You toss a basketball toward the basket. At the highest point of its arc,

a. The horizontal component of the velocity is zero.
b. The vertical component of the velocity is zero.
c. Both the horizontal and vertical components of the velocity are zero.

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]

Answer:

F = 32.2m

Explanation:

The force of gravity on an object is given by:

[tex]F=mg[/tex]

where

m is the mass of the object

g is the acceleration due to gravity

Here we have:

- An object of mass m

- The acceleration of gravity is expressed as [tex]g=32.2 ft/s^2[/tex]

Therefore, substituting into the formula above, we find that the force of gravity on the object is

[tex]F=m\cdot 32.2 = 32.2m[/tex]

Answer:

F = 32.2m

Explanation:

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

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.

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:

Ffs = 251 N

Fn = 691 N

Explanation:

Take the y direction to be normal to the ramp and the x direction to be parallel to the ramp.

The angle of the ramp is 20°, so the angle that the weight vector makes with the normal is also 20°.  Therefore:

Fgx = Fg sin 20°

Fgy = Fg cos 20°

Sum of the forces in the x direction (parallel to the ramp):

∑F = ma

Ffs − Fgx = 0

Ffs = Fgx

Ffs = Fg sin 20°

Ffs = 735 sin 20°

Ffs ≈ 251

Sum of the forces in the y direction (normal to the ramp):

∑F = ma

Fn − Fgy = 0

Fn = Fgy

Fn = Fg cos 20°

Fn = 735 cos 20°

Fn ≈ 691

To answer that question we need to apply equations of movement ( from Newton´s laws )

In equilibrium:

∑F  = 0         or    ∑Fx = 0     ;  ∑Fy = 0

Solution is:

a) F(sf) = 251 [N]

b) Fn = 691 [N]

From the attached drawings we can see:  ( Body free diagram)

∑ Fₓ  = F(sf)  - Pₓ  = 0           where P = m×g  = 735 [N] ( the weigth)

and Pₓ = P× cos20°

Then     F(sf)  = Pₓ × sin 20°

F(sf) = m×g×cos20°  =  735× 0.34202 [N]

F(sf) = 251.3847

F(sf) = 251 [N]

rounding to the nearest number

F(sf) = 691 [N]

∑ Fy = 0

∑ Fy = Fn - Py  = 0                     Py = P×cos20°      Py = m×g×cos20°

Py = 735×0.939693 [N]     Py =   [N]

Fn = Py = 690.674 [N]

rounding to the nearest whole number

Fn = 691 [N]

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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.

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:

[tex]\lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}[/tex]

Explanation:

the velocity as a function of time is

[tex]v(t)=\frac{mg}{c}(1-e^{\frac{-ct}{m}})[/tex]

[tex]\therefore v(t)=\frac{mg}{c}(1-\frac{1}{e^{\frac{ct}{m}}})[/tex]

[tex]\therefore v(t)=\frac{mg}{c}(1-\frac{1}{e^{\frac{ct}{m}}})\\\\\therefore \lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}(1-\frac{1}{\infty })\\\\\therefore \lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}(1-0)\\\\\therefore \lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}[/tex]

The limit as t approaches infinity of the speed v for an object with mass m dropped from rest with air resistance considered, is the terminal velocity mg/c. The exponential term in the equation approaches zero as time becomes very large, leading to the object reaching a constant terminal velocity.

To calculate the limit as t approaches infinity of the speed v for an object with mass m dropped from rest with air resistance taken into account, we use the provided equation v = mg/c (1 − e^{-ct/m}), where g is the acceleration due to gravity, which averages 9.80 m/s², and c is a positive constant representing air resistance. The limit represents the object's terminal velocity, which is the constant speed an object reaches when the force of gravity is balanced by the drag force of air resistance.

As t → ∞ (increases towards infinity), the exponential term e^{-ct/m} approaches zero. Hence, the limit of the speed v becomes:

∑ lim t→∞ v = lim t→∞ mg/c (1 − e^{-ct/m}) = mg/c.

The value mg/c is known as the terminal velocity, which is the maximum speed that the object will reach as it continues to fall.

