A particle travels in a circular orbit of radius 21 m. Its speed is changing at a rate of 23.1 m/s2 at an instant when its speed is 37.2 m/s. What is the magnitude of the acceleration (in m/s?) of the particle?

Answer:

1.23 J/(g °C)

Explanation:

[tex]m_{w}[/tex] = mass of water = 50 g

[tex]c_{w}[/tex] = specific heat of water = 4.186 J/(g °C)

[tex]T_{wo}[/tex] = Initial temperature of water = 22.00 °C

[tex]m_{a}[/tex] = mass of alloy = 25 g

[tex]c_{a}[/tex] = specific heat of alloy = ?

[tex]T_{ao}[/tex] = Initial temperature of alloy = 93.00 °C

[tex]T_{f}[/tex] = Final equilibrium temperature = 31.10 °C

Using conservation of heat

Heat Lost by alloy = Heat gained by water

[tex]m_{a}c_{a}(T_{ao}-T_{f})[/tex] = [tex]m_{a}c_{a}(T_{f} - T_{ao})[/tex]

(25) [tex]c_{a}[/tex] (93 - 31.10) = (50) (4.186) (31.10 - 22)

[tex]c_{a}[/tex] = 1.23 J/(g °C)

Answer:

The work done will be 57.15 J

Explanation:

Given that,

Mass = 12 kg

Distance = 3 m

Force = 22 N

Angle = 30°

We need to calculate the work done  

The work done is defined as,

[tex]W = Fd\cos\theta[/tex]

Where, F = force

d = displacement

Put the value into the formula

[tex]W=22\times3\times\cos30^{\circ}[/tex]

[tex]W=22\times3\times\dfrac{\sqrt{3}}{2}[/tex]

[tex]W = 57.15\ J[/tex]

Hence, The work done will be 57.15 J

Answer:

Speed of particle, v = 51.6 m/s

Explanation:

It is given that,

Initial speed of the particle which is moving towards east, u = 20 m/s

Acceleration of the particle, a = 3.95 m/s²

Time taken, t = 8 s

We have to find the speed of the particle after this acceleration. Let the speed is v. It can be calculated using first equation of motion as :

v = u + at

[tex]v=20\ m/s+3.95\ m/s^2\times 8\ s[/tex]

v = 51.6 m/s

So, the speed of the particle after this acceleration is 51.6 m/s

a.Breccia contains angular clasts and conglomerate contains rounded clasts.

Breccia and conglomerate are rocks that can primarily be distinguished by the shape of their clasts; breccia has angular fragments while the fragments in conglomerate are rounded.

One can distinguish between breccia and conglomerate based on the nature of their clasts. Breccia contains angular clasts, meaning its fragments have not been smoothed or rounded and retain sharp, rough edges. On the other hand, conglomerate contains rounded clasts.

The fragments within a conglomerate have been weathered and eroded over time, resulting in more rounded and smooth shapes. Both of these sedimentary rocks are formed from fragments of igneous rock or the shells of living organisms, but their difference lies in the shape and wear of their components.

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Answer:

Frequency

Explanation:

Photons are the packet of energy. They are massless and chargeless particles. They travel in the vacuum with the speed of light. The energy of photon is given by :

[tex]E=h\nu[/tex]

Where

h = Planck's constant

[tex]\nu[/tex] = frequency of photon

Or [tex]E=\dfrac{hc}{\lambda}[/tex]

c = speed of light

[tex]\lambda[/tex] = wavelength of photon

From the above equation, it is clear that the energy of photon is directly proportional to its frequency.

Answer:

Part a)

U = 31.25 J

Part b)

U = 312.5 J

Explanation:

Part A)

A spring that requires a force of 50 N to be stretched 0.2 m from its equilibrium position.

So here we have

[tex]F = kx[/tex]

[tex]50 = k(0.2)[/tex]

k = 250 N/m

now the energy stored in the spring is given by

[tex]U = \frac{1}{2}kx^2[/tex]

[tex]U = \frac{1}{2}(250)(0.5)^2[/tex]

[tex]U = 31.25 J[/tex]

Part B)

A spring that requires 50 J of work to be stretched 0.2 m from its equilibrium position.

