Tribes of the Sioux Nation (among other Plains Indians) maintained historical calendars composed of winter counts. Tribe historians would_____________________.

Answer:

The impulse needed to stop the child = -200 Ns

Explanation:

Impulse: This can be defined as the product of force and time of a body. The S.I unit of impulse is kgm/s or Ns

From newton's second law,

I = mΔv ................... Equation 1.

Where I = Impulse needed to stop the child, m = mass of the child, Δ = change in velocity of the child.

Given: m = 20.0 kg,

Δv = v-u where v =0 m/s, u = 10.0 m/s

Δv = 0-10 = -10 m/s.

Substituting into equation 1

I = 20(-10)

I = -200 Ns.

Thus the impulse needed to stop the child = -200 Ns

Note: The negative sign shows that the impulse act against the motion of the child

The correct impulse needed to stop the child is 200 N·s.

The impulse needed to stop the child is 16 N·s.To find the impulse needed to stop the child, we can use the formula for impulse, which is the change in momentum of an object. The momentum of an object is given by the product of its mass and velocity. Impulse is the product of the average force applied to the object and the time interval during which the force is applied. The impulse is equal to the change in momentum.

The initial momentum of the child (p_initial) is given by the mass of the child (m) multiplied by the initial velocity of the car (v_initial), since the child's velocity is assumed to be the same as that of the car. The final momentum of the child (p_final) is zero because the child comes to a stop.

Given:

- Mass of the child, m = 20.0 kg

- Initial velocity of the car (and the child), v_initial = 10.0 m/s

- Time taken to stop, Aat = 0.050 s

- The change in momentum (Aap) is:

Aap = [tex]p_final - p_initial[/tex]

Aap = [tex]0 - (m * v_initial)[/tex]

Aap = - (20.0 kg * 10.0 m/s)

Aap= - 200 kgA·m/s

The negative sign indicates that the momentum is decreasing (the child is stopping).

 The impulse (J) is equal to the change in momentum:

J = Aap

J = - 200 kgA·m/s

Since impulse is a vector quantity and we are interested in the magnitude of the impulse needed to stop the child, we take the absolute value:

J = | - 200 kgA·m/s |

J = 200 kgA·m/s

Now, we can express the impulse in Newtons-seconds (NA·s) by noting that 1 kgA·m/s is equivalent to 1 NA·s:

J = 200 NA·s

However, the answer provided initially (16 NA·s) seems to be incorrect. Let's re-evaluate the calculation to ensure accuracy.

The impulse J is the change in momentum Aap over the time interval Aat. The average force F_avg applied to the child is then the impulse divided by the time interval:

F_avg = J / Aat

 Rearranging for J gives:

J = F_avg / Aat

Since we know the change in momentum Aap is equal to the impulse J, we can write:

J = Aap = m * Aav

The change in velocity Aav of the car (and thus the child) is the final velocity [tex]v_final[/tex]minus the initial velocity [tex]v_initial[/tex]. Since the car stops, [tex]v_final[/tex]= 0, and Aav is simply [tex]-v_initial[/tex].

Aav = v[tex]_final - v_initial[/tex]

Aav = 0 - 10.0 m/s

Aav= - 10.0 m/s

Now we can calculate the impulse needed to stop the child:

J = m * Aav

J = 20.0 kg * (- 10.0 m/s)

J = - 200 kgA·m/s

Taking the magnitude:

J = | - 200 kgA·m/s |

J = 200 kgA·m/s

Since 1 kgA·m/s is equivalent to 1 NA·s, the impulse in Newtons-seconds is:

J = 200 NA·s

This confirms that the initial answer provided (16 N·s) is indeed incorrect, and the correct impulse needed to stop the child is 200 N·s.

Answer:

Explanation:

It is dangerous to jump from rowboat to a dock because you may end up hurting yourself.

As momentum is conserved, the momentum of the boat after you jump off from boat is equal and opposite to your momentum. So you end up landing far behind than you expected and hurt yourself.

Explains why jumping from a rowboat to a dock violates the conservation of momentum principle, leading to potential boat instability.

In terms of conservation of momentum, when you jump from a rowboat to a dock, your momentum changes suddenly as you leave the boat, but the boat still retains its momentum. This violates the principle of conservation of momentum, which states that in the absence of external forces, the total momentum of a system remains constant.

For instance, if the rowboat (system) and you initially have zero momentum while at rest, the moment you jump, your momentum changes, but the boat will move in the opposite direction to compensate, due to Newton's third law of motion, causing instability and potentially capsizing the boat.

Answer:

-1.03 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity. The S. I unit of acceleration is m/s².

Mathematically, acceleration is expressed as

a = (v-u)/t ........................ Equation 1

Where a = acceleration, v = final velocity, u = initial velocity, t  = time.

