10. A tube in which the
cross-sectional area gradually decreases to a minimum diameter in its center section is called a “___”

Answer:

It is D

Explanation:

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Answer:

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Answer:

The tension in \(T_1\) is 0.1024 Newtons.

The tension in \(T_2\) is 0.1157 Newtons.

Explanation:

Lets create a free body diagram showing all of the forces; we need to show the vertical and horizontal components of the tension.

I will attach a picture of my free body diagram. Notice I created 2 new triangles with the adjacent angles of angle A and C from the original picture.

Lets make a list of all the variables we have now. Also lets write down the information we are given.

\(A=27\\C=38\\m=12\\g=9.81\\W\\T_{1} \\T_{1x} \\T_{1y}\\T_{2} \\T_{2x} \\T_{2y} \\\)

In this situation the sum of of the vertical tension components must support the weight. To find the vertical components we can use the SIN function.

\(sin(x)=\frac{O}{H} \\O=Hsin(x)\)

Therefore we can write that the sum of the forces in the y direction is

\(\sum F_y=T_1sin(A)+T_2sin(C)=W\)

This system is in equilibrium; the object should not move along the x-axis. Therefore, the horizontal components of \(T_1\) and \(T_2\) must then equal each other. To find the horizontal components we can use the COS function.

\(cos(x)=\frac{A}{H} \\A=Hcos(x)\)

Therefore we can write that the sum of the forces in the x direction is

\(\sum F_x=T_1cos(A)=T_2cos(C)\)

Now we have to equations to help us solve the problem.

\(T_1cos(A)=T_2cos(C)\)

\(T_1sin(A)+T_2sin(C)=W\)

We do not know the numerical values of \(T_1\) and \(T_2\) so we will have to manipulate algebraically to solve them.

In the first equation lets solve for \(T_1\).

\(T_1cos(A)=T_2cos(C)\)

Divide both sides by \(cos(A)\).

\(T_1=\frac{T_2cos(C)}{cos(A)}\)

Separate the right side into two fractions.

\(T_1=\frac{T_2}{1} *\frac{cos(C)}{cos(A)}\)

Use the reciprocal trig identity for cosine.

\(sec(x)=\frac{1}{cos(x)}\)

\(T_1=\frac{T_2}{1} *cos(C)}*sec(A)\)

\(T_1=T_2*cos(C)}*sec(A)\)

Now insert our answer for \(T_1\) into the second equation.

\(T_2*cos(C)}*sec(A)*sin(A)+T_2*sin(C)=W\)

Solve for \(T_2\). Lets replace each trig function with its own variable to make this easier.

\(T_2*cos(C)}*sec(A)*sin(A)+T_2*sin(C)=W\)

\(cos(C)=x\\sec(A)=y\\sin(A)=z\\sin(C)=u\)

\(T_2*x*y*z+T_2*u=W\\T_2xyz+T_2u=W\)

Now lets solve for \(T_2\).

Factor \(T_2\) out of each term.

\(T_2(xyz)+T_2(u)=W\)

Factor \(T_2\) out of each term.

\(T_2(xyz+u)=W\)

Divide each side by \(xyz+u\).

\(T_2=\frac{W}{xyz+u}\)

Lets substitute the trig functions back in for the variables

\(T_2=\frac{W}{cos(C)*sec(A)*sin(A)+sin(C)}\)

\(W\) is the weight.

The formula for weight is \(W=mg\). Where \(m\) is the mass in kilograms.

12 grams is 0.012 kilograms.

\(W=0.012*9.81\)

\(W=0.11772\)

Numerical Evaluation

Lets evaluate \(T_2\).

\(T_2=\frac{0.11772}{cos(38)*sec(27)*sin(27)+sin(38)}\)

\(T_2=0.11573\)

Lets evaluate \(T_1\).

\(T_1cos(A)=T_2cos(C)\)

\(T_1*cos(27)=0.11573*cos(38)\\T_1=0.10235443\)

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Answer:

T₁ = 102.25 N (2 d.p.)

T₂ = 115.61 N (2 d.p.)

