Lake Erie contains roughly 4.00 ✕ 1011 m3 of water. (a) How much energy is required to raise the temperature of that volume of water from 17.8°C to 21.6°C? (Assume the density of this water to be equal to that of water at 20°C and 1 atm.) J (b) How many years would it take to supply this amount of energy by using the 1,400-MW exhaust energy of an electric power plant? yr

Answer:

Density of unknown liquid is [tex]2047 kg/m^3[/tex]

Explanation:

When rock is suspended in air then the weight of the rock is counter balanced by the tension force in the string

So here we have

[tex]T = mg = 43.8 N[/tex]

now when the rock is immersed in water then the tension in the string is and buoyancy force due to water is counter balanced by the weight of the object

so here we have

[tex]T_1 + F_b = mg[/tex]

[tex]31.3 + F_b = 43.8[/tex]

[tex]F_b = 43.8 - 31.3 = 12.5 N[/tex]

now we have

[tex](1000)V(9.81) = 12.5[/tex]

[tex]V = 1.27 \times 10^{-3} m^3[/tex]

now when the rock is immersed into other liquid then we have

[tex]T_2 + F_b' = mg[/tex]

[tex]18.3 + F_b' = 43.8[/tex]

[tex]F_b' = 25.5 N[/tex]

now we have

[tex]\rho(1.27 \times 10^{-3})(9.81) = 25.5[/tex]

[tex]\rho = 2047 kg/m^3[/tex]

Explanation:

1. The entire span of possible sound waves is called the acoustic spectrum. It is subdivided into infrasonic sounds, audible sounds, and ultrasonic sounds.

2. The difference between a musical note and another note at twice the frequency is called an octave.

3. Sound intensity varies with the inverse square of distance.

The possibility of sound frequencies is called the sound spectrum. Notes at double the frequency are an octave higher, and the perception of sound varies with distance from the source.

The total possibility of sound frequencies is referred to as the sound spectrum. The difference between a single musical note and another note at twice the frequency of the first is known as an octave in music. The note that is twice the frequency appears higher in pitch.

The relationship of sound to the distance between the source and receiver affects the perceived loudness of the sound. This is due to the way sound waves spread out and weaken as they travel away from the source. As a result, the further somebody is from the source of the sound, the quieter it will appear.

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Answer:

The speed of the cylinder has decreased

Explanation:

Given data:

Distance between the center of the turn table and the cylinder = R

Distance between the center of the turn table and the cylinder when it is moved to a new point, R' = R/2

the tangential velocity in circular motion is given as:

[tex]V=\frac{2\pi R}{T}[/tex]

where T is the time

now for R = R' = R/2

the new velocity V' comes as:

[tex]V'=\frac{2\pi \frac{R}{2}}{T}[/tex]

or

[tex]V'=\frac{\pi R}{T}[/tex]

thus the velocity becomes half or we can say the velocity decreases

Now, the acceleration (a) in circular motion is given as:

a = V²/R

now for

R = R' = R/2

the velocity V' = V/2

thus,

new acceleration a' comes as:

[tex]a = \frac{(\frac{V}{2})^2}{R/2}[/tex]

or

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

thus,

the acceleration decreases by 2 times

The statement that accurately describes the motion of the cylinder at the new location is : ( 1 )  and ( 3 )

Given data :

Distance between cylinder and center of the turn table = R

Distance between cylinder and turntable when moved = R/2

where ;

Tangential Velocity/speed =  [tex]\frac{2\pi R}{T}[/tex]  

R = distance between objects

T = time

∴ The initial Tangential velocity of the cylinder at rest ( V ) = [tex]\frac{2\pi R}{T}[/tex]  ----- ( 1 )

 while the

Tangential velocity/speed of the cylinder when it moves ( V' ) = [tex]\frac{2\pi R/2}{T}[/tex]

                                                                                                       = [tex]\frac{\pi R}{T}[/tex] ----- ( 2 )

Therefore comparing equations ( 1 ) and ( 2 )  the speed of the cylinder has decreased as the cylinder moves from R to R/2 .

