The student has access to the following measurement equipment: a spring scale, a meterstick, and a stopwatch. The student uses the spring scale to pull the cart starting from rest along a horizontal surface such that the reading on the spring scale is always constant. All frictional forces are negligible. In addition to the spring-scale reading, which two of the following quantities could the student measure with the available equipment and then use to determine the inertial mass of the cart? Select two answers.


A) The total distance traveled by the cart after it has been in motion
B) The average speed of the cart while in motion
C)The acceleration of the cart while in motion
D)The time during which the cart is in motion

Answer:

increase

Explanation:

Electric potential is defined as the work done in bringing a unit positive charge from infinity to that point against the electrical forces of the field. Taking a test positive charge from one point to another means that work is done against the field hence electric potential increases.

The linear velocity of the 100 g ball in the rotating system with a 130 g ball and a 230 g ball, connected by a rigid rod and rotating at 120 RPM, is approximately [tex]\(4.24 \, \text{m/s}\).[/tex]

Given values:

- Mass of the 130 g ball [tex](\(m_1\))[/tex]: 130 g

- Mass of the 230 g ball [tex](\(m_2\))[/tex]: 230 g

- Length of the rod [tex](\(r\)):[/tex] 34 cm = 0.34 m

- Initial angular velocity [tex](\(ω_{\text{initial}}\))[/tex]: [tex]\( ω_{\text{initial}} = \frac{2π \times 120}{60} \)[/tex]

Now, let's calculate the initial angular velocity:

[tex]\[ ω_{\text{initial}} = \frac{2π \times 120}{60} = 4π \, \text{rad/s} \][/tex]

Next, calculate the moment of inertia for each ball:

[tex]\[ I_1 = \frac{2}{5}m_1r^2 = \frac{2}{5} \times 0.13 \times (0.34)^2 \][/tex]

[tex]\[ I_2 = \frac{2}{5}m_2r^2 = \frac{2}{5} \times 0.23 \times (0.34)^2 \][/tex]

Now, calculate the total moment of inertia:

[tex]\[ I_{\text{total}} = I_1 + I_2 \][/tex]

Substitute the values into the conservation of angular momentum equation:

[tex]\[ I_{\text{total}} \times ω_{\text{initial}} = I_{\text{total}} \times ω_{\text{final}} \][/tex]

Solve for [tex]\(ω_{\text{final}}\)[/tex] and then use it to calculate the linear velocity [tex](\(v\))[/tex] of the 100 g ball:

[tex]\[ v = ω_{\text{final}} \times r \][/tex]

After performing these calculations, we can determine the speed of the 100 g ball. Let me provide you with the numerical results in the next response.

After performing the calculations, the final angular velocity [tex](\(ω_{\text{final}}\))[/tex] is determined to be the same as the initial angular velocity [tex](\(4π \, \text{rad/s}\))[/tex]. Now, we can calculate the linear velocity [tex](\(v\))[/tex] of the 100 g ball using the formula [tex]\(v = ω_{\text{final}} \times r\).[/tex]

[tex]\[ v = 4π \times 0.34 \][/tex]

[tex]\[ v \approx 4 \times 3.14 \times 0.34 \][/tex]

[tex]\[ v \approx 4.24 \, \text{m/s} \][/tex]

Therefore, the speed of the 100 g ball in this rotating system is approximately [tex]\(4.24 \, \text{m/s}\).[/tex]

Answer:

I think the answer is c

Explanation:

But it is kinda of opinionated

Statements C best describe a leader. A leader is a person who inspires other team members to reach the organizational goals

A leader is someone who motivates others to achieve the organization's objectives.

A leader is the person who owns the company and manages it on a daily basis. Any member of the management team might be considered a leader, but only a leader would be able to own the company.

A leader is a person who inspires other team members to reach the organizational goals

Hence,statements C best describe a leader.

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Answer:

We say that the object's spectrum is shifted.

Explanation:

Spectral lines will be shifted to the blue part of the spectrum1 if the source of the observed light is moving toward the observer, or to the red part of the spectrum when is moving away from the observer (that is known as the Doppler effect).

That shift can be used to find the velocity of the object by means of the Doppler velocity.

[tex]v = c\frac{\Delta \lambda}{\lambda_{0}}[/tex]  (1)

Where [tex]\Delta \lambda[/tex] is the wavelength shift, [tex]\lambda_{0}[/tex] is the wavelength at rest, v is the velocity of the source and c is the speed of light.

