For general projectile motion, which of the following best describes the horizontal component of a projectile's acceleration? Assume air resistance is negligible.

-The horizontal component of a projectile's acceleration continually decreases.
-The horizontal component of a projectile's acceleration is zero.
-The horizontal component of a projectile's acceleration remains a nonzero constant.
-The horizontal component of a projectile's acceleration continually decreases.
-The horizontal component of a projectile's acceleration initially decreases and then increases.

Final answer:

There are at least two fundamental action-reaction pairs of forces when a girl stands on a sofa, including the force of gravity and the normal force, as well as any frictional forces if there is lateral movement.

Explanation:

Action-Reaction Pairs

In the scenario where a girl stands on a sofa, there are several action-reaction pairs of forces at play. According to Newton's third law, every action has an equal and opposite reaction. When the girl exerts a force on the sofa (due to her weight), the sofa exerts an equal and opposite force back onto the girl. This is one pair. Additionally, if the girl were to push on the sofa to stand up or adjust her position, the sofa would push back with an equal force in the opposite direction, forming another action-reaction pair.

The number of action-reaction pairs can vary based on what specific interactions are being considered. Generally, in this scenario, we consider at least two fundamental pairs: the force of gravity on the girl (action) and the normal force from the sofa on the girl (reaction), and if there's any lateral movement while standing, there would be a frictional force between the sofa and the girl's feet (action), and an equal and opposite frictional force from the girl's feet on the sofa (reaction).

The semimajor axis is equal to 270 Km. The eccentricity of the orbit is 0.33 Km.

An elliptic orbit or elliptical orbit can be described as a Kepler orbit with an eccentricity of less than 1. It is a circular orbit with an eccentricity equal to 0. It is a Kepler orbit with an eccentricity greater than 0 and less than 1 and a Kepler's orbit with negative energy. The radial elliptic orbit has an eccentricity equal to 1.

Given the satellite, moving in an elliptical orbit has the farthest point from the  above earth's surface, [tex]R_a[/tex] = 360 Km

The closest point from the  above earth's surface, [tex]R_p[/tex] = 180 Km

Semi-major axis can be calculated as , [tex]{\displaystyle a = \frac{R_a+R_p}{2}[/tex]

[tex]{\displaystyle a = \frac{360 + 180}{2}[/tex]

a = 270 km

The eccentricity of the elliptical orbit, we can calculate as shown below:

[tex]{\displaystyle e = \frac{a- R_p}{a}[/tex]

e = (270 - 180)/270

e = 1/3

e = 0.33 km

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Answer:a) a force acting on an object in motion causes it to stop moving

Explanation: gradpoint

Answer: Option (c) is the correct answer.

Explanation:

Compounds which contain carbon and hydrogen atoms are known as organic compounds.

As general chemical formula of alcohols is R-OH, where R = any alkyl or aryl group.

For example, [tex]CH_{2}CH_{2}OH[/tex] is known as ethanol and it is an alcohol as it contains the functional group "-OH".

So, alcohols contains elements carbon, hydrogen and oxygen.

Thus, we can conclude that alcohols are organic compounds that contain carbon, oxygen, and hydrogen.

The force does 195 Joules of work on the object.

To calculate the work done by a force, we use the formula:

Work = Force x Distance x cos(theta)

Given that the force is 25 N, the angle is 30 degrees, and the distance is 3 m, we can substitute these values into the formula to calculate the work:

Work = 25 N x 3 m x cos(30 degrees) = 195 J

Therefore, the force does 195 Joules of work on the object.

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The expression for the linear expression of the two objects will be

[tex]a=\dfrac{m_1g}{m_2+m_1-M}[/tex]

Acceleration is defined as the change of the velocity with the time. Acceleration is a vector quantity and is defined by both the magnitude and the direction.

The mass of sphere is m1 radius is R and mass m2 is connected by light chord.

Now by using free body diagram

The tension in the chord will be

[tex]m_1g-T_1=m_1a[/tex]

[tex]T_1=m_1(g-a)[/tex]

For tension T2 we will have

[tex]T_2=m_2a[/tex]

Now the torque will be

[tex]T_{net}=I\alpha[/tex]

[tex]T_2R-T_1R=I(\dfrac{a}{R})[/tex]

[tex]T_2R-T_1R=MR^2(\dfrac{a}{R})[/tex]

[tex]T_2-T_1=Ma[/tex]

[tex]m_2a-m_1(g-a)=Ma[/tex]

[tex]m_2a-m_1g+m_1a=Ma[/tex]

[tex]a(m_2+m_1-M)=m_1g[/tex]

[tex]a=\dfrac{m_1g}{m_2+m_1-M}[/tex]

Thus the expression for the linear expression of the two objects will be

[tex]a=\dfrac{m_1g}{m_2+m_1-M}[/tex]

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In Texas, the speed limit in urban districts is generally set at 30 mph unless otherwise posted. This means that unless there are signs indicating a different speed limit, drivers are expected to adhere to a maximum speed of 30 miles per hour when traveling through urban areas. Therefore, the correct answer is A. 30 mph.

