Given the position vector of the particle
r(t)=(t+1)i+(t^2−1)j+2t k, find the particle's velocity and acceleration vectors at t=1

Explanation:

The angle of the handle relative to the horizontal is 35°.  The angle of the ramp to the horizontal is 7°.  So the angle of the handle relative to the ramp is 28°.

cos 28° = 50 / F

F = 50 / cos 28°

F = 56.6 lbs

56.6 lbs force must be applied in the direction of the handle so that the component of the force parallel to the ramp is 50 lbs.

Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change in momentum.

As given in the problem if a ramp leading to the entrance of a building is inclined upward at an angle of 7 degrees. A suitcase is to be pulled up the ramp by a handle that makes an angle of 35 degrees horizontal.

The angle of the handle relative to the horizontal is 35°.  The angle of the ramp to the horizontal is 7°.  So the angle of the handle relative to the ramp is 28°.

cos 28° = 50 / Force

Force  = 50 / cos 28°

Force  = 56.6 lbs

Thus, the 56.6 lbs force must be applied in the direction of the handle so that the component of the force parallel to the ramp is 50 lbs

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The magnitude of linear momentum can be calculated using the equation p=mv. The momentum for the different objects given are: (a) Proton: 8.517 * 10^-21 kg*m/s, (b) Bullet: 7.2 kg*m/s, (c) Sprinter: 761.25 kg*m/s, and (d) Earth: 1.78324 * 10^29 kg*m/s.

The magnitude of the linear momentum of an object is calculated using the equation p=mv, where 'p' is the momentum, 'm' is the mass of the object, and 'v' is its velocity.

(a) Using the given values, m = 1.67 * 10^-27 kg and v = 5.10 * 10^6 m/s, the momentum of the proton is p = m*v = (1.67 * 10^-27 kg)*(5.10 * 10^6 m/s) = 8.517 * 10^-21 kg*m/s.

(b) The bullet's mass is 15.0 g which is 0.015 kg. Its velocity is 480 m/s. Thus, its momentum is p = m*v = (0.015kg)*(480m/s) = 7.2 kg*m/s.

(c) The sprinter's mass is 72.5 kg and velocity is 10.5 m/s. So, his momentum is p = m*v = (72.5kg)*(10.5m/s) = 761.25 kg*m/s.

(d) The Earth's mass is 5.98 * 10^24 kg and its orbital speed is 2.98 * 10^4 m/s. Hence, its momentum is p = m*v = (5.98 * 10^24 kg)*(2.98 * 10^4 m/s) = 1.78324 * 10^29 kg*m/s.

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The magnitude of the linear momentum for a proton, a bullet, a sprinter, and the Earth is approximately 8.52 * 10^-21 kg*m/s, 7.2 kg*m/s, 760.5 kg*m/s, and 1.78 * 10^29 kg*m/s, respectively.

To calculate the magnitude of the linear momentum, we need to multiply the mass of the object by its velocity, which follows the formula: momentum = mass * velocity.

For the given cases:

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Answer:

A. The starting height of the ball

Explanation:

When we talk about controlled variables, we refer to the variable that should be kept the same throughout the experiment. The reason why we do this, is to limit anything else that is not being tested, that may affect the results of the experiment.

In the scenario given, the experiment is to see the relationship between the initial height of a basketball and the height of its rebound bounce.

So you the starting height of the ball should vary, meaning it is NOT controlled.

Answer:

Starting height of the ball

Explanation:

To achieve a temperature independent resistance of 3.80 kΩ, you need to use a carbon resistor and Nichrome wire-wound resistor that counterbalance each other. This is possible because carbon and Nichrome have opposite temperature coefficients of resistance.

In order for the resistance to remain constant with temperature, the carbon resistor and the Nichrome wire-wound resistor must counterbalance each other. Meaning, when one's resistance increases with temperature, the other's resistance decreases, keeping the total resistance the same. Given that carbon and Nichrome have opposite temperature coefficients of resistance, they can accomplish this task.

Generally, the resistance R of a resistor is given by the formula R = R0(1 + α(T-T0)), where α is the temperature coefficient, T is the temperature and R0 is the resistance at reference temperature T0. As the temperature increases, a positive α will increase the resistance while a negative α will decrease it.

