The Huka Falls on the Waikato River is one of New Zealand's most visited natural tourist attractions. On average the river has a flow rate of about 300,000 L/s. At the gorge, the river narrows to 18 m wide and averages 22 m deep.(a) What is the average speed (in m/s) of the river in the gorge?_______m/s.(b) What is the average speed (in m/s) of the water in the river downstream of the falls when it widens to 63 m and its depth increases to an average of 42 m________m/s.

Answer:

20 metric tons

Explanation:

Using the Principle of Conservation of Linear Momentum which states that the total momentum of a system is constant in any direction in which no external force acts.

momentum before collision= momentum after collision

mathematically expressed as

m1u1 + m2u2 = m1v1+m2v2 =0

where:

m1= mass of rolling freight car = 10 metric tons = 10Mg

1 ton = 1000kg= 1Mg

u1 =initial velocity of freight car = 108km/h,

m2 = mass of stationary freight car = ?, this the unknown we are finding

u2 = initial velocity of

m2 = 0, because it is at rest,

v1= final velocity of m1,

v2= final velocity of m2

since they move together in the same direction they share a common final velocity as such

v1=v2= 36km/h

substituting valves in expression give

m1u1 + m2u2 = m1v1 + m2v2= 0

10Mgx 108km/hr + m2 x 0 = 10Mg x 36km/h + m2x36km/h

1080( Mg x km/hr) + 0 = 360 (Mg x Km/h) + 36m2 (km/h)

1080 (Mgx km/hr) - 360(Mg x km/h) = 36m2 (km/h)

720(Mg x km/h) = 36m2 (km/h)

divide both side by  36 (km/h) gives

m2 = 20Mg =20 metric tons

Answer: B). 20 metric tons

Explanation: If a freight car with a mass of 10 metric tons is going at 108 km/h collides with another freight car, that is parked. If the speed of the cars, after they crash together, is 36 km/h, the mass of the second car is 20 metric tons.

Answer:

[tex]SE_{gel} = 3.75[/tex]

Explanation:

First, we have to calculate the gel's column height using the cylinder's volume, as follows:

[tex]V=\pi\times r \times h\\h=\frac{V}{\pi \times r}\\h=\frac{4.1m^3}{\pi \times 1m}= 1.30 m[/tex]

Then, as the pressure given at the bottom of the tank is the sum of the surface pressure and the gel's column pressure, we need to calculate only the gel's column pressure:

ft of water is a unit of pressure, but we need to convert it to atm and then to Pa, in order to calculate our results in the correct units. Therefore, the conversion factor is:

1 ft of water (4°C) = 0.0295 atm

[tex]60 ft water \times \frac{0.0295 atm}{1 ft water}= 1.77  atm\\P_{bottom}=P_{surface}+P_{gel}\\P_{gel}=P_{bottom}-P_{surface}=1.77 atm - 1.3 atm\\P_{gel}= 0.47 atm\times \frac{101325Pa}{1 atm}=47622.75 Pa[/tex]

Now, to calculate the specific gravity, we need to find first the gel's density:

[tex]P_{gel} = \rho gh\\\rho = \frac{P_{gel}}{gh}=\frac{47622.75 Pa}{9.8 m/s^2 \times 1.30m}= 3738.04 \frac{kg}{m^3}[/tex]

[tex]SE_{gel} = \frac{\rho_{gel}}{\rho_{water}}= \frac{3738.04 kg/m^3}{997 kg/m^3} = 3.75\\SE_{gel} =3.75[/tex]

The specific gravity of the gel is 3.75.

Final answer:

To calculate the specific gravity of a gel, convert the pressure reading from feet of water to pascals, add the tank's pressurized atmosphere, then use the relation between pressure, density, gravity, and height to solve for the density of the gel, which is then compared to the density of water.

