In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop.

(c)How much work does the gravitational force do on the ball while it is compressing the mattress?

(d)How much work does the mattress do on the ball?

(e)Now model the mattress as a single spring with an unknown spring constant k, and consider the whole system formed by the ball, the earth and the mattress. By how much does the potential energy of the mattress increase as it compresses?

(f)What is the value of the spring constant k?

Answer:

The scale reading at the top = 565.8N

The scale reading at the bottom = 610.44N

Explanation:

We are given:

t = 32seconds

R = 9.7meters

mass (m) = 60Kg

We take g as gravitational field = 9.8

Therefore, to find the scale reading (N) at the top, let's use the formula:

[tex] N_t = m (g - w^2 R) [/tex]

Where w = 2π/t

w = 2π/32 = 0.197

Substituting the figures into the equation, we have

[tex] N = 60 (9.8 - 0.197^2 * 9.7) [/tex]

N = 60 (9.8 - 0.37)

N = 60 * 9.43

Na = 565.8N

To find scalar reading at the bottom, we use;

[tex] N_b = m(g + w^2 R) [/tex]

= 60 (9.8 + 0.37)

= 60 * 10.17

[tex] N_b = 610.44 [/tex]

Answer:

The scale reading at the top is [tex]z_{top} = 565.6\mu N[/tex]

The scale reading at the bottom is [tex]z_{bottom} = 610.356\ N[/tex]

Explanation:

From the question

       The radius is [tex]r = 9.7 m[/tex]

       The period is [tex]T = 32sec[/tex]

      The mass is [tex]m =60kg[/tex]

Generally the mathematical representation for angular velocity of the wheel is

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

                    [tex]= \frac{2*3.142}{32}[/tex]

                    [tex]= 0.196 \ rad/sec[/tex]

The velocity at which the point scale move can be obtained as

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

                        [tex]= 9.7* 0.196[/tex]

                       [tex]= 1.9 m/s[/tex]

Considering the motion of the 60kg mass as shown on the first and second uploaded image

Let the z represent the reading on the scale which is equivalent to the normal force acting on the mass.

          Now at the topmost the reading of the scale would be

                       [tex]mg - z_{top} =\frac{mv^2}{r} =m\omega^2r[/tex]

Where mg is the gravitational force acting on the mass and [tex]\frac{mv^2}{r}[/tex] is the centripetal force keeping the mass from spiraling out of the circle

Now making z the subject of the formula

                    [tex]z_{top} = mg - \frac{mv^2}{r}[/tex]

                       [tex]= m(g- \frac{v^2}{r})[/tex]

                       [tex]= 60(9.8 - \frac{1.9^2}{9.7} )[/tex]

                      [tex]= 565.6\mu N[/tex]

Now at the bottom the scale would be

                      [tex]z_{bottom}-mg = \frac{mv^2}{r} = m \omega^2r[/tex]

This is because in order for the net force to be in the positive y-axis (i.e for the mass to keep moving in the Ferris wheel) the Normal force must be greater than the gravitational force.  

Making  z the subject

                  [tex]z_{bottom}= mg +m\omega^2r[/tex]

                  [tex]z =m(g+\omega^2r)[/tex]

                  [tex]z= 60(9.8 + (0.196)^2 *(9.7))[/tex]

                     [tex]= 610.356\ N[/tex]

Answer:

[tex]d=2.4\ miles[/tex]

Explanation:

Given:

According to given condition:

[tex]t_w=t_b+\frac{36}{60}[/tex]

where:

[tex]t_w=[/tex] time taken to reach the building by walking

[tex]t_b=[/tex] time taken to reach the building by biking

We know that,

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

so,

[tex]\frac{d}{v_w} =\frac{d}{v_b} +\frac{36}{60}[/tex]

[tex]\frac{d}{3}=\frac{d}{12} +\frac{3}{5}[/tex]

[tex]d=2.4\ miles[/tex]

Answer:

The distance from her apartment to the classroom building is 2.4 miles.

Explanation:

Given that,

Time  = 36 min

Walking average speed of her = 3 m/h

biking average speed of her = 12 m/h

If she takes n minutes to ride, then if the distance is d miles,

We need to calculate the distance

Using formula of time

[tex]t=t_{1}+t_{2}[/tex]

[tex]t=\dfrac{d}{v_{1}}+\dfrac{d}{v_{2}}[/tex]

Put the value into the formula

[tex]\dfrac{36}{60}=\dfrac{d}{3}-\dfrac{d}{12}[/tex]

[tex]d=\dfrac{36\times12}{60\times3}[/tex]

[tex]d=2.4\ miles[/tex]

Hence, The distance from her apartment to the classroom building is 2.4 miles.

Answer:

7.08 kg

Explanation:

Given:

Length of scale (x) = 11.0 cm = 0.110 m [1 cm = 0.01 m]

Range of scale is from 0 to 200 N.

Frequency of oscillation of fish (f) = 2.55 Hz

Mass of the fish (m) = ?

Now, range of scale is from 0 to 200 N. So, maximum force, the spring can hold is 200 N. For this maximum force, the extension in the spring is equal to the length of the scale. So, [tex]x = 0.11\ m[/tex]

Now, we know that, spring force is given as:

[tex]F=kx\\\\k=\frac{F}{x}[/tex]

Where, 'k' is spring constant.