To make the combined resistance temperature independent, the sum of the change in resistance of the carbon resistor and the Nichrome resistor should be zero. Therefore, you would set up the equation where the increase of the carbon resistance equals the decrease of the Nichrome resistance. Solving this equation will give you the exact values required for the resistances of carbon and Nichrome at 0ºC in order to have a total resistance of 3.80 kΩ.

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Answer:

[tex]I = 94.33 kg m^2[/tex]

Explanation:

Let say the rod is slightly pulled away from its equilibrium position

So here net torque on the rod due to its weight is given as

[tex]\tau = mg dsin\theta[/tex]

since rod is pivoted at distance of 1.4 m from centre of gravity

so its moment of inertia about pivot point is given as

[tex]Inertia = I[/tex]

now we have

[tex]I \alpha = mg d sin\theta[/tex]

now for small angular displacement we will have

[tex]\alpha = \frac{mgd}{I}\theta[/tex]

so angular frequency of SHM is given as

[tex]\omega = \sqrt{\frac{mgd}{I}}[/tex]

now we will have

[tex]\frac{2\pi}{9} = \sqrt{\frac{4.8(9.8)1.4}{I}[/tex]

[tex]I = 94.33 kg m^2[/tex]

Final answer:

The moment of inertia from the pivot is approximately [tex]72.4 kg\(\cdot\)m2.[/tex]

Explanation:

To generate an accurate answer for the moment of inertia of the stick about the pivot when it is used as a physical pendulum, we can use the formula for the period of a physical pendulum, which is:

[tex]T = 2\(\pi\)\(\sqrt{I/(mgd)}\),[/tex]

where T is the period, I is the moment of inertia, m is the mass of the rod, g is the acceleration due to gravity (approximately 9.8 m/s2), and d is the distance from the pivot to the center of mass. Given that T = 9 seconds, m = 4.8 kg, and d = 1.4 m, we can rearrange the formula to solve for I:

[tex]I = T2mgd/(4\(\pi\)2).[/tex]

Plugging the given values into this equation:

[tex]I = 92 * 4.8 kg * 9.8 m/s^2 * 1.4 m / (4 * \(\pi\)2).[/tex]

Calculating this yields the moment of inertia I:

[tex]I \(\approx\) 72.4 kg\(\cdot\)m2.[/tex]

Answer:

[tex]E=3.3\times 10^4N/C[/tex]

Option D is the correct answer.

Explanation:

Electric field, E is the ratio of potential difference and distance between them.

Potential difference, V = 227 V

Distance between plates = 6.8 mm = 0.0068 m

Substituting,

         [tex]E=\frac{V}{d}=\frac{227}{0.0068}=3.3\times 10^4N/C[/tex]

Option D is the correct answer.

Answer:

3.8 x 10⁴ Ω

Explanation:

C = Capacitance = 165 μF

R = resistance of the filament = ?

t = time taken to charge the capacitor = 10 s

Q₀ = maximum charge stored by capacitor

Q = charge stored by capacitor at time "t" = 0.80 Q₀

T = Time constant

Charge stored by capacitor at any time is given as

[tex]Q = Q_{o}(1 - e^{\frac{-t}{T}})[/tex]

[tex]0.80 Q_{o} = Q_{o}(1 - e^{\frac{-10}{T}})[/tex]

T = 6.21 s

Time constant is given as

T = RC

6.21 = R (165 x 10⁻⁶)

R = 3.8 x 10⁴ Ω

The resistance of the filament in the bulb R = 3.8 x 10⁴ Ω

It is given that:-

Capacitance  C= 165 μF

Resistance of the filament = R=?