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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:

B. False

Explanation:

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

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

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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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: .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:

1.034 m above the floor

Explanation:

The location of center of body for a compound body, when the weights are given is calculated as:

[tex]\bar x = \frac{W_1x_1+W_2x_2+W_3x_3+W_4x_4+W_5x_5+.......+Wnx_n}{W_1+W_2W_3W_4W_5+.....+W_n}[/tex]

where,

[tex]\bar x[/tex] is the center of gravity of the entire body

W = weight of the individual body

x = center of gravity of the individual body

Thus on substituting the values we get,

[tex]\bar x = \frac{438\times 1.28+144\times 0.760+87\times 0.250}{438+144+87}[/tex]

or

[tex]\bar x = \frac{691.83}{669}[/tex]

or

[tex]\bar x =1.034m[/tex]

Hence, the center of gravity of the entire body lies 1.034 m above the floor

Answer:

Required mass of sand is 20 kg

Explanation:

Given:

Mass of the plank = 25 kg

Distance of the Center of gravity of the Plank from the fulcrum = [tex]\frac{2}{2}-0.50 = 0.5m[/tex]

Distance of the Center of gravity of the sand box from the fulcrum = [tex]\frac{2}{2}-\frac{0.75}{2}= 0.625m[/tex]

Balancing the torque due to the plank and the sand box with respect to the fulcrum

Torque = Force × perpendicular distance

thus, we get

(25 × g) × 0.5 = weight of sand × 0.625

where, g is the acceleration due to gravity

or

(25 × g) × 0.5 = (mass of sand × g) × 0.625

or

mass of sand = 20 kg

Hence, the required mass of the sand is 20 kg

Answer:

20 Kg mass of sand should be put into the box so that the plank balances horizontally on a fulcrum placed horizontally on a fulcrum placed just below its midpoint.

Explanation:

Use the second condition of equilibrium.

[tex]$\sum} \tau=0$[/tex]

[tex]MgL-$M g x_{c m}=0$[/tex]

[tex]$M=\frac{m x_{c m}}{L}[/tex]

[tex]=\frac{25(0.50)}{0.625}[/tex]

[tex]=20 \mathrm{~kg}$[/tex]

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Answer:

Speed of A = 891 km/h

Speed of B = 804 km/h

Explanation:

Let the speed of aeroplane B is v, the speed of aeroplane is A is 87 km/h faster than B.

So, the speed of aeroplane A is 87 + v.

Distance traveled by A after 7 hours, d1 = (87 + v) x 7

Distance traveled by B after 7 hours, d2 = v x 7

Total distance traveled = 11865 km

So, d1 + d2 = 11865

(87 + v) x 7 + v x 7 = 11865

609 + 14 v = 11865

14 v = 11256

v = 804 km/h

So, the speed of A = 87 + 804 = 891 km/h

Speed of B = 804 km/h

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.

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:

Kinetic Energy

Explanation:

Heat energy is another name for thermal energy. Kinetic energy is the energy of a moving object. As thermal energy comes from moving particles, it is a form of kinetic energy.

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:

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.

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.

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:

Rate of change of area is [tex]75.398ft^{2}/sec[/tex]

Explanation:

[tex]Area=\pi r^{2}\\\\\frac{d(Area)}{dt}=\frac{\pi r^{2}}{dt}=2\pi r\frac{dr}{dt}[/tex]

Applying values we get [tex]\frac{d(Area)}{dt}=2\pi 4\times 3=75.398ft^{2}/sec[/tex]

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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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.

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.

Answer:

Boat speed = 8 miles/hr

Explanation:

Let the speed of the boat be U

Let the speed of the current be V.

Therefore, the downstream speed is U+V

and the upstream speed is U-V

Now we know that Speed = distance / time

Therefore, downstream speed, U+V = 63 / 7

                                                    U+V = 9 miles/hr   ------(1)

                  Upstream speed, U-V = 63 / 9

                                                U-V = 7 miles/hr      --------(2)

Therefore subtracting (2) from (1), we get

( U+V) - ( U-V ) = 9-7

2V = 2

V = 1 miles/hr

Therefore the speed of the current is V = 1 mile/hr

Now from (1) we get

U+V = 9

U+1 = 9

U = 8

Therefore, the speed of the boat is U = 8 miles/hr

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.