Now, plug in the given values and solve for 'k'. This gives,

[tex]k=\frac{200\ N}{0.11\ m}=1818.18\ N/m[/tex]

Now, the oscillation of the fish represents simple harmonic motion as it is attached to the spring.

So, the frequency of oscillation is given as:

[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]

Squaring both sides and expressing it in terms of 'm', we get:

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

Now, plug in the given values and solve for 'm'. This gives,

[tex]m=\frac{1818.18\ N/m}{4\pi^2\times (2.55\ Hz)^2}\\\\m=\frac{1818.18\ N/m}{256.708\ Hz^2}\\\\m=7.08\ kg[/tex]

Therefore, the mass of the fish is 7.08 kg.

Answer:

0.5099 T

Explanation:

We are given that

Number of turns=N=1785

Resistance of circuit, R=[tex]43.9\Omega[/tex]

Area of each turn,[tex]A=4.5\times 10^{-4} m^2[/tex]

Charge , q=[tex]9.33\times 10^{-3} C[/tex]

We have to find the magnitude of the magnetic field.

We know that magnetic field, [tex]B=\frac{qR}{NA}[/tex]

Substitute the values

Magnetic field, B=[tex]\frac{9.33\times 10^{-3}\times 43.9}{1785\times 4.5\times 10^{-4}}=0.5099 T [/tex]

Hence, the magnitude of the magnetic field=0.5099 T

Answer:

D) the masses of the atoms and the strength of the bond.

Explanation:

Mostly in diatomic and triatomic molecules, bonds of the molecules experience vibrations and rotations. When there is a continuous change in the bond distance of the two atoms then these vibrations are termed as stretching vibrations.

As these vibrations exist in the bonds between the atoms. So they depend upon the masses of the atoms and strength of the bonds. Greater masses of the atoms and strong bond strength will result in reduction of vibration. Thats why we don't observe such stretching vibrations in larger, massive molecules. They mostly exists in the diatomic and triatomic molecules where the bond strength is not that much stronger and the masses of the atoms are small.

For a square in an electric field, the potential difference between points C and D is the same as that between A and B, and the potential difference between A and D is zero.

To answer these questions, we first need to understand the concept of electric potential difference, or voltage. It's defined as the change in potential energy of a charge moved between two points, divided by the charge. In the case of a square area in an electric field, if the potential difference between points A and B (ΔV_AB) and between B and D (ΔV_BD) is x volts, then the potential difference between A and C (ΔV_AC) is -x volts, as it's negative of ΔV_AB. Due to this, the potential difference between C and D (ΔV_CD) would be also x volts as it's same as that between A and B and equal to ΔV_BD.

For Part C of the question, thinking about the square as a cycle, if we traverse from A to B to D (or vice versa) the total potential difference would be x - x = 0 volts. Therefore, the potential difference between A and D (ΔV_AD) is zero.

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a) 37.9 m/s

b) 1.35 rad below horizontal

c) 7.56 s

Explanation:

a-c)

At the beginning, both the owl and the vole are travelling in a horizontal direction at a speed of

[tex]v_x=8.3 m/s[/tex]

After the vole wiggles free, the owl takes 2 seconds to react; the horizontal distance covered by the owl during this time is

[tex]d_x = v_x t =(8.3)(2)=16.6 m[/tex]

The vertical motion of the wiggle is a free fall motion, so it is a uniformly accelerated motion with constant acceleration

[tex]g=9.8 m/s^2[/tex] in the downward direction

The wiggle falls from a height of h' = 282 m to a height of h = 2 m, so the vertical displacement is

s = h' - h = 282 - 2 = 280 m

The time it takes the wiggle to cover this distance is given by the suvat equation:

[tex]s=u_y t - \frac{1}{2}gt^2[/tex]

where [tex]u_y = 0[/tex] is the initial vertical velocity. Solving for t,

[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(280)}{9.8}}=7.56 s[/tex]

The owl must cover the vertical distance of 280 m in this time interval, so its vertical speed must be:

[tex]v_y=\frac{s}{t}=\frac{280}{7.56}=37.0 m/s[/tex]

Therefore, the speed of the owl during the dive is the resultant of the velocities in the two directions:

[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{8.3^2+37.0^2}=37.9 m/s[/tex]

b)

In part a-c, we calculated that the components of the velocity of the owl in the horizontal and vertical direction, and they are

[tex]v_x=8.3 m/s\\v_y=37.0 m/s[/tex]

This means that the angle of the owl's dive, with respect to the original horizontal direction, is

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

And substituting these values, we find:

[tex]\theta=tan^{-1}(\frac{37.0}{8.3})=77.4^{\circ}[/tex]

And this angle is below the horizontal direction.

Converting into radians,

[tex]\theta=77.4\cdot \frac{2\pi}{360}=1.35 rad[/tex]

Final answer:

The owl's dive speed is 7.15 m/s, dive angle is 30 degrees below the horizontal, and it takes about 1.76 seconds for the mouse to fall.

Explanation:

The owl's dive speed: Using the equation of motion for constant acceleration, we can find that the owl's dive speed is 7.15 m/s.

The owl's dive angle below the horizontal: The owl's dive angle is 30 degrees above the horizontal when diving.