Given: u = 13.60 m/s, v = 7.20 m/s t = 6.2 s.

Substituting into equation 2

a = (7.20-13.60)/6.2

a = -6.4/6.2

a = -1.03 m/s²

Note: a is negative because, the hockey puck is decelerating.

Hence the average acceleration = -1.03 m/s²

Answer: have "cis C=C double bonds" and "liquid" at room temperature.

Explanation:

The unsaturated fatty acids have one or more C=C double bonds in the cis formation. Thus, this results in the molecules not been as stable as the saturated fats. They have weaker intermolecular bonds thus resulting in lower melting point . The consequently results in it being liquid at room temperature.

Unsaturated fats have one or more double bonds in their carbon chain and are usually liquid at room temperature. In contrast, saturated fats, which have no double bonds, are typically solid at room temperature.

When comparing saturated and naturally occurring unsaturated fats, the unsaturated fats have one or more double bonds in their carbon chain and are typically liquid at room temperature. Saturated fats, on the other hand, have no double bonds and are usually solid at room temperature. This is due to the straight, closely packed chains of saturated fat molecules, whereas the double bonds in unsaturated fats cause bends, preventing them from packing closely together and remaining fluid at room temperature.

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Answer:To ensure significant dialogue which will guarantee that the needed Products meets specifications

It will help to develop a better Operational model for the users.

Explanation: The design phase is one of the most critical phase in product development,it is in this phase that the product structure, contents and specifications are put into action. It is necessary that the end users have a significant time to dialogue with the project or product development team to guarantee that the end product meets specifications.

It is also necessary to spent adequate time in order to make the product user friendly (easy to operate and maintain).

Answer:

A) [tex] Z = 0.577 C +112.931[/tex]

[tex] Z = 0.577*(100) +112.931=170.631 Z[/tex]

B) [tex] C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C[/tex]

C) [tex] K = C +273.15[/tex]

[tex] K = -22.41 +273.15 =250.739 K[/tex]

Explanation:

For this case we want to create a function like this:

[tex] Z = a C + b[/tex]

Where Z represent the degrees for the Z scale C the Celsius grades and  tha valus a and b parameters for the model.

The boiling point of nitrogen is -195,8 °C

The melting point of iron is 1538 °C

We know the following equivalences:

-195.8 °C = 0 °Z

1538 °C = 1000 °Z

Let's say that one point its (1538C, 1000 Z) and other one is (-195.8 C, 0Z)

So then we can calculate the slope for the linear model like this:

[tex] a = \frac{z_2 -z_1}{c_2 -c_1}= \frac{1000 Z- 0Z}{1538C -(-195.8 C)}=\frac{1000 Z}{1733.8 C}=0.577 \frac{Z}{C}[/tex]

And now for the slope we can use one point let's use for example (-195.8C, 0Z), and we have this:

[tex] 0 = 0.577 (-195.8) + b[/tex]

And if we solve for b we got:

[tex] b = 0.577*195.8 =112.931 Z[/tex]

So then our lineal model would be:

[tex] Z = 0.577 C +112.931[/tex]

Part A

The boiling point of water is 100C so we just need to replace in the model and see what we got:

[tex] Z = 0.577*(100) +112.931=170.631 Z[/tex]

Part B

For this case we have Z =100 and we want to solve for C, so we can do this:

[tex] Z-112.931 = 0.577 C[/tex]

[tex] C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C[/tex]

Part C

For this case we know that [tex] K = C +273.15[/tex]

And we can use the result from part B to solve for K like this:

[tex] K = -22.41 +273.15 =250.739 K[/tex]

Final answer:

The boiling point of water on the Z scale is 500°Z. To convert 100°Z to the Celsius scale, the equivalent temperature is 10°C. To convert 100°Z to the Kelvin scale, the equivalent temperature is 283.15 K.

Explanation:

The Z scale is a new temperature scale created by the demented scientist. According to the Z scale, the boiling point of nitrogen is 0°Z and the melting point of iron is 1000°Z.

A) To find the boiling point of water on the Z scale, we need to look at the temperature range between the boiling point of nitrogen and the melting point of iron. Since the boiling point of nitrogen is 0°Z and the melting point of iron is 1000°Z, the range between them is 1000-0 = 1000°Z. Therefore, the boiling point of water on the Z scale is 500°Z, which is halfway between 0°Z and 1000°Z.

B) To convert 100°Z to the Celsius scale, we can use the formula: Celsius = (Z - 0) * (100 - 0) / (1000 - 0). Plugging in the values, we get Celsius = (100 - 0) * (100 - 0) / (1000 - 0) = 10000 / 1000 = 10°C.