Also Given that Tangential acceleration = v² / R .

Since velocity decreases after the cylinder moves, the magnitude of the acceleration of the cylinder will decrease as well.

Hence we can conclude that the acceleration and speed of the cylinder has decreased with the movement of the cylinder.

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The frequency of a wave is measured in hertz (Hz). Hence the correct answer is option E

The frequency of a wave is measured in hertz (Hz). Hertz is the unit used to measure the number of waves passing by a point in one second. This means that option B) 'measured in hertz (Hz)' is the correct answer. Frequency is not equal to the speed of the wave divided by the wavelength of the wave, as stated in option D). While option C) 'the number of peaks passing by any point each second' is related to frequency, it does not encompass the entire definition. Therefore, option E) 'all of the above' is also incorrect.

Answer: If it has ions, it is an electrolyte

Explanation:

Let's start by explaining that electrolytes are compounds that contain charged particles or ions, which can be cations (positive ions) or anions (negative ions).

So, it is this composition that makes an electrolytic material conduct electricity.

In this sense, the way to identify if a material is an electrolyte or not, is knowing whether it is composed of ions or not.

Explanation:

The equation of motion of an object is given by :

[tex]h(t)=-16t^2+112t+128[/tex]

Where

t is the time in seconds

We need to find the time when the object hits the ground. When the object hits the ground, h(t) = 0

So,

[tex]-16t^2+112t+128=0[/tex]

[tex]-t^2+7t+8=0[/tex]

On solving above equation using online calculator, t = 8 seconds. So, the object hit the ground after 8 seconds. Hence, this is the required solution.

Answer:

Part a)

[tex]EMF = 5.6 \times 10^{-3} V[/tex]

Part b)

Since the radius is decreasing so induced current will increase the flux through the coil

So it would be clockwise in direction

Explanation:

As we know that magnetic flux linked with the coil is given as

[tex]\phi = \pi r^2 B[/tex]

now the rate of change in flux is given as

[tex]\frac{d\phi}{dt} = 2\pi r \frac{dr}{dt} B[/tex]

now we know that circumference is decreasing at rate of 15 cm/s

so here we know the length of circumference as

[tex]C = 2\pi r[/tex]

So rate of change in circumference is

[tex]\frac{dC}{dt} = 2\pi \frac{dr}{dt}[/tex]

[tex]\frac{1}{2\pi}(15 cm) = \frac{dr}{dt}[/tex]

final length of circumference at t = 8 s

[tex]C = 167 - (15)(8) = 47[/tex]

Part a)

Now the induced EMF is given as

[tex]EMF = (2\pi r)(\frac{1}{2\pi})(0.15)(0.5)[/tex]

[tex]EMF = (0.47)(\frac{1}{2\pi})(0.15)(0.5)[/tex]

[tex]EMF = 5.6 \times 10^{-3} V[/tex]

Part b)

Since the radius is decreasing so induced current will increase the flux through the coil

So it would be clockwise in direction

The problem is solved using Faraday's Law of Electromagnetic Induction: the induced emf equals to the change in magnetic flux divided by change in time. Then, Lenz's law is used to determine the direction of the induced current.

The problem you're dealing with is an application of Faraday's Law of Electromagnetic Induction. This law states that the electromotive force (or emf) induced in a circuit is equivalent to the rate of change of magnetic flux through that circuit.

For Part A, you know that the rate of change of the circumference is constant and you can convert that to a rate of change of radius (since circumference = 2πr). Using these formulas, you can determine the rate of change of area as πr(dr/dt), where dr/dt is the rate of change of radius. The induced emf is proportional to the rate of change of magnetic flux, which is the product of magnetic field strength and area. Thus, induced emf equals to (change in flux/change in time), or -Bπr(dr/dt).