The object's spectrum is affected by the Doppler effect due to its motion away from us.

The apparent change in the wavelength of a spectral line from a distant object compared to the laboratory measurement is known as the Doppler effect. In this case, the spectral line appears at a longer wavelength (328 nm) in the spectrum of the distant object compared to the laboratory measurement (321 nm), indicating that the object is moving away from us.

The Doppler effect occurs because when an object moves away from an observer, the wavelength of the light it emits appears to increase, causing a shift toward the red end of the spectrum. On the other hand, if the object was moving toward us, the spectral line would appear at a shorter wavelength and shift toward the blue end of the spectrum.

This phenomenon is important in astronomy as it allows us to determine whether distant objects like stars are moving toward or away from us, which helps us understand their motion and the expansion of the universe.

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Answer:

The new electric field strength of this charge is half of the strength of the other charge.

Explanation:

The electric force acting on the charge particle is given by :

F = q E

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

Where

q is the charged particle

E is the electric field

If the magnitude of a charge is twice as much as another charge, q' = 2q, but the force experienced is the same, then the new electric field is given by :

[tex]E'=\dfrac{F}{q'}[/tex]

[tex]E'=\dfrac{F}{(2q)}[/tex]

[tex]E'=\dfrac{1}{2}\times \dfrac{F}{q}[/tex]

[tex]E'=\dfrac{E}{2}[/tex]

So, the new electric field strength of this charge is half of the strength of the other charge. Hence, this is the required solution.

Final answer:

The electric field strength of the charge with twice the magnitude is half the strength of the electric field of the other charge because the force experienced by both is equal.

Explanation:

If the magnitude of a charge is twice as much as another charge, but the force experienced by each is the same, then the electric field strength that the larger charge is in is half the strength of the other charge’s electric field. The electric field strength (E) at a point is defined as the force (F) experienced by a positive test charge (q) placed at that point divided by the magnitude of the charge itself: E = F/q.

If we have two charges, q1 and q2, where q2 is twice q1 (“q2 = 2 × q1”), and the forces on both charges are equal (“F1 = F2”), we can set up the following equations: E1 = F1/q1 and E2 = F2/q2. Combining our known relationships, we get E1 = F/q1 and E2 = F/(2 × q1), which simplifies to E1 = 2 × E2. Therefore, E2 is half the magnitude of E1, meaning that the electric field strength of the charge with twice the magnitude is half the strength of the electric field of the other charge.

Answer:

Weber's Law

Explanation:

Weber’s law is an important law in psychology.

This law quantifies the perception of change within a provided stimulus.

This law is also known as Weber-Fechner law as it relates two hypotheses in the field of psychophysics which are Weber law and the Fechner law.

Weber’s law describes that an observed change in a stimulus is a constant ratio of the original stimulus.

Final answer:

Weber's Law describes the relationship between a stimulus and the resulting sensation, stating that the Just Noticeable Difference (JND) is a constant fraction of stimulus intensity. It is a key principle in psychophysics, useful for experiments such as determining the JND for different weights of rice in bags.

Explanation:

The concept you're inquiring about is Weber's Law, which is a fundamental principle in the field of psychophysics, a branch of psychology. Weber's Law describes the relationship between a stimulus and its resulting sensation by proposing that the Just Noticeable Difference (JND) is a constant fraction of the stimulus intensity.

To put this into practice, for example, if one were testing the JND of various weights of rice in bags, they could use incremental percentage increases of the weight to determine the smallest difference that can be perceived. By choosing increments like 10 percent or 20 percent between certain weight thresholds, a researcher can apply Weber's Law to determine the JND in a standardized way.

This law helps explain why when you have a cup of coffee with little sugar, adding a teaspoon can make a noticeable difference, but in a cup with much more sugar, that same teaspoon becomes less detectable, requiring a proportionally larger amount of sugar to notice a change in sweetness.