So here we know the formula of spring energy as

[tex]U = \frac{1}{2}kx^2[/tex]

[tex]50 = \frac{1}{2}k(0.2)^2[/tex]

[tex]k = 2500 N/m[/tex]

now by the formula of energy stored in spring

[tex]U = \frac{1}{2}kx^2[/tex]

[tex]U = \frac{1}{2}(2500)(0.5)^2[/tex]

[tex]U = 312.5 J[/tex]

Final answer:

The work required to stretch the first spring 0.5 m from its equilibrium position, given a spring constant of 250 N/m, is 31.25 J. For the second spring, with a spring constant of 2500 N/m, the work required is 312.5 J.

Explanation:

To calculate the work required to stretch the springs mentioned in the student's question, we use Hooke's law which is given by W = ½ k x², where W is the work done, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

Part a

Given that a force of 50 N stretches the spring 0.2 m, we first determine the spring constant using F = k x:
k = F / x = 50 N / 0.2 m = 250 N/m.
Then, we calculate the work done to stretch the spring 0.5 m from its equilibrium position using W = ½ k x² = ½ × 250 N/m × (0.5 m)² = 31.25 J.

Part b

If 50 J of work is required to stretch the spring 0.2 m, the spring constant can be obtained by rearranging the work formula:
50 J = ½ k (0.2 m)², solving for k, yields k = 50 J / (0.5 × (0.2 m)²) = 2500 N/m.
Finally, calculating the work needed to stretch the spring 0.5 m from its equilibrium using W = ½ k x² = ½ × 2500 N/m × (0.5 m)² = 312.5 J.

Answer:

The speed of the plank is 81.68 m/s

Explanation:

Given that,

Speed of bullet = 152 m/s

Speed of wood = 128 m/s

Speed of another bullet = 97 m/s

We need to calculate the speed of plank

Using conservation of momentum

[tex]m_{1}u_{1}=(m_{1}+m_{2})v[/tex]

Where,

u = initial velocity

v = final velocity

[tex]152m_{1}=(m_{1}+m_{2})128[/tex]....(I)

[tex]m_{1}97=(m_{1}+m_{2})v[/tex]....(II)

From equation(I) and equation(II)

[tex]\dfrac{152}{97}=\dfrac{128}{v}[/tex]

[tex]= \dfrac{128\times97}{152}[/tex]

[tex]v=81.68\ m/s[/tex]

Hence, The speed of the plank is 81.68 m/s

Using the physics of projectile motion, it can be calculated that a tennis ball shot vertically upward with an initial speed of 20 m/s will take approximately 2.04 seconds to reach its maximum height.

The subject of this question is the physics of motion, specifically projectile motion. The tennis ball is being shot vertically upward, so its motion can be considered as projectile motion. The time it takes for the ball to reach its maximum height can be calculated by using the vertical motion properties of projectiles. We know that the upwards velocity of the ball when released is 20 m/s and the acceleration due to gravity is -9.8 m/s² (indicating the ball is slowing as it rises, eventually down to 0 m/s at its highest point).

Now, to calculate time, we can use this formula for acceleration: 'v = u + gt'. Here, v is final vertical velocity, u is initial vertical velocity, g is acceleration due to gravity and t is the time. When the ball reaches its maximum height, its final vertical velocity will be 0. So we transform the formula to 't= (v-u)/g', and inputting the known values, '-(0 - 20 m/s)/-9.8 m/s², gives us a time of approximately 2.04 seconds

Thus, it will take approximately 2.04 seconds for the tennis ball to reach its maximum height.

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Answer:

p = 20, q = 50

Explanation:

p² + 16q = 1200

300 - p² + 2q = 0

By adding the two equations, we can eliminate p²:

16q + 300 + 2q = 1200

18q = 900

q = 50

Solving for p:

p² + 16(50) = 1200

p = 20

Final answer:

The equilibrium quantity is approximately 64.29 units, and the equilibrium price is approximately 13.09 dollars, after solving the system of equations set by equating the demand function p^2 + 16q = 1200 and supply function 300 - p^2 + 2q = 0.

Explanation:

To find the equilibrium quantity and equilibrium price for the given demand function p2 + 16q = 1200 and supply function 300 − p2 + 2q = 0, we need to set the quantity demanded equal to the quantity supplied and solve the system of equations.

Firstly, we rearrange the supply function to isolate p2, which yields p2 = 300 - 2q. Next, we substitute this expression into the demand function equation:

(300 - 2q) + 16q = 1200

Now, we solve for q:

300 + 14q = 1200

14q = 900

q = 900 / 14

q = 64.29 (rounded to two decimal places)

Then, we substitute q back into either the original demand or supply function to find p. Using the demand function:

p2 + 16(64.29) = 1200

p2 + 1028.64 = 1200

p2 = 171.36

[tex]p = \sqrt{171.36}[/tex]

p = 13.09 (rounded to two decimal places)

The equilibrium quantity is approximately 64.29 units, and the equilibrium price is approximately 13.09 dollars.