For part B, you apply Lenz's law, which states that the direction of induced current is such that the magnetic field due to it opposes the change in the initial magnetic field. Since the field is getting smaller (since the loop is contracting), then the current will act in such a way as to try to keep the size of the field the same. This means going in the counterclockwise direction.

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1. 8.11 m/s

In order to find the speed of the bird, we need to combine both the horizontal and the vertical component of its velocity.

The horizontal component is equal to the tangential velocity of the circular motion of the bird. We have:

r = 6.00 m radius of the circle

T = 5.00 s period of the circular motion

So the tangential velocity (=horizontal component of the velocity) is

[tex]v_x=\frac{2\pi r}{T}=\frac{2\pi (6.00 m)}{5.00 s}=7.54 m/s[/tex]

The vertical component of the velocity is given by the problem

[tex]v_y = 3.00 m/s[/tex]

So the speed of the bird is the magnitude of its velocity:

[tex]v=\sqrt{v_x^2 + v_y^2}=\sqrt{(7.54 m/s)^2+(3.00 m/s)^2}=8.11 m/s[/tex]

2. 9.48 m/s^2

The only component of the acceleration of the bird is the one that determine the circular motion in the horizontal plane - so it is the centripetal acceleration.

The centripetal acceleration of the bird is given by:

[tex]a=\frac{v_x^2}{r}[/tex]

where

[tex]v_x = 7.54 m/s[/tex] is the tangential velocity of the bird

r = 6.00 m is the radius of the circular trajectory

Substituting,

[tex]a=\frac{(7.54 m/s)^2}{6.00 m}=9.48 m/s^2[/tex]

3. Towards the centre of the circle

The bird is moving upward at constant velocity and in a straight line, so there is no component of the acceleration in the vertical direction.

In the horizontal plane, however, the bird is moving in a uniform circular motion: this means that the tangential acceleration is zero, while there is a centripetal acceleration (calculated in the previous part of the exercise). The direction of the centripetal acceleration is always towards the centre of the circular trajectory: so, the acceleration of the bird has exactly the same direction as the centripetal acceleration, i.e. towards the centre of the circle.

4. [tex]21.7^{\circ}[/tex]

We said that the components of the velocity of the bird are:

[tex]v_x = 7.54 m/s[/tex] in the horizontal direction

[tex]v_y = 3.00 m/s[/tex] in the vertical direction

So we can find the angle of the velocity with the horizontal by using the equation:

[tex]\theta = tan^{-1} (\frac{v_y}{v_x})[/tex]

Substituting the values in, we find

[tex]\theta = tan^{-1} (\frac{3.00 m/s}{7.54 m/s})=21.7^{\circ}[/tex]

Answer: 3 seconds

Explanation:

Since they provide the same impulse,

Impulse= Ft

F1 = 6N

t1 = 2s

F2= 4N

t2= ?

F1= Force of first engine

t1= time elapsed by first engine

F2= force of second engine

t2= time elapsed by second engine

F1t1 = F2t2

6 × 2 = 4t2

t2= 12/4

t2 = 3 seconds

This second engine fires for 3 s

[tex]\texttt{ }[/tex]

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

thrust of first engine = F₁ = 6 N

elapsed time of first engine = t₁ = 2 s

thrust of second engine = F₂ = 4 N

Asked:

elapsed time of second engine = t₂ = 2 s

Solution:

We will use Newton's Law of Motion to solve this problem as follows:

[tex]\Sigma F = ma[/tex]

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

[tex]\Sigma F = I \div t[/tex]

[tex]\boxed {I = \Sigma F \times t}[/tex] → Impulse Formula

[tex]\texttt{ }[/tex]

Two rocket engines provide the same impulse :

[tex]I_1 = I_2[/tex]

[tex]\Sigma F_1 \times t_1 = \Sigma F_2 \times t_2[/tex]

[tex]6 \times 2 = 4 \times t_2[/tex]

[tex]12 = 4 \times t_2[/tex]

[tex]t_2 = 12 \div 4[/tex]

