's second law is Mathematically given as:

[tex]F=\frac{dp}{dt}[/tex]

Momentum is constant and the reaction time is equal, so the force exerted will also be equal.

The correct answer is C. both forces have the same magnitude.

According to Newton's third law of motion, for every action, there is an equal and opposite reaction. This means that when the rifle fires the bullet, the force exerted by the rifle on the bullet is equal in magnitude to the force exerted by the bullet on the rifle.

 Let's denote the force exerted by the rifle on the bullet as [tex]\( F_{rb} \)[/tex]and the force exerted by the bullet on the rifle as[tex]\( F_{br} \).[/tex] According to Newton's third law:

[tex]\[ F_{rb} = -F_{br} \][/tex]

 The magnitudes of these forces are equal, even though they are in opposite directions.

 To calculate the magnitude of these forces, 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.

 Before the rifle is fired, both the bullet and the rifle are at rest, so their initial momenta are zero. After the rifle is fired, the bullet has a velocity of 300 m/s, and we can assume the rifle has a much smaller recoil velocity due to its much larger mass.

 The change in momentum of the bullet is:

[tex]\[ \Delta p_{bullet} = m_{bullet} \times v_{bullet} \][/tex]

[tex]\[ \Delta p_{bullet} = 0.00500 \, \text{kg} \times 300 \, \text{m/s} \][/tex]

[tex]\[ \Delta p_{bullet} = 1.5 \, \text{kg} \cdot \text{m/s} \][/tex]

 The change in momentum of the rifle is equal in magnitude and opposite in direction to the change in momentum of the bullet, assuming no other forces are acting on the system (like air resistance or friction):

[tex]\[ \Delta p_{rifle} = -\Delta p_{bullet} \][/tex]

[tex]\[ \Delta p_{rifle} = -1.5 \, \text{kg} \cdot \text{m/s} \][/tex]

 Since the time interval [tex]\( \Delta t \)[/tex] over which the forces are applied is the same for both the bullet and the rifle, the forces can be calculated using the impulse-momentum theorem:

[tex]\[ F_{rb} = \frac{\Delta p_{bullet}}{\Delta t} \][/tex]

[tex]\[ F_{br} = \frac{\Delta p_{rifle}}{\Delta t} \][/tex]

 Since [tex]\( \Delta p_{bullet} = -\Delta p_{rifle} \)[/tex], it follows that:

[tex]\[ F_{rb} = -F_{br} \][/tex]

 Therefore, the forces are equal in magnitude and opposite in direction, which is consistent with Newton's third law. The correct choice is C, both forces have the same magnitude.

Answer:168 W

Explanation:

Given

Power needed [tex]P=42 W[/tex]

initial Launch velocity is v

Energy of projectile when it is launched [tex]E=\frac{1}{2}mv^2[/tex]

[tex]Power=\frac{Energy}{time}[/tex]

[tex]Power=\frac{E}{t}[/tex]

[tex]42=\frac{\frac{1}{2}mv^2}{t}--------1[/tex]

Power when it is launched with 2 v

[tex]E_2=\frac{1}{2}m(2v)^2=\frac{4}{2}mv^2[/tex]

[tex]P=\frac{2mv^2}{t}---------2[/tex]

Divide 1 & 2 we get

[tex]\frac{42}{P}=\frac{1}{2\times 2}[/tex]

[tex]P=42\times 4=168 W[/tex]    

To accelerate the projectile to twice its launch speed, four times the power is needed.

To find the power needed to accelerate the projectile from rest to a launch speed of 2v in a time of t, we need to recognize that power is directly proportional to the change in kinetic energy. The change in kinetic energy from rest to launch speed v is given by KE = (1/2)mv^2, and the change in kinetic energy from rest to launch speed 2v is given by KE' = (1/2)m(2v)^2 = 4(1/2)mv^2 = 4KE.

Since power is directly proportional to the change in kinetic energy, the power needed to accelerate the projectile to a launch speed of 2v is four times the power needed to accelerate it to a launch speed of v. Therefore, the power needed is 4(42.0 W) = 168.0 W.