C) To convert 100°Z to the Kelvin scale, we use the formula: Kelvin = Celsius + 273.15. Plugging in the value, we get Kelvin = 10 + 273.15 = 283.15 K.

Answer:

A) The ball's gravitational potential energy is greatest at the instant when the ball is at its highest point.

Explanation:

Suppose you kick a soccer ball straight up to a height of 10 meters. Which of the following is true about the gravitational potential energy of the ball during its flight?

A) The ball's gravitational potential energy is greatest at the instant when the ball is at its highest point. (true)

B) The ball's gravitational potential energy is always the same. (false)

    because the gravitational potential energy is changed as the height changed.

C) The ball's gravitational potential energy is greatest at the instant the ball leaves your foot. (false)

    Because The ball's gravitational potential energy is greatest at the instant when the ball is at its highest point.

D) The ball's gravitational potential energy is greatest at the instant it returns to hit the ground. (false)

     The ball's gravitational potential energy is greatest at the instant when the ball is at its highest point. not when return to the ground

Final answer:

The gravitational potential energy of a soccer ball kicked straight up is A) The ball's gravitational potential energy is greatest at the instant when the ball is at its highest point.

Explanation:

If you kick a soccer ball straight up to a height of 10 meters, the gravitational potential energy of the ball is greatest at the instant when the ball is at its highest point. At the highest point, all the energy in the system is gravitational potential energy.

This is because gravitational potential energy is calculated by the formula PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height above the reference point, which in this case is the ground. At the ball's highest point, the h value is at its maximum of 10 meters. Therefore, the correct answer to the question is A) The ball's gravitational potential energy is greatest at the instant when the ball is at its highest point.

Answer:

Explanation:

Gravitational Potential Energy at earth surface [tex]U_1=\frac{GM_em}{R_e}[/tex]

Gravitational Potential Energy at height h is [tex]U_2=\frac{GM_em}{R_e+h}[/tex]

Energy required to lift the satellite [tex]E_1=U_1-U_2[/tex]

[tex]E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}[/tex]

Now Energy required to orbit around the earth

[tex]E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}[/tex]

[tex]\Delta E=E_1-E_2[/tex]

[tex]\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}[/tex]

[tex]E_1=E_2[/tex]  (given)

[tex]\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0[/tex]

[tex]\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0[/tex]

[tex]h=\frac{R_e}{2}[/tex]

[tex]h=3.19\times 10^6\ m[/tex]

(b)For greater height [tex]E_1[/tex]  is greater than [tex]E_2[/tex]

thus energy to lift the satellite is more than orbiting around earth

Answer:

Tech A is right.

Explanation:

Its the ability of quartz crystal that if we apply a mechanical stress on them by an applied force then they can produce the electricity or voltage. This phenomenon of producing electricity by squeezing a crystal is termed as piezoelectricity or a piezoelectric effect.

Tech A is correct. Quartz crystals under mechanical stress can produce a voltage. The piezoelectric principle is used in coolant temperature sensors.

Tech A is correct. Quartz crystals, when subjected to mechanical stress, exhibit piezoelectric behavior, which means they can generate a voltage. This phenomenon is based on the piezoelectric effect, where the crystal structure of quartz generates an electric charge when deformed. This property is extensively used in various applications, including sensors, resonators, and oscillators.

Tech B is also correct. The piezoelectric principle is indeed used in coolant temperature sensors. These sensors utilize a piezoelectric material, such as quartz, to convert temperature changes into an electric signal. When the temperature of the coolant changes, it causes the piezoelectric material to deform, resulting in a change in voltage or current that can be measured and interpreted to determine the coolant temperature.

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Answer:

(a) velocity at 9 rev/s is 34.3m/s and the velocity at 7rev/s is 40.4 m/s. These were values can be gotten by using the formula v = 2(pi)R/T. Where T is the period. T = 1/w(angular velocity).

(b) acceleration at 9.0rev/s is 1961 m/s²

(c) acceleration at 7.0rev/s is 1814 m/s²

These values can be gotten by using the formula a = v²/R

Explanation:

The full solution can be found in the attachment below.

Thank you for reading.

Final answer:

To find the speed of the stone for each rate of rotation, use the formula Speed = (2 * pi * radius * rate of rotation). The centripetal acceleration at 9.00 rev/s is approximately 3.1 m/s^2 and at 7.00 rev/s is approximately 3.9 m/s^2.

Explanation:

To find the speed of the stone for each rate of rotation, we can use the formula:

Speed = (2 * pi * radius * rate of rotation)

Using a length of 0.600 m and a rate of rotation of 9.00 rev/s, the speed of the stone is approximately 3.4 m/s. Using a length of 0.900 m and a rate of rotation of 7.00 rev/s, the speed of the stone is approximately 4.9 m/s.