[tex]\boxed {t_2 = 3 \texttt{ s}}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Dynamics

Answer:

Maximum distance of image from mirror is equal to focal length of the mirror

Explanation:

As we know by the equation of mirror we have

[tex]\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}[/tex]

here we know for convex mirror

object position is always negative as it will be placed behind the mirror always

while the focal length of the convex mirror is always taken positive

So here we have

[tex]\frac{1}{d_i} + \frac{1}{-d_o} = \frac{1}{f}[/tex]

[tex]\frac{1}{d_i} = \frac{1}{d_o} + \frac{1}{f}[/tex]

so here maximum value of image distance is equal to focal length of the mirror

Answer:

Part a)

2)  >  1)  >  3)

Part b)

1)  =  2)  =  3)

Explanation:

Due to charge induction the magnitude of charge on the inner surface of the outer shell is having same charge as that of the small sphere inside but the sign of charge must be opposite.

So here we can say

1)+ 4q, 0

so inner surface has charge - 4q and outer surface charge is +4q

2) -6q , +10q

so inner surface charge is +6q, outer surface charge is +4q

3) +16q , -12q

so inner surface charge is -16q, outer surface charge is +4q

Part a)

situations in which inner surface charge is arranged in decreasing order is given as

2)  >  1)  >  3)

Part b)

Situations in which outer surface charge is arranged in decreasing order is given as

1)  =  2)  =  3)

Situation (1) has the most positive charge on the inner surface of the shell, while situation (2) has the most positive charge on the outer surface.

To rank the situations according to their charge on the inner and outer surfaces of the metallic spherical shell, we need to consider the net charges on the small charged ball and the shell. Let's analyze each situation:

Therefore, ranking the situations according to the charge on the inner surface would be: (1), (3), (2) - from most positive to least positive. For the outer surface, the ranking would be: (2), (1), (3) - from most positive to least positive.

The tension in each rope supporting the child on the swing can be calculated by summing the forces (weight of the child and the pulling force) and dividing by 2, because the force is being equally distributed among the ropes due to the symmetry of the situation. This gives a tension of approximately 130N in each rope. The closest answer to this is B: 120N.

This problem relates to forces and tension. As the child is at rest, the total net force on the child is zero. This means that the sum of the upward forces (tension in the ropes) is equal to the downward forces (the weight of the child and the horizontal force pulling him backwards).

The weight of the child can be calculated using the formula Weight = m*g, where m is the mass and g is the acceleration due to gravity. In this case, the weight of the child is 160 N. As for the horizontal pull, it is given as 100N.

The combination of these two forces is then equally distributed across the two ropes holding the swing, due to the symmetry of the situation. Therefore, we sum the forces (160N for weight and 100N for pull) and divide by 2 to find the tension in each rope.

The calculation is, therefore, (160N + 100N) / 2 = 130N.

The answer closest to this value is B: 120N. Without more precise numbers, we choose the closest value.

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Answer:

Decreasing rate of thickness when x=3 in is [tex]\dfrac{dx}{dt}= -7.28\times 10^{-3} in/min[/tex]

Explanation:

Lets assume that thickness of ice over the spherical iron ball =x

So radius diameter of Sphere=5+x in

    Inner radius=5 in

So volume V=[tex]\dfrac{4}{3}\pi [(5+x )^3-5^3][/tex]

         V=[tex]\dfrac{4}{3}\pi [x^3+75x^2+15x][/tex]

Now given that ice melts rate=[tex] 15in^3/min[/tex]

[tex]\dfrac{dV}{dt}= -15in^3/min[/tex]

[tex]\dfrac{dV}{dt}=\dfrac{4}{3}\pi [3x^2+150x+15]\dfrac{dx}{dt}[/tex]

When x=3 in [tex]\dfrac{dV}{dt}= -15in^3/min[/tex]

[tex]-15=\dfrac{4}{3}\pi [3\times 3^2+150\times 3+15]\dfrac{dx}{dt}[/tex]