Time taken to charge the capacitor = t = 10 s

Now,

Q₀ = maximum charge which can be stored by the capacitor

Since the capacitor to 80% of its maximum charge

Then,

Q = charge stored by capacitor at time "t" = 0.80 Q₀

T = Time constant

The charge stored by the capacitor at any time is given as

[tex]Q=Q_{0} (1-e^{\dfrac{-t}{T} } )[/tex]

[tex]0.80Q_{0} =Q_{0} (1-e^{\dfrac{-10}{T} } )[/tex]

[tex]T=6.21 sec[/tex]  

Now the Time constant is given as

[tex]T=R\times C[/tex]

[tex]6.21= R\times (165\times 10^{-6} )[/tex]

[tex]R=3.8\times 10^{4}[/tex]Ω

The resistance of the filament in the bulb R = 3.8 x 10⁴ Ω

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Answer:

27.22 K

Explanation:

T₁ = initial temperature in fahrenheit = - 4.00 ⁰F

T₂ = final temperature in fahrenheit = 45.0 ⁰F

To convert the temperature from fahrenheit to kelvin, we can use the relation

[tex]K = \frac{F - 32}{1.8} + 273.15[/tex]

where F = Temperature in fahrenheit  and K = temperature in kelvin

T'₁ =  initial temperature in kelvin = (- 4.00 - 32)/1.8 + 273.15 = 253.15 K

T'₂ =  final temperature in kelvin = (45.0 - 32)/1.8 + 273.15 = 280.37 K

ΔT = Change in temperature

Change in temperature on kelvin scale is given as

ΔT = T'₂ - T'₁

ΔT = 280.37 - 253.15

ΔT = 27.22 K

To find the temperature change on the Kelvin scale, convert the given temperatures from Fahrenheit to Kelvin and subtract them.

The temperature change on the Kelvin scale can be determined by converting the given temperatures from Fahrenheit to Kelvin and then finding the difference between them.

First, convert -4.00°F to Kelvin:
273.15 K + (-4.00°F + 459.67 °F) × (5/9)

Next, convert 45.0°F to Kelvin:
273.15 K + (45.0°F + 459.67 °F) × (5/9)

Finally, subtract the two Kelvin temperatures to find the temperature change.

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Explanation:

The speed [tex]v[/tex] of a wave is given by the following equation:

[tex]v=\frac{\lambda}{T}[/tex] (1)

Where [tex]\lambda[/tex] is the wavelength and [tex]T[/tex] the period of the wave.

In addition we know there is an inverse relation between the period of a wave and its frequency [tex]f[/tex] :

[tex]f=\frac{1}{T}[/tex]  (2)

Substituting (2) in (1):

[tex]v=\lambda.f[/tex]  (3)

Now we can find the velocity of the wave with the known given values applying equation (3):

[tex]v=(10m)(2Hz)[/tex]  (4)

Finally:

[tex]v=20 \frac{m}{s}[/tex]  

Answer:

d = 15 mm

Explanation:

Force of repulsion between two current carrying wire is given by

[tex]F = \frac{\mu_0 i_1 i_2 L}{2\pi d}[/tex]

now this force of repulsion is counterbalanced by the weight of the wire

so we have

[tex]mg = \frac{\mu_0 i_1 i_2 L}{2\pi d}[/tex]

now we have

[tex]d = \frac{\mu_0 i_1 i_2 L}{2\pi mg}[/tex]

so here we can say that

[tex]d = \frac{\mu_0 i_1 i_2}{2\pi (m/L)g}[/tex]

now plug in all values in it

[tex]d = \frac{4\pi \times 10^{-7} (72.8)(72.7)}{2\pi (0.071)}[/tex]

[tex]d = 0.015 m[/tex]

d = 15 mm

Answer:

I think it will be half of the initial charge

Explanation:

Because we know, the resulting charge will be q1+q2/2, since one is neutral so the charge will be half q/2

Answer:

Final velocity will be 1.7 m/s in the initial direction of running back

Explanation:

Since there is no external force on the system of two line man so here we can say that momentum of the two person will remain constant

So we can say

initial momentum of the two persons = final momentum of them

[tex]m_1v_{1i} + m_2v_{2i} = (m_1 + m_2) v[/tex]

now from the above equation we will have

[tex]v = \frac{m_1v_{1i} + m_2v_{2i}}{m_1 + m_2}[/tex]

now plug in all the values in the above equation

[tex]v = \frac{118(1) + 74(-6)}{118 + 74}[/tex]

[tex]v = - 1.7 m/s[/tex]

Answer:

Power required, P = 175 watts

Explanation:

It is given that,

Weight of a student, F = mg = 200 N

The student climbs a flight of stairs of height, h = 7 m

Time taken, t = 8 s

We have to find the power required to perform this task. Work done per unit time is called the power required. Mathematically, it is given by :

[tex]P=\dfrac{W}{t}[/tex]

W = work done

t = time taken

[tex]P=\dfrac{mgh}{t}[/tex]

[tex]P=\dfrac{200\ N\times 7\ m}{8\ s}[/tex]

P = 175 watts

Hence, the power required to complete this task is 175 watts.