The rationale behind setting a default speed limit for urban districts is to promote safety for both motorists and pedestrians. Urban areas typically have higher population densities, increased pedestrian activity, and more complex traffic patterns compared to rural or suburban areas. As a result, lower speed limits help reduce the risk of accidents, particularly those involving pedestrians and cyclists.

By setting a default speed limit of 30 mph in urban districts, Texas law aims to create safer environments for all road users. However, it's essential for drivers to remain vigilant and observant of posted speed limit signs, as speed limits may vary depending on factors such as road conditions, proximity to schools or residential areas, and construction zones.

Overall, the default speed limit of 30 mph in urban districts reflects a balance between ensuring efficient traffic flow and prioritizing the safety of all individuals on the road. Therefore, the correct answer is A. 30 mph.

The complete question is:

In Texas, the speed limit in urban districts is _______ unless otherwise posted.

A. 30 mph

B. 50 mph

C. 40 mph

D. 25 mph

Assuming pulley is frictionless. Let the tension be ‘T’. See equation below.

The expression for the acceleration of the red block after it is released from the rest is

[tex]\fbox{\begin{minispace}\\{a = \dfrac{{\left( {{m_g} - \mu {m_r}} \right)g}}{\left( {{m_g} + {m_r}} \right)}}\end{minispace}}[/tex]

Further Explanation:

The friction force acting on the red block placed on the table is acting opposite to the direction of motion of block. The motion of the system, of red block and grey block, is vertically downward direction. It implies that the tension on the string is less than the weight on the grey block.

Given:

The mass of the red block is [tex]{m_r}[/tex].

The mass of the grey block is [tex]{m_g}[/tex].

The coefficient of friction between red block and table is [tex]\mu[/tex].

Concept:

The net force on the red block is acting along the direction of motion of the system and is equal to the difference between the tension in the string and friction force.

The net force on the red block is:

[tex]{m_r}a = T - f[/tex]

Rearrange the above expression.

[tex]T = {m_r}a + f[/tex]

Substitute [tex]\mu {m_r}g[/tex] for [tex]f[/tex] in above equation.

[tex]\fbox{\begin\\T = {m_r}a + \mu {m_r}g\end{minispace}}[/tex]                                                           …… (1)

Here,  [tex]T[/tex] is the tension on the string, [tex]{m_r}[/tex] is the mass of the red block, [tex]a[/tex] is the acceleration of the system and [tex]g[/tex] is the acceleration due to gravity.

The net force on the grey block is along the direction of motion of the system and it is equal to the difference between the weight on grey block and tension on the string.

The net force on the grey block is:

[tex]{m_g}a = {m_g}g - T[/tex]                                                                        …… (2)

Here, [tex]{m_g}[/tex] is the mass of the grey block.

Substitute [tex]{m_r}a + \mu {m_r}g[/tex] for [tex]T[/tex] in equation (2).

[tex]{m_g}a = {m_g}g - \left( {{m_r}a + \mu {m_r}g} \right)[/tex]

Rearrange the above expression.

[tex]{m_g}a + {m_r}a = {m_g}g - \mu {m_r}g[/tex]

Simplify the above expression.

[tex]\fbox{\begin{minispace}\\{a = \dfrac{{\left( {{m_g} - \mu {m_r}} \right)g}}{\left( {{m_g} + {m_r}} \right)}}\end{minispace}}[/tex]

Learn more:

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2.  The motion of a body under friction brainly.com/question/4033012

3. A ball falling under the acceleration due to gravity brainly.com/question/10934170

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Acceleration, friction, tension, system of two blocks, pulley mass system, motion, net force, string, block placed on the table, block hanging through a pulley, block is released from rest.