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Answer:45.24 m/s

Explanation:

Given

Height of Building h=24 m

Range of ball R=100 m

Considering Vertical motion of ball

using [tex]y=u_yt+\frac{a_yt^2}{2} [/tex]

initial vertical velocity is zero therefore [tex]u_y=0[/tex]

[tex]24=0+\frac{9.8\times t^2}{2}[/tex]

[tex]t=\sqrt{\frac{48}{9.8}}[/tex]

[tex]t=2.21 s[/tex]

Now considering Horizontal Motion

[tex]R=u_xt+\frac{a_xt^2}{2}[/tex]

[tex]100=u_x\times 2.21+0[/tex]  , as there is no horizontal acceleration

[tex]u_x=45.24 m/s[/tex]

The physics problem can be addressed by using the principles of projectile motion. The time of flight determined by the vertical motion is used to calculate the horizontal initial velocity. The initial velocity of the ball is approximately 45.24m/s.

This is a problem in Physics based on the principles of Projectile Motion. We need to determine the initial velocity of the ball. The key point in this problem is that the horizontal motion of the projectile (in this case, the ball) is determined purely by the initial horizontal velocity, and is unaffected by the vertical motion. This is called the independence of the horizontal and vertical motions.

The time the ball is in the air is governed entirely by its vertical motion. Thus, we can find the time of flight by using the equation for vertical motion: y = 1/2gt², where y is the vertical displacement (24m in this case), g is the acceleration due to gravity (9.81 m/s²), and t is the time. So, t = sqrt(2y/g) = sqrt(2*24/9.81) = 2.21s.

Using this time, we can find the initial horizontal velocity using the equation for horizontal motion: x = vxt where x is the horizontal displacement (100m in this case), vx is the horizontal velocity, and t is the time. Rearranging the equation we get: vx = x/t which is approximately 45.24m/s . So, the initial velocity of the ball is around 45.24m/s

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Answer:

P=144.74W

Explanation:

We can model the power output of a resistance by using the following formula:

[tex]P=\frac{V^{2}}{R}[/tex]

Wehre P is the power output, V is the rms voltage and R is the resistance. The resistance of the space heater will remain the same, so we can calculate it from the power output in Germany and its rms voltage. So when solving for R, we get:

[tex]R=\frac{V^{2}}{P}[/tex]

and we can now use the provided data:

[tex]R=\frac{(230V)^{2}}{531.7W}[/tex]

which yields:

R= 99.49 Ω

Once we know what the heater's resistance is, we can now go ahead and calculate the power outpor of the heater in the U.S.

[tex]P=\frac{V^{2}}{R}[/tex]

so

[tex]P=\frac{(120V)^{2}}{99.49\Omega}[/tex]

P=144.74W

Answer:

183.75641 Joules

Explanation:

F = Force of the vacuum cleaner = 84.5 N

s = Displacement of the vacuum cleaner = 2.62 m

[tex]\theta[/tex] = Angle the force makes with the horizontal = 33.9°

Work done is given by

[tex]W=F\times scos\theta\\\Rightarrow W=84.5\times 2.62\times cos33.9\\\Rightarrow W=183.75641\ J[/tex]

The work done by the force of the vacuum cleaner is 183.75641 Joules

The work done by the man pushing the vacuum cleaner with a force of 84.5 N at an angle of 33.9° over a distance of 2.62 m is approximately 184.8 joules.

To calculate the work done by a force, you can use the formula W = F × d × cos(θ), where W is the work done, F is the magnitude of the force, d is the distance the object moves, and θ is the angle the force makes with the horizontal direction of movement. In this case, the man pushes a vacuum cleaner with a force of 84.5 N at an angle of 33.9° over a distance of 2.62 m. We multiply the force by the distance and the cosine of the angle to find:

W = 84.5 N × 2.62 m × cos(33.9°)

Calculating cosine of 33.9 degrees and multiplying with the force and distance, we get:

W = 84.5 × 2.62 × 0.8326 ≈(joules)

The man does approximately 184.8 joules of work pushing the vacuum cleaner.

Answer:

β = 16.7°

Explanation:

The sum of forces on the x-axis are:

[tex]T*sin\beta=m*a[/tex]

The sum of forces on the y-axis are:

[tex]T*cos\beta=m*g[/tex]

By dividing x-axis by the y-axis equation:

[tex]tan\beta=a/g[/tex]

Solving for β:

[tex]\beta=atan(a/g)[/tex]

β = 16.7°