The centripetal acceleration of the stone at 9.00 rev/s can be found using the formula:

Centripetal acceleration = (speed)^2 / radius

Substituting the values, we get a centripetal acceleration of approximately 3.1 m/s^2.

The centripetal acceleration at 7.00 rev/s can be found using the same formula. Substituting the values, we get a centripetal acceleration of approximately 3.9 m/s^2.

Answer:

A. 3.1m/s²

Explanation:

Since the body is sliding down the plane, the frictional force (Ff) being a force of opposition will be acting upwards.

The forces acting on the body in the vertical direction will be the weight (W) and the normal reaction(R) acting in the opposite direction.

Taking the sum of the forces along the plane,

Fm - Ff = mass × acceleration (newton's law)

Since Fm = Wsin(theta)

Ff = nR where n is the coefficient of friction we will have;

Wsin(theta) - nR = ma... (1)

W = mg = 6×10 = 60N

n = 0.40

R = Wcos (theta) {resolving weight to the vertical)

R = 60cos39°

R = 46.6N

Substituting this values into eqn 1 to the get the acceleration,

60sin39° - 0.4(46.6) = 6a

37.8 - 18.64 = 6a

19.16 = 6a

a = 19.16/6

a = 3.19m/s²

The acceleration of the body down the plane will be 3.1m/s²

Answer:

-3.769 m/min

Explanation:

[tex]\dfrac{dA}{dt}[/tex] = Rate of change of area = -49 m²/min (negative due to shrinking)

s = Side length = 13 m

[tex]\dfrac{ds}{dt}[/tex] = Rate of change of side

Area of a square is given by

[tex]A=s^2[/tex]

Differentiating with respect to time

[tex]\dfrac{dA}{dt}=\dfrac{ds^2}{dt}\\\Rightarrow \dfrac{dA}{dt}=2s\dfrac{ds}{dt}\\\Rightarrow \dfrac{ds}{dt}=\dfrac{dA}{dt}\times \dfrac{1}{2s}\\\Rightarrow \dfrac{ds}{dt}=-49\times \dfrac{1}{13}\\\Rightarrow \dfrac{ds}{dt}=-3.769\ m/min[/tex]

The rate of change of the sides of the square is -3.769 m/min

Answer:

3. the mass of the atom is concentrated in a very small area.

Explanation:

As the Rutherford experiment has shown that majority of particles has passed the foil with slightest deviations. It shows that the most of the part of atom is empty which doesn't effect the deviation of alpha particles. Few particles are deflected at significant angels shows that there must be a mass occupying the smaller area of atom which is later on called as nucleus.

Answer:

a. [tex]v'=8.925\ m.s^{-1}[/tex]

b. [tex]I=1585397.5872\ N.s[/tex]

c. [tex]s=265.4125\ m[/tex]

Explanation:

a.

Velocity of the two cars just after the collision:

Using the law of conservation of

[tex]m_m.v+m_s\times 0=(m_m+m_s)\times v'[/tex]

[tex]2100\times 17+ 0=(2100+1900)\times v'[/tex]

[tex]v'=8.925\ m.s^{-1}[/tex]

b.

Frictional force acting on the combined mass of cars:

[tex]f=\mu_k\times N[/tex]

where:

N = reaction normal to the contact surface by the ground

[tex]f=0.68\times (2100+1900)\times 9.8[/tex]

[tex]f=26656\ N[/tex] is the resistance force to the motion of the wheels

Now, by Newton's second law of motion:

[tex]f=\frac{dp}{t}[/tex]

[tex]f=\frac{(m_m+m_s)\times v'}{t}[/tex]

[tex]26656=\frac{(2100+1900)\times 8.925}{t}[/tex]

[tex]t=59.4762\ s[/tex]

Now impulse is the impact load acting acting for certain time:

[tex]I=f\times t[/tex]

[tex]I=26656\times 59.4762[/tex]

[tex]I=1585397.5872\ N.s[/tex]

c.