Explanation:

The diagram shows a body of mass 12 kg (weight = 12g N) held in equilibrium by two light, inextensible strings. One string makes an angle of 27° with the positive horizontal and the other string makes an angle of 38° with the negative horizontal.

The body is in equilibrium, so both the horizontal and vertical components of the forces must sum to zero.

Resolving horizontally, taking (→) as positive:

\(\implies -T_1 \cos (27^{\circ})+T_2 \cos (38^{\circ})=0\)

\(\implies T_1 \cos (27^{\circ})=T_2 \cos (38^{\circ})\)

\(\implies T_1=\dfrac{T_2 \cos (38^{\circ})}{ \cos (27^{\circ})}\)

Resolving vertically, taking (↑) as positive:

\(\implies T_1 \sin(27^{\circ})+T_2 \sin(38^{\circ})-12\text{g}=0\)

\(\implies T_1 \sin(27^{\circ})+T_2 \sin(38^{\circ})=12\text{g}\)

Substitute the found expression for T₁ into the second equation and take g = 9.8 ms⁻²:

\(\implies \left(\dfrac{T_2 \cos (38^{\circ})}{ \cos (27^{\circ})}\right) \sin(27^{\circ})+T_2 \sin(38^{\circ})=12\text{g}\)

\(\implies T_2 \cos (38^{\circ})\tan(27^{\circ})+T_2 \sin(38^{\circ})=12\text{g}\)

\(\implies T_2 \left(\cos (38^{\circ})\tan(27^{\circ})+ \sin(38^{\circ})\right)=12\text{g}\)

\(\implies T_2 =\dfrac{12\text{g}}{\cos (38^{\circ})\tan(27^{\circ})+ \sin(38^{\circ})}\)

\(\implies T_2=115.614550...\:\text{N}\)

Substitute the found value of T₂ into the equation for T₁ and take g = 9.8 ms⁻²:

\(\implies T_1=\dfrac{\left(\dfrac{12\text{g}}{\cos (38^{\circ})\tan(27^{\circ})+ \sin(38^{\circ})}\right) \cos (38^{\circ})}{ \cos (27^{\circ})}\)

\(\implies T_1=102.250103...\text{N}\)

Therefore, the value of T₁ and T₂ in the given diagram is:

Answer:

Below

Explanation:

weight = mass * g  <==g is less on the moon, so your weight is less on the                      moon

         to calculate weight, you have to use the g of the planet you are on

The MASS does not change....it will be 6kg on the moon or earth

Weight of 30 kg on earth =    30 * g = 30 * 9.81 = 294 N

The circuit in which lamps Q and R light but not lamp P when switch S is open is circuit B.

An electric circuit is a path for transmitting electric current.

Given the circuits below, when switch S is open, we want to determine the circuit in which lamps Q and R light but not lamp P.

To determine the circuit, we proceed as follows.

To determine the circuit in which lamps Q and R light but not lamp P, it must satisfy this condition

Looking at all the circuits, the circuit which satisfy these condition is circuit B

So, the circuit in which lamps Q and R light but not lamp P is circuit B.

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When you increase your speed from 35 mph to 70 mph, the force of impact you experience will be increased 4 times.

Hence, the correct option is B.

The force of impact in a collision is directly proportional to the square of the velocity. When the speed is doubled (increased from 35 mph to 70 mph), the force of impact will increase by a factor of four (\(2^2\)). This is a fundamental concept in physics known as the kinetic energy formula, which is given by:

Kinetic Energy = (1/2) * mass * \(velocity^2\)

Since kinetic energy is directly proportional to the square of velocity, doubling the speed results in four times the kinetic energy and, consequently, four times the force of impact.

Therefore, When you increase your speed from 35 mph to 70 mph, the force of impact you experience will be increased 4 times.

Hence, the correct option is B.

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Answer:

the answer to this question is Accelerated motion

1.00 kg of ice at -24.0°C is placed in contact with a 1.00 kg block of a metal at 5.00°C. They come to equilibrium at -8.88°C.