Answer:

2.19 m/s

Explanation:

The momentum (p) is the product of the mass (m) and velocity (v) of an object.

p = m × v

The momentum of a 0.0075 kg bullet traveling at a speed of 365 m/s is:

p(bullet) = 0.0075 kg × 365 m/s = 2.74 kg.m/s

We want the 1.25 kg-coffee cup to have the same momentum than the bullet.

p(cup) = p(bullet) = 2.74 kg.m/s

p (cup) = m(cup) × v(cup)

2.74 kg.m/s = 1.25 kg × v(cup)

v(cup) = 2.19 m/s

As per the statement, Manny walks about 3 miles and the reference point for calculating the distance is the same as the starting point.  

Hence the option B is correct.

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Answer:

The difference in the temperatures between the hot and cold reservoirs determined how well a heat engine would work

Explanation:

I got a 100% on my test

To make cars and airplanes lighter, engineers use materials with high strength-to-weight ratios and apply aerodynamic principles to reduce drag. The balance of strength versus mass is resolved through materials selection and careful design optimization, including computer simulations and wind tunnel tests to create vehicles that are strong, lightweight, and efficient.

When designing cars or airplanes, the challenge is to reduce the mass of the vehicle while maintaining structural integrity. One way to achieve a lighter car is to incorporate plastics instead of metal in parts of its construction. For airplanes, advanced composite materials such as carbon fiber are often used because of their strength-to-weight ratios. Aerodynamics play a critical role in this process, as reducing drag through aerodynamic shaping can not only decrease fuel consumption but also reduce the required structural mass.

The balance between strength and mass is resolved through the use of advanced materials and design strategies. Modern design approaches involve analyzing aerodynamic forces and optimizing the shape and structures to endure these forces. With the help of computer simulations and wind tunnel testing, engineers can fine-tune designs to achieve the desired balance. The selection of materials follows a criterion that ensures they are lightweight yet able to withstand the high stresses experienced during operation. Often, the design includes failsafes that account for the worst expected loads.

Ultimately, the goal is to meet specific design objectives that include performance metrics like range, payload capacity, and fuel efficiency. Designers look at comparator aircraft or vehicles and use their data as references for optimizing wing loading, thrust-to-weight ratio, and other design parameters. Thus, the optimal design is often a trade-off between various competing factors including mass, strength, and aerodynamic efficiency.

Remember that the total velocity of the motion is the vector sum of the velocity you would have in still water and the stream. Always place the vectors carefully to be able to come up with an accurate sum vector.

Final answer:

When the ice on a mountainside melts, dirt and rocks that were held in place by the ice become unstable and may move downhill, altering the landscape. Saturated soil can cause mud flows and contribute to erosion. Glaciers also play a significant role in sculpting landscapes by carrying rock and sediment as they move.

Explanation:

When ice on a mountainside melts, the dirt and rocks that were previously held in place by the ice may become less stable. Erosion processes, such as water runoff and gravity, can lead to increased movement of these materials. As the water from the melting ice saturates the mountain's soil and rock, it can trigger debris flows or mud flows. These events have the potential to transport large amounts of material downslope, often resulting in significant changes to the landscape. The movement of this material can lead to the formation of new features, such as valleys or depressions, and can also contribute to the reshaping or smoothing of existing mountainous terrain.

It's also worth noting that glaciers, which are large bodies of ice that move slowly over land, have historically been powerful agents of erosion. As they advance and retreat, glaciers grind down rocks and carry sediment far from its origin, which is evidence of their strong impact on shaping the landscapes.

Final Answer:

The daredevil must have a minimum speed of approximately 14.7 m/s to clear the canyon.

Explanation:

To clear the canyon, the daredevil's vertical displacement at the edge of the canyon must be equal to the width of the canyon. Using the kinematic equation for vertical motion [tex]\( y = v_0t + \frac{1}{2}gt^2 \)[/tex], where y is the vertical displacement, [tex]\( v_0 \)[/tex] is the initial vertical velocity, g is the acceleration due to gravity, and t is the time of flight, and solving for [tex]\( v_0 \)[/tex], we get [tex]\( v_0 = \sqrt{2gy} \)[/tex]. Substituting the given values [tex]\( g = 9.8 \, \text{m/s}^2 \) and \( y = 8.88 \, \text{m} \)[/tex], we find [tex]\( v_0 \approx 14.7 \, \text{m/s} \)[/tex], which is the minimum speed required to clear the canyon.

It takes 0.322 s to give the merry-go-round an angular velocity of 1.43 rad/s

[tex]\texttt{ }[/tex]

Centripetal Acceleration can be formulated as follows:

[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

[tex]\texttt{ }[/tex]

Centripetal Force can be formulated as follows:

[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

moment of inertia = I = 84.4 kg.m²

initial angular velocity = ωo = 0 rad/s

angular acceleration = α = 4.44 rad/s²

final angular velocity = ω = 1.43 rad/s

Asked:

time taken = t = ?