The distance the the cars skid before coming to rest:

using equation of motion:

[tex]v'_f=v'+a.t[/tex]

we've [tex]v'_f=0\ m.s^{-1}[/tex] since final velocity is zero

[tex]0=8.925+a\times 59.4762[/tex]

[tex]a=0.1501\ m.s^{-2}[/tex]

now since we have:

[tex]s=v'.t+\frac{1}{2} a.t^2[/tex]

[tex]s=8.925\times 59.4762-0.5\times 0.1501\times 59.4762^2[/tex]

[tex]s=265.4125\ m[/tex]

A. The velocity of the two cars just after the collision is 8.925 m/s

B. The impulse that acts on the skidding cars just after the collision is 35700 Ns

C. The distance travelled by the skidding cars before coming to rest is 5.98 m

A. Determination of the velocity the two cars

Mass of moving car (m₁) = 2100 Kg

Initial velocity of moving car (u₁) = 17 m/s

Mass of stationary car (m₂) = 1900 Kg

Initial velocity of stationary car (u₂) = 0 m/s

[tex]v = \frac{u_1m_1 + u_2m_2}{m_1m_2} \\ \\ v = \frac{(2100 \times 17) + (1900 \times 0)}{2100 + 1900} \\ \\ v = 8.925 \: m/s[/tex]

Thus, the velocity of the two cars is 8.925 m/s

B. Determination of the impulse

Mass of moving car (m₁) = 2100 Kg

Mass of stationary car (m₂) = 1900 Kg

Velocity of the two cars (v) = 8.925 m/s

I = Ft = mv

I = (2100 + 1900) × 8.925

Thus, the impulse on the skidding cars is 35700 Ns

C. Determination of the distance travelled by the skidding cars.

Mass of moving car (m₁) = 2100 Kg

Mass of stationary car (m₂) = 1900 Kg

Total mass = 2100 + 1900 = 4000 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Normal reaction (N) = mg = 4000 × 9.8 = 39200 N

Coefficient of kinetic friction (μ) = 0.68

F = μN

F = 0.68 × 39200

Impulse (I) = 35700 Ns

Force (F) = 26656 N

I = Ft

35700 = 26656 × t

Divide both side by 26656

t = 35700 / 26656

Initial velocity of the two cars (u) = 8.925 m/s

Final velocity of the two cars (v) = 0 m/s

Time (t) = 1.34 s

[tex]s = \frac{(v + u)t}{2} \\ \\ s = \frac{(8.925 + 0)1.34}{2} \\ \\ s = \frac{8.925 \times 1.34}{2} \\ \\ s = 5.98 \: m[/tex]

Therefore, the distance travelled by the skidding cars before coming to rest is 5.98 m

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Answer:

speed of sound in unknown gas will be 360m/sec

Explanation:

We have given length of the pipe l = 0.770 m

It is given that pipe is open at one end and closed at one end

Frequency of third harmonic f = 750

Third harmonic frequency of pipe which one end is open and one end is closed is given by

[tex]f=\frac{5v}{4L}[/tex], here v is the speed of the sound, L is the length of the pipe

So  [tex]750=\frac{5v}{4\times 0.770}[/tex]

v = 360 m /sec

So speed of sound in unknown gas will be 360m/sec

The given data is incomplete. The complete question is as follows.

At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coefficient of friction was estimated to be 0.36.  Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes. (why does the car's mass not matter?)

Explanation:

Let us assume that v is the final velocity and u is the initial velocity of the car. Let s be the skid marks and [tex]\mu[/tex] be the friction coefficient and m be the mass of car.

Hence, the given data is as follows.

                v = 0,     s = 84 m,     [tex]\mu[/tex] = 0.36

According to Newton's law of second motion the expression for acceleration is as follows.

                      F = ma

                 [tex]-\mu N[/tex] = ma

                 [tex]-\mu mg[/tex] = ma

                      a = [tex]-\mu g[/tex]

Also,    

               [tex]v^{2} = u^{2} + 2as[/tex]

              [tex](0)^{2} = u^{2} + 2(-\mu g)s[/tex]

                  [tex]u^{2} = 2(\mu g)s[/tex]

                            = [tex]\sqrt{2(0.36)(9.81 m/s^{2})(84 m)}[/tex]

                            = 24.36 m/s

Thus, we can conclude that the speed of the car when the driver slammed on (and locked) the brakes is 24.36 m/s.

A baseball bat hits a ball is an example of contact force.

Option A.

Contact force is defined as those forces which are acting between two bodies by actually touch between two bodies.

In case of the book falling to ground, the force acting is the gravitational force. Gravitational force acts even when two bodies aren't in contact. So its a non contact force.

Similar is the condition of the leaf and ground where the gravitational force acts.

The magnetic force between the paper clip and the magnet is also non contact force, which acts even when two bodies aren't in contact.

But the bat hitting the ball is having direct contact between the two bodies, and their contact is what makes the ball fly off. So its a contact force.

Contact force is the type of force that occurs when the external force act on the system act physically. A baseball bat hitting a ball is the right example of contact forces.

Any force that requires touch is referred to as a contact force. The majority of apparent interactions between macroscopic groupings of matter are caused by contact forces.

Contact forces are used in everyday circumstances. For example, a baseball bat hits a ball.

Hence a baseball bat hitting a ball is the right example of contact forces.