Explanation:

The question is asking to calculate the specific gravity of a gel based on the pressure reading from a probe at the bottom of a cylindrical tank. We are given the pressure as 60 ft of water. To convert this to a pressure that we can use to find density, we need to convert feet of water to pascals:

1 ft H2O = 2989.07 Pa

60 ft H2O = 60 * 2989.07 Pa = 179344.2 Pa

Since the tank is pressurized to 1.3 atmospheres at the surface, we must also consider this in our pressure calculation:

1 atm = 101325 Pa

1.3 atm = 1.3 * 101325 Pa = 131722.5 Pa

The total pressure at the bottom of the tank is the sum of the pressure due to the gel and the pressurized air:

Total pressure = 179344.2 Pa + 131722.5 Pa

Total pressure = 311066.7 Pa

To find the specific gravity, we use the following relation where the specific gravity is the ratio of the density of the gel (ρgel) to the density of water (ρH2O):

Specific gravity = ρgel / ρH2O

We know that pressure is also the product of density (ρ), gravity (g=9.81 m/s²), and height (h=60 ft * 0.3048 m/ft) for the fluid:

Pressure = ρgel * g * h

ρgel = Pressure / (g * h)

ρgel = 311066.7 Pa / (9.81 m/s² * 60 ft * 0.3048 m/ft)

After calculating ρgel, we divide that by the density of water (1000 kg/m³) to find the specific gravity.

Answer: A hazard is a condition that has the potential to cause harm.

A natural hazard is a potentially harmful situation, where a person places himself in a naturally unsafe zone.

A natural disaster is a large scale destruction of life and properties by the forces of nature.

A natural hazard can become a natural disaster in some cases where the natural harmful situation a person places himself in, is acted upon by the forces of nature on a large scale.

Answer:

Explanation:

Given

Each student exert a force of [tex]F=400 N[/tex]

Let mass of car be m

there are 18 students who lifts the car

Total force by 18 students [tex]F=18\times 400=7200 N[/tex]

therefore weight of car [tex]W=7200[/tex]

mass of car [tex]m=\frac{W}{g}[/tex]

[tex]m=\frac{7200}{9.8}=734.69 kg[/tex]

(b)[tex]7200 N \approx 1618.624\ Pound-force[/tex]

[tex]734.69 kg\approx 1619.71 Pounds[/tex]                  

The mass of the car lifted by eighteen students is 734.69 kg, and its weight is approximately 1620.29 pounds.

To determine the mass of the car that eighteen students lift in a college homecoming competition, given that each student exerts an upward force of 400 N, first calculate the total force exerted by the students: 18 students × 400 N per student = 7200 N.

Next, we use the equation for weight (W = mg) to find the mass (m), where W is the weight, m is the mass, and g is the acceleration due to gravity (9.8 m/s²). Thus, the mass of the car is m = W/g = 7200 N / 9.8 m/s² ≈ 734.69 kg.

To find the weight in pounds, we can use the conversion factor 1 kg ≈ 2.20462 lbs. Therefore, the weight of the car in pounds is approximately 734.69 kg × 2.20462 lbs/kg ≈ 1620.29 lbs.

Answer:

Current divider

Explanation:

To calculate current flow through any branch of a circuit by substituting the values of [tex]I_X\ and\ R_X[/tex] for the branch  values when total circuit current and the resistance are known, use the current divider formula.

A current divider is a circuit that produces output current as a function of input current. It is a rule to find the splitting of the current in all branches of the circuit. Hence, the correct option is (B).  

The formula to calculate the current flow through any branch of a circuit when the total circuit current and resistance are known is the current divider formula.

To calculate the current flow through any branch of a circuit by substituting the values of IX and RX for the branch values when total circuit current and the resistance are known, use the current divider formula. The current divider rule is particularly useful in parallel resistor circuits and it allows for the easy calculation of the current flowing through a resistor in parallel.

In a parallel circuit, the voltage across each branch is the same, but the current through each branch can be different, depending on the resistance of that branch. The current divider rule states that the current through a branch is the ratio of the total parallel resistance to the branch resistance, times the total current entering the parallel combination.

For example, if we want the current through resistor R1 (I1), and we know the total current (Itot) and the total parallel resistance (Rp), as well as the resistance of R1 (R1), we can use the formula:

I1 = Itot * (Rp / R1)

By knowing the total resistance and the total current, one can deduce the individual branch currents using Ohm's Law and the principles of parallel circuits.