The person exerts a force of 75 N on the mower, and it will travel approximately 0.529 meters before coming to a stop.

To find the force (F) exerted by the person on the mower, we can use Newton's second law, which states:

[tex]\[ \text{Net force} = \text{Mass} \times \text{Acceleration} \][/tex]

Given:

- Mass of the mower (m) = 24 kg

- Net external force (F_net) = 51 N

- Force of friction (F_friction) = 24 N

We know that net external force is the sum of all forces acting on the object, so:

[tex]\[ F_{\text{net}} = F - F_{\text{friction}} \][/tex]

[tex]\[ 51 \, \text{N} = F - 24 \, \text{N} \][/tex]

[tex]\[ F = 51 \, \text{N} + 24 \, \text{N} \][/tex]

[tex]\[ F = 75 \, \text{N} \][/tex]

So, the person exerts a force of [tex]\( 75 \, \text{N} \)[/tex] on the mower.

Now, to find how far the mower will go before stopping, we can use the equation of motion:

[tex]\[ v^2 = u^2 + 2as \][/tex]

Where:

- \( v \) = final velocity (0 m/s, since the mower stops)

- \( u \) = initial velocity = 1.5 m/s

- \( a \) = acceleration

- \( s \) = distance traveled

We know that the net force acting on the mower is responsible for the acceleration, and it's given by Newton's second law:

[tex]\[ F_{\text{net}} = m \times a \][/tex]

So:

[tex]\[ a = \frac{F_{\text{net}}}{m} \][/tex]

[tex]\[ a = \frac{51 \, \text{N}}{24 \, \text{kg}} \][/tex]

[tex]\[ a ≈ 2.125 \, \text{m/s}^2 \][/tex]

Now, using the equation of motion:

[tex]\[ 0^2 = (1.5 \, \text{m/s})^2 + 2 \times 2.125 \, \text{m/s}^2 \times s \][/tex]

[tex]\[ 0 = 2.25 + 4.25s \][/tex]

[tex]\[ 4.25s = -2.25 \][/tex]

[tex]\[ s = \frac{-2.25}{4.25} \][/tex]

[tex]\[ s ≈ -0.529 \, \text{m} \][/tex]

The negative sign indicates that the mower moves in the opposite direction to its initial motion before stopping. So, the mower will travel approximately [tex]\( 0.529 \, \text{m} \)[/tex] before coming to a stop.

The force exerted by the person on the mower is 75 N. The mower will travel approximately 1.125 meters before stopping.

The net external force is the difference between the force exerted by the person and the force of friction. Thus, we can write:

[tex]F_{net} = F - F_{friction}[/tex]

Substituting the given values:

[tex]51\,\text{N} = F - 24\,\text{N}[/tex]

Solving for F:

[tex]F = 51\,\text{N} + 24\,\text{N} = 75\,\text{N}[/tex]

Finding the Distance the Mower Travels before Stopping:

When the force F is removed, the mower will decelerate due to the frictional force. First, we find the acceleration (a) using Newton's second law:

[tex]F_{friction} = m \cdot a[/tex]

Solving for a gives:

[tex]a = \frac{F_{friction}}{m} = \frac{24\,\text{N}}{24\,\text{kg}} = 1\, \text{m/s}^2[/tex]

Since friction opposes the motion, this is a deceleration (a is negative):

[tex]a = -1\, \text{m/s}^2[/tex]

Next, we use the kinematic equation to find the distance (d):

[tex]v^2 = v_0^2 + 2 \cdot a \cdot d[/tex]

Substituting the known values:

[tex]0 = (1.5 \frac{m}{s})^2 + 2 \cdot (-1 \frac{m}{s^2}) \cdot d[/tex]

Solving for d:

[tex]0 = 2.25 - 2d[/tex]
[tex]2d = 2.25[/tex]
[tex]d = 1.125 \text{m}[/tex]

(a) 6.04 rev/s

The speed of the ball is given by:

[tex]v=\omega r[/tex]

where

[tex]\omega[/tex] is the angular speed

r is the distance of the ball from the centre of the circle

In situation 1), we have

[tex]\omega=8.13 rev/s \cdot 2\pi = 51.0 rad/s[/tex]

r = 0.600 m

So the speed of the ball is

[tex]v=(51.0 rad/s)(0.600 m)=30.6 m/s[/tex]

In situation 2), we have

[tex]\omega=6.04 rev/s \cdot 2\pi = 37.9 rad/s[/tex]

r = 0.900 m

So the speed of the ball is

[tex]v=(37.9 rad/s)(0.900 m)=34.1 m/s[/tex]

So, the ball has greater speed when rotating at 6.04 rev/s.