Answer:

Force exerted, F = 2.64 × 10⁷ Newton

Explanation:

It is given that,

Mass of the artillery shell, m = 1100 kg

It is accelerated at, [tex]a=2.4\times 10^4\ m/s^2[/tex]

We need to find the magnitude of force exerted on the ship by the artillery shell. It can be determined using Newton's second law of motion :

F = ma

[tex]F=1100\ kg\times 2.4\times 10^4\ m/s^2[/tex]

F = 26400000 Newton

or

F = 2.64 × 10⁷ Newton

So, the force exerted on the ship by the artillery shell is 2.64 × 10⁷ Newton.

Answer: The force exerted on the artillery shell is [tex]2.64\times 10^6N[/tex]  and the magnitude of force exerted on the ship by artillery shell is [tex]2.64\times 10^6N[/tex]

Explanation:

Force is defined as the push or pull on an object with some mass that causes change in its velocity.

It is also defined as the mass multiplied by the acceleration of the object.

Mathematically,

[tex]F=ma[/tex]

where,

F = force exerted on the artillery shell

m = mass of the artillery shell = 1100 kg

a = acceleration of the artillery shell = [tex]2.40\times 10^4m/s^2[/tex]

Putting values in above equation, we get:

[tex]F=1100kg\times 2.40\times 10^4m/s^2\\\\F=2.64\times 10^6N[/tex]

Now, according to Newton's third law, every action has an equal and opposite reaction.

So, the force exerted on the artillery shell will be equal to the force exerted on the ship by artillery shell acting in opposite direction.

Hence, the force exerted on the artillery shell is [tex]2.64\times 10^6N[/tex]  and the magnitude of force exerted on the ship by artillery shell is [tex]2.64\times 10^6N[/tex]

Answer:

The work is required to stretch it from 8 m to 16 m is 192 N-m

Explanation:

Given that,

Natural length = 8 m

Force F = 12 N

After stretched,

length = 10 m

We need to calculate the elongation

[tex]x = 10-8=2\ m[/tex]

Using hook's law

The restoring force is directly proportional to the displacement.

[tex]F\propto (-x)[/tex]

[tex]F = -kx[/tex]

Where, k = spring constant

Negative sign shows the displacement in opposite direction

Now, The value of k is

[tex]k = \dfrac{F}{x}[/tex]

[tex]k = \dfrac{12}{2}[/tex]

[tex]k = 6[/tex]

When stretch the string from 8 m to 16 m.

Then the elongation is

[tex]x=16-8=8\ m[/tex]

Now, The work is required to stretch it from 8 m to 16 m

[tex]W = \dfrac{1}{2}kx^2[/tex]

Where, k = spring constant

x = elongation

[tex]W=\dfrac{1}{2}\times6\times8\times8[/tex]

[tex]W=192\ N-m[/tex]

Hence, The work is required to stretch it from 8 m to 16 m is 192 N-m

Answer:

The distance is 8.25 m.

Explanation:

Given  that,

Speed = 2 m/s

Acceleration = 0.5 m/s^2

Time = 3 sec

We need to calculate the distance

Using equation of motion

[tex]s = ut+\dfrac{1}{2}at^2[/tex]

Where, u = initial velocity

a = acceleration

t = time

Put the value in the equation

[tex]s=2\times3+\dfrac{1}{2}\times0.5\times3^2[/tex]

[tex]s=8.25\ m[/tex]

Hence, The distance is 8.25 m.

Julie's distance at the end of the 3 seconds is 8.25 m.