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Answer:

2.5 seconds

Explanation:

s(t) = -16t^2 + 80t + 384

for

0≤t≤8

First we differentiate s(t) to get s'(t)

s'(t) = -32t + 80

Let us then find the critical point; thus we will equate s'(t) to zero and then search for values where s'(t) is undefined

s'(t) = -32t + 80 = 0

t = 80/32

t = 2.5 sec

Let us evaluate s at the critical points and end points

s(0) = -16(0)^2 + 80(0) + 384 = 384

s(2.5) = -16(2.5)^2 + 80(2.5) + 384 = 684

s(8) = -16(8)^2 + 80(8) + 384 = 0

Thus, the stone attains it maximum height of 684ft at at t=2.5s

Answer:

ball D will fall toward the ground at the same time as ball C

Explanation:

both balls experience the same downward (vertical) force of gravity as such they will both fall down at the same time, given that all other factors are equal.

although the ball were through with different forces,

those forces where in the horizontal direction but the force of gravity (downward force) will act on them equally to bring them down at the same time

Answer:

e). [tex]a = 0.066 m/s^2[/tex]

Explanation:

As we know that wheel is turned by 90 degree angle

so the angular speed of the wheel is given as

[tex]\omega_f^2 - \omega_i^2 = 2\alpha \theta[/tex]

now we have

[tex]\omega_f^2 - 0 = 2(0.01)(\frac{\pi}{2})[/tex]

[tex]\omega = 0.177 rad/s[/tex]

now the centripetal acceleration of the point P is given as

[tex]a_c = \omega^2 R[/tex]

[tex]a_c = (0.177)^2(2)[/tex]

[tex]a_c = 0.063 m/s^2[/tex]

tangential acceleration is given as

[tex]a_t = R\alpha[/tex]

[tex]a_t = 2(0.01)[/tex]

[tex]a_t = 0.02 m/s^2[/tex]

now net acceleration is given as

[tex]a = \sqrt{a_t^2 + a_c^2}[/tex]

[tex]a = \sqrt{0.02^2 + 0.063^2}[/tex]

[tex]a = 0.066 m/s^2[/tex]

The question involves the calculation of linear acceleration of a point on a wheel undergoing angular acceleration. The linear acceleration involves both tangential and radial components, and these are combined to provide the total acceleration. The closest answer is 0.066 m/s^2.

The subject of this question is physics, specifically the concepts of angular acceleration and linear acceleration. Given an angular acceleration, we can find the linear acceleration by using the formula a = rα, where a is linear acceleration, r is the radius, and α is angular acceleration. Substituting the given values, we get a = 2.0 m * 0.01 rad/s2 = 0.02 m/s2. This value is the tangential acceleration of point P.

Since the point P is on the y-axis, the linear acceleration, a, is given by the radial acceleration which is equal to ω2r, where ω is the angular velocity and r is the radius of the circle. But we also have ω2 = 2αr, where ρ is the angular acceleration and r is the radius. So, the radial acceleration is (2αr)2r = 4α2r3. Substituting for α and r we get: 4*(0.01 rad/s2)2*(2.0 m)3 = 0.0016 m/s2.

The total linear acceleration is the vector sum of the radial and tangential accelerations. Since these vectors are perpendicular, we calculate their resultant by Pythagoras' theorem: √(at2 + ar2) = √(0.022 m/s2 + 0.00162 m/s2) = 0.02016 m/s2. Hence, the choice that's closest is (a) 0.066 m/s2.

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Answer:

Explanation:

Facts about magnetic flux;

(1). Change in magnetic flux is equal to the current generated.

(2). The magnetic field moves from the north to the south.

(3). Magnetic flux is equal to how much magnetic field goes through a direction.

(4). Magnetlc direction can be found found by using Lenz's law.

Faraday's law= emf= N(BA/ ∆t.

Where N= number of loop

∆B/∆T = rate at which flux changes.

The magnitude φ(initial) of the magnetic flux through one turn of the coil before it is rotated is

φ(initial) = BA while φ(final)= zero(0).

The initial magnetic flux through one turn of the coil, with the coil initially perpendicular to Earth's magnetic field, is 6.10 × 10⁻⁹ T×m².

To calculate the magnitude of the initial magnetic flux (φinitial) through one turn of the coil, we use the formula:

φ = B × A × cos(θ)

where:

Since initially, the plane of the coil is perpendicular to Earth's magnetic field, θ = 0°, and hence cos(θ) = 1. First, we must convert the area from cm² to m²:

A = 12.2 cm² × (1 m² / 10,000 cm²) = 12.2 × 10⁻⁴ m²

Now, we can calculate the magnetic flux:

φinitial = 5.00 × 10⁻⁵ T × 12.2 × 10⁻⁴ m² × cos(0°)

φinitial = 5.00 × 10⁻⁵ T × 12.2 × 10⁻⁴ m²

φinitial = 6.10 × 10⁻⁹ T×m²

This is the magnitude of the initial magnetic flux through one turn of the coil before it is rotated.