(b) [tex]1561 m/s^2[/tex]

The centripetal acceleration of the ball is given by

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

where

v is the speed

r is the distance of the ball from the centre of the trajectory

For situation 1),

v = 30.6 m/s

r = 0.600 m

So the centripetal acceleration is

[tex]a=\frac{(30.6 m/s)^2}{0.600 m}=1561 m/s^2[/tex]

(c) [tex]1292 m/s^2[/tex]

For situation 2 we have

v = 34.1 m/s

r = 0.900 m

So the centripetal acceleration is

[tex]a=\frac{v^2}{r}=\frac{(34.1 m/s)^2}{0.900 m}=1292 m/s^2[/tex]

Final answer:

The rate of rotation that gives the greater speed for the ball is 8.13 rev/s. The centripetal acceleration of the ball at 8.13 rev/s is 39.43 m/s^2, while the centripetal acceleration at 6.04 rev/s is 32.90 m/s^2.

Explanation:

The rate of rotation that gives the greater speed for the ball is 8.13 rev/s. The speed of the ball is directly proportional to the rate of rotation. So, a higher rate of rotation will result in a greater speed for the ball.

The centripetal acceleration of the ball at 8.13 rev/s can be calculated using the formula:
centripetal acceleration = (angular velocity)^2 * radius
Plugging in the values:
centripetal acceleration = (8.13 rev/s)^2 * 0.6 m = 39.43 m/s^2

The centripetal acceleration of the ball at 6.04 rev/s can be calculated in the same way:
centripetal acceleration = (6.04 rev/s)^2 * 0.9 m = 32.90 m/s^2

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

[tex]p_i = 0[/tex]

The final total momentum is instead:

[tex]p_f = m_a v_a + m_c v_c[/tex]

where

[tex]m_a = 125 kg[/tex] is the mass of the astronaut

[tex]v_a = 2.50 m/s[/tex] is the velocity of the astronaut

[tex]m_c = 1900 kg[/tex] is the mass of the capsule

[tex]v_c[/tex] is the velocity of the capsule

Since the total momentum must be conserved, we have

[tex]p_i = p_f = 0[/tex]

so

[tex]m_a v_a + m_c v_c=0[/tex]

Solving the equation for [tex]v_c[/tex], we find

[tex]v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s[/tex]

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

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

The change in momentum of the astronaut is

[tex]\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s[/tex]

And the duration of the push is

[tex]\Delta t = 0.600 s[/tex]

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

[tex]F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N[/tex]

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

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

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

[tex]K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J[/tex]

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

[tex]K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J[/tex]

The astronaut's push off leads to a change in the speed of the space capsule of -0.164 m/s. The average force exerted is approximately 416.67N. The final kinetic energy of the astronaut and the space capsule are 156.25J and 25628.1J respectively.

This problem covers the principle of conservation of momentum. The astronaut and the space capsule constitute a closed system, where the total momentum before and after the push must be equal.

(a) Using the principle of conservation of momentum (initial momentum = final momentum), we can calculate the change in speed of the space capsule when the astronaut pushes off. We start with the equation m1V1 + m2V2 = 0, where m1 is the astronaut's mass and V1 is her speed, and m2 is the space capsule's mass and V2 is its velocity. Solving for V2 gives us a change in speed of the space capsule of -0.164m/s (which is in the opposite direction to the astronaut's motion).

(b) The average force exerted can be calculated by changing momentum over time (Force = Change in momentum / Time). Here we obtain approximately 416.67N.

(c) The final kinetic energy of each object is K = 1/2*m*v^2. For the astronaut, this is approximately 156.25J and for the space capsule, this is 25628.1J.

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Answer:

D. 39 N m

Explanation:

m = mass of the weight used in crossfit workout = 7.0 kg

Force due to the weight used is given as

F = mg

F = (7.0) (9.8)

F = 68.6 N

d = distance of point of action of weight from shoulder joint = 0.57 m

τ = Torque about the shoulder joint due to the weight

Torque about the shoulder joint due to the weight is given as

τ = F d

Inserting the values

τ = (68.6) (0.57)

τ = 39 Nm

Answer:

Wavelength of light is 750 nm.