To calculate the distance traveled by Julie at the end ]of 3 seconds, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these given values into equation 1

Hence, Julie's distance at the end of the 3 seconds is 8.25 m.
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Answer:

24.14 ⁰C

Explanation:

T₀ = initial temperature = 3 °C

R₀ = initial resistance of thermometer at initial temperature = 217.62 Ω

R = Final resistance of thermometer at final temperature = 215.32 Ω

T = final temperature = ?

α = temperature coefficient of resistivity for carbon = - 5.00 x 10⁻⁴ C⁻¹

Final resistance of thermometer at final temperature is given as

R = R₀ (1 + α (T - T₀ ))

Inserting the values

215.32 = 217.62 (1 + (- 5.00 x 10⁻⁴) (T - 3))

T = 24.14 ⁰C

Answer:

15.Radiowave

16.laser is device that generates an intense beam of other electromagnetic radiation by emission of photons from excited atoms.

17.this is a laboratory instrument commonly used to display and analyse the waveformof electronic signals.

19. this is a space entirely devoid of matter.

Answer:

Induced emf of the wire is 6.36 Volts.

Explanation:

It is given that,

Length of the wire, l = 75 cm = 0.75 m

Magnetic field, B = 0.53 T

Velocity, v = 16 m/s

The wire is moving straight up in the magnetic field. So, an emf is induced in the wire. It is given by :

[tex]\epsilon=Blv[/tex]

[tex]\epsilon=0.53\ T\times 0.75\ m\times 16\ m/s[/tex]

[tex]\epsilon=6.36\ V[/tex]

So, the induced emf of the wire is 6.36 V. Hence, the correct option is (b) "6.36 V".

Answer:

6.36

Explanation:

If a 75 cm straight wire moves straight up through a 0.53 T magnetic field with a velocity of 16 m/s, the induced emf in the wire is 6.36.

The shell has horizontal position [tex]x[/tex] and vertical position [tex]y[/tex] at time [tex]t[/tex] according to

[tex]x=\left(310\dfrac{\rm m}{\rm s}\right)\cos50.0^\circ t[/tex]

[tex]y=\left(310\dfrac{\rm m}{\rm s}\right)\sin50.0^\circ t-\dfrac g2t^2[/tex]

where [tex]g=9.80\dfrac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity.

The shell explodes after 39.0 s, at which point its coordinates are

[tex]x=7770\,\mathrm m[/tex]

[tex]y=1810\,\mathrm m[/tex]

Start with 2 arbitrary vectors, [tex]\vec v_1[/tex] and [tex]\vec v_2[/tex]. (pic 1)

Vectors are determined by their lengths and direction. This means that translating the vector (i.e. sliding it left/right and up/down in the plane) doesn't fundamentally change that vector. To this end, we could just as easily represent [tex]\vec v_2[/tex] as if it had originated from the tip of [tex]\vec v_1[/tex]. This "new" [tex]\vec v_2[/tex] and the "old" [tex]\vec v_2[/tex] are the same vector. (pic 2)

If we connect the origin of [tex]\vec v_1[/tex] with the tip of "new" [tex]\vec v_2[/tex], we get a new vector, and this we define as the vector sum [tex]\vec v_1+\vec v_2[/tex]. (pic 3)

We can do this other way, by first traslating [tex]\vec v_1[/tex] to the tip of [tex]\vec v_2[/tex], then connecting the origin of [tex]\vec v_2[/tex] with the tip of "new" [tex]\vec v_1[/tex]. This demonstrates that vector addition is commutative (order of the vectors being added doesn't matter - you always end up at the same terminus). The "parallelogram method" refers to how a parallelogram is traced out. (pic 4)

Multiplying a vector by -1 reverses its direction. (pic 5)

Adding [tex]\vec v_1[/tex] and [tex]-\vec v_2[/tex] works the same way as standard vector addition, giving us the new vector [tex]\vec v_1-\vec v_2[/tex]. (pic 6)

We can do the same in the reverse order, but now we get a different vector, [tex]\vec v_2-\vec v_1[/tex]. (pic 7)

These vectors have the same length but point in opposite directions. (pic 8)

But notice that we can translate the vectors [tex]\vec v_1-\vec v_2[/tex] and [tex]\vec v_2-\vec v_1[/tex] so that we get a vector that either starts at the tip of [tex]\vec v_2[/tex] and ends at the tip of [tex]\vec v_1[/tex] (pic 9), or starts at the tip of [tex]\vec v_1[/tex] and ends at the tip of [tex]\vec v_2[/tex] (pic 10). The "triangle method" refers to the triangles that are traced out by either vector sum [tex]\vec v_1-\vec v_2[/tex] and [tex]\vec v_2-\vec v_1[/tex] together with [tex]\vec v_1[/tex] and [tex]\vec v_2[/tex].