Complete Question:

A coil has 230 turns enclosing an area of 12.2 cm³. In a physics laboratory experiment, the coil is rotated during the time interval 3.00 × 10⁻² s from a position in which the plane of each turn is perpendicular to Earth's magnetic field to one in which the plane of each turn is parallel to the field. The magnitude of Earth's magnetic field at the lab location is 5.00 × 10⁻⁵ T.

What is the magnitude φinitial of the magnetic flux through one turn of the coil before it is rotated?

Answer:

4400 K approximately

Explanation:

Stefan-Boltzmann law establish that the power radiated from a black body in terms of its temperature which is proportional to [tex]T^4[/tex]

[tex]i = aT^4[/tex]  with a Stefan-Boltzmann constant

Also we know that [tex]i_{sunspot}[/tex] is three times [tex]i_{sun}[/tex]

[tex]i_{sun}=3i_{sunspot}[/tex] or [tex]\frac{i_{sun}}{i_{sunspot}} = 3[/tex]

Using the Stefan-Boltzmann we can write

[tex]\frac{i_{sun}}{i_{sunspot}} = 3 = \frac{aT_{sun}^4}{aT_{sunspot}^4}[/tex]

solving for [tex]T_{sunspot}[/tex]

[tex]T_{sunspot} = \left(\frac{T_{sun}^4}{3}\right)^{1/4}[/tex]

Replacing the value of [tex]T_{sun}[/tex] (5800 K) it is obtained that [tex]T_{sunspot}[/tex] is 4407.05

The magnitude of the resultant force on the race car and driver is 357.375 kN.

First, we need to find the centripetal acceleration of the race car traveling around the circular track. The centripetal acceleration can be found using the formula:

a = v² / r

where v is the velocity of the car and r is the radius of the circular track. Plugging in the values, we get:

a = (99 m/s)² / 52 m = 188.25 m/s²

Next, we can use Newton's second law of motion to find the magnitude of the resultant force acting on the car. The formula is:

F = m * a

where F is the force, m is the mass of the car, and a is the centripetal acceleration. Plugging in the values, we get:

F = (1900 kg) * (188.25 m/s²) = 357,375 N

Finally, we can convert the force to kilonewtons (kN) by dividing by 1000:

F = 357,375 N / 1000 = 357.375 kN

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Answer:

a= 0.063 m

b = 0.116 m

Explanation:

First of all, we need the spring constant in order to solve this problem. You are not giving that data, but I will tell you how to solve this assuming a value of k, In this case, let's assume the value of k is 1600 N/m. (I solved an exercise like this before, using this value).

Now, we need to use the expressions to calculate the distance of the spring.

The elastic potential energy (Uel) is given with the following formula:

Uel = 1/2 kx²

Solving for x:

x = √2*Uel/k

Replacing the data in the above formula (And using the value of k os 1,600):

x = √2 * 3.2 / 1600

x = 0.063 m

b) For this part, we need to apply the work energy theorem which is:

K1 + Ugrav1 + Uel1 + Uo = K2 + Ugrav2 + Uel2

Since in this part, the exercise states that the book is dropped, we can say that the innitial and the end is 0, therefore, K1 = K2 = 0.

The spring at first is not compressed, so Uel1 = 0, and Uo which is the potential energy of other factors, is also 0, because there are no other force or factor here. Therefore, our theorem is resumed like this:

Ugrav1 = Uel2

The potential energy from gravity is given by:

Ug = mgy

And as the spring is placed vertically, we know the height which the book is dropped, so the distance y is:

y = x + h

And this value of x, is the one we need to solve. Replacing this in the theorem we have:

mg(h+x) = 1/2kx²

g would be 9.8 m/s²

Now, replacing the data:

1.2*9.8(0.8 + x) = 1/2*1600x²

Rearranging and solving for x we have:

1.2*9.8*2(0.8 + x) = 1600x²

18.82 + 23.52x = 1600x²

1600x² - 23.52x - 18.82 = 0

Now we need to solve for x, using the general formula:

x = - (-23.52) ± √(-23.52)² - 4 * 1600 * (-18.82) / 2*1600

x = 23.52 ± √553.19 + 120,448 / 3200

x = 23.52 ± 347.85 / 3200

x1 = 23.52 + 347.85 / 3200 = 0.116 m

x2 = 23.52 - 347.85 / 3200 = -0.101 m

Using the positive value, we have that the distance is 0.116 m.