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Answer:

3 effect of oxidized lipids

4 an inflammatory response

5 the formation of plaques

Explanation:

Destruction of cells due to oxidation of lipids whereby free radicals steal electrons in cell membrane.

This occurs when tissues get injured by trauma, bacteria or toxins, thereby causing damages cells to release chemicals like histamine, brakykinn that cause vessels to leak fluid into the injured tissues causing swelling.

-plaques are regions of destroyed cells which are visible structures formed inside a cell culture.

Answer:

[tex]dQ=-500J[/tex]

Explanation:

We have given work is done on the gas is 1200 J

So work done will be [tex]dW=-1200J[/tex] ( as work is done on the gas )

It is given that internal energy is increases by exactly 700 J

So [tex]dU=700J[/tex]

From thermodynamic equation [tex]dQ=dU+dW[/tex]

So [tex]dQ=700-1200=-500J[/tex]

Here negative sign indicates that heat flow out of the gas

If heat was negative then heal was flowing in the gas

Answer:

392307.6923 m/s²

Explanation:

t = Time taken

u = Initial velocity = 560 m/s

v = Final velocity = 460 m/s

s = Displacement = 13 cm

a = Acceleration

From the equation of motion we have

[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{460^2-560^2}{2\times 13\times 10^{-2}}\\\Rightarrow a=-392307.6923\ m/s^2[/tex]

The acceleration of the bullet as it passes through the board is -392307.6923 m/s²

Answer:

Susie has strabismus.

Explanation:

Strabismus is a medical condition in which the patient can not keep his/her eyes aligned under the normal conditions. In strabismus one or both of eyes rolls into up, down, in or out direction. Physicians usually recommend the exercises for eyes or placing a patch on the unaffected eye or they also do surgery where it's required like in severe cases.  

Answer: The astronomer who, at the turn of the century, measured the spectra of hundreds of thousands of stars (leaving a catalog that astronomers used for the rest of the century) was: Annie Jump Cannon.

Annie Jump Cannon was born on 11th December 1863 in Dover, Delaware. She was an american astronomer whose work contributes in the expansion of current stellar classification. She made the first accurate attempt to organize and classify stars on the basis of their temperature and spectral form. She suffered from hearing loss during her childhood. She was almost deaf during her career. She was member of National women's party and by her nature, she was a extreme suffragist.

Annie Jump was the eldest of the three daughters of Wilson Cannon (shipbuilder and state senator) and Mary Jump (second wife of Wilson Cannon). It was Annie's mother who encouraged her to peruse her education and career in her interests. She suggested her to study in Wellesley college, perusing in mathematics, biology and chemistry. Cannon's mother teaching like household economics, helped her to organize her research later on.

She took her early education and was brilliant student of mathematics, then went to Wellesley College of Massachusetts in 1880 and studied physics and astronomy, later she went to work at Harvard Observatory.

Astronomer Annie Jump was the pioneer, who develop a simple spectral classification system. She categorized nearly 400,000 stars in total. She for the very first time found a double star, three hundred variable stars and 5 novas. She worked as astronomer for more then forty years till 1940, until she got retired. During her career she also work as suffragist, and helped many women to attain respect and reputation in scientific society. She was a hard working woman with a calm nature who build the ways for upcoming women in the field of scientific research and astronomy.

Annie Jump Canon died on 13th April, 1941 in Cambridge, Massachusetts. she died at the age of 72. She was hospitalized prior to her death due extreme illness. After her death American Astronomical Society announce an award named after her (The Annie Jump Cannon Award), yearly to those female astronomers, who worked remarkably in astronomy.

The astronomer who, at the start of the 20th century, measured and catalogued the spectra of hundreds of thousands of stars was Annie Jump Cannon. She worked under Edward C. Pickering at the Harvard Observatory and her detailed catalogue served as a major reference for astronomers for the rest of the century.

The astronomer you're referring to is Annie Jump Cannon who worked at the Harvard Observatory under Edward C. Pickering. In the late 1800s, Pickering started an ambitious project of classifying stellar spectra, the foundation of this work was a remarkable collection of close to one million photographic spectra of stars. Cannon was hired by Pickering to assist in the classification of spectra and her efficiency was such, that she could visually identify and determine the spectral types of several hundred stars per hour.

This relentless work resulted in Cannon creating a comprehensive catalog containing the spectral types for hundreds of thousands of stars. This seminal piece of work became the blueprint for much of twentieth-century astronomy, and served to influence and inform a whole generation of astronomers. Therefore, Cannon's detailed and systematic study was instrumental in developing a deeper understanding of the universe as we know it today.

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a) The jogger has run approximately 7.978 miles after 1 hour. b) The jogger's acceleration at t = 0 is approximately 0.41 ft/s^2. c) The equation does not provide a direct relationship between distance and time, so we cannot determine the time required to run 6 miles.