Answer:

Velocity of rocket will be equal to 7.81 m/sec

Explanation:

We have given mass of the rocket [tex]m_1=4kg[/tex]

Mass of fuel is given [tex]m_2=50gram=0.05kg[/tex] ( As 1 kg is equal to 1000 gram )

Velocity of fuel [tex]v_2[/tex] = 625 m/sec

We have to find the velocity of rocket [tex]v_1[/tex]

From conservation of momentum we know that initial momentum is equal to final momentum '

So [tex]m_1v_1=m_2v_2[/tex], here [tex]m_1[/tex] is mass of rocket [tex]v_1[/tex] is velocity of rocket [tex]m_2[/tex] is mass of fuel and [tex]v_2[/tex] is velocity of fuel

So [tex]4\times v_1=625\times 0.05[/tex]

[tex]v_1=7.81m/sec[/tex]

So velocity of rocket will be equal to 7.81 m/sec

Answer:

1050 possible selections

Explanation:

Number of red balls = 7

Number of blue balls = 10

Let blue balls be B and red balls be R

four balls are to be selected. At least three must be blue.

the first combination is 4blue 0red

second combination is 3blue 1red

(10C4*7C0 + 10C3*7C1)

210 *1 +120 *7

210+840

=1050 possibilities

There are 71 different selections possible if at least 3 balls must be blue.

To find the number of different selections possible, we can consider the different scenarios where at least 3 balls are blue:

Summing up the number of selections from both scenarios, we get 70 + 1 = 71 different selections possible.

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Answer:

a.  [tex]W_w=313.5\ J[/tex]

b. [tex]W_g=-155.312\ J[/tex]

c. [tex]F_N=222.045\ N[/tex]

d.  [tex]W_t=313.5\ J[/tex]

Explanation:

Given:

a.

Work done by the worker who applied the force:

[tex]W_w=F.s\ cos 0^{\circ}[/tex] since the direction of force and the displacement are the same.

[tex]W_w=209\times 1.5[/tex]

[tex]W_w=313.5\ J[/tex]

b.

Work done by the gravitational force:

[tex]W_g=m.g\times h[/tex]

where:

g = acceleration due to gravity

h = the vertically downward displacement

Now, we find the height:

[tex]h=s\times sin\ \theta[/tex]

[tex]h=1.5\times sin\ 25^{\circ}[/tex]

[tex]h=0.634\ m[/tex]

So, the work done by the gravity:

[tex]W_g=25\times 9.8\times (-0.634)[/tex]  ∵direction of force and displacement are opposite.

[tex]W_g=-155.312\ J[/tex]

c.

The normal reaction force on the crate by the inclined surface:

[tex]F_N=m.g.cos\ \theta[/tex]

[tex]F_N=25\times 9.8\times cos\ 25[/tex]

[tex]F_N=222.045\ N[/tex]

d.

Total work done on crate is with respect to the worker: [tex]W_t=313.5\ J[/tex]

Answer:

[tex]MA=\frac{F_l}{F_e}>1[/tex]

Explanation:

The mechanical advantage with a shorter lever will always be greater than 1.

It is so because with a longer effort arm we need to apply lesser force to lift a unit mass which is at a shorter distance from the fulcrum. This is in accordance with the conservation of moments.

[tex]F_e\times d_e=F_l\times d_l[/tex]

where:

[tex]F_e=[/tex] force on the effort arm

[tex]F_l=[/tex] force on the load arm

[tex]d_e\ \&\ d_l[/tex] are the lengths of load effort arm and load arm respectively.

So, one factor balances the other keeping the product of the two constant.

And we know that mechanical advantage is :

[tex]MA=\frac{F_l}{F_e}[/tex]

Answer:

262 kN/C

Explanation:

If the electrons is moving parallel, thus it has a retiline movement, and because the velocity is varing, it's a retiline variated movement. Thus, the acceleration can be calculated by:

v² = v0² + 2aΔS

Where v0 is the initial velocity (2.0x10⁷ m/s), v is the final velocity (4.0x10⁷ m/s), and ΔS is the distance (1.3 cm = 0.013 m), so:

(4.0x10⁷)² = (2.0x10⁷)² + 2*a*0.013

16x10¹⁴ = 4x10¹⁴ + 0.026a

0.026a = 12x10¹⁴

a = 4.61x10¹⁶ m/s²

The electric force due to the electric field (E) is:

F = Eq

Where q is the charge of the electron (-1.602x10⁻¹⁹C). By Newton's second law:

F = m*a

Where m is the mass, so:

E*q = m*a

The mass of one electrons is 9.1x10⁻³¹ kg, thus, the module of electric field strenght (without the minus signal of the electron charge) is:

E*(1.602x10⁻¹⁹) = 9.1x10⁻³¹ * 4.61x10¹⁶

E = 261,866.42 N/C

E = 262 kN/C

Answer:

c. the tilt of the axis of rotation with respect to the Plane of the Ecliptic

Explanation:

The inclination of the ecliptic (or known only as obliqueness) refers to the angle of the axis of rotation with respect to a perpendicular to the plane of the eclipse. He is responsible for the seasons of the year that the planet Earth lends. It is not constant but changes through the movement of nutation. The terrestrial plane of Ecuador and the ecliptic intersect in a line that has an end at the point of Aries and at the diametrically opposite point of Libra.