Answer:

The time taken to rotate the sphere one time is,  t = 22 s

Explanation:

Given data,

The mass of the sphere, m = 8200 kg

The radius of the sphere, r = 90 cm

                                             = .9 m

The force applied by the girl, F = 75 N

The moment of inertia of the sphere is,

                            I = 2/5 mr²

                              = (2/5) 8200 x (.9)²

                              = 2657 kg·m²

The torque,

                            τ = I α

                             75 x 0.9 = 2657 x α

                              α = 0.0254 rad/s²

The angular displacement,

                            θ = ½αt²

                             2π =  ½ x 0.0254 rad/s² x t²

                                t = 22 s

Hence, the time taken to rotate the sphere one time is,  t = 22 s

Answer:

80% (Eighty percent)

Explanation:

The material has a refractive index (n) of 1.25

Speed of light in a vacuum (c) is 2.99792458 x 10⁸  m/s

We can find the speed of light in the material (v) using the relationship

n = c/v, similarly

v = c/n

therefore v = 2.99792458 x 10⁸  m/s ÷ (1.25) = 239 833 966 m/s

v = 239 833 966 m/s

Therefore the percentage of the speed of light in a vacuum that is the speed of light in the material can be calculated as

(v/c) × 100 = (1/n) × 100 = (1/1.25) × 100 = 0.8 × 100 = 80%

Therefore speed of light in the material (v) is eighty percent of the speed of light in the vacuum (c)

Answer:

The percentage of speed of light in vacuum to the speed of light in the said material is 80%

Explanation:

The common values of refractive index are between 1 and 2 since nothing can travel faster than the speed of light, therefore, no material has a refractive index lower than 1.

According to the formula [tex]n=\frac{c}{v}[/tex]

where [tex]n[/tex] is the index of refraction

[tex]c[/tex] is the speed of light in vacuum

and [tex]v[/tex] is the speed of light in the material, it can be seen that n and v are inversely proportional which means greater the refractive index lower is the speed  of light.

Since we know that speed of light in vacuum is 300,000 km/s using the formula we get,

[tex]v=\frac{c}n}[/tex]

[tex]v=\frac{300000}{1.25}=240,000  km/s[/tex]

for finding percentage,

[tex]=\frac{240000}{300000}*100 = 80[/tex] %

Based on discoveries to date, the conclusion as “Planetary systems are common and planets similar in size to Earth are also common” is justified.

Answer: Option C

Explanation:

Some studies show that on average, each star has at least single planet. This means that most stars, such as the Solar System, possess planets (otherwise exoplanets). It is known that small planets (more or less Earthly or slightly larger) are more common than giant planets. The mediocrity principles state that planet like Earth should be universal in the universe, while the rare earth hypothesis says they are extremely rare.

Size is often considered an important factor, because planets the size of the Earth are probably more terrestrial and can hold the earth's atmosphere. The planetary system is a series of gravitational celestial objects orbiting a star or galaxy. Generally, planetary systems describe systems with one or more planets, although such systems may also consist of bodies such as dwarf planets, asteroids and the like.

Answer:

[tex]t_2=\sqrt{2}(t_1)[/tex]

Explanation:

The kinetic energy of a body is that energy it possesses due to its motion. In classical mechanics, this energy depends only on its mass and speed, as follows:

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

The speed in an uniformly accelerated motion is given by:

[tex]v=at[/tex]

Replacing this expression in the formula for the kinetic energy, we have:

[tex]K=\frac{ma^2t^2}{2}\\[/tex]

So, if we have [tex]K_2=2K_1[/tex]:

[tex]K_1=\frac{ma^2t_1^2}{2}(1)\\K_2=\frac{ma^2t_2^2}{2}\\2K_1=\frac{ma^2t_2^2}{2}\\K_1=\frac{ma^2t_2^2}{4}(2)\\[/tex]

Equaling (1) and (2) and solving for [tex]t_2[/tex]:

[tex]\frac{ma^2t_1^2}{2}=\frac{ma^2t_2^2}{4}\\t_2=\frac{4t_1^2}{2}\\t_2=\sqrt{2t_1^2}\\t_2=\sqrt{2}(t_1)[/tex]

Answer:

Earth attract the less massive moon with a force that is the same as the force with which the moon attracts the Earth.