Explanation:

It is given that,

Frequency of light, [tex]\nu=4\times 10^{14}\ Hz[/tex]

The relationship between the wavelength and the frequency of light is given by :

[tex]c=\nu\times \lambda[/tex]

Where

c = speed of light

[tex]\nu[/tex] = frequency of light

[tex]\lambda[/tex] = wavelength of light

[tex]\lambda=\dfrac{c}{\nu}[/tex]

[tex]\lambda=\dfrac{3\times 10^8\ m/s}{4\times 10^{14}\ Hz}[/tex]

[tex]\lambda=7.5\times 10^{-7}\ m[/tex]

[tex]\lambda=750\ nm[/tex]

Hence, the correct option is (d) "750 nm".

The wavelength corresponding to light with a frequency of 4 x 10^14 Hz is found using the formula λ = c/f and equals 750 nm, falling within the visible spectrum.

The wavelength corresponding to a frequency of 4 x 10^14 Hz can be calculated using the equation c = λf, where c is the speed of light (3.0 × 10^8 m/s), λ is the wavelength in meters, and f is the frequency in hertz (Hz). To find the wavelength, we rearrange the equation to λ = c/f. Plugging in the values, we get λ = (3.0 × 10^8 m/s) / (4 x 10^14 Hz) which equals 750 nm. Therefore, the wavelength corresponding to light with a frequency of 4 x 10^14 Hz is 750 nm, which falls within the range of visible light wavelengths (400 nm to 750 nm).

Answer:

r² / 110² − 3z² / 1375² = 1

Explanation:

The equation of a hyperboloid (which is a hyperbola rotated about the z axis or conjugate axis) that is centered at the origin is:

x² / a² + y² / b² − z² / c² = 1

If the cross sections are circular rather than elliptical, then a = b.

(x² + y²) / a² − z² / c² = 1

Or, if you prefer cylindrical coordinates:

r² / a² − z² / c² = 1

We know that at z = 0, r = 110.  And at z = -500, r = 130.

110² / a² − 0 = 1

130² / a² − (-500)² / c² = 1

Solving:

a² = 110²

c² = 1375² / 3

Plugging in:

r² / 110² − 3z² / 1375² = 1

Answer:

(a) The motion of rocket is in upward direction and reaches to maximum height then the rocket starts falling freely.

(b) The maximum height attained by the rocket from the ground is 348.65 m

(c) The time taken by the rocket to maximum height after lift off is 8.85 s.

(d) The total time taken by the rocket in air is 17.3 second.

Explanation:

u = 57 m/s, a = 3 m/s^2, h = 140 m

let the rocket attains a velocity v after covering 140 m and it takes t time to reach upto 140 m.

Use III equation of motion

V^2 = u^2 + 2a h

v^2 = 57^2 + 2 x 3 x 140

v = 63.95 m/s

Now use I equation of motion

v = u + at

t = (63.95 - 57) / 3 = 2.32 s

(a) The motion of rocket is in upward direction and reaches to maximum height then the rocket starts falling freely.

(b) Let H be the maximum height reached by the rocket after the engine stops.

Use III equation of motion

v^2 = u^2 + 2aH

here, v = 0, u = 63.95 m/s, a = - 9.8 m/s^2

0 = 63.95^2 - 2 x 9.8 x H

H = 208.65 m

The maximum height attained by the rocket from the ground is h + H = 140 + 208.65 = 348.65 m

(c) Let t' be the time in which rocket reaches to maximum height after engine is stopped.

Use I equation of motion

v = u + a t'

0 = 63.95 - 9.8 x t'

t' = 6.53 s

The time taken by the rocket to maximum height after lift off is t + t' = 2.32 + 6.53 = 8.85 s.

(d) let t'' be the time taken by the rocket to fall freely

Use II equation of motion

H' = ut'' + 1/2 gt''^2

Here, H' = 348.65 m, u = 0

348.65 = 0 + 0.5 x 9.8 x t''^2

t''^ = 8.44 s

The total time taken by the rocket in air is t + t' + t'' = 2.32 + 6.53 + 8.44 = 17.3 second.

Answer:

The net force on him is zero.

Explanation:

The velocity of skydiver is constant.

As we know that the acceleration is rate of change in velocity. So, here velocity os constant it means acceleration of skydiver is zero.

According to Newton's second law

Force acting on a body is equal to the product of mass and velocity of the body.

As acceleration is zero that means the net force acting on the body is zero.