When the Sun crosses the Aries, the spring equation occurs (between March 20 and 21, the beginning of spring in the northern hemisphere and the early autumn of the southern hemisphere), and from which the Sun is in the North Hemisphere; Pound until you reach the point of the autumn equinox (around September 22-23, beginning fall in the northern hemisphere and spring in the southern hemisphere).

Answer:

Tribe historians would: Depict a significant event for each year on a buffalo or deer skin

Tribes of the Sioux Nation maintained historical calendars composed of winter counts, using pictorial representations to record significant events. Winter counts served as a visual record of the tribe's history and were passed down through generations.

Tribes of the Sioux Nation (among other Plains Indians) maintained historical calendars composed of winter counts. Tribe historians would use pictorial representations to record significant events that occurred during each year. Each year, a new pictorial symbol would be added to the count. For example, if a significant battle occurred during a year, the historian would draw a symbol representing that battle. Winter counts served as a visual record of the tribe's history and were often passed down from generation to generation.

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Answer: 0.006in/s

Explanation:

Let the rate at which air is being blown into a spherical balloon be dV/dt which is 1.68in³/s

Also let the rate at which the radius of the balloon is increasing be dr/dt

Given r = 4.7in and Π = 3.14

Applying the chain rule method

dV/dt = dV/dr × dr/dt

If the volume of the sphere is 4/3Πr³

V = 4/3Πr³

dV/dr = 4Πr²

If r = 4.7in

dV/dr = 4Π(4.7)²

dV/dr = 277.45in²

Therefore;

1.68 = 277.45 × dr/dt

dr/dt = 1.68/277.45

dr/dt = 0.006in/s

Answer:

Explanation:

There are two prevailing theories: One is that the Earth held onto some water when it formed, as there would have been ice in the nebula of gas and dust (called the proto-solar nebula) that eventually formed the sun and the planets about 4.5 billion years ago. Some of that water has remained with the Earth, and might be recycled through the planet's mantle layer, according to one theory.

The second theory holds that the Earth, Venus, Mars and Mercury would have been close enough to that proto-solar nebula that most of their water would have been vaporized by heat; these planets would have formed with little water in their rocks. In Earth's case, even more water would have been vaporized when the collision that formed the moon happened. In this scenario, instead of being home-grown, the oceans would have been delivered by ice-rich asteroids, called carbonaceous chondrites.

Two theories explain the formation of Earth's hydrosphere: water from interstellar grains and water from comets and asteroids.

There are two competing scientific theories supported by evidence that explain the formation of Earth's hydrosphere:

Water from comets and asteroids: Another theory suggests that the water may have been brought to Earth when comets and asteroids impacted it. Scientists estimate that comet impacts during Earth's early years could have contributed enough water to account for what we see today.

Traction loss in a vehicle's rear wheels is most likely due to either acceleration-induced traction loss or rear wheel traction loss (skid). The former is caused by rapid acceleration, while the latter can be due to turning or braking.

The traction loss described in your question occurs in the rear wheels of a vehicle and is likely due to acceleration-induced traction loss or rear wheel traction loss (skid). Acceleration-induced traction loss happens when rapid acceleration causes the tires to lose grip on the road. This is common in high-powered, rear-wheel drive vehicles. Rear wheel traction loss or a rear wheel skid, on the other hand, generally occurs when the rear wheels lose grip during turning or braking. In this scenario, the vehicle's rear end can swing out in either direction, causing the vehicle to spin.

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Final answer:

Rear wheel traction loss (skid) occurs when the rear tires lose grip on the road surface. In a front-wheel-driven car, friction at the rear wheel is in the opposite direction of motion. This concept is important for analysis in vehicle dynamics and skid prevention.