Explanation:

This can be explained by Newton's third law:

[tex]F_{12}=-F_{21}[/tex]

The force exerted by body 1 on body 2 is the same as that exerted by body 2 on body 1, only with the opposite sign.

In this case that force is the gravitational force, but the law still applies.

So the moon and the earth are attracted with the same magnitude of force.

Answer:

Evaporation is the physical change in which a substance converts from its liquid state to its gaseous state. Condensation is the physical change in which a substance converts from its gaseous state to its liquid state.

Explanation:

Evaporation and condensation are opposite processes to each other. Evaporation changes a liquid to a gas and condensation is the reverse.

Answer:

f = 63.8 N

Explanation:

initial angle to the vertical = 5 degrees

initial holding force = 40 N

final angle to the vertical = 8 degrees

final holding force = ?

find the final holding force

        m = mass  and  g = acceleration due to gravity

        40 = mgSin5

        mg = [tex]\frac{40}{sin5}[/tex] ....equation 1

        f = mgSin8  ......equation 2

   substituting the value of mg from equation 1 (where mg = [tex]\frac{40}{sin5}[/tex] ) into equation 2

    f = mgSin8 = [tex]\frac{40}{sin5}[/tex] x Sin8

   f = 63.8 N

Answer:

64 N

Explanation:

Given that the angles are very small, the following approximation can be made:

F ≈ m*g*α

where F is the force needed to hold the ball, m is the ball mass, g is the acceleration of gravity and α is the angle between the cord and the vertical.

Let's call F1 the force needed to hold the ball at 5° (α1) and F2 the force needed to hold the ball at 8° (α2).

F1 ≈ m*g*α1

F2 ≈ m*g*α2

Dividing the equations:

F1/F2 = α1/α2

F2 = (α2/α1)*F1

F2 = (8/5)*40

F2 = 64 N

As the gases ( O₂ & CO₂ ) are most abundant here and supports life.

Hope it helps!

Happy Learning!!

:D

Answer:

Speed of the ambulance is 32.7 metres per second.

Explanation:

Let the actual frequency of the siren be [tex]f_{0}[/tex].

Frequency observed by me when ambulance is approaching = 540 Hz

Frequency observed by me when ambulance is moving away = 446 Hz

Let [tex]v_{s}[/tex] be the speed of sound and [tex]v_{a}[/tex] be the speed of ambulance.Then according to Doppler effect:

When source is moving towards observer,frequency observed is given as

[tex]f_{0} \times \frac{v_{s} }{v_{s} - v_{a} }[/tex] = 540 Hz    

When source is moving away from observer,frequency observed is given as

[tex]f_{0} \times \frac{v_{s} }{v_{s} + v_{a} }[/tex] = 446 Hz    

Taking [tex]v_{s} = 343 \frac{m}{s}[/tex] and solving the above two equations by eliminating [tex]f_{0}[/tex],

we get [tex]v_{a} = 32.7 \frac{m}{s}[/tex]

Answer:

The tension in the string connecting block 50 to block 51 is 50 N.

Explanation:

Given that,

Number of block = 100

Force = 100 N

let m be the mass of each block.

We need to calculate the net force acting on the 100th block

Using second law of newton

[tex]F=ma[/tex]

[tex]100=100m\times a[/tex]

[tex]ma=1\ N[/tex]

We need to calculate the tension in the string between blocks 99 and 100

Using formula of force

[tex]F_{100-99}=ma[/tex]

[tex]F_{100-99}=1[/tex]

We need to calculate the total number of masses attached to the string

Using formula for mass

[tex]m'=(100-50)m[/tex]

[tex]m'=50m[/tex]

We need to calculate the tension in the string connecting block 50 to block 51

Using formula of tension

[tex]F_{50}=m'a[/tex]

Put the value into the formula

[tex]F_{50}=50m\times a[/tex]

[tex]F_{50}=50\times1[/tex]

[tex]F_{50}=50\ N[/tex]

Hence, The tension in the string connecting block 50 to block 51 is 50 N.

