Learning goal: to practice problem-solving strategy 6.1 work and kinetic energy. your cat "ms." (mass 8.50 kg ) is trying to make it to the top of a frictionless ramp 2.00 m long and inclined 19.0 ∘ above the horizontal. since the poor cat can't get any traction on the ramp, you push her up the entire length of the ramp by exerting a constant 41.0 n force parallel to the ramp. if ms. is moving at 1.90 m/s at the bottom of the ramp, what is her speed when she reaches the top of the incline?

Answer:

Our answer is 3380kg

Explanation:

The force required to move the truck at constant speed in the given case is.

F=Mg sin∅ =(9100kg)(9.8m/s²)sin15° =2.31×10⁴ N

The net force on the truck after the mass is fell down from the truck is

Fnet =F- mg sin∅

ma= F-mg sin∅

m(1.5m/s²) =( 2.31 × 10⁴N) -m(9.8m/s²)sin 15°

Solve for m.

m((1.5m/s²) +(9.8m/s²)sin 15°) =(2.31 ×10⁴ N))

m = 5720kg

Mass of load is.

Δm =M -m =(9100kg) -(5720kg) =3380kg

Answer:

-496.37 J

Explanation:

P(V2-V1) = 10(.5-.01)

10(.49) =4.9

L x ATM = 4.9 x 101.3= 496.37 J

External pressure means negative therefore its -496.37J

The work done in this scenario can be calculated by multiplying the change in volume, external pressure, and a conversion factor. In this case, the work done is 494.9 J.

The work done in this scenario can be calculated using the formula:
Work = change in volume * external pressure * conversion factor

Given:
Initial volume (V1) = 0.0100 l
Final volume (V2) = 0.500 l
External pressure = 10.00 atm

First, we need to find the change in volume:
Change in volume = V2 - V1 = 0.500 l - 0.0100 l = 0.490 l

Next, we can calculate the work done:
Work = change in volume * external pressure * conversion factor
= 0.490 l * 10.00 atm * 101.3 J/l atm
= 494.9 J

Therefore, the work done in joules is 494.9 J.

50 mph east,because Velocity is a physical vector quantity both magnitude and direction are needed to define it.

When an item is moving, its velocity is the rate at which its direction is changing as seen from a certain point of view and as measured by a specific unit of time.

Uniform motion an object is said to have uniform motion when object cover equal distance in equal interval of time within exact fixed direction. For a body in uniform motion, the magnitude of its velocity remains constant over time.

50 mph east,because Velocity is a physical vector quantity both magnitude and direction are needed to define it.

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The kinetic energy of the object at t = 0 is approximately 66.27 Joules.

To calculate the kinetic energy of the object, we can use the equation: KE = 0.5 * m * v^2. Given the mass of the object is 5.22 kg and the x-component of velocity is 5.10 m/s, we can substitute these values into the equation:

KE = 0.5 * 5.22 kg * (5.10 m/s)^2

Calculating this gives us a kinetic energy of approximately 66.27 Joules.

To determine the density of states in silicon or GaAs at specific energies, one needs to use the formula related to effective mass and semiconductor band structure. The question does not provide enough information to perform these calculations, and additional data is required.

The question implies determining the number of energy states within a certain energy range in silicon and gallium arsenide (GaAs) semiconductors at different temperatures. The challenge is to understand the concept of density of states (DoS) and how it varies with energy and temperature. The density of states is a function that describes the number of states per interval of energy at each energy level available to be occupied by electrons or holes. At T = 300 K and T = 400 K, we would use the DoS formula, which depends on effective mass and energy of the semiconductor material. However, the question as provided does not include enough information or specific parameters to calculate the density of states for silicon and GaAs between given energy levels and at specific temperatures.

To find the density of states at E = 0.80 eV, E = 2.2 eV, and E = 5.0 eV, you would use a formula related to the effective mass of the electrons and the structure of the semiconductor band. However, without the actual formulas or values specific for silicon and GaAs, it is not possible to calculate the exact density of states at these energy levels. Furthermore, the additional information provided in the challenge problems discusses concepts like the free electron gas model and the Fermi factor but is not directly applicable to calculating the density of states without further context.

The coefficient of kinetic friction depends on the materials of the interacting surfaces and their microscopic characteristics, not on the speed of motion. The experimental data in Tables 6.1 and 5.2 indicate this by showing that frictional coefficients are about materials, not speed.

The coefficient of kinetic friction is a factor that determines the amount of frictional force between two objects that are sliding against each other. It depends on the nature of the materials in contact, rather than on the speed of motion. This concept can be demonstrated by the data in Tables 6.1 and 5.1, which show coefficients of kinetic friction that are less than their static counterparts and do not correspond to speed. This indicates that kinetic friction is more about the materials' interactions at the microscopic level.

For instance, through a simple experiment with a cup sliding on a table, the coefficient of kinetic friction can be determined without considering the speed of the cup's motion. Instead, the frictional force is calculated using the normal force, which is based on the weight of the cup plus any added load. Similarly, different surfaces have different coefficients of friction, as shown in Table 5.2, but this is about surface characteristics and not the speed of motion.

To sum up, the coefficient of kinetic friction does not depend on speed. Instead, it's about the materials in contact and their microscopic interactions. The direction of friction is always opposite that of motion, illustrated in Equations 6.1 and 6.2 which showcase the dependence of friction on materials and the normal force.

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Gravity is not a direct force causing plate motion in the context of plate tectonics; instead, key forces include ridge push, slab pull, and basal drag, with slab pull being the predominant mechanism.