[tex]\dfrac{dx}{dt}= -7.28\times 10^{-3} in/min[/tex]

So decreasing rate of thickness when x=3 in is [tex]\dfrac{dx}{dt}= -7.28\times 10^{-3} in/min[/tex]

Answer:

7.0 m

Explanation:

We can solve the problem by using the work-energy theorem, which states that the work done on the block is equal to the block's change in kinetic energy:

[tex]W=K_f - K_i[/tex]

where:

W is the work done (by gravity, since it is the only force acting on the block)

[tex]K_i[/tex] is the initial kinetic energy

[tex]K_f = 0[/tex] is the final kinetic energy, which is zero since the block comes to a stop

The work done by gravity against the block is:

[tex]W=-(mgsin \theta) d[/tex]

where

m is the mass of the block

g = 9.8 m/s^2 is the acceleration due to gravity

[tex]\theta=20.5^{\circ}[/tex] is the angle of the incline

[tex](mg sin \theta)[/tex] is the component of the force of gravity that acts along the plane parallel to the incline, and the negative sign is due to the fact that this force acts opposite to the displacement of the block, d

d is the displacement of the block

Instead, the initial kinetic energy is

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

where

v = 4.90 m/s is the initial speed of the block

Substituting everything into the first equation, we can solve to find d, the displacement covered by the block up the incline:

[tex]-mgsin \theta d = -\frac{1}{2}mv^2\\d = \frac{v^2}{g sin \theta}=\frac{(4.90)^2}{(9.8)(sin 20.5^{\circ})}=7.0 m[/tex]

Final answer:

To find how far a block slides up an incline given an initial velocity and angle, we use energy conservation and kinematic equations, considering the acceleration due to gravity and the incline's angle.

Explanation:

The question involves calculating the distance a block slides up a frictionless incline given an initial velocity, making use of concepts from kinematics and energy conservation in physics. Using the energy conservation principle, the initial kinetic energy of the block is converted into potential energy at the highest point of its trajectory on the incline. Given an initial velocity of 4.90 m/s and an incline angle of 20.5°, we can calculate the distance using the formula v^2 = u^2 + 2as, where v is the final velocity (0 m/s at the highest point), u is the initial velocity, a is the acceleration (negative due to gravity acting opposite to the motion), and s is the distance travelled. The acceleration component along the incline can be calculated as a = -g*sin(θ), where g is the acceleration due to gravity (9.81 m/s²) and θ is the incline angle. Hence, the distance s can be found by rearranging the formula. This approach demonstrates the application of kinematic equations and the concept of work-energy theorem in solving problems related to motion on an incline.

Answer: .2) The final temperature will be exactly midway between the initial temperatures of substances A and B.

Explanation:

[tex]heat_{absorbed}=heat_{released}[/tex]

As we know that,  

[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]

[tex]m_A\times c_A\times (T_{final}-T_A)=-[m_B\times c_B\times (T_{final}-T_2)][/tex]         .................(1)

where,  

q = heat absorbed or released

[tex]m_A[/tex] = mass of A = 2x

[tex]m_B[/tex] = mass of B = x

[tex]T_{final}[/tex] = final temperature = z

[tex]T_A[/tex] = temperature of A

[tex]T_2[/tex] = temperature of B

[tex]c_A[/tex] = specific heat capacity of A = y

[tex]c_B[/tex] = specific heat capacity of B = 2y

Now put all the given values in equation (1), we get

[tex]2x\times y\times (z-T_A)=-[x\times 2y\times (z-T_B)][/tex]

[tex]2z=T_B+T_A[/tex]

[tex]z=\frac{T_B+T_A}{2}[/tex]

Therefore, the final temperature of the mixture will be exactly midway between the initial temperatures of substances A and B.