Answer:

The observer detects light of wavelength is 115 nm.

(b) is correct option

Explanation:

Given that,

Wavelength of source = 500 nm

Velocity = 0.90 c

We need to calculate the wavelength of observer

Using Doppler effect

[tex]\lambda_{o}=\sqrt{\dfrac{1-\beta}{1+\beta}}\lambda_{s}[/tex]

Where, [tex]\beta=\dfrac{c}{v}[/tex]

[tex]\lambda_{o}=\sqrt{\dfrac{c-0.90c}{c+0.90c}}\times500\times10^{-9}[/tex]

[tex]\lambda_{o}=115\ nm[/tex]

Hence, The observer detects light of wavelength is 115 nm.

Answer:

[tex]2.5 rad/s^2, 688^{\circ}[/tex]

Explanation:

The angular acceleration is given by:

[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]

where

[tex]\omega_f = 74.0 rev/min \cdot (\frac{2\pi rad/rev}{60 s/min})=7.75 rad/s[/tex] is the final angular speed

[tex]\omega_i = 0[/tex] is the initial angular speed

t = 3.10 s is the time interval

Solving the equation,

[tex]\alpha = \frac{7.75 rad/s - 0}{3.10 s}=2.5 rad/s^2[/tex]

Now we can find the angular displacement by using:

[tex]\theta = \omega_i t + \frac{1}{2}\alpha t^2[/tex]

Substituting,

[tex]\theta=0+\frac{1}{2}(2.5 rad/s^2)(3.10 s)^2=12.0 rad[/tex]

In degrees:

[tex]\theta = \frac{12.0 rad}{2\pi}\cdot 360^{\circ}=688^{\circ}[/tex]

Answer:

2.23 s

Explanation:

For a second-order reaction:

1 / [A] = 1 / [A]₀ + kt

Given [A] = 0.330 M, [A]₀ = 0.750 M, and k = 0.760 M⁻¹s⁻¹:

1 / 0.330 = 1 / 0.750 + 0.760t

t = 2.23

It would take 2.23 seconds.

Final answer:

To find the time it takes for a second-order reaction concentration to decrease from 0.750 M to 0.330 M at the given rate constant, we use the integrated second-order rate law. After substituting the given values and solving for time, we find that it takes approximately 1.20 seconds.

Explanation:

To calculate the time taken for the concentration of A to decrease from 0.750 M to 0.330 M in this second-order reaction, we can use the integrated second-order rate law which is given by:

[tex]\( \frac{1}{[A]_t} - \frac{1}{[A]_0} = k \cdot t \)[/tex]

Where [tex]\([A]_t\)[/tex] is the concentration of A at time t, [tex]\([A]_0\)[/tex] is the initial concentration of A, k is the rate constant, and t is the time.

Plugging in the values, we get:

[tex]\( \frac{1}{0.330 \ M} - \frac{1}{0.750 \ M} = 0.760 \ M^{-1}s^{-1} \cdot t \)[/tex]

After solving for t, we find that the time required is approximately 1.20 seconds. Note that the units of k given [tex](\( M^{-1}s^{-1} \))[/tex] are consistent with a second-order reaction, which confirms the correctness of the approach.

Answer:

Option B is the correct answer.

Explanation:

Shear stress is the ratio of shear force to area.

We have

       Shear stress = 3 N/mm² = 3 x 10⁶ N/m²

       Area = Area of rectangle = 10 x 10⁻² x d = 0.1d

       Shear force = 50000 N

Substituting

        [tex]\texttt{Shear stress}=\frac{\texttt{Shear force}}{\texttt{Area}}\\\\3\times 10^6=\frac{50000}{0.1d}\\\\d=0.1667m=16.67cm[/tex]

Width of beam = 16.67 cm

Option B is the correct answer.

To calculate the depth of the beam, shear stress is divided by shear force to get the cross-sectional area, which is then divided by the width to obtain the depth. The depth is found to be 16.67 cm.

The subject of the question relates to the calculation of the depth of a rectangular beam based on the given maximum shear force and shear stress. The formula to be used involves shear stress (τ), shear force (V), and the cross-sectional area (A), which is the product of the beam's width and depth (b × d).