How long the mouse falls: Applying the kinematic equation for vertical motion, the time it takes for the mouse to fall completely is approximately 1.76 seconds.

Answer:

They both have the same speed when hitting the ground below

Explanation:

Conservation of the Mechanical Energy

In the absence of external non-conservative forces, the total mechanical energy of a particle is conserved or is kept constant.

The mechanical energy is the sum of the potential gravitational and kinetic energies, i.e.

[tex]\displaystyle M=mgh+\frac{mv^2}{2}[/tex]

When the first ball is launched at the top of the building of height h, at a speed vo, the mechanical energy is

[tex]\displaystyle M=mgh+\frac{mv_o^2}{2}[/tex]

When the ball reaches ground level (h=0) the mechanical energy is

[tex]\displaystyle M'=\frac{mv_f^2}{2}[/tex]

Equating M=M'

[tex]\displaystyle mgh+\frac{mv_o^2}{2}=\frac{mv_f^2}{2}[/tex]

We could solve the above equation for vf but it's not necessary because we have derived this relation regardless of the direction of the initial speed. It doesn't matter if it's launched with an angle above the horizontal, directly horizontal, or even directly downwards, the final speed is always the same.

It can also be proven with the exclusive use of the kinematic equations.

Note: The speed is the same for both balls, but not the velocity since the direction of the final velocity will be different in each case. Its magnitude is the same for all cases.

Answer: They both have the same speed when hitting the ground below

Explanation:

Below is an attachment containing the solution.

The magnitude of the magnetic force experienced by the bottom side of the rectangular coil, placed in a uniform magnetic field and carrying a current, is 21.16 N.

The subject of this question relates to the interaction of a current-carrying wire in a magnetic field, a fundamental concept in physics. In this particular setup, the magnetic force on each segment of the rectangular coil can be determined by the formula: F = I (L × B), where F is the magnetic force, I is the current, L is the length of the wire and B is the magnetic field. But in this instance, we're specifically interested in the force exerted on the bottom side of the loop, for which the magnetic field and current are perpendicular to each other.

Therefore, the force is given by F = ILB. By substituting the given values into the equation—the length of the bottom side (b = 0.46 m), the magnitude of the current (I = 10 A), and the strength of the magnetic field (B = 4.60 T)—we obtain: F = (10 A)(0.46 m)(4.60 T) = 21.16 N.

So, the magnitude of the magnetic force on the bottom side of the loop is 21.16 N.

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Answer: a) r = 281m, b) r = 5.13×10^-7m

Explanation: From the question, we realised that the particle enters the earth magnetic field with it velocity perpendicular to the magnetic field thus making the particle have a circular motion.

The force exerted on a charge in a magnetic field perpendicular to the velocity of the particle is responsible for the centripetal force required to give the object it circular motion.

Magnetic force = centripetal force.

qvB = mv²/r

By dividing "v" on both sides, we have that

qB = mv/r

Above is the formulae that defines the circular motion of a particle in the earth's magnetic field

Where q = magnitude of electronic charge.

B = strength of magnetic field = 1.62×10^-7 T

v = speed of particle = 8.03×10^6 m/s

A) If the particle where to be an electron, q ( magnitude of electron) =1.609×10^-19c.

m = mass of electronic charge = 9.11×10^-31 kg.

By substituting these parameters into the formulae we have that

1.609×10^-19× 1.62×10^-7 = 9.11×10^-31 × 8.03×10^6/ r

By cross multiplying, we have that

1.609×10^-19× 1.62×10^-7 × r = 9.11×10^-31 × 8.03×10^6

r = 9.11×10^-31 × 8.03×10^6 / 1.609×10^-19× 1.62×10^-7

r = 7.32*10^(-24)/ 2.61×10^-26

r = 2.81×10²

r = 281m

B)

If the particle is proton, q = magnitude of a proton charge = 1.609×10^-19c,

m = mass of proton = 1.673×10^-27 kg

By substituting these parameters into the formulae we have that

1.609×10^-19× 1.62×10^-7 = 1.673×10^-27×8.03×10^6/ r

By cross multiplying, we have that

1.609×10^-19× 1.62×10^-7 × r = 1.673×10^-27× 8.03×10^6

r = 1.673×10^-27 × 8.03×10^6 / 1.609×10^-19× 1.62×10^-7

r = 1.34*10^(-32)/ 2.61×10^-26

r = 0.513×10^-6 m

r = 5.13×10^-7m

Answer:

Explanation:

The dimness of two light source, which have the same intensify depends on their distance from the observer. That is why some stars appear brighter than the others, this is due to the distance of each of them from the earth.

The closer a light source, the brighter they appear

Now, if the right lamp has a dim light actually, it shows that the bicycle is closer than it appears.

Answer a) impulse is 1360 kg.m/s

Explanation: impulse is impact force times the time of impact

I = 3400 * 0.4 = 1360kg.m/s

Answer b) final velocity is 10.1m/s

Explanation: impulse is the change of momentum

I = m(v-u)

V and u are final and initial velocities respectively.