Answer:

Maximum height reached by the bullet = 4.01 m

Explanation:

Horizontal displacement = 92 m

Time taken = 6.4 s

Horizontal velocity

           [tex]=\frac{92}{6.4}=14.375m/s[/tex]

We have angle of projection = 32°

Horizontal velocity = u cos 32 = 14.375

                      u = 16.95 m/s

Vertical velocity = u sin θ = 16.95 x sin 32 = 8.98 m/s

Time of flight till it reaches maximum height = 0.5 x 6.4 = 3.2s

Now we have vertical motion of bullet

        S = ut + 0.5 at²

       Vertical velocity = u = 16.95 m/s

       a = -9.81 m/s²

        t = 3.2s

Substituting

        S = 16.95 x 3.2 - 0.5 x 9.81 x 3.2² = 4.01 m

Maximum height reached by the bullet = 4.01 m

Answer:

Magnitude of force, F = 11 Newtons

Explanation:

Charge 1, [tex]q_1=5.3\ \mu C=5.3\times 10^{-6}\ C[/tex]

Charge 2, [tex]q_2=11.2\ \mu C=11.2\times 10^{-6}\ C[/tex]

The separation between the charges is, r = 0.22 m

We have to find the magnitude of the force between two point charges. It can be calculated using the formula as :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

[tex]F=9\times 10^9\times \dfrac{5.3\times 10^{-6}\ C\times 11.2\times 10^{-6}\ C}{(0.22\ m)^2}[/tex]

F = 11.03 N

or

F = 11 Newtons

Hence, this is the required solution.

Answer:

Centripetal acceleration of the car is (4.96 g) m/s²

Explanation:

It is given that,

Radius of circle, r = 95 m

Speed of the car, v = 68 m/s

We need to find the centripetal acceleration. It is given by :

[tex]a_c=\dfrac{v^2}{r}[/tex]

So, [tex]a_c=\dfrac{(68\ m/s)^2}{95\ m}[/tex]

[tex]a_c=48.67\ m/s^2[/tex]

Since, g = 9.8 m/s²

So,

[tex]a_c=(4.96\ g)\ m/s^2[/tex]

So, the magnitude of the centripetal acceleration is (4.96 g) m/s². Hence, this is the required solution.

Answer:

The number of protons 6.19 more than electron.

Explanation:

Given that,

Charge [tex]Q=+9.9\times10^{-6}\ C[/tex]

We know that,

formula of charge

[tex]Q = nc[/tex]

Where,

Q = total charge

n = number of protons

e = charge of electron

Put the value into the formula

[tex]n = \dfrac{Q}{e}[/tex]

[tex]n=\dfrac{9.9\times10^{-6}}{1.6\times10^{-19}}[/tex]

[tex]n=6.1875\times10^{13}\ C[/tex]

According to statement of question

Divide the answer by [tex]10^{13}[/tex]

[tex]n=\dfrac{6.1875\times10^{13}}{10^{13}}[/tex]

[tex]n=6.19\ C[/tex]

Hence, The number of protons 6.19 more than electron.

Answer:

Minimum elastic modulus of fiber = 455.64 GPa

Explanation:

Contents of composite material = Epoxy and Unidirectional fibers

Elastic modulus of epoxy = 3.5 GPa

Elastic modulus of composite material = 320 GPa

Volume fraction of fiber = 70 %

Volume fraction of epoxy = 100 - 70 = 30%

Elastic modulus of composite material = 3.5 x 0.3 + Elastic modulus of fiber x 0.7 = 320

0.7 x Elastic modulus of fiber = 320 - 1.05 = 318.95

Elastic modulus of fiber = 455.64 GPa

Minimum elastic modulus of fiber = 455.64 GPa

Explanation:

The boat's velocity relative to the water is 6.10 m/s, and the water's velocity relative to the shore is 1.95 m/s.  We want the boat's velocity relative to the shore to be perpendicular to the shore.