Solution:

[tex]\omega = \omega_o + \alpha t[/tex]

[tex]1.43 = 0 + 4.44t[/tex]

[tex]1.43 = 4.44t[/tex]

[tex]t = 1.43 \div 4.44[/tex]

[tex]t = 0.322 \texttt{ s}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Circular Motion

[tex]\texttt{ }[/tex]

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

First let us calculate the acceleration of the car.

The net force is equivalent to the frictional force:

m a = µk  m g

a = µk g

a = 0.550 * 9.8 m/s^2

a = 5.39 m/s^2 (negative direction) = -5.39 m/s^2

Then we use the equation:

v^2 = vi^2 + 2 a d

where v is final speed or velocity, vi is initial speed = 60 mph = 26.8224 m/s, d is distance

v^2 = (26.8224 m/s)^2 + 2 (-5.39 m/s^2) * 57 m

v^2 = 104.98 m^2/s^2

v = 10.25 m/s       (ANSWER)

The compression forces acting on both sides of the center board = 70.338 N

The force acting on a system with static equilibrium is 0

[tex] \large {\boxed {\bold {\sum F = 0}} [/tex]

(forces acting as translational motion only, not including rotational forces)

[tex] \displaystyle \sum F_x = 0 \\\\\ sum F_y = 0 [/tex]

For objects undergoing rotation, the equilibrium must be met

[tex] \large {\boxed {\bold {\sum \tau = 0}} [/tex]

The force acting on the touchpad between the two fields is called the normal force  (N)

While the frictional force arises because of 2 objects that come into direct contact, especially when there is motion between two objects that are in direct contact.

There are 2 friction forces

The board sandwiched between two other boards will not fall if

a downward force = an upward force

or ΣFy = 0 (idle / equilibrium)

so that

A downward force = weight of the board sandwiched = 87.5 N

An upward force = arises from two static friction forces namely: us.N (N =  reaction force of the compression forces=F) ---> us.F

then the force that works can be stated:

EFy = 0

W = 2fs

W = 2 (us.F)

87.5 N = 2 (0.622. F)

F = 70,338 N

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Answer:

Work done, [tex]W=1.03\times 10^6\ J[/tex]

Explanation:

It is given that,

Mass of the car, m = 950 kg

Distance up to the incline, d = 710 m

Angle of inclination of the car is 9 degrees.

When the car is push in the inclined path, the vertical component of the force will act on it such that the work done is given by :

[tex]W=mg\ sin\theta\times d[/tex]

[tex]W=950\times 9.8\times \ sin(9) \times 710[/tex]  

[tex]W=1.03\times 10^6\ J[/tex]

So, the minimum work needed to push a car is [tex]1.03\times 10^6\ J[/tex]. Hence, this is the required solution.

The minimum work needed to push a car up an incline can be found by calculating the work done against gravity. This is done by multiplying the mass of the car by the gravitational acceleration and the height, which can be deduced from the sin of the incline angle and the distance.

The question is asking for the minimum work done to push a car up an inclined plane. To solve this, you use the concept of work done against gravity. The formula for work done against gravity is Work = m * g * h, where m is the mass, g is the gravitational acceleration and h is the height. Since the distance and incline angle are provided, we can calculate the height with the trigonometric function sine, h = sin(angle) * distance. Substituting these values, we get Work = 950kg * 9.81m/s² * sin(9.0°) * 710m. The required calculation will then provide the answer to the work done in pushing the car up the incline.

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Final answer:

Despite their different masses, two objects with equal speeds moving on a frictionless incline reach the same height because mechanical energy is conserved and potential energy at a given height is not dependent on mass.

Explanation:

In a scenario where two objects with different masses but equal speeds slide up a frictionless incline, the laws of physics tell us that both objects should reach an equivalent height. This is because the mechanical energy (kinetic plus potential energy) in a closed system without any external forces (like friction) is conserved. Since both objects start with the same kinetic energy (due to equal speeds) and potential energy is directly related to height (not mass), they should convert their kinetic energy into the same amount of potential energy, meaning they should rise to the same height before coming to a stop and sliding back down.

It's important to note that if there were friction or some other external force acting on the objects, the outcomes could be different as the mechanical energy would no longer be conserved in the same way, but in a frictionless situation, mass does not affect the height an object reaches if it starts with the same kinetic energy.