We can use the principle of conservation of heat to solve this problem. The heat lost by the metal must equal the heat gained by the ice.

The heat lost by the metal is given by

Q1 = m1c1ΔT1

Where m1 is the mass of the metal, c1 is its specific heat, and ΔT1 is the change in temperature.

The heat gained by the ice is given by

Q2 = m2c2ΔT2

Where m2 is the mass of the ice, c2 is its specific heat, and ΔT2 is the change in temperature.

Since the two objects come to thermal equilibrium, we can set Q1 equal to Q2

m1c1ΔT1 = m2c2ΔT2

Solving for c1, we get

c1 = m2c2ΔT2 / (m1ΔT1)

By putting these values we get

c1 = (1.00 kg)(2.06 kJ/kg·K)(-24.0°C - (-8.88°C)) / [(1.00 kg)(5.00°C - (-8.88°C))]

c1 = 0.902 kJ/kg·K

Hence, the specific heat of the metal is 0.902 kJ/kg·K.

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Answer: D

Explanation:

they are all odorless, colorless, monatomic gases with very low chemical reactivity.

Brainliest?

The elements in group 18 that they rarely react with any other elements. The correct option is D.

The periodic table, also known as the periodic table of the elements, is an arrangement of chemical elements in rows and columns.

It is widely used in chemistry, physics, and other sciences, and is widely regarded as a chemistry icon.

The periodic table is worked upon by scientists to quickly refer to information about an element, such as its atomic mass and chemical symbol.

The arrangement of the periodic table also allows scientists to detect trends in element properties such as electronegativity, ionization energy, and atomic radius.

The noble gases are found at the far right of the periodic table and were previously known as "inert gases" due to their filled valence shells, which make them extremely nonreactive. In comparison to other element groups, noble gases were discovered relatively late.

Thus, the correct option is D.

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Answer:

Experiments showed that increasing the light frequency increased the kinetic energy of the photoelectrons, and increasing the light amplitude increased the current.

Explanation:

The photocurrent increases linearly with the intensity of the incident light but is independent of its frequency. The stopping potential increases linearly with the frequency of the incident light but is independent of its intensity.

They should use a large marble with the ramp at a 75 degree angle

The tension force in the string is 98.72 N.

The angular speed of the ball is calculated as follows;

ω = 2πf

ω = 2π x 5

ω = 31.42 rad/s

The tension force in the string is calculated as follows;

F = T = mω²r

T = 0.04 x (31.42)² x 2.5

T = 98.72 N

Thus, the tension force in the string is 98.72 N.

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The speed of the truck just before braking is 23.24 m/s.

The speed of the truck before braking is calculated by applying the third kinematic equation as shown below.

v² = u² - 2as

where;

When the truck stops, the final velocity = 0

0 = u² - 2as

u²  = 2as

u = √2as

u = √ ( 2 x 6 x 45 )

u = 23.24 m/s

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Answer:

The value is      \(A = 39315 \ m^2\)

Explanation:

From the question we are told that

    The velocity which the rover is suppose to land with is  \(v = 1 \ m/s\)

    The  mass of the rover and the parachute is  \(m = 2270 \ kg\)

     The  drag coefficient is  \(C__{D}} = 0.5\)

      The atmospheric density of Earth  is  \(\rho = 1.2 \ kg/m^3\)

     The acceleration due to gravity in Mars is  \(g_m = 3.689 \ m/s^2\)

Generally the Mars  atmosphere density is mathematically represented as

          \(\rho_m = 0.71 * \rho\)

=>        \(\rho_m = 0.71 * 1.2\)

=>        \(\rho_m = 0.852 \ kg/m^3\)

Generally the drag force on the rover and the parachute  is mathematically represented as

          \(F__{D}} = m * g_{m}\)

=>       \(F__{D}} = 2270 * 3.689\)  

=>       \(F__{D}} = 8374 \ N\)  

Gnerally this drag force is mathematically represented as

         \(F__{D}} = C__{D}} * A * \frac{\rho_m * v^2 }{2}\)

Here A is the frontal area

So  

         \(A = \frac{2 * F__D }{ C__D} * \rho_m * v^2 }\)

=>       \(A = \frac{2 * 8374 }{ 0.5 * 0.852 * 1 ^2 }\)

=>       \(A = 39315 \ m^2\)

The skater's final angular velocity is approximately 9.86 rad/s.