If we say that θ is the angle the boat makes with the shore, then:

6.10 cos θ - 1.95 = 0

6.10 cos θ = 1.95

cos θ = 1.95 / 6.10

θ = 71.4°

The boat should be steered 71.4° relative to the shore, or 18.6° relative to the vertical.

The angle at which the boat must be steered can be determined by using the trigonometric function sine (sin), taking the river current speed as the opposite side and the boat's speed in still water as the hypotenuse in a right triangle. Calculate θ = arcsin(1.95 m/s / 6.10 m/s) to find the angle.

To find the angle the boat must be steered, we need to consider the velocity of both the boat and the river current. Firstly, consider that the boat's velocity (6.10 m/s) and the river current velocity (1.95 m/s) form a right triangle. The speed in still water is the hypotenuse while the river current speed forms the opposite side of the triangle.

To find the angle θ which is the angle between the boat's direction and the river current, we use the trigonometric function sine (sin), defined as the length of the opposite side divided by the length of the hypotenuse. Therefore, sin(θ) = (river current speed) / (boat speed in still water), sin(θ) = 1.95 m/s / 6.10 m/s. To find the angle θ, find the inverse of sin(θ).

Therefore, θ = arcsin(1.95 m/s / 6.10 m/s), which can be calculated using a scientific calculator to get an accurate degree measurement for the angle.

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The droplet has the charge of 9.6 ×[tex]10^{-19}[/tex] C with 6 negatively charged electrons on it.

If the charge on any negatively charged oil droplet is always a whole-number multiple of the fundamental charge of a single electron [tex]10^{-19[/tex] C.  and Millikan was measuring the charge on an oil droplet with 6 negatively charged electrons.

Given that :

Number of electrons = 6

Given charge = −1.60× [tex]10^{-19}[/tex]C

The total charge can be calculated as:

Total charge = Number of electrons × Fundamental charge

Total charge  = 6 × −1.60× [tex]10^{-19}[/tex]C

                       = 9.6 ×[tex]10^{-19}[/tex]

The droplet has the charge of 9.6 ×[tex]10^{-19}[/tex] C.

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Using the findings of Millikan's oil drop experiment, the charge on an oil droplet with 6 negatively charged electrons would be -9.60×10^−19 Coulombs.

According to the data from Millikan’s oil drop experiment, the charge of a single electron is identified to be −1.60×10−19 C. It was found that the charge on any negatively charged oil droplet is always a whole-number multiple of this fundamental charge. Therefore, if Millikan was measuring the charge on an oil droplet with 6 negatively charged electrons on it, he would calculate the charge as

-1.60×10−19 C (charge of a single electron) multiplied by 6 (number of electrons) resulting in a total charge of -9.60×10^−19 C.

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(a) [tex]4.3\cdot 10^9 J[/tex]

The kinetic energy of an object is given by:

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

where

m is the mass of the object

v is its speed

For the ship in the problem, we have

[tex]m=6.40\cdot 10^7 kg[/tex] is the mass

[tex]v=11.6 m/s[/tex] is the speed

So its kinetic energy is

[tex]K=\frac{1}{2}(6.40\cdot 10^7 kg)(11.6 m/s)^2=4.3\cdot 10^9 J[/tex]

(b) [tex]-4.3\cdot 10^9 J[/tex]

According to the work-kinetic energy theorem, the work done on an object is equal to the change in kinetic energy of the object:

[tex]W= \Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2[/tex]

where

W is the work done

m is the mass of the object

v is the final speed of the object

u is the initial speed of the object

Here we want to find the work done to stop the ship, so the final speed of the ship is v=0, while the initial speed is u=11.6 m/s. So the work done will be

[tex]W= 0 -\frac{1}{2}(6.40\cdot 10^7 kg)(11.6 m/s)^2=-4.3\cdot 10^9 J[/tex]

(c) [tex]1.3\cdot 10^6 N[/tex]

The work done on an object can be also written as follows

[tex]W=Fd[/tex]

where

F is the magnitude of the force

d is the displacement of the object

Here we know:

[tex]W=-4.3\cdot 10^9 J[/tex] is the work done

d = 3.20 km = 3200 m is the displacement of the ship

So we can solve the formula to find F, the force exerted on the ship to stop it:

[tex]F=\frac{W}{d}=\frac{-4.3\cdot 10^9 J}{3200 m}=-1.3\cdot 10^6 N[/tex]

and the negative sign simply means that the force is opposite to the displacement of the ship (in fact, the force acts against the motion of the ship).

