A car traveling 92 km/h is 130 m behind a truck traveling 75 km/h. how long will it take the car to reach the truck?

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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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

law of motion

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displacement of a skateboarder

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The distance of the elevator

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Answer:

Acceleration of the train is 0.07 m/s²            

Explanation:

Initial speed of the train, u = 22 m/s

Final speed of the train, v = 32 m/s

Time taken by the train to do the process is 142 seconds

We have to find the acceleration of the train. It is given by the rate of change of speed of the train i.e.

[tex]a=\dfrac{v-u}{t}[/tex]

[tex]a=\dfrac{32\ m/s-22\ m/s}{142\ s}[/tex]

[tex]a=0.07\ m/s^2[/tex]

So, the acceleration of the car is 0.07 m/s². Hence, the correct option is (d) 0.07 m/s²

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:

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

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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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.

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)

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:

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

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.

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

Answer:

Speed is slightly greater than 14 m/s.          

Explanation:

The ball will under go projectile motion. The horizontal velocity remains constant while there would be increase in vertical velocity due to acceleration due to gravity in downward direction.

Using first equation of motion, after 1 s, vertical velocity will be:

v = u +at

v=0+(9.8 m/s²)(1 s) = 9.8 m/s

Horizontal velocity = 10 m/s

Net velocity:

[tex]V= \sqrt{(9.8)^2+(10)^2}=\sqrt{196.04}=14.001 m/s[/tex]

Speed is slightly greater than 14 m/s.

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.

The loss in kinetic energy in an inelastic collision is equal to 251104 J.

According to the law of conservation of linear momentum, the sum of the momentum before and after an inelastic collision must be equal.

m₁u₁ + m₂ u₂ =   m₁ v₁ + m₂ v₂

where m₁ and m₂ is the mass of the collided bodies, u₁ & u₂ are their initial speed while v₁ & v₂ is their final speed.

Given the initial velocity of the truck is v₁ = 27 m/s and the other car is at rest, u₂ = 0 m/s.

The mass of the truck, m₁ = 1490 Kg

The mass of the stationary car, m₂ = 1280 Kg

From the law of conservation of momentum, find the final speed after a collision:

m₁ u₁ + m₂ u₂ = (m₁ + m₂)v'

1490 × 27 + 1280 ×0 = (1490 + 1280) v'

v' = 14.52 m/s

The loss in the kinetic energy = [tex]=\frac{1}{2} mv_1^2-\frac{1}{2} (m_1+m_2)^2v^'^2[/tex]

[tex]==\frac{1}{2} \times 1490\times(27)^2-\frac{1}{2} (2770)^2(14.52)^2[/tex]

[tex]= 543105 -292000\\=251104 J[/tex]

Therefore, the loss in the kinetic energy in the collision is equal to 251104 J.

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The shortest time in which a car could accelerate from 0 mph to 60 mph is [tex]13.7s[/tex].

Further explanation:

The opposite force acting on the body is known as frictional force. It always acts in the opposite direction of motion of body.

Concept used:

The force applied to a body to keep it at rest is known as the static friction force. It always acts opposite to the direction of motion of body. It is defined as the product of coefficient of friction and the normal force acting on the body.

The expression for the normal reaction of the body is given as.

[tex]N = mg[/tex]  

The expression for the net force is given as.

[tex]{F_{net}} = ma[/tex]                                 …… (1)

The expression for the static friction is given as.

[tex]{F_s} = {\mu _s}N[/tex]

The expression for the balanced forces is given as.

[tex]{F_{net}} = {F_s} - {F_r}[/tex]

Substitute[tex]{\mu _s}N[/tex] for [tex]{F_s}[/tex] and  for[tex]{F_r}[/tex] in the above expression.

[tex]\begin{aligned}{F_{net}}&={\mu _s}N-{\mu _r}N\\&= \left( {{\mu _s} - {\mu _r}} \right)N \\ \end{aligned}[/tex]

Substitute [tex]mg[/tex] for [tex]N[/tex] in above expression.

[tex]{F_{net}}=\left({{\mu _s}-{\mu _r}}\right)\left( {mg}\right)[/tex]       …… (2)

Compare equation (1) and (2) we get.

[tex]a=g\left({{\mu _s}-{\mu _r}}\right)[/tex]                              …… (3)

Here, [tex]a[/tex] is the acceleration of the body, g is the acceleration due to gravity, [tex]{\mu _s}[/tex] is the coefficient of static friction and  is the coefficient of reactive force.

The expression for the first equation of motion is given as.