The skater's final angular velocity can be calculated using the principle of conservation of angular momentum. The equation for angular momentum is given by:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Initially, the skater has an angular momentum of:

L_initial = I_initial * ω_initial

Substituting the given values:

L_initial = 2.12 kg m² * 3.25 rad/s

The skater's final angular momentum remains the same, as angular momentum is conserved:

L_final = L_initial

The final moment of inertia is given as 0.699 kg m². Therefore, the final angular velocity can be calculated as:

L_final = I_final * ω_final

0.699 kg m² * ω_final = 2.12 kg m² * 3.25 rad/s

Solving for ω_final:

ω_final = (2.12 kg m² * 3.25 rad/s) / 0.699 kg m²

Hence, the skater's final angular velocity is approximately 9.86 rad/s.

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We have the next information

T=2.00 x 10^-1 s=0.2s

r=3.5 inch

For the angular speed

where omega is the angular speed, T is the period

We substitute

For the linear speed on the rim of the disc, we will use the next formula

in this case r= 3.5/2=1.75 inch

Then for the linear speed on the point at 0.750 inches from the center of the disk.

ANSWER

ω=31.41 rad/sec

v on the rim= 54.97 inches/sec

v on the point=23.56 inches/sec

Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

\(k = \frac{1}{2} m {v}^{2} \\ \)

m is the mass

v is the velocity

From the question we have

\(k = \frac{1}{2} \times 8 \times {4}^{2} \\ = 4 \times 16\)

We have the final answer as

Hope this helps you

the answer to your question is 200 J

Answer:

99.9999 m

Explanation:

\(a_{x} =0\) \(v_{xo} =14\) \(a_{y} =-g\) \(v_{yo} =0\)

X-direction             | Y-direction

\(x=x_{o}v_{xo}t+\frac{1}{2} a_{x}t^2\)  | \(y=y_{o}v_{yo}t+\frac{1}{2} a_{y}t^2\)

\(x=0+14(7.14285)\) | \(0=250-\frac{1}{2} (-9.8)t^2\)

x = 99.9999 m       | \(-250=-4.9t^2\)

                               | \(\sqrt{\frac{-250}{-4.9} } =\sqrt{t^2}\)

                              | 7.14285s = t

Analysing the Question:

We are given:

height of the hot air balloon (h) = 250 m

horizontal velocity of hot air balloon (p) = 14 m/s

Since the balloon had no vertical velocity, the basket will drop vertically at a velocity of 0 m/s and gain its velocity due to acceleration

whereas the horizontal velocity of the balloon is 14 m/s, so the basket will have a horizontal velocity of 14 m/s when falling

there will be no change in the horizontal velocity of the basket since there is no force against the horizontal velocity of the falling basket

Time taken by the basket to reach the ground:

from the last section, we deduced that:

initial vertical velocity (u) = 0 m/s

time taken by the basket to reach the ground (t) = t seconds

acceleration of the basket due to gravity (a) = 10 m/s²

from the second equation of motion:

h = ut + 1/2*at²

replacing the variables

250 = (0)t + 1/2 * (10)(t)²

250 = 5t²

t² = 50

t = √50

t = 5√2 seconds   OR     7 seconds

Horizontal distance travelled by the basket:

Distance travelled = horizontal speed * time taken to reach the ground

Distance travelled = 14 * 7

Distance travelled = 98 m

The potential energy of the student standing on the building floor, 15 m above the ground, is 30,000 Joules.

To calculate the potential energy (P.E) of the student standing on a building floor, use the formula P.E = mgh, where m = mass of the student, g = acceleration due to gravity, and h = height above the ground.