When two substances with different temperatures come into contact and reach thermal equilibrium, heat flows until they reach the same final temperature. In this case, the final temperature will be closer to the initial temperature of substance B because substance B requires more heat to raise its temperature and has a smaller mass. The specific heat capacity ratio determines the proportion of heat absorbed or released by the substances.

When two substances at different temperatures come into contact with each other and reach thermal equilibrium, heat flows from the hotter substance to the cooler substance until they reach the same final temperature. In this case, substance A has twice the mass of substance B but substance B has twice the specific heat capacity of substance A. The specific heat capacity determines how much heat is needed to raise the temperature of a substance. Since substance B requires more heat to raise its temperature and has a smaller mass, it will be able to absorb more heat from substance A than substance A can absorb from substance B.

This means that the final temperature will be closer to the initial temperature of substance B than substance A. The specific heat capacity ratio determines the proportion of heat absorbed or released by the substances.

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Answer:

At light intensity I = 3, is P a maximum

Explanation:

Given:

[tex]P=\frac{90I}{I^2+I+9}[/tex]

now differentiating the above equation with respect to Intensity 'I' we get

[tex]\frac{dp}{dI}=\frac{(I^2+I+9).\frac{d(90I)}{dI}-90I.\frac{d((I^2+I+9)}{dI}}{(I^2+I+9)^2}[/tex]

or

[tex]\frac{dp}{dI}=\frac{(I^2+I+9).90-90I.(2I+1)}{(I^2+I+9)^2}[/tex]

or

[tex]\frac{dp}{dI}=\frac{90I^2+90I+810)-(180I^2+90I)}{(I^2+I+9)^2}[/tex]

or

[tex]\frac{dp}{dI}=\frac{-90I^2+810)}{(I^2+I+9)^2}[/tex]

Now for the maxima [tex]\frac{dP}{dI}=0[/tex]

thus,

[tex]0=\frac{-90I^2+810)}{(I^2+I+9)^2}[/tex]

or

[tex]-90I^2+810=0[/tex]

or

[tex]I^2=\frac{810}{90}[/tex]

or

[tex]I^2=9[/tex]

or

I = 3

thus, for the value of intensity I = 3, the P is maximum

at I = 3

[tex]P=\frac{90\times3}{3^2+3+9}[/tex]

or

[tex]P=\frac{270}{21}[/tex]

or

[tex]P=12.85[/tex]

Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

Mass of object 2, m₂ = 6 kg

Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

[tex]E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2[/tex]

[tex]E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2[/tex]

E = 9 J

So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.

Answer:given below

Explanation:

Cart is pulled 14 cm from mean position

and its position is given by

[tex]x=14cos\left ( \pi t\right )[/tex]

therefore its velocity is acceleration is given by

[tex]v=-14\pi sin\left ( \pi t\right )[/tex]

[tex]a=-14\pi ^2cos\left ( \pi t\right )[/tex]

[tex]\left ( a\right ) cart\ max.\ speed\ is[/tex]

[tex]v_{max}=14\pi at\ t=0.5sec[/tex]

and its position is x=0

acceleration at t=0.5sec

a=0

[tex]\left ( b\right )[/tex]

[tex]a_{max}=14\pi ^2 cm/s^2[/tex]

at t=0,1,2 sec

at t=0

x=14 cm

v at t=0

v=0 cm/s

for t=1 sec

x=-14 cm i.e. 14 cm behind mean position

v=0 m/s

The cart's maximum speed is 14π m/s and occurs at the equilibrium point in its oscillation at integer multiples of the half period. At these moments, the magnitude of the acceleration is 14π² m/s². The magnitude of the acceleration is greatest when the cart is at the extreme points of its oscillation, where its speed is zero.

The cart, undergoing simple harmonic motion, has a maximum speed when it passes through equilibrium - the midpoint of its oscillating path. From the equation for the straightforward harmonic motion, we know that the speed v of the cart is given by the derivative of s(t) = 14cos(πt). The derivative of this function is v(t) = -14πsin(πt), and the maximum speed occurs when sin(πt) = ±1, which gives a maximum speed of ±14π m/s.