We have that for the Question "What is the tension in the string connecting block 100 to block 99? What is the tension in the string connecting block 50 to block 51?"

it can be said that

From the question we are told

Each of 100 identical blocks sitting on a frictionless surface is connected to the next block by a massless string. The first block is pulled with a force of 100 N.

Assuming mass of each block is 1 kg

The equation for the force is given as

[tex]F = ma\\\\a = \frac{F}{m}\\\\ = \frac{100}{100*1}\\\\ = 1 m/s^2[/tex]

Now, between block 100 and 99,

[tex]F = ma\\\\F = 1*1 \\\\= 1 N[/tex]

Now between block 50 and 51. There are 50 blocks behind 51 st block,  so,

[tex]m = 50 Kg\\\\a = 1 m/s2 (assuming all blocks accelerate at same rate)\\\\F = 50 * 1\\\\F= 50 N[/tex]

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Termination of translation requires a termination signal, and a release factor

Explanation:

There are 3 stops codons of the 64 possible codons. These are UAA, UAG, or UGA. These do not code for amino acids and are therefore not recognized by any anticodons for any of the ‘charged’ tRNA. These codons, are recognized by release factors that ‘knock off’ the newly synthesized peptide from the ribosome through peptidyl-tRNA hydrolysis. There are several release factors (RF1 and RF2) in bacteria but in eukaryotes only one RF has been discovered to recognize the 3 stop codons.

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Translation termination is signaled by a stop codon and facilitated by release factors and ribosomes, which lead to the release of the newly synthesized protein and the disassembly of the ribosomal complex.

Termination of translation requires a termination signal, which is a stop codon (UAA, UAG, or UGA) that doesn't code for an amino acid but rather signals the end of the translation process. During this stage, release factors recognize the stop codon and prompt the addition of a water molecule to the carboxyl end of the peptidyl-tRNA in the P site. This action leads to the release of the newly synthesized polypeptide chain. Ribosomes, which consist of a large and a small subunit, dissociate from the mRNA and from each other upon the completion of translation.

Ribosomes, release factors, and the mRNA template are the key participants in the termination process. Initiator tRNA and elongation factors with charged tRNAs are involved in the initiation and elongation stages of translation, but not in termination.

Answer:

2.35 x 10^{5} J

Explanation:

force (f) = 1080 N

distance (d) = 218 m

find the work done

work is said to be done when force applied to an object moves the object, therefore work done = force x distance

work done = 1080 x 218 = 235,440 J = 2.35 x 10^{5} J

Answer:

111.6 g

Explanation:

Given

m + M = 2.58 / 9.8

= 0.2632 kg

When the can of fruit juice is balance from scale , we get following relation

M x ( 50 - 21.2 ) = m x 21.2 ( balancing the torque due to weight of scale and can about the balancing point )

M x 28.8 = 21.2 m

= 21.2 ( 0.2632 - M )

= 5.58 - 21.2 M

M ( 28.8 + 21.2 ) = 5.58

M = .1116 kg

= 111.6 g

Answer:

it takes the sun about 230 million years to orbit the milky way.

Explanation:

we're moving at an average velocity of 828,000 km/hr but it still takes us a long time to orbit the milky way.

Answer:

[/tex] 408.8[/tex]

Explanation:

[tex]t[/tex] = thickness of the floor = 0.159 m

[tex]k[/tex] = thermal conductivity of wooden floor = 0.141 Wm⁻¹ ⁰C⁻¹

[tex]T_{i}[/tex] = Temperature of the interior of the cabin = 18.0 ⁰C

[tex]T_{o}[/tex] = Temperature of the air below the cabin = - 16.4 ⁰C

Difference in temperature is given as

[tex]\Delta T[/tex] = Difference in temperature = [tex]T_{i} - T_{o} = 18 - (- 16.4) = 34.4[/tex]⁰C

[tex]A[/tex] = Area of the floor = 13.4 m²

[tex]Q[/tex] = Rate of heat conduction

Rate of heat conduction is given as

[tex]Q = \frac{kA \Delta T}{t}[/tex]

[tex]Q = \frac{(0.141) (13.4) (34.4)}{0.159}\\Q = 408.8[/tex] W

Answer:

both technicians are correct

Explanation:

both technicians are correct because technician A says the first external component to be installed is the intake manifold and technician B says the exhaust manifold should be installed after the intake manifold has been installed, this implies the same thing because installing the exhaust manifold after installing the intake manifold means you are installing the intake manifold first.