1360 = 200 (v - 3.3)

1360 = 200v - 660

V = (1360+660)÷200

V = 10.1m/s

Q2 answer is: astronaut has KE approximately 208 times that of the truck

Explanation :

KE = 0.5mv^2

For truck

KE = 11000 * 33.3^2 = 12197790J

For astronaut

KE = 42 * 7777.7^2 = 2540689926J

Comparing

2540689926/12197790 = 208.2

NB: speed has been converted to m/s by multiplying with 0.28 I.e 1000/3600

To calculate the radial acceleration of a ball in circular motion, derive its velocity using the horizontal distance traveled upon string break and use the centripetal acceleration formula.

The question provided involves calculating the radial acceleration of a ball in circular motion before the string breaks. To solve this, we must understand that the radial or centripetal acceleration formula is a = v^2 / r, where a is the centripetal acceleration, v is the velocity of the object in circular motion, and r is the radius of the circle.

However, the information given directly does not include the velocity v. We can derive the velocity using the horizontal distance the ball traveled after the string broke. Since the only force acting on the ball after the string breaks is gravity, the horizontal motion can be considered uniform. The formula distance = velocity x time (d = vt) can be rearranged to find the velocity (v = d/t). Using the principle of projectile motion, the time (t) it takes for the ball to hit the ground can be found using the formula derived from the vertical motion due to gravity: t = sqrt(2h/g), where h is the height above the ground and g is the acceleration due to gravity (9.81 m/s2).

Coming back to finding the radial acceleration, once we have the velocity, we simply substitute values into the centripetal acceleration formula. Note that the actual calculations were not performed as the goal here is to elucidate the strategy for solving the problem.

Final answer:

The radial acceleration of the ball during circular motion was calculated to be approximately 47.7 m/s².

Explanation:

To find the radial acceleration of the ball during its circular motion, we'll use the information provided about the ball's horizontal displacement after the string breaks.

In a situation where the string breaks and an object follows a projectile motion, the horizontal component of its initial velocity (vx) can be calculated using the horizontal displacement (d) and the time of flight (t). The formula for horizontal displacement is d = vx * t.

The time of flight (t) can be found using the vertical motion equations, considering that the ball drops 1.50 m. Since the ball is released from rest in the vertical direction, we have:

Vertical displacement (y): 1.50 m

Acceleration due to gravity (g): 9.81 m/s2

Initial vertical velocity (vy0): 0 m/s

Using the equation y = vy0 * t + 0.5 * g * [tex]t^2[/tex], we can solve for the time of flight (t), which will also be the same time the ball is moving horizontally since horizontal and vertical motions are independent.

Applying this equation to calculate the time (t):

1.50 = 0 + 0.5 * 9.81 * [tex]t^2[/tex]
t = sqrt(2 * 1.50 / 9.81)
t ≈ 0.553 s (rounded to three significant figures)

Now we can find the horizontal velocity (vx) using the horizontal distance (d):

2.10 = vx * 0.553
vx ≈ 3.80 m/s

The horizontal velocity of the ball at the instant the string breaks is the same as the tangential velocity of the ball while it was in circular motion. Therefore, we can calculate the radial or centripetal acceleration (ar) using the formula for centripetal acceleration:

ar = v2 / r
where v is the tangential velocity and r is the radius.

So:

ar = [tex](3.80 m/s)^2[/tex] / 0.300 m
ar ≈ 47.7 m/s2

The radial acceleration of the ball during its circular motion was approximately 47.7 [tex]m/s^2[/tex].

Answer:

1.42×10⁻¹⁶ N.

Explanation:

The force on a charge moving in a magnetic field is given as

F = qBvsin∅..................... Equation 1

Where F = Force on the charge, q = charge, B = Magnetic Field, v = speed, ∅ = angle between the magnetic field and the speed

Given: B = 0.78 T,  v = 1140 m/s, ∅ = 90° ( Perpendicular) q = 1.60 × 10⁻¹⁹ C.

Substitute into equation 1

F = 0.78(1140)(1.60 × 10⁻¹⁹)sin90°

F = 1.42×10⁻¹⁶ N.

Hence the force on the charge = 1.42×10⁻¹⁶ N.

Final answer:

During the elevator's deceleration, the scale reading would increase to approximately 96.3 N as the passenger experiences an upward acceleration.

Explanation:

During the stopping process of the elevator that is descending at 4 m/s and comes to a rest in 2 seconds, the reading on the scale will change due to the acceleration the passenger experiences while the elevator decelerates. The initial velocity v0 is -4 m/s (downwards) and the final velocity v is 0 m/s, as the elevator comes to a stop. The time t taken to stop is 2 seconds, so to find the acceleration, we use the formula a = (v - v0)/t.

This gives us an acceleration of (0 - (-4))/2 = 2 m/s2 upwards, since deceleration is a positive acceleration in the opposite direction of movement. A person's apparent weight on a scale is the normal force exerted by the scale, which can be calculated from Newton's second law, F = ma, where m is mass and a is the acceleration. In this case, the person's actual mass (m) can be deduced from their weight (W = mg), where g is the acceleration due to gravity (approximately 9.81 m/s2).

By rearranging the weight formula, m = W/g, we find m = 80 N / 9.81 m/s2 ≈ 8.16 kg. Now, we add the elevator's upward acceleration to the gravity, so the total acceleration the person feels is (g + a) = 9.81 m/s2 + 2 m/s2 = 11.81 m/s2. Applying F = ma gives us the scale reading: F = 8.16 kg × 11.81 m/s2 ≈ 96.3 N.