Explanation:

The traction loss referred to in the question occurs in the rear wheels of a vehicle and is specifically known as rear wheel traction loss (skid). This is a type of skid that happens when the rear tires of a vehicle lose grip on the road surface, often as a result of oversteering, sudden acceleration, or slippery conditions. It's essential to distinguish this from other types of traction loss, such as braking-induced traction loss, which occurs when the tires can no longer grip the road during heavy braking, or driver-induced skids, which are the result of driver error.

With respect to the problem provided, in a front-wheel-driven car, the friction at the rear wheel is generally in the opposite direction of motion. This is because the rear wheels are not powered in this configuration; instead, they follow the rotation induced by the car's movement, which is primarily driven by the front wheels. Therefore, while the front wheels are pulled forward by the engine, creating friction in the direction of motion, the rear wheels experience a kind of rolling resistance and the friction present there acts in the opposite direction of the car's movement.

Understanding the dynamics of friction and traction is critical for analyzing cases like where a vehicle skids and comes to rest after traveling a certain distance or determining the force required to maintain a constant speed in the presence of kinetic friction. These analyses involve concepts like travel reduction ratio, torque transferred to the wheel axle, tractive efficiency of the wheel, and the actual velocity of the wheel.

The angular velocity of a solid uniform sphere rolling without slipping down an inclined plane does not depend on the mass or the radius of the sphere; it depends on the height from which it rolls down and gravitational acceleration only.

The question asks what determines the angular velocity of a solid uniform sphere as it rolls without slipping down an inclined plane. It specifically queries whether this angular velocity depends on the mass (M) of the sphere, the radius (R) of the sphere, both, or neither.

The answer is that the angular velocity of a rolling sphere at the bottom of the incline does not depend on the mass of the sphere or the radius of the sphere. According to the principles of conservation of energy and the dynamics of rolling motion, all objects regardless of their mass or radius, will roll down an incline and reach the bottom with the same angular velocity if they start from rest, provided that they do not slip and there is no air resistance. This is because the potential energy lost is completely converted into kinetic energy (both translational and rotational), and the mechanical energy of the system remains constant.

The angular velocity of the sphere at the bottom of the incline depends on d) neither the mass nor the radius of the sphere.

When the sphere rolls without slipping, there is a relationship between its linear velocity (v) and angular velocity (ω) given by the equation v = ωR.

Total mechanical energy of the sphere = translational kinetic energy + rotational kinetic energy + potential energy.

At the bottom of the incline, all potential energy will have converted into kinetic energy:

By energy conservation:

Substituting I and the relation v = ωR gives us:

Solving for ω:

From the above steps, it's evident that the angular velocity ω depends on the height h the sphere rolls down, and the acceleration due to gravity g, but not on the mass M or radius R of the sphere.

Therefore, the correct answer is: d) neither the mass nor the radius of the sphere.

Answer:

A) Increasing,  Increasing,  Increasing

Explanation:

The greater the rate of fuel ejection higher will be the kinetic energy of the rocket. As rocket is fire upward its fuel rejection rate increases and hence it's kinetic energy increases.

Gravitational potential energy,

as the rocket moves further it's r from the Earth increases. Hence the Gravitational potential energy increases as its height from the Earth increases.

Therefore,  mechanical energy of the rocket must also increase as it is sum of kinetic and potential energy.

Hence Option A is correct.

For the first kilometer of the rocket's ascent, the kinetic energy, gravitational potential energy, and mechanical energy of the Earth-rocket system increase.

The changes in the kinetic energy, gravitational potential energy, and mechanical energy of the Earth-rocket system for the first kilometer of the rocket's ascent can be determined.

Since the mass of fuel ejected is small compared to the mass of the rocket, the rocket's kinetic energy will increase as it accelerates away from Earth. The gravitational potential energy of the Earth-rocket system will also increase as the rocket moves higher against the pull of gravity. Therefore, the changes in the three quantities are as follows:

Answer:

a) 200m

b) 40 m/s

c) 81.55m

d) 31.62m

Explanation:

Solution

a)

y = y0 + u×t+⅟2×a×t2 =

y0 = 0

u = 0

y = unknown

a = 4m/s2

t = time = 10 seconds

y = 0.5×4×102 = 200m

b) v = u + at

v = 0 + 4×10 = 40 m/s

c) v2 = u2 - 2×g×y

at maximum height  v = 0

we have

402 = 2×9.81×y

Y =81.55m

d)

for the stage 2 we haace

 y = y0 + u×t+⅟2×a×t2 =

y = 0 + 4×2+0.5×6×22 = 92m

for the stage one we have

y = 0+40×2-0.5×9.81×4= 60.38m

distance between the first and second stage  2s aftee separation = 92-60.38 = 31.62m

Answer:

374.39 J/K

Explanation:

Entropy: This can be defined as the degree of disorder or randomness of a substance.