The ship's kinetic energy is 5.77 × 10^9 J. The work required to stop it is -5.77 × 10^9 J. The magnitude of the force required to stop it is 1.80 × 10^3 N.

(a) The kinetic energy of the ship can be calculated using the formula:

Ek = (1/2)mv2

where m is the mass of the ship and v is its velocity. Plugging in the given values, we get a kinetic energy of 5.77 × 109 J.

(b) To stop the ship, work needs to be done to bring it to a complete stop. The work done is equal to the change in kinetic energy of the ship. Initially, the ship had a kinetic energy of 5.77 × 109 J. When it comes to a stop, its kinetic energy becomes zero. Therefore, the work required to stop the ship is -5.77 × 109 J (negative since work is done on the ship).

(c) The magnitude of the constant force required to stop the ship can be calculated using the work-energy theorem:

Work = Force × Distance

Given that the displacement of the ship is 3.20 km, or 3.20 × 106 m, and the work done on the ship is -5.77 × 109 J, we can solve for the force:

Force = Work / Distance = -5.77 × 109 J / (3.20 × 106 m) = -1.80 × 103 N

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(a) The pitcher must throw the ball at 27.7 m/s

The momentum of an object is given by:

[tex]p=mv[/tex]

where

m is the mass of the object

v is the object's velocity

Let's calculate the momentum of the bullet, which has a mass of

m = 2.70 g = 0.0027 kg

and a velocity of

[tex]v=1.50\cdot 10^3 m/s[/tex]

Its momentum is:

[tex]p=mv=(0.0027 kg)(1.50\cdot 10^{3} m/s)=4.05 kg m/s[/tex]

The pitcher must throw the baseball with this same momentum. The mass of the ball is

m = 0.146 kg

So the velocity of the ball must be

[tex]v=\frac{p}{m}=\frac{4.05 kg m/s}{0.146 kg}=27.7 m/s[/tex]

So, the pitcher must throw the ball at 27.7 m/s.

(b) a. The bullet has greater kinetic energy

The kinetic energy of an object is given by

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

where m is the mass of the object and v is its speed.

For the bullet, we have:

[tex]K=\frac{1}{2}(0.0027 kg)(1.50\cdot 10^3 m/s)^2=3037.5 J[/tex]

For the ball:

[tex]K=\frac{1}{2}(0.146 kg)(27.7 m/s)^2=56.0 J[/tex]

So, the bullet has greater kinetic energy.

Final answer:

The baseball's speed must be 27.74 m/s to match the momentum of the bullet, and the bullet has greater kinetic energy than the baseball.

Explanation:

Calculating Baseball's Speed and Comparing Kinetic Energies

To validate the pitcher's claim that a 0.146-kg baseball can have the same momentum as a 2.70-g bullet traveling at 1.50 x 10³ m/s, we must use the formula for momentum (p = mv), which gives us:

Momentum of bullet = (2.70 g) x (1.50 x 10³ m/s) = (0.0027 kg) x (1500 m/s) = 4.05 kg m/s.

Therefore, the baseball's speed v needed to have the same momentum is calculated by rearranging the formula to v = p/m, which gives us:

Baseball's speed v = 4.05 kg m/s / 0.146 kg = 27.74 m/s.

To determine which has the greater kinetic energy, we use the kinetic energy formula KE = (1/2)mv². Calculating the kinetic energy of both the bullet and the baseball:

KE of bullet = (1/2)(0.0027 kg)(1500 m/s)² = 3037.5 J.

KE of baseball = (1/2)(0.146 kg)(27.74 m/s)² = 55.90 J.

Comparing the results shows that the bullet has greater kinetic energy than the baseball.