Calculation:

Shear Stress (τ) = Shear Force (V) / Area (A)

Given that the maximum shear stress is 3 N/mm² and the shear force is 50000 N, we can rearrange the formula to solve for the cross-sectional area (A):

A = V / τ = 50000 N / 3 N/mm²

Convert 50000 N to mm² by multiplying by 1000 to maintain the correct units:

A = 50000 N × 1000 mm²/N = 50000000 mm²

A = 50000000 mm² / 3 N/mm² = 16666666.67 mm²

Next, we find the depth, remembering the width of the beam is 10 cm (which is 100 mm because 1 cm = 10 mm):

Depth (d) = A / width (b) = 16666666.67 mm² / 100 mm

Depth (d) = 166666.67 mm which is equal to 16.67 cm.

Therefore, the correct answer is (c) 16.67 cm.

To find an integer solution to given linear equations, iterate through integer values for x and y within the range of -10 to 10 and substitute them into the equations to check for solutions.

To find an integer solution for the set of given linear equations using brute force, you would substitute integer values for x and y within the specified range of -10 to 10. The equations given as examples, such as 7y = 6x + 8, 4y = 8, and y + 7 = 3x, are all linear equations that can be rearranged into the standard form y = mx + b, where m is the slope and b is the y-intercept.

In practice, you would start with -10 for x, solve for y in each equation, and check if y is an integer. Repeat incrementally until you reach 10 for x. If you find a pair of x and y values that satisfy both equations, you've found a solution. Remember to verify your solution by substiting the found values back into the original equations to ensure they hold true.

For example, if the equations were 2x + 3y = 6 and 4x - y = 10, you would try each pair of x and y between -10 and 10 until you find a pair that works for both equations. This brute force method is not the most efficient, but it will eventually yield an integer solution, if one exists within the given range.

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The solution that satisfies both equations is x = 3 and y = 2.

In many engineering and scientific applications, finding solutions to systems of linear equations is crucial. Given the linear equations 8x + 7y = 38 and 3x - 5y = -1, we are tasked with finding integer solutions for x and y in the range of -10 to 10.

Steps to Solve Using Brute Force :

We will systematically check all integer pairs (x, y) within the specified range:

After performing these steps, we find the solution: x = 3 and y = 2. Both equations are satisfied by this pair because:

8(3) + 7(2) = 24 + 14 = 38 and 3(3) - 5(2) = 9 - 10 = -1

Thus, x = 3 and y = 2 is the correct solution.

Answer:

Torque is defined as the force that produces rotation around the axis of an object. Torque is also known as the moment of force. It's S.I unit is Nm.

Formula of torque is:

                τ= rxF(sinθ)

Given Data:

Force = 20 N

r = 0.25 m

θ = 120°

Solution:

              τ= rxF(sinθ)

              = 0.25 x 20 (sin 120)

              =  2.903 Nm

Thus, the total torque applied on the wrench is 2.903 Nm.

The magnitude of the torque when the force is applied at an angle of 120 degrees to the wrench is 4.33 N.m.

To determine the magnitude of the torque when a force is applied at an angle to a wrench, we can use the formula:

Torque = Force x Distance x sin(angle)

In this case, the force is 20 N, the distance is 0.25 meters, and the angle is 120 degrees. Plugging in these values, we get:

Torque = 20 N x 0.25 m x sin(120°)

Torque = 20 N x 0.25 m x 0.866

Torque = 4.33 N.m

Therefore, the magnitude of the torque when the force is applied at an angle of 120 degrees to the wrench is 4.33 N.m.

Answer:

2/(3π) ft/min ≈ 0.212 ft/min

Explanation:

Volume of a cone is:

V = ⅓ π r² h

The diameter is three times the altitude, so:

2r = 3h

r = 3/2 h

Substituting:

V = ⅓ π (3/2 h)² h

V = ⅓ π (9/4 h²) h

V = ¾ π h³

Taking derivative with respect to time:

dV/dt = 9/4 π h² dh/dt

Given dV/dt = 6 and h = 2:

6 = 9/4 π (2)² dh/dt

6 = 9π dh/dt

dh/dt = 2/(3π)

dh/dt ≈ 0.212 ft/min

Explanation:

This problem can be solved using the Radioactive Half Life Formula:

[tex]A=A_{o}.2^{\frac{-t}{h}}[/tex] (1)

Where:

[tex]A=3.13mg[/tex] is the final amount of the material

[tex]A_{o}=50mg[/tex] is the initial amount of the material

[tex]t=19.7days[/tex] is the time elapsed

[tex]h[/tex] is the half life of the material (the quantity we are asked to find)

Knowing this, let's substitute the values and find [tex]h[/tex] from (1):

[tex]3.13mg=(50mg)2^{\frac{-19.7days}{h}}[/tex] (2)