The maximum speed is attained at integer multiples of the half period (1/2, 3/2, 5/2 seconds, and so on). At this moment, the cart is at the equilibrium position, s=0. Acceleration a(t) is given by the second derivative of the position function s(t), which is a(t) = -14π²cos(πt). The magnitude of the acceleration at the time of maximum speed is 14π² m/s².

In the case of the maximum magnitude acceleration, this is attained at the turning points of the oscillation when cos(πt) = ±1. At these moments, the speed of the cart is zero.

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Answer:

intensity.

Explanation:

when the light collected by the lens is focused into a small spot it tends to increase the intensity of the light.

as different path of light with different intensity combines from passing through the lens it tends to make the light path and intensity coherent and after being coherent there intensity increases.

Based on the provided information, the correct statement is:

b. The bottom of your front bumper must not be more than 28 inches above the pavement.

The modification to increase ground clearance in the pick-up truck involves specific regulations for the front bumper height.

The statement "The bottom of your front bumper must not be more than 28 inches above the pavement" indicates a legal restriction to maintain a certain elevation for safety and compliance reasons.

This limitation aims to prevent potential hazards, such as underride collisions, and ensures that the modified truck adheres to regulatory standards. By specifying the maximum height of the front bumper, authorities aim to strike a balance between vehicle customization and road safety.

This regulation emphasizes the importance of maintaining a reasonable height for front bumpers to mitigate risks and uphold public safety standards, reflecting the broader considerations in vehicular modifications within the legal framework.

The probable question may be:

You have a pick-up truck that weighed 4,000 pounds when it was new. You are modifying it to increase its ground clearance. When you are finished

a. The bottom of your front bumper can be up to 32 inches above the pavement.

b. The bottom of your front bumper must not be more than 28 inches above the pavement.

c. You will no longer be required to have a rear bumper.

The magnitude of the electric field is found by using the components of tension in the string, related to the horizontal electric force and the vertical gravitational force, and applying trigonometry to solve for E.

To find the magnitude and direction of the electric field, we must recognize that there are two forces acting on the plastic ball: the force of gravity and the electric force. The force of gravity acts downward and has a magnitude of mg, where m is the mass of the ball and g is the acceleration due to gravity. The electric force acts horizontally and has a magnitude of Eq, where E is the electric field strength and q is the charge on the ball.

These two forces result in a tension in the string that holds the ball in equilibrium. The electric force provides the horizontal component of this tension, and the force of gravity provides the vertical component. As the string makes an angle of 17.4° with the wall, we can use trigonometry to relate the forces. From the vertical component, we have:

Tcos(θ) = mg

And from the horizontal component, we have:

Tsin(θ) = Eq

Dividing these two equations gives us:

tan(θ) = Eq/mg

Substituting the values provides:

tan(17.4°) = E(1.10 × 10-3 C)/(0.013 kg × 9.81 m/s2)

Solving for E yields:

E = mg(tan(θ))/q

By calculating this, we can find the electric field's magnitude.

Answer:

1. 1800 lb ft, 5.27 J

2. 905

Explanation:

1. Work is the change in energy.

W = mgh

W = (120 lb) (15 ft)

W = 1800 lb ft

W = mgh

W = (0.250 kg) (9.8 m/s²) (2.15 m)

W = 5.27 J

2. The IMA (ideal mechanical advantage) of a screw is the circumference of the screw divided by the pitch.

IMA = 2πr / p

The IMA of a handle (lever) is the circumference of the handle divided by the circumference of the screw.

IMA = 2πR / 2πr

The total IMA is the product:

IMA = (2πR / 2πr) (2πr / p)

IMA = 2πR / p

IMA = 2π (36) / (0.25)

IMA = 905

Answer:

a). same as

b). less than

Explanation:

a). When a bicycle is moving, the linear speed at the top of the rear wheel is same as the linear speed at the top of the front wheel. Since the clown's bicycle is a rigid body, both the wheels that is the front wheel and the rear wheel will move with the same linear speed.

b). Since we know that angular speed varies inversely to the radius of the wheel.