The correct answer is that the reading on the scale during the stopping process is 160 N.

When the elevator comes to a stop, it decelerates, which means there is an upward acceleration. According to Newton's second law, the net force acting on an object is equal to the mass of the object times its acceleration ([tex]F_net = m * a[/tex]). In this case, the net force is the difference between the normal force and the gravitational force [tex](F_net = N - mg)[/tex], where mg is the weight of the person.

Given that the elevator decelerates at a rate of 4 m/s in 2 seconds, we can calculate the deceleration (a) as follows:

[tex]\[ a = \frac{\Delta v}{\Delta t} = \frac{4 \text{ m/s}}{2 \text{ s}} = 2 \text{ m/s}^2 \][/tex]

Now, we can set up the equation for the net force during deceleration:

[tex]\[ N - mg = m * a \][/tex]

Since the weight of the person (mg) is 80 N, we can substitute this value into the equation:

[tex]\[ N - 80 \text{ N} = m * 2 \text{ m/s}^2 \][/tex]

To find the normal force (N) during deceleration, we need to solve for N:

[tex]\[ N = m * 2 \text{ m/s}^2 + 80 \text{ N} \][/tex]

We know that when the elevator was moving at a constant rate, the normal force was equal to the weight of the person [tex](N_constant = 80 N).[/tex] During deceleration, the normal force must be twice the weight of the person to provide the additional force required to decelerate the person:

[tex]\[ N = 2 * 80 \text{ N} = 160 \text{ N} \][/tex]

Therefore, the reading on the scale during the stopping process is 160 N.

Explanation:

Below is an attachment containing the solution .

Answer:

The induced emf 1.43 s after the circuit is closed is 4.19 V

Explanation:

The current equation in LR circuit is :

[tex]I=\frac{V}{R} (1-e^{\frac{-Rt}{L} })[/tex]    .....(1)

Here I is current, V is source voltage, R is resistance, L is inductance and t is time.

The induced emf is determine by the equation :

[tex]V_{e}=L\frac{dI}{dt}[/tex]

Differentiating equation (1) with respect to time and put in above equation.

[tex]V_{e}= L\times\frac{V}{R}\times\frac{R}{L}e^{\frac{-Rt}{L} }[/tex]

[tex]V_{e}=Ve^{\frac{-Rt}{L} }[/tex]

Substitute 6.05 volts for V, 0.655 Ω for R, 2.55 H for L and 1.43 s for t in the above equation.

[tex]V_{e}=6.05e^{\frac{-0.655\times1.43}{2.55} }[/tex]

[tex]V_{e}=4.19\ V[/tex]

Based on the calculations, the induced emf is equal to 4.19 Volts.

Given the following data:

In a RL circuit, current is given by this mathematical expression:

[tex]I=\frac{V}{R} (1-e^{\frac{Rt}{L} })[/tex]

Where:

For an induced emf in a circuit, we have:

[tex]E=L\frac{dI}{dt} \\\\E=L \times I \\\\E=L \times \frac{V}{R} (1-e^{\frac{-Rt}{L} })\\\\E=V e^{\frac{-Rt}{L} }\\\\E=6.05 e^{\frac{-0.655 \times 1.43}{2.55} }[/tex]

E = 4.19 Volts.

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The temperature of the oven is 200°F

Explanation:

Given-

We have to apply Newton's law of cooling or heating

[tex]\frac{dT}{dt} = k ( T - Tm)\\\\\frac{dT}{T - Tm} = kdt[/tex]

On Integrating both sides, we get

[tex]T = Tm + Ce^k^t[/tex]

On putting the value,

T(0) = 65°F

[tex]65 = Tm + C\\C = 65 - Tm\\\\T = Tm + (65 - Tm) e^k^t[/tex]

After 1/2 minute, thermometer reads 110°F. So,

[tex]110 = Tm + (65 - Tm)e^0^.^5^k[/tex]                - 1

After 1 minute, thermometer reads 140°F. So,

[tex]140 = Tm + (65 - Tm)e^k[/tex]                     - 2

Dividing equation 2 by 1:

[tex]e^k^-^0^.^5^k = \frac{140 - Tm}{110 - Tm} \\\\e^0^.^5^k = \frac{140 - Tm}{110 - Tm}[/tex]                                - 3

From 1 we have,

[tex]e^0^.^5^k = \frac{110 - Tm}{65 - Tm}[/tex]

Putting this vale in eqn 3. We get,

[tex]\frac{110 - Tm}{65 - Tm} = \frac{140 - Tm}{110 - Tm}[/tex]

[tex](110-Tm)^2 = (140-Tm) (65-Tm)\\\\12100 + Tm^2 - 220Tm = 9100 - 140Tm - 65Tm + Tm^2\\\\3000 - 220Tm = -205Tm\\\\3000 = 15Tm\\\\Tm = 200[/tex]

Therefore, the temperature of the oven is 200°F

The Temperature ( hotness )  of the oven is ; 200°F

Given data :

Initial thermometer reading ( To ) = 65°F

Thermometer reading after 1/2 minute ( T1/2 ) = 110°F

Thermometer reading after 1 minute ( T1 ) = 140°F

Temperature of oven ( Tm ) = ?