(a) To find the distance the jogger has run when t = 1 h, we need to substitute t = 1 into the given equation. Plugging in t = 1, we have: v = 7.5(1 + 2(0.04(1)))^0.3. Simplifying this expression, we get v ≈ 7.5(1.08)^0.3. Evaluating this expression, we find v ≈ 7.978 mi/h. To find the distance, we multiply the speed by the time: d = v × t = 7.978 × 1 = 7.978 mi.

(b) The acceleration can be found by taking the derivative of the velocity equation with respect to time. Differentiating the given equation, we have dv/dt = 7.5(1 + 2(0.04x))^(-0.7) × 2(0.04). Substituting t = 0 into this expression, we get: a = 7.5(1 + 2(0.04(0)))^(-0.7) × 2(0.04). Simplifying this expression, we find a ≈ 0.28 mi/h^2. To convert this to ft/s^2, we multiply by 5280 ft/mi and divide by 3600 s/h: a = 0.28 × 5280 / 3600 ≈ 0.41 ft/s^2.

(c) To find the time required for the jogger to run 6 mi, we need to solve the equation v = 7.5(1 + 2(0.04x))^0.3 for t. Plugging in d = 6 and v = 7.5(1 + 2(0.04x))^0.3, we can solve for t. However, the given equation does not provide a direct relationship between distance and time, so we cannot solve it.

Answer:

When a converging lens is placed under water then its focal length increases.

Explanation:

When any lens, be it converging or diverging is immersed into water then the speed of light before and after the refraction through the lens has less speed and due to less difference in speed it shows less deviations as a result the focal length is increased for the lenses.

Lesser the difference between the refractive index of the medium and the lens glass greater becomes its focal length.

The focal length of the lens immersed in a medium is given by:

[tex]\frac{1}{f} =\frac{n_l-n_m}{n_m} \times (\frac{1}{R_1} +\frac{1}{R_2} )[/tex]

where:

[tex]n_l=[/tex] refractive index of the glass material with resp. to air

[tex]n_m=[/tex] refractive index of the medium with resp. to air

[tex]R_1\ \&\ R_2=[/tex] radii of curvature of the two surfaces of the lens

Answer:

Explanation:

According to the lens maker's formula

The focal length of the converging lens in air is given by

[tex]\frac{1}{f_{a}}=\left ( \mu _{a}-1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )[/tex]       .... (1)

Where, R1 and R2 be the radius of curvature of the lens and μa is the refractive index of glass in air.

When the lens is placed in water. the focal length is given by

[tex]\frac{1}{f_{w}}=\left ( \frac{\mu _{a}}{\mu _{w}}-1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )[/tex]        .... (2)

Where, R1 and R2 be the radius of curvature of the lens and μa is the refractive index of lens in air and μw is the refractive index of glass in water.

Dividing equation (1) by (2), we observe that the value of focal length,

the focal length of lens in water increases.

Final answer:

The speed of the ram at the instant of contact is 8.413 m/s.

Explanation:

To find the speed of the ram at the instant of contact, we can use the principle of conservation of mechanical energy. The initial potential energy stored in the compressed spring is converted into the kinetic energy of the ram as it hits the staple. Since the mass of the spring is ignored, we can consider the system as consisting of only the ram.

Using the equation for the potential energy of the spring (PE = 0.5kx^2), we can calculate the initial potential energy. The compressed length of the spring is 3.20 x 10^-2 m, so the initial potential energy is 0.5(37107 N/m)(3.20 x 10^-2 m)^2 = 192.128 J.

At the instant of contact, all of the potential energy has been converted into kinetic energy. Therefore, we can equate the initial potential energy to the final kinetic energy of the ram. The kinetic energy is given by KE = 0.5mv^2, where m is the mass of the ram (0.179 kg) and v is the speed of the ram at the instant of contact.

Solving for v, we have: 192.128 J = 0.5(0.179 kg)v^2. Rearranging the equation and solving for v, we find that v = 8.413 m/s. Therefore, the speed of the ram at the instant of contact is 8.413 m/s.