Answer:

[tex]T_H = 1335\ J[/tex]

Explanation:

Given,

Work done by engine = 19900 J

rejected energy of heat =  5300 J

temperature = 285 K

efficiency of the engine

[tex]\eta = \dfrac{W}{Q_H}[/tex]

where

[tex]W = Q_H - Q_C[/tex]

[tex]19900 = Q_H - 5300[/tex]

[tex]Q_H = 25200\ J[/tex]

efficiency in terms of temperature

[tex]\eta = \dfrac{T_H-T_C}{T_H}[/tex]

from the above two equation

[tex]\dfrac{Q_H-Q_C}{Q_H}= \dfrac{T_H-T_C}{T_H}[/tex]

[tex]1-\dfrac{Q_C}{Q_H}= 1- \dfrac{T_C}{T_H}[/tex]

[tex]\dfrac{Q_C}{Q_H}= \dfrac{T_C}{T_H}[/tex]

[tex]T_H=T_C\ \dfrac{Q_H}{Q_C}[/tex]

[tex]T_H=285\times \dfrac{25200}{5300}[/tex]

[tex]T_H = 1335\ J[/tex]

The smallest possible temperature of the hot reservoir, given the work done by the engine, heat rejected into the cold reservoir, and temperature of the cold reservoir, is approximately 1357.14 Kelvin.

The problem is based on the Carnot heat engine, a theoretical model used for understanding the thermodynamics of heat conversion into work. The efficiency of Carnot's engine depends on the temperatures of the hot and cold reservoirs (Th and Tc respectively). The efficiency, represented as e, is calculated as e = (Th-Tc) / Th.

Before we do the calculation, it's important to note that temperatures must be expressed in Kelvin for this formula to be accurate. Tc, the temperature of the cold reservoir, is given as 285 K. The work done by the engine, represented as W, is 19900 J, and the heat rejected by the engine into the cold reservoir, represented as Qc, is 5300J. The heat absorbed by the engine from the hot reservoir, represented as Qh, can be calculated using the first law of thermodynamics, which states Qh = W + Qc. Therefore, Qh = 19900 J + 5300 J = 25200 J.

Now, to find Th, we rearrange the efficiency formula to Th = Tc / (1 - e), where e is now W / Qh = 19900 J / 25200 J ≈ 0.79. Substituting Tc = 285 K and e ≈ 0.79 into the formula, Th = 285 K / (1 - 0.79) ≈ 1357.14 K. This is the smallest possible temperature of the hot reservoir for the given conditions.

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Answer:

[tex]c_q=0.7409\ J.g^{-1}.^{\circ}C^{-1}[/tex]

Explanation:

Given:

We have:

Assuming that heat loss is neither from any of the components before mixing nor from the mixture just after mixing. Also the container does not absorbs any heat.

Therefore,

heat gained by the water = heat lost by the quartz

[tex]Q_w=Q_q[/tex]

[tex]m_w.c_w. (T_f-T_{wi})=m_q.c_q.(T_{qi}-T_f)[/tex]

    where: [tex]c_q=[/tex] specific heat capacity of quartz

[tex]250\times 4.186\times (27.9-25)=58.1\times c_q\times (98.4-27.9)[/tex]

[tex]c_q=0.7409\ J.g^{-1}.^{\circ}C^{-1}[/tex]

To calculate the specific heat capacity of quartz, use the equation q = mcΔT and set it equal to the equation for the heat transferred to the water. Solve for c to find the specific heat capacity of quartz.

The specific heat capacity of quartz can be calculated using the equation:

q = mcΔT

where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

The heat transferred to the water can be calculated using the equation:

q = mcΔT

where q is the heat transferred, m is the mass, c is the specific heat capacity of water (4.184 J/g °C), and ΔT is the change in temperature.

By setting the two equations equal to each other and solving for c, the specific heat capacity of quartz can be determined.

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