Determine the Temperature  of the Oven

To determine the temperature of the oven we will apply Newton's law of heating and cooling.

[tex]\frac{dT}{dt} = K(T -Tm )[/tex]    (  Integrating the expression we will have )

T = Tm + [tex]Ce^{kt}[/tex]   ---- ( 1 )

where k = time

at T = 0 equation becomes

65 = Tm + C

∴ C = 65 - Tm

Back to equation ( 1 )

T = Tm + ( 65 -Tm )[tex]e^{kt}[/tex]  ---- ( 2 )

After 1/2 minute equation 2 becomes

110 = Tm + ( 65 - Tm )[tex]e^{0.5k}[/tex] ---- ( 3 )

After 1 minute equation 2 becomes

140 = Tm + ( 65 - Tm )[tex]e^{k}[/tex]  ---- ( 4 )

Next step : Divide equation 4 by equation 3

[tex]e^{0.5k} = \frac{140 - Tm}{110 - Tm}[/tex]  ----- ( 5 )

Also From equation 3

[tex]e^{0.5k} = \frac{110 - Tm}{65 - Tm}[/tex]  --- ( 6 )

Next step : Equating equations ( 5 ) and ( 6 )

( 110 - Tm )² = ( 140 - Tm )( 65 - Tm )

3000 - 220 Tm = -205Tm

∴ Tm = 3000 / 15 = 200°F

Hence we can conclude that the temperature ( hotness ) of the oven is 200°F.

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Answer:

Part (a) current in the wire is 1.144 A

Part (b) the drift velocity of electrons in the wire is 1.028 x 10⁻⁴ m/s

Explanation:

Given;

diameter d  = 1.02 mm

current density J = 1.40×10⁶ A/m²

number of electron = 8.5×10²⁸ electrons

Part (a) Current in the wire

I = J×A

Where A is area of the wire;

[tex]A = \frac{\pi d^2}{4} \\\\A = \frac{\pi (1.02X10^{-3})^2}{4} = 8.1723 X10^{-7} m^2[/tex]

I = 1.40 x 10⁶ x 8.1723 x 10⁻⁷

I = 1.144 A

Part (b) the drift velocity of electrons in the wire

[tex]V = \frac{J}{nq} = \frac{1.4X10^6}{8.5X10^{28} X 1.602X10^{-19}} = 1.028 X10^{-4} m/s[/tex]

We were given the

diameter = 1.02 mm

current density = 1.40×10⁶ A/m²

number of electron = 8.5×10²⁸ electrons

We can use the formula:

I = J×A

where I is current, J is density and A is area.

A = π d²

        4

  = π (1.02ₓ 10⁻³)² = 8.1723 x 10⁻⁷

              4

I = J×A

I = 1.40 x 10⁶ x 8.1723 x 10⁻⁷

I = 1.144 A

V = J/ nq

    =   1.4 ₓ 10⁶ / (8.5ₓ 10²⁸ₓ 1.602ₓ 10⁻¹⁹)

   = 1.028ₓ 10⁻⁴ m/s

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Answer:

At the very bottom, whnere the sag is the greatest, Jay’s acceleration is upward.

Explanation:

As Jay lands on the trampoline, Jay’s motion was dowward,  the trampoline was opposing his motion and hence, caused him to reach an initial halt position. Afterwards, the trampoline causes Jay to move back into the  air and therefore, the change in velocity vector act in upward direction. The acceleration vector is always align towards the change in velocity vector's direction.

Answer:

The required constant friction force for the last 20 m is 6,862.8 N

Explanation:

Energy Conversion

There are several ways the energy is manifested in our physical reality. Some examples are Kinetic, Elastic, Chemical, Electric, Potential, Thermal, Mechanical, just to mention some.

The energy can be converted from one form to another by changing the conditions the objects behave. The question at hand states some types of energy that properly managed, will make the situation keep under control.

Originally, the m=220 kg car is at (near) rest at the top of a h=101 m tall track. We can assume the only energy present at that moment is the potential gravitational energy:

[tex]E_1=mgh=220\cdot 9.8\cdot 101=217,756\ J[/tex]

For the next x1=230 m, a constant friction force Fr1=350 N is applied until it reaches ground level. This means all the potential gravitational energy was converted to speed (kinetic energy K1) and friction (thermal energy W1). Thus

[tex]E_1=K_1+W_1[/tex]

We can compute the thermal energy lost during this part of the motion by using the constant friction force and the distance traveled:

[tex]W_1=F_{r1}\cdot x_1=350\cdot 230=80,500\ J[/tex]

This means that the kinetic energy that remains when the car reaches ground level is

[tex]K_1=E_1-W_1=217,756\ J-80,500\ J=137,256\ J[/tex]

We could calculate the speed at that point but it's not required or necessary. That kinetic energy is what keeps the car moving to its last section of x2=20 m where a final friction force Fr2 will be applied to completely stop it. This means all the kinetic energy will be converted to thermal energy:

[tex]W_2=F_{r2}\cdot x_2=137,256[/tex]

Solving for Fr2

[tex]\displaystyle F_{r2}=\frac{137,256}{20}=6,862.8\ N[/tex]

The required constant friction force for the last 20 m is 6,862.8 N

To determine the required constant friction force for the last 20 m of the roller coaster ride, we consider the energy changes that occur. The work done by the friction force is equal to the loss in potential energy. By using the equation -mgh = -f x d, we can solve for the friction force.