Answer:

speed of eight ball speed after the collision is 3.27 m/s

Explanation:

given data

initially moving v1i = 3.4 m/s

final speed is v1f = 0.94 m/s

angle = θ w.r.t. original line of motion

solution

we assume elastic collision

so here using conservation of energy

initial kinetic energy = final kinetic energy .............1

before collision kinetic energy = 0.5 × m× (v1i)²

and

after collision kinetic energy =  0.5 × m× (v1f)²  + 0.5 × m× (v2f)²

put in equation 1

0.5 × m× (v1i)² =  0.5 × m× (v1f)²  + 0.5 × m× (v2f)²

(v2f)² = (v1i)² - (v1f)²

(v2f)² = 3.4² - 0.94²

(v2f)² = 10.68

taking the square root both

v2f = 3.27 m/s

speed of eight ball speed after the collision is 3.27 m/s

Final answer:

To find the eight ball's speed after the collision, we can use the principle of conservation of momentum and the fact that the collision is elastic. By solving equations for momentum and kinetic energy, we can determine the final velocity of the eight balls. Given the initial velocity and final velocity of the cue ball, and assuming equal masses for both balls, we can substitute these values and solve for the final velocity of the eight balls.

Explanation:

To find the eight ball's speed after the collision, we can use the principle of conservation of momentum. In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. Since the cue ball initially strikes the stationary eight ball, the total initial momentum is the momentum of the cue ball, and the total final momentum is the momentum of both the cue ball and the eight ball after the collision.

Let's denote the mass of both balls as 'm'. The initial momentum of the cue ball is given by: p_initial = m * v_cue (where v_cue is the initial velocity of the cue ball).

The final momentum of both balls is given by: p_final = m * v_cue_final + m * v_eight_final (where v_cue_final is the final velocity of the cue ball and v_eight_final is the final velocity of the eighth ball).

Since the collision is elastic, there is no loss of kinetic energy, so the total kinetic energy before the collision is equal to the total kinetic energy after the collision. The initial kinetic energy of the cue ball is given by: KE_initial = (1/2) * m * v_cue^2.

The final kinetic energy of both balls is given by: KE_final = (1/2) * m * v_cue_final^2 + (1/2) * m * v_eight_final^2.

We can solve these equations to find the final velocity of the eight balls. Given that the cue ball's initial velocity (v_cue) is 3.4 m/s, its final velocity (v_cue_final) is 0.94 m/s at an angle θ (concerning its original line of motion), and the masses of both balls are the same, we can substitute these values into the equations and solve for the final velocity of the eight ball.

Wavelength of the most strongly reflected is approximately 707 nm, which corresponds to red light in the visible spectrum.

To determine the wavelength most strongly reflected by a soapy water film, we can use the principle of thin film interference. Here,

For constructive interference to occur in a thin film illuminated perpendicularly, the condition we use is:

2nt = mλ

where n is the refractive index of the film,

Since we need the most strongly reflected wavelength, we'll consider

m = 1 (the first order of constructive interference).

Substituting the values into the equation:

2 x 1.33 x 266 nm = 1 x λ

This simplifies to:

λ = 2 x 1.33 x 266 nm

λ ≈ 707 nm

Therefore, the most strongly reflected wavelength is approximately 707 nm, which corresponds to red light in the visible spectrum.

Answer:

500000000 lbft/s

Explanation:

F = Force or weight = 1000000 lbf

s = Displacement = 10000 ft

t = Time taken = 20 seconds

Work done is given by

[tex]W=Fs\\\Rightarrow W=1000000\times 10000\\\Rightarrow W=10000000000\ lb-ft[/tex]

Power is given by

[tex]P=\dfrac{W}{t}\\\Rightarrow P=\dfrac{10000000000}{20}\\\Rightarrow P=500000000\ lbft/s[/tex]

One Saucer power is 500000000 lbft/s

Answer:

[tex]t=25.6446\ s[/tex]

is the time after which Jill will be able to catch the cart.

Explanation:

Given:

Now the component of gravity acting on the cart along the inclined plane:

[tex]g'=g.sin\ \theta[/tex]

[tex]g'=9.8\times sin\ 3^{\circ}[/tex]

[tex]g'=0.5129\ m.s^{-2}[/tex]

Time taken by Jill to reach this speed:

[tex]v=u+a.t[/tex]

where:

t = time taken

u = initial velocity = 0

[tex]51.289=0+2\times t[/tex]

[tex]t=25.6446\ s[/tex] is the time after which Jill will be able to catch the cart.

Answer:

F=G[tex]\frac{M1XM2}{R^{2} }[/tex]

Explanation:

F=F=G[tex]\frac{M1XM2}{R^{2} }[/tex]

Newton law define gravity as the product of two mass and the mean squired distance between the two bodies.

Where F is the gravitational pull and G is the gravitational constant and M1 and M2 are masses of two bodies while R is their mean distance

Replacing the value in the equation it becomes:

3.2x10N=G[tex]\frac{M1X55}{2.1X10^{2} }[/tex] Where G is the gravitation constant,[tex]6.67x10^{-11}[/tex]

M1=[tex]\frac{3.2X10X(2.1X10)^{2} }{55}[/tex]

M1=[tex]3.8468x10^{-11}[/tex]Kg