[tex]v = u + at[/tex]      

Rearrange the above expression for time is given as.

[tex]\fbox{\begin\\t = \dfrac{{\left( {v - u} \right)}}{a}\end{minispace}}[/tex]                              …… (4)

Here, [tex]v[/tex] is the final velocity, [tex]u[/tex] is the initial velocity and [tex]t[/tex] is the time.

Substitute [tex]1[/tex] for [tex]{\mu _s}[/tex], [tex]0.8[/tex] for[tex]{\mu _r}[/tex] and [tex]9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}[/tex] for [tex]g[/tex] in equation (3).

[tex]\begin{aligned}a&=\left( {9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}}\right)\left( {1 - 0.8}\right)\\&=1.96\,{\text{m/}}{{\text{s}}^{\text{2}}} \\ \end{aligned}[/tex]

Substitute [tex]0\,{\text{mph}}[/tex] for [tex]u[/tex], [tex]60\,{\text{mph}}[/tex] for [tex]v[/tex] and [tex]1.96\,{\text{m/}}{{\text{s}}^{\text{2}}}[/tex] for [tex]a[/tex] in equation (4).

[tex]\begin{aligned}t&=\frac{{\left( {60\,{\text{mph}}-0\,{\text{mph}}}\right)}}{{1.96\,{\text{m/}}{{\text{s}}^{\text{2}}}}}\\&=\frac{{60\,{\text{mph}}\left( {\frac{{0.447\,{\text{m/s}}}}{{1\,{\text{mph}}}}} \right)}}{{1.96\,{\text{m/}}{{\text{s}}^{\text{2}}}}}\\&= 13.7\,{\text{s}} \\ \end{aligned}[/tex]

Thus, the time required to accelerate a car is [tex]\fbox{\begin\\13.7\, {\text{s}}\end{minispace}}[/tex].

Learn more:

1.  Conservation of momentum https://brainly.com/question/9484203.

2.  Motion under friction https://brainly.com/question/7031524.

3. Net force on the body https://brainly.com/question/4033012.

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Force, friction, Acceleration, acceleration due to gravity, normal, weight, mass, motion, sliding, sled, hill, inclined, plane, coefficient of friction, angle of inclination, 13.68 s, 13.69 s.  

Final Answer:

So, under ideal conditions with the assumption of maximum static friction being available throughout the acceleration, the shortest time in which a typical car could accelerate from 0 to 60 mph on rubber-on-concrete would be approximately 2.734 seconds.

Explanation:

To solve this problem, we need to understand how the force of friction allows a car to accelerate and then apply the equations of motion to determine the shortest time it would take for the car to accelerate from 0 to 60 miles per hour (mph).

First, we need to define our variables and constants:
- The static friction coefficient between rubber and concrete, μs, which is 1.00. This is the factor that will limit our acceleration since it represents the maximum frictional force that can be exerted before sliding begins.
- The acceleration due to gravity, g, which is 9.81 m/s^2.
- The initial speed of the car, 0 mph, which needs to be converted to meters per second (m/s).
- The final speed of the car, 60 mph, which also needs to be converted to meters per second.

The force of static friction (f_friction) responsible for the acceleration can be calculated by multiplying the static friction coefficient (μs) with the normal force (N). The normal force, in this case, is equal to the weight of the car, which is the mass (m) times the acceleration due to gravity (g). Since we are looking for the highest possible acceleration, we can assume that the force of static friction is at its maximum value.

f_friction = μs * m * g

However, we don't need to know the mass of the car because it will cancel out in the equation of motion. The maximum possible acceleration (a_max) happens when the force of static friction is at its highest, and it can be calculated using:

a_max = f_friction / m = μs * g

Now we have the maximum possible acceleration the tires can provide on concrete, which is:

a_max = 1.00 * 9.81 m/s^2
a_max = 9.81 m/s^2

Next, let's convert the final velocity from mph to m/s:

v_final(mph) = 60 mph
v_final(m/s) = v_final(mph) * 0.44704 (conversion factor from mph to m/s)
v_final(m/s) = 60 * 0.44704
v_final(m/s) ≈ 26.8224 m/s

With the initial velocity (v_initial) being 0 m/s (since we start from rest), we can use one of the kinematic equations to find the time (t):

v_final = v_initial + a_max * t
26.8224 m/s = 0 m/s + 9.81 m/s^2 * t

Solving for t, we get:

t = v_final / a_max
t ≈ 26.8224 m/s / 9.81 m/s^2

t ≈ 2.734 seconds

So, under ideal conditions with the assumption of maximum static friction being available throughout the acceleration, the shortest time in which a typical car could accelerate from 0 to 60 mph on rubber-on-concrete would be approximately 2.734 seconds.