To determine the required constant friction force for the last 20 m of the roller coaster ride, we need to consider the energy changes that occur. As the car rolls downhill, it loses potential energy and gains kinetic energy. The work done by the friction force is equal to the loss in potential energy. Using the equation -m g h = - f x d, where m is the mass of the car, g is the acceleration due to gravity, h is the height of the track, f is the friction force, and d is the distance over which the force acts, we can solve for the friction force.

In this case, the mass of the car is 220 kg, the height of the track is 101 m, and the distance over which the force acts is 20 m. Plugging these values into the equation, we get:

-220 kg * 9.8 m/s² * 101 m = -f * 20 m

Solving for f, we find that the required constant friction force for the last 20 m of the ride is approximately 2151 N.

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Answer:

for the people of the Earth traveling they last much more than 70 years

Explanation:

In order to answer this answer we must place ourselves in the context of special relativity, which are the expressions for time and displacement since the speed of light has a finite speed that is the same for all observers.

 The life time of the person is 70 years in a fixed reference system in the person this time we will call their own time (t₀), when the person is placed in a ship that moves at high speed, very close to the speed of the light the time or that an observer measures on Earth, the expression for this time is

              t = t₀ 1 / √(1 - (v / c)²)

We see that if the speed of the ship is very close to the speed of light the  

the value of the root of the denominator is very high, for which for the person on Earth it measures a very large time even when the person on the ship travels has a time within its 70 years of life

In concussion, for the people of the Earth traveling they last much more than 70 years

Answer:

577g

Explanation:

Given parameters:

Temperature change = 5.9°C

Amount of heat lost = 427J

Unknown:

Mass of the block = ?

Solution:

The heat capacity of a body is the amount of heat required to change the temperature of that body by 1°C.

                H =  m c Ф

  H is the heat capacity

 m  is the mass of the block

  c is the specific heat capacity

   Ф is the temperature change

Specific heat capacity of lead is 0.126J/g°C

   m = H / m Ф

   m = [tex]\frac{427}{0.126 x 5.9}[/tex]  = 577g

Mass of the lead block is 577g

Answer: 0.5654

Explanation:

Answer:

The current in the primary coil would be [tex]0.5\ A[/tex].

Explanation:

Given the power supplied by a transformer is 60 watts.

And the voltage in the primary coil is 120 Volts.

We need to find the current supply in the primary coil.

We will use the formula

[tex]P=V\times I[/tex]

Where,

[tex]P[/tex] is the power in Watts.

[tex]V[/tex] is the voltage in Volts.

[tex]I[/tex] is the current supply in Ampere.

[tex]I=\frac{P}{V}\\\\I=\frac{60}{120}\\ \\I=0.5\ A[/tex]

So, the current in the primary coil would be [tex]0.5\ A[/tex].

Answer:

4 A

Explanation:

We are given that

[tex]R_1=R_2=R_3=4\Omega[/tex]

I=12 A

We have to find the current flowing through each resistor.

We know that in parallel combination current flowing through different resistors are different and potential difference across each resistor is same.

Formula :

[tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}[/tex]

Using the formula

[tex]\frac{1}{R}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}[/tex]

[tex]R=\frac{4}{3}\Omega[/tex]

[tex]V=IR[/tex]

Substitute the values

[tex]V=12\times \frac{4}{3}=16 V[/tex]

[tex]I_1=\frac{V}{R_1}=\frac{16}{4}=4 A[/tex]

[tex]I_1=I_2=I_3=4 A[/tex]

Hence, current flows through any one of the resistors is 4 A.

The current flows through a resistor is equal to 4A.

To understand more, check below explanation.

When n number of resistors are connected in parallel to a battery  and current I flowing in the circuit.

Then, current flows in each resitor [tex]=\frac{I}{n}[/tex]

It is given that, three identical resistors are connected in parallel to a battery and total current of 12A flows through the circuit.

So that, current in each resistor [tex]=\frac{12}{3}=4A[/tex]

Hence, the current flows through a resistor is equal to 4A.

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Answer:

Force is needed to begin to move the book and  force needed to keep the book moving at a constant velocity is 10.78 N and 5.85 N.

Explanation:

Given :

Mass of book , M = 2.03 kg.

Coefficient of static friction , [tex]\mu_s=0.542 \ .[/tex]

Coefficient of kinetic friction , [tex]\mu_k=0.294\ .[/tex]

Force, required needed to begin to move the book ,

[tex]F=\mu_sN=\mu_s(mg)=0.542\times 2.03\times 9.8=10.78\ N.[/tex]

Now, We know kinetic friction acts when object is in motion .

Therefore , Force, required o keep the book moving at a constant velocity

[tex]F=\mu_kN=\mu_k(mg)=0.294\times 2.03\times 9.8=5.85\ N.[/tex]

Hence, this is the required solution.