That is ω = 1 / r

Since the rear wheel has twice the radius of that of the front wheel, therefore real wheel will have less angular speed than the front wheel.

Therefore, the angular speed of the rear wheel is less than the angular speed of the front wheel.

Answer:

It is called single transformation

Answer:

Single transformation

Explanation:

Energy can neither be created nor be destroyed, it can only covert from one form to another. Sometimes, to get a work done one form of energy only require to be transformed into another form then it is known as single transformation.

For Example, cell phone transforming electrical energy into electromagnetic energy that makes it's working feasible.

Moving vehicle converting chemical energy to kinetic energy.

Potential energy can be found using this formula:

PE= m * g * h

where:

PE= potential energy

m=mass

g=gravitational acceleration constant (9.8 m/s^2)

h= height

So your answer is height because you also use the gravitational constant.

Answer:

We know, U = m * g * h

So, seema needs the value of g & h

In short, Your Answer would be Option C) acceleration due to gravity and the height the ball reaches

Explanation:

Acceleration is the rate of the change in velocity.

The answer would be the rate at which velocity changes.

Answer:

The speed is 20 m/s, and the direction is "downward". Acceleration is the rate of change of velocity. Usually, acceleration means the speed is changing, but not always. When an object moves in a circular path at a constant speed, it is still accelerating, because the direction of its velocity is changing.

Explanation:

Answer:

B. False

Explanation:

Acceleration is the slope of a velocity vs. time graph.

Displacement is the area under a velocity vs. time graph.

Answer:

d

velocity multiplied to mass

Answer:

Momentum of the person, p = 75 kg-m/s

Explanation:

It is given that,

Mass of the person, m = 15 kg

The person is moving with a velocity of, v = 5 m/s

We need to find the walker's momentum. It is equal to the product of mass and velocity with which it is moving. Mathematically, it is given by :

p = mv

[tex]p=15\ kg\times 5\ m/s[/tex]

p = 75 kg-m/s

So, the momentum of the person is 75 kg-m/s. Hence, this is the required solution.

Answer: [tex]3.97(10)^{-19}J[/tex]

Explanation:

The energy [tex]E[/tex] of a photon is given by:

[tex]E=h\nu[/tex]   (1)

Where:

[tex]h=6.626(10)^{-34}\frac{m^{2}kg}{s}[/tex] is the Planck constant

[tex]\nu=6(10)^{14}s^{-1}[/tex] is the frequency of green light

Solving (1):

[tex]E=(6.626(10)^{-34}\frac{m^{2}kg}{s})(6(10)^{14}s^{-1})[/tex]   (2)

[tex]E=3.9756(10)^{-19}\frac{m^{2}kg}{s^{2}}=3.9756(10)^{-19}J[/tex]   (3)

Explanation:

The electro magnetic force is given by

F = [tex]\frac{k.q_{1}.q_{2}}{r^{2}}[/tex]

where [tex]q_{1}[/tex] and [tex]q_{2}[/tex] are charged particles

           k =Coulombs constant

          r = distance between two charges

And gravitational force is given by

F = [tex]\frac{G.m_{1}.m_{2}}{r^{2}}[/tex]

where [tex]m_{1}[/tex] and [tex]m_{2}[/tex] are masses

           G =Garvitation constant

          r = distance between two masses

Now since the planets, stars and galaxies are electrically neutral, therefore they have zero electrical charge and so electro magnetic forces have no affect on  these planets, stars and heavenly bodies.

     Whereas the masses of the heavenly bodies are very large, so they are largely affected by the gravitational force since Gravitational force is directly proportional to the product of the masses of a body.

Therefore, though the electromagnetic force is stronger than the gravitational force, the electromagnetic force does not dominate the forces in the heavenly bodies as they as not electrically charged.