A star with an initial mass of 10 solar masses will likely form a neutron star instead of a black hole after a supernova due to the mutual repulsion between densely packed neutrons that can support the core against its own gravitational weight.

Given a star with initial 10 solar masses that ends its life in a supernova explosion, it is more likely to form a neutron star rather than a black hole. Most stars end their life as white dwarfs or neutron stars. In the case of a very massive star, if the core's remaining mass is more than about three times that of the Sun, it may collapse into a black hole. However, if the mass is less than about three solar masses, the mutual repulsion between densely packed neutrons can support the core against its own weight, preventing it from becoming a black hole.

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To solve this problem, we use the formula:

I100 / I1 = [P / 4π(100m)^2] / [P / 4π(1m)^2]

I100 / I1 = 1 / 100^2

I100 / I1 = 10^-4

Therefore the change in intensity from 1m to 100m in decibels is:

B100 – B1 = 10 log(10^-4) dB = -40 dB

So the intensity at 100m is calculated as:

B100 = B1 – 40 dB = 140 dB – 40 dB = 100 dB

Answer:

100 dB

The intensity of the sound produced due to the firecracker at the distance of [tex]100\text{ m}[/tex] is [tex]\fbox{\begin\\100\text{ dB}\end{minispace}}[/tex].

Further Explanation:

The vibration, produced in the medium, which travels as an wave of pressure or density through a medium is known as sound. The sound is a longitudinal wave and it requires a medium for its propagation.  

Given:

The distance between the place where firecracker is set off by a child and the family house is [tex]100\text{ m}[/tex].  

The intensity of the sound produced by the firecracker at the distance [tex]1\text{ m}[/tex] is[tex]140\text{ dB}[/tex].

Concept:

The intensity of the sound wave is defined as the power carried by the sound waves in the direction perpendicular to the direction of propagation per unit time.

The intensity of the sound wave is:

[tex]\fbox{\begin\\I=\dfrac{P}{4\pi r^2}\end{minispace}}[/tex]                                                        ...... (1)

Here, [tex]I[/tex] is the intesnity of the sound wave, [tex]P[/tex] is the power carried by the sound wave and [tex]r[/tex] is the distance between source and the listener.

The intensity of the sound in decible at the house is:

[tex]\fbox{\begin\\\beta _{100}=\beta _{1}+10\log \dfrac{I_{100}}{I_{1}}\end{minispace}}[/tex] ...... (2)

Here, [tex]\beta _{100}[/tex] is the intensity of the sound at the distance of [tex]100\text{ m}[/tex], [tex]\beta _{1}[/tex] is the intensity of sound produced by the firecracker at the distance of [tex]1\text{ m}[/tex], [tex]I_{100}[/tex] is the intensity of sound produced by firecracker at [tex]100\text{ m}[/tex] and [tex]I_{1}[/tex] is the intensity of sound produced by firecracker at [tex]1\text{ m}[/tex].

The ratio of the intensity of sound at distance [tex]100\text{ m}[/tex] and [tex]1\text{ m}[/tex] is:

[tex]\fbox{\begin\\\dfrac{{{I_{100}}}}{{{I_1}}}=\frac{{r_1^2}}{{r_2^2}}\end{minispace}}[/tex]                                                                                     …… (3)

Calculation:

Substitute the values in equation (3).

[tex]\begin{aligned}\frac{{{I_{100}}}}{{{I_1}}}&=\frac{{{{\left( 1 \right)}^2}}}{{{{\left( {100} \right)}^2}}}\\&={10^{ - 4}}\\\end{aligned}[/tex]

Substitute the values in equation (2).

[tex]\begin{aligned}{\beta _{100}}&=140\,{\text{dB}} - 40\,{\text{dB}} \\&=100\,{\text{dB}}\\\end{aligned}[/tex].

Thus, the intensity of the sound produced due to the firecracker at the distance of [tex]100\text{ m}[/tex] is [tex]\fbox{\begin\\100\text{ dB}\end{minispace}}[/tex]

Learn more:

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

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

3. Conservation of energy brainly.com/question/3943029

Answer Details:

Grade: College

Subject: Physics

Chapter: Waves and Oscillation

Keywords:

Intensity of sound, sound waves, vibration, firecracker, power carried by sound waves, 100db, 100 dB, 100dB, 100 m, bursting of firecrackers, sound inside the house, 140 db, 140 dB.