The S.I unit of entropy is J/K

ΔS = ΔH/T ..................................... Equation 1

Where ΔS = entropy change, ΔH = Heat change, T = temperature.

ΔH = cm................................... Equation 2

Where,

c = specific latent heat of fusion of water = 333000 J/kg, m = mass of ice = 0.3071 kg.

Substitute into equation 2

ΔH = 333000×0.3071

ΔH = 102264.3 J.

Also, T = 273.15 K

Substitute into equation 1

ΔS = 102264.3/273.15

ΔS = 374.39 J/K

Thus, The change in entropy = 374.39 J/K

The change in entropy of the gas is 374.39 J/K.

The given parameters:

The heat of fusion of the ice is calculated as follows;

[tex]\Delta H = mL\\\\\Delta H = 0.3071 \times 333,000\\\\\Delta H = 102,264.3 \ J[/tex]

The change in entropy of the gas is calculated as follows;

[tex]\Delta S = \frac{\Delta H}{T} \\\\\Delta S = \frac{102,264.3}{273.15} \\\\\Delta S = 374.39 \ J/K[/tex]

Thus, the change in entropy of the gas is 374.39 J/K.

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Answer:

Range, [tex]R = MV²/2QE[/tex]

Explanation:

The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.

Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses  all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.

So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops

Therefore {tex}R = MV²/2QE{/tex}

Answer:

The Tension T is 42120N

The Horizontal force component is 18322.2N

The Vertical force component is - 4729N

Explanation:

First, you have to find the angle between the drawbridge and the cable using sine and cosine rule. This will result in angle 44.2°. Hence, the angle between the horizontal axis and the cable will be 64.2° (44.2° + 20°).

Having done that, you apply two conditions of equilibrium.

1. THE VECTOR SUM OF ALL FORCES EQUAL ZERO.

∑Fx = 0

∑Fx = Rx - Tcos64.2 = 0

Rx = 0.435T

∑Fy = 0

∑Fy = Ry + Tsin64.2 - W - w = 0

W = 2000kg × 9.8 = 19600N

w =1000kg × 9.8 = 9800N

Ry + 0.9T = 29400N

Ry = 29400 - 0.9T

2. THE SUM TOTAL OF TORQUES EQUALS ZERO

Rx: τ = 0

Ry: τ = 0

T: τ = 5 × Tsin44.2

= 3.49T m

W: τ = 4 × 19600sin90

= 78400Nm

w: τ = 7 × 9800sin9

= 68600Nm

Note:

Rx = x component of Reaction force

Ry = y component of Reaction force.

T = Tension

W = weight of bridge

w = weight of Sir Lance a Lost and his steed

τ = torque

Note: The torque of Tension is counter clockwise while that of the weights is clockwise.

Hence,

∑τccw = ∑τcw

3.49T = 78400 + 68600

3.49T = 14700Nm

T = 147000/3.49

T = 42120N

Rx = 0.435 × 42120

Rx = 18322.2N

Ry = 29400N - (0.9×42120)N

Ry = 29400 - 34129

Ry = -4729N

Note: Ry being negative means that the hinge of the drawbridge exerts a downward force.

In this exercise we have to use the knowledge of tension and force, so we can say that this will result in:

A)The Tension T is 42120N

B)The Horizontal force component is 18322.2N

C)The Vertical force component is - 4729N

Before starting the calculations we have to remember some concepts such as:

First we will add all the force vectors so that it results in zero and that means that the body is in equilibrium, so:

[tex]\sum F_x = 0\\ \sum F_x = R_x - Tcos(64.2) = 0\\ R_x = 0.435T\\ \sum F_y = 0\\ \sum F_y = R_y + Tsin(64.2) - W - w = 0\\ W = 2000kg * 9.8 = 19600N\\ w =1000kg * 9.8 = 9800N\\ R_y + 0.9T = 29400N\\ R_y = 29400 - 0.9T[/tex]

Secondly, we will add up all the torques so that it results in zero and that means that the body is in equilibrium, so:

[tex]T: T = 5 * Tsin(44.2) = 3.49T m\\ W: T = 4 * 19600sin(90)= 78400Nm\\ w: T = 7 * 9800sin(90) = 68600Nm\\ \sum Tccw = \sum Tcw\\ 3.49T = 78400 + 68600\\ 3.49T = 14700Nm\\ T = 147000/3.49\\ T = 42120N R_x = 0.435 * 42120\\ R_x = 18322.2N\\ R_y = 29400N - (0.9*42120)N\\ R_y = 29400 - 34129\\ R_y = -4729N[/tex]

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Answer:

[tex]u_x=8.5454\ m.s^{-1}[/tex]

[tex]d=9.5782+16=25.5782\ m[/tex] was the her position ahead from her window when she caught the object.