Answer:

ΔS = - 3.74 × 10⁻²³ J/K = - 3.7 × 10⁻²³ to 2 s.f

Explanation:

The change in entropy for a system with changing allowable Microsystems is given as

ΔS = K In (W/W₀)

K = Boltzmann's constant = 1.381 × 10⁻²³ J/K

W₀ = initial number of microstates

To calculate this, how many ways can 2 atoms occupy 16 available microstates with order important?

That is, ¹⁶P₂ = 16!/(16 - 2)! = 16 × 15 = 240 microstates.

W = the number of microstates for Chlorine atom when Bromine atom desorps = 16 microstates.

ΔS = K In (W/W₀)

ΔS = (1.381 × 10⁻²³) In (16/240)

ΔS = (1.381 × 10⁻²³) × 2.7081

ΔS = - 3.74 × 10⁻²³ J/K

Entropy change is positive when an atom desorbs from a surface and drifts away, allowing for more freedom of movement and increasing the system's entropy.

Entropy change is a measure of disorder in a system. When an atom desorbs from a surface and drifts away, the change in entropy is positive as it increases the freedom of movement of the atom. This occurs because there are more possible locations for the atom to occupy, leading to an increase in entropy.

Answer:

a) [tex]F_H=776.952\ N[/tex]

b) [tex]F_g=706.32\ N[/tex]

c) [tex]v=5.4249\ m.s^{-1}[/tex]

d) [tex]KE=1059.48\ J[/tex]

Explanation:

Given:

a)

Now the force of lift by the helicopter:

Here the lift force is the resultant of the force of gravity being overcome by the force of helicopter.

[tex]F_H-F_g=m.a[/tex]

where:

[tex]F_H=72\times 0.981+72\times9.81[/tex]

[tex]F_H=776.952\ N[/tex]

b)

The gravitational force on the astronaut:

[tex]F_g=m.g[/tex]

[tex]F_g=72\times 9.81[/tex]

[tex]F_g=706.32\ N[/tex]

d)

Since the astronaut has been picked from an ocean we assume her initial velocity to be zero, [tex]u=0\ m.s^{-1}[/tex]

using equation of motion:

[tex]v^2=u^2+2a.h[/tex]

[tex]v^2=0^2+2\times 0.981\times 15[/tex]

[tex]v=5.4249\ m.s^{-1}[/tex]

c)

Hence the kinetic energy:

[tex]KE=\frac{1}{2} m.v^2[/tex]

[tex]KE=0.5\times 72\times 5.4249^2[/tex]

[tex]KE=1059.48\ J[/tex]

Answer:

Explanation:

mass of helicopter, m = 72 kg

height, h = 15 m

acceleration, a = g/10

(a) Work done by the force

Work, W = force due to helicopter x distance

W = m x ( g + a) x h

W = 72 ( 9.8 + 0.98) x 15

W = 11642.4 J

(b) Work done by the gravitational force

W = - m x g x h

W = - 72 x 9.8 x 15

W = - 10584 J

(c) Kinetic energy = total Work done

K = 11642.4 - 10584

K = 1058.4 J

(d) Let the speed is v.

K = 0.5 x m v²

1058.4 = 0.5 x 72 x v²

v = 5.42 m/s

Now the force of lift by the helicopter:

where:

force by the helicopter

force of gravity

b)

The gravitational force on the astronaut:

d)

Since the astronaut has been picked from an ocean we assume her initial velocity to be zero,

using equation of motion:

c)

Hence the kinetic energy:

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Answer:

e. it is anywhere inside (the force is zero everywhere)

Explanation:

The relation between the force and electric field is given as follows

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

where q is the charge inside, and E is the electric field inside the balloon created by the charge on the surface.

By Gauss' Law, the electric field inside the balloon created by the charges on the surface (excluding the charge q, since the electric field of the same charge cannot apply a force on the same charge) is zero.

[tex]\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E4\pi r^2 = 0\\E = 0[/tex]

Since the external electric field inside the sphere is zero, then the force on the point charge is zero everywhere.

It is anywhere inside (the force is zero everywhere).

The electrical field on the surface of the charged sphere is calculated by applying Coulomb law;

[tex]E = \frac{Qr}{4\pi \varepsilon _0R^3}[/tex]

Electric force is calculated as follows;

F = Eq

The electric field inside the charged sphere is zero.

F = 0 x q = 0

Thus, we can conclude that, it is anywhere inside (the force is zero everywhere).

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Answer:

[tex]W = 10.28\ mm[/tex]

Explanation:

Given,

Red light wavelength = 633 nm

width of slit = 0.320 mm

distance,d = 2.60 m

Condition of first maximum

[tex]a sin \theta_1 = m\lambda [/tex]

[tex]\theta_1 =sin^{-1}(\dfrac{m\lambda}{a})[/tex]

m = 1

[tex]\theta_1 =sin^{-1}(\dfrac{633\times 10^{-9}}{0.32\times 10^{-3}})[/tex]

[tex]\theta_1 = 0.1133^\circ[/tex]

Width of the first minima

[tex]y_1 = L tan \theta_1[/tex]

[tex]y_1 = 2.60\times tan( 0.11331)[/tex]

[tex]y_1 = 5.14 \ mm[/tex]

Now, width of the central region

[tex]W = 2 y_1[/tex]

[tex]W = 2\times 5.14[/tex]

[tex]W = 10.28\ mm[/tex]