Given:

Mass of the student (m) = 200 kg

Acceleration due to gravity (g) = 10 m/s^2

Height above the ground (h) = 15 m

Using the formula, calculate the potential energy as follows:

P.E = mgh

= 200 kg x 10 m/s² x 15 m

= 30,000 kg·m²/s²

= 30,000 J (Joules)

Therefore, the potential energy of the student standing on the building floor, 15 m above the ground, is 30,000 Joules.

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The net force in Newtons of the object is 10N.

Solution:

m1 = 4kg

a = 2.5m/s²

Force m = ma

F = 4 * 2.5 = 10N.

The greater the net force acting on an object the greater its acceleration. Newton's second law of motion summarizes these relationships. We can rewrite the acceleration equation above to find the net force. The net force on an object is the combined effect of all compressive and tensile forces actually acting on the object.

When the forces pushing or pulling an object are unbalanced the object accelerates in the direction of the net force. The net force is the sum of all the individual forces acting on an object. Newton's first law can be viewed as a special case of Newton's second law when the net force F is zero. The acceleration must also be zero.

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Answer:

sorry hindi q po alm hahahahahah

Given:

Mass of truck, m1 = 3200 kg

Mass of trailer, m2 = 2400 kg

acceleration = 0.45 m/s²

Let's determine the following:

(a) The tension force between the truck and the trailer.

To find the tension between the truck and the trailer, apply the formula:

Where:

Fnet is the tension between the truck and the trailer

a is the acceleration = 0.45 m/s²

m1 is the mass if the truck = 3200 kg

m2 is the mass of trailer = 2400 kg

Hence, we have:

Therefore, the tension between the truck and the trailer is 2520N

(b) The force with which the truck is pulling.

The tension between the truck and the trailer is equivalent to the force on the trailer.

thus, we have:

Therefore, the force with which the truck is pulling is 1080N.

ANSWER:

(a) 2520N

(b) 1080 N

Change of state due to cooling is due to the removal of thermal energy from a substance. A substance changes states, such as from a gas to a liquid or from a liquid to a solid, when its particles lose kinetic energy as it loses heat and moves more slowly. Eventually, the particles reorganize into a more ordered form with less energy. The term "solidification" or "freezing" refers to this process.

The net work must be done on the particle to cause it to move to the right at 47 m/s is 59.4J, The spring constant of spring is 281.75N/m and the length of spring stretched is 0.02m.

(1.)Given the mass of particle (m) = 75g = 0.075kg

The velocity of particle (Vl) towards left = 25m/s

The velocity of particle towards right (Vr) = 47m/s

The work done to move the particle from left to right = W

Here work done can be considered as change in Kinetic energy then:

W = (KEf - KEi)

\(W = 1/2*m*Vr^2 - 1/2*m*Vl^2\)

\(W = (1/2)*75 g*(47 m/s)^2 - (1/2)*75 g*(25 m/s)^2\)

\(W = (1/2)*75 g*(1584 m^2/s^2)\)

W = 59.4J

Therefore, the net work required to cause the particle to move from 25 m/s to 47 m/s is 59.4 J.

(2.) Given the length of spring = 8cm = 0.08m

The mass of block = 2.3kg

The extension in spring = 16cm = 0.16m

Let the spring constant = k

(A) We know that the force exerted in stretching the spring = F = kx where

F = mg then

mg = kx

2.3 * 9.8 = k * (0.16 - 0.08)

k = 22.54/0.08 = 281.75N/m

(B) The mass of new block = 3kg

Let the extended spring = x'

Then mg = kx' such that:

3 * 9.8 = 281.75 * x'

x' = 29.4/281.75 = 0.10m

the extension in spring = 0.10 - 0.08 = 0.02m

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Answer: 430297.6828056 light milliseconds

Explanation: the distance between  earth and venus is

184835061.608992 kilometers based on the model. the spacecraft would be at the halfway point which would be 92417530.804496 kilometers from earth. then, you find the midpoint of the 2 planets if they were aligned in a straight path. that would be 129000000 kilometers. then you find how fast it would take light to travel that distance. therefore you would get 430297.6828056 light milliseconds as the answer. Hope the answer was helpful. -Stephen

Explanation:

Here is the first one....you should try the second one...