Explanation:

Given:

[tex]r=3.2\times 5[/tex]

[tex]r=16\ m[/tex]

Since the projectile was thrown horizontally therefore the vertical component of velocity is zero.

Now the time taken for the object to reach to Henrietta ignoring the height where she catches:

[tex]h=u_x.t+\frac{1}{2} \times g.t^2[/tex]

[tex]43.9=0+\frac{1}{2} \times 9.8\times t^2[/tex]

[tex]t=2.9932\ s[/tex] is the time taken by the projectile to reach Henrietta.

Now the distance further walked by Henrietta in the above time from the point where she was when the projectile was launched:

[tex]\Delta d=v_w\times t[/tex]

[tex]\Delta d=9.5782\ m[/tex]

Now the total  horizontal distance from the window to Henrietta when the projectile reached her:

[tex]d=\Delta d+r[/tex]

[tex]d=9.5782+16=25.5782\ m[/tex] was the her position ahead from her window when she caught the object.

Now the initial horizontal velocity of launch of the projectile:

[tex]u_x=\frac{d}{t}[/tex]

[tex]u_x=\frac{25.5782}{2.9932}[/tex]

[tex]u_x=8.5454\ m.s^{-1}[/tex]

Answer:

b. fluid friction occurs

Explanation:

Water is a liquid and both going through air gas and liquid cause fluid friction.

Answer:

Initial concentration of the reactant = 3.34 × 10^(-2)M

Explanation:

Rate of reaction = 2.30×10−4 M/s,

Time of reaction = 80s

Final concentration = 1.50×10−2 M

Initial concentration = Rate of reaction × Time of reaction + Final concentration

= 2.30×10−4 M/s × 80s + 1.50×10−2 M = 3.34 × 10^(-2)M

Initial concentration = 3.34 × 10^(-2)M

Explanation:

Formula for final volume of chamber if the partition is ruptured will be as follows.

        [tex]V_{2}[/tex] = 1.5 + 1.5

                   = 3.0 [tex]ft^{3}[/tex]

As mass remains constant then the specific volume at this state will be as follows.

             [tex]\nu_{2} = \frac{V_{2}}{m}[/tex]

                          = [tex]\frac{3.0}{2}[/tex]

                          = 1.5 [tex]ft^{3}/lbm[/tex]

Now, at final temperature [tex]T_{2}[/tex] = 300 F according to saturated water tables.

   [tex]\nu_{f} = 0.01745 ft^{3}/lbm[/tex]

   [tex]\nu_{fg} = 6.4537 ft^{3}/lbm[/tex]  

   [tex]\nu_{g} = 6.47115 ft^{3}/lbm[/tex]

Hence, we obtained [tex]\nu_{f} < \nu_{2} < \nu_{g}[/tex] and the state is in wet condition.

       [tex]\nu_{2} = \nu_{f} + x_{2}\nu_{fg}[/tex]

             1.5 = [tex]0.01745 + x_{2} \times 6.4537[/tex]

        [tex]x_{2}[/tex] = 0.229

Now, the final pressure will be the saturation pressure at [tex]T_{2}[/tex] = 300 F

and,   [tex]P_{2}[/tex] = [tex]P_{sat}[/tex] = 66.985 psia

Formula to calculate internal energy at the final state is as follows.

         [tex]U_{2} = m(u_{f}_{300 F} + x_{2}u_{fg_{300 F}}[/tex]

                   = [tex]2(269.51 + 0.229 \times 830.45)[/tex]

                   = 920.56 Btu

Therefore, we can conclude that the final pressure of water, in psia is 66.985 psia and total internal energy, in Btu, at the final state is 920.56 Btu.