[tex]\frac{3.13mg}{50mg}=2^{\frac{-19.7days}{h}}[/tex] (3)

Applying natural logarithm in both sides:

[tex]ln(\frac{3.13mg}{50mg})=ln(2^{\frac{-19.7days}{h}})[/tex] (4)

[tex]-2.77=-\frac{19.7days}{h}ln(2)[/tex] (5)

Clearing [tex]h[/tex]:

[tex]h=\frac{-19.7days}{-2.77}(0.693)[/tex] (6)

Finally:

[tex]h=4.928days[/tex]

The half-life is the period required for the quantity of a radioactive substance to become half its original quantity. By using given information and the half-life formula, we can calculate the half-life of the unknown radioactive substance.

The half-life of a radioactive substance is the time period required for the quantity of the substance to reduce to half its initial amount due to radioactive decay. This problem involves using the half-life formula: T = (t/log_2) * (log(N0/N)), where T is the half-life, t is the elapsed time, N0 is the initial quantity of the substance, and N is the quantity after time t.

From the question, we know N0 = 50.0mg, N = 3.13mg, and t = 19.7 days. Substituting these values into the formula allows us to calculate the half-life, which will give the answer.

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Answer:

Coefficient of static friction required = 0.94

Explanation:

Coefficient of static friction required [tex]=\frac{v^2}{gR}[/tex]

Velocity, v = 120 km/h = 33.33 m/s

g = 9.81 m/s²

R = 120 m

Coefficient of static friction required[tex]=\frac{33.33^2}{9.81\times 120}=0.94[/tex]

Coefficient of static friction required = 0.94

Answer:

The process in which heat flows by the mass movement of molecules from one place to another is C) convection.

Hope this helps :)

The process in which heat flows by the mass movement of molecules from one place to another is convection.

Heat transfer by convection is a method of heat transfer that involves the mass movement of molecules from one place to another.

Heat can also be transfered in the following methods;

Thus, the process in which heat flows by the mass movement of molecules from one place to another is convection.

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Final answer:

Magnetic flux is a representation of the magnetic field passing through an area—it's calculated using the field strength, area vector, and the cosine of their angle. Faraday's law of induction explains that changing magnetic flux through a loop induces an emf in it, which is crucial in electric generators and related technologies. Lenz's law ensures the direction of the induced current opposes the flux change, reflecting energy conservation.

Explanation:

Magnetic Flux and Faraday's Law of Induction

Magnetic flux is essentially a measure of the quantity of magnetic field lines passing through a given area. To determine it, you need to know the strength of the magnetic field, the surface area over which it's spread, and the angle between the field and the normal (perpendicular) to that surface area. Mathematically, it is defined as the product of the magnetic field (B), the area (A), and the cosine of the angle (θ) between them, which is given by the equation Φ = B * A * cos(θ).

Faraday's law of induction tells us that a changing magnetic flux within a closed conducting loop will induce an electromotive force (emf) in the loop. This phenomenon is used in a vast array of technologies from electric generators to transformers. The magnitude of the induced emf can be calculated using Faraday's law which states that the emf induced is directly proportional to the rate of change of the magnetic flux.

Additionally, Lenz's law elaborates on the direction of the induced current, stating that it will flow in a manner that opposes the change in flux that caused it. This is a manifestation of the conservation of energy.

Application Example

An example of Faraday's law would be a coil of wire exposed to a changing magnetic field. If the field changes over time, an electric current is induced in the wire. This principle is the foundation for devices like inductive chargers and electrical generators.

Answer:

Electric field, E = 2 N/C

Explanation:

It is given that,

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

Electric force, [tex]F=3.2\times 10^{-19}\ N[/tex] (in +x direction)

We have to find the electric field at this point. The electric field is given by the electric force acting on a charged particle per unit electric charge. Mathematically, it is given by :

[tex]E=\dfrac{F}{q}[/tex]

q = charge on an electron, [tex]1.6\times 10^{-19}\ C[/tex]

[tex]E=\dfrac{3.2\times 10^{-19}\ N}{1.6\times 10^{-19}\ C}[/tex]

E = 2 N/C

The electric field at this point is 2 N/C.

Answer:

Beats is a phenomenon which occurs when two sound waves with different frequencies overlap each other.

Explanation:

When these two sound waves overlap, a third frequency is formed. The places where the two waves reinforces each other, constructive interference takes place and a louder sound is produced whereas, at points where destructive interference takes place a softer sound is produced.

These alternative loud and soft sounds are called beats.