Answer:

Ep= 3.8 10⁵ N/C

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Equivalence

1nC= 10⁻⁹C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

q₁ =+7.5 nC = +7.5*10⁻⁹C  

q₂ =  -2.0 nC = -2.0*10⁻⁹C

d₁ =d₂ = 1.5cm = 1.5 *10⁻²m  = 0.015 m

Calculation of the electric fieldsat the midpoint (P) between the two charges

Look at the attached graphic:

E₁: Electric Field at point ;Due to charge q₁. As the charge q₁ is positive negative (q₁+), the field leaves the charge .

E₂: Electric Field at point : Due to charge q₂. As the charge q₂ is negative (q₂-) ,the field enters the charge

E₁ = k*q₁/d₁² = 9*10⁹ *7.5  *10⁻⁹/ ( 0.015 )² = 3*10⁵ N/C

E₂ = k*q₂/d₂²= 9*10⁹ *2*10⁻⁹/( 0.015 )² = 0.8*10⁵ N/C

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Ep= E₁ + E₂  

Ep= 3*10⁵ N/C +  0.8*10⁵ N/C

Ep= 3.8 10⁵ N/C

The electric field strength at the midpoint between a +7.5 nC point charge and a -2.0 nC point charge that are 3.0 cm apart is 1.56 *10^6 N/C (Newtons per Coulomb) away from the positive charge.

The question is asking for the electric field at the midpoint between two point charges. The formula for the electric field created by a point charge is given by E=kQ/r^2, where E is the electric field, k is Coulomb's constant (approximately 9.0 * 10^9 N*m^2/C^2), Q is the charge, and r is the distance from the charge. In this scenario we have two charges, so we calculate the electric field created by each charge and sum the results.

Applying this to both point charges and summing the electric fields we obtain: E_total = |E1| + |E2| = k*|Q1|/d^2 + k*|Q2|/d^2 = ((9.0*10^9 N*m^2/C^2) * 7.5 *10^-9 C/0.015m^2) + ((9.0*10^9 N*m^2/C^2) * 2 *10^-9 C/ 0.015m^2) = 1.8 *10^6 N/C and -0.24 * 10^6 N/C respectively. The total electric field strength at the midpoint is the sum which equals 1.56 *10^6 N/C away from the positive charge.

Learn more about Electric Field Strength here:

https://brainly.com/question/1812671

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Answer:

39.8 m ≈ 40 m

Explanation:

power (P) = 50 W

sound intensity level ([tex]p[/tex]) = 94 dB

the distance (r) can be gotten from the equation I = [tex]\frac{power}{4nr^{2} }[/tex] (take not that π is shown as [tex]n[/tex])

making r the subject of the formula we have r = [tex]\sqrt{\frac{power}{4nI} }[/tex]   (take not that π is shown as [tex]n[/tex])

But to apply this equation we need to get the value of the intensity (I)

now that we have the value of the intensity (I)  we can substitute it into the formula for the distance (r)

distance (r) = [tex]\sqrt{\frac{power}{4nI} }[/tex]

r = [tex]\sqrt{\frac{50}{4x3.142x0.00251} }[/tex] = 39.8 m ≈ 40 m

Answer:

57.19461 m/s²

-9.20833 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Equation of motion

[tex]v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{\dfrac{663}{3.6}-0}{3.22}\\\Rightarrow a=57.19461\ m/s^2[/tex]

The acceleration during the race is 57.19461 m/s²

[tex]v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{0-\dfrac{663}{3.6}}{20}\\\Rightarrow a=-9.20833\ m/s^2[/tex]

The acceleration while stopping is -9.20833 m/s²

The assumptions made here are:

The initial velocity of the dragster at the start of the race is zero

The acceleration is constant in both cases of acceleration.

The final velocity when it stops is zero

Motion is in a straight line path

Answer: a) 42Nm b) 8.4m/s

Explanation:

Impulse is defined as object change in momentum.

Since Force = mass × acceleration

F = ma

Acceleration is the rate of change in velocity.

F = m(v-u)/t

Cross multiply

Ft = m(v-u)

Since impulse = Ft

and Ft = m(v-u)... (1)

The object change in velocity (v-u) = Ft/m from eqn 1

Going to the question;

a) Impulse = Force (F) × time(t)

Given force = 14N and time = 3seconds

Impulse = 14×3

Impulse = 42Nm

b) The object change in velocity (v-u) = Ft/m where mass = 5kg

v-u = 14×3/5

Change in velocity = 42/5 = 8.4m/s