The question pertains to the forces that are responsible for plate tectonics and plate motion. It is a common misconception that gravity is a direct force causing plate motion; however, in the context of tectonic plates, gravity is not a force that directly initiates their movement. In tectonic plate motion, the key driving forces are ridge push, slab pull, and basal drag. Ridge push occurs at mid-ocean ridges, where newly formed lithosphere pushes plates apart. Slab pull is the force exerted by a subducting plate that pulls the rest of the plate after it as it descends into the mantle due to its higher density. Basal drag is the force exerted by the mantle's convection currents on the base of the tectonic plates.

Gravity, while essential in providing the overall setting by influencing the density and buoyancy of rocks, does not directly move the plates on its own. Rather, gravity contributes indirectly to gravitational sliding, which pulls lithospheric plates down from the elevated mid-ocean ridges due to the height difference. However, current evidence supports slab pull as the predominant mechanism over ridge push and gravitational sliding.

The current required in the loop to experience the given torque is [tex]\boxed{6.366\times{{10}^2}\,{\text{A}}}[/tex]  or [tex]\boxed{636.6\,{\text{A}}}[/tex] .

Further Explanation:

Given:

The diameter of the circular loop is [tex]20\,{\text{cm}}[/tex] .

The torque experienced by the circular loop is [tex]1.0\times{10^{-3}}\,{\text{N}}\cdot{\text{m}}[/tex] .

Concept:

Since the circular loop is kept in the effect of the Earth’s Magnetic field, it will experience a magnetic torque due to the magnetic lines of force passing through the area of cross-section of the loop.

The torque experienced by the loop is expressed as:

[tex]\boxed{\tau =BIA}[/tex]

Here, [tex]\tau[/tex]  is the torque experienced, [tex]B[/tex]  is the magnetic field, [tex]I[/tex]  is the current in the loop and [tex]A[/tex]  is the area of cross-section of the loop.

The strength of the Earth’s magnetic field is [tex]5\times{10^{-5}}\,{\text{T}}[/tex] .

Substitute the values in the above expression.

[tex]\begin{aligned}1.0\times{10^{-3}}&=\left({5\times{{10}^{-5}}}\right)\timesI\times\left({\pi \times{{\left({\frac{d}{2}}\right)}^2}}\right)\\I&=\frac{{1.0\times{{10}^{-3}}}}{{5\times{{10}^{-5}}\left({\pi {{\left({\frac{{0.20}}{2}}\right)}^2}}\right)}}\\&=6.366\times{10^2}\,{\text{A}}\\\end{aligned}[/tex]

The current required in the loop to experience the given torque is [tex]\boxed{6.366\times{{10}^2}\,{\text{A}}}[/tex]  or [tex]\boxed{636.6\,{\text{A}}}[/tex] .

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Answer Details:

Grade: College

Subject: Physics

Chapter: Electromagnetism

Keywords:

Earth’s magnetic field, torque, maximum torque, maximum current, through the loop, experience a modest torque, T=BIA, 636 A, wire is oriented.

Final answer:

The outside air temperature is approximately 3375 °C.

Explanation:

The outside air temperature can be calculated using the density altitude and pressure altitude. The formula to calculate the temperature is:

T = T0 + (d × 120)

Where:

T is the outside air temperature

T0 is the temperature at sea level

d is the difference between the pressure altitude and the density altitude, divided by 1,000 (d = (PA - DA) / 1000)

Plugging in the given values, we have:

T = 15 + (28 × 120)

T = 15 + 3360

T = 3375

So, the outside air temperature is approximately 3375 °C.

Final answer:

The rate at which the length of the person's shadow is growing is \(\frac{{48}}{{5}}\) feet per second.

Explanation:

In this problem, we can use similar triangles to determine the rate at which the length of the person's shadow is growing. Let's consider the situation at a particular moment when the person is a certain distance away from the pole. At this moment, the length of the person's shadow is the distance from the pole to the person multiplied by the ratio of the height of the pole to the height of the person. Let's call this length S. The rate of change of S, which represents the speed at which the length of the person's shadow is growing, can be determined using derivatives. To find this rate, we need to differentiate the expression for S with respect to time (t), since the person is moving and therefore the distance from the pole is changing over time.

The length of the person's shadow (S) can be expressed as:

S = \(\frac{{12}}{{5}}x\)

where x is the distance from the pole to the person at a particular moment.

To find the rate of change of S with respect to time (\(\frac{{dS}}{{dt}}\)), we differentiate the expression for S:

\(\frac{{dS}}{{dt}} = \frac{{12}}{{5}}\frac{{dx}}{{dt}}\)

Since the person is moving away from the pole at a rate of 4 feet per second, \(\frac{{dx}}{{dt}} = 4\). Plugging this value into the equation, we can calculate the rate at which the length of the person's shadow is growing:

\(\frac{{dS}}{{dt}} = \frac{{12}}{{5}} \cdot 4 = \frac{{48}}{{5}}\) feet per second

By using similar triangles and differentiating with respect to time, we find that the length of the person's shadow is growing at a rate of approximately 2.86 feet per second.

The question involves calculating the rate at which the length of a person's shadow grows as they walk away from a street light. To start, we can use similar triangles to relate the heights and shadows of the pole and the person. Given:

Let the distance of the person from the pole be x, and the length of the shadow be s. By similar triangles, we have:

(Height of the pole)/(Total distance from the pole to the tip of the shadow) = (Height of the person)/(Length of the shadow)

This converts to:

12/(x + s) = 5/s

Cross multiplying gives us:

12s = 5(x + s)

Which simplifies to:

12s = 5x + 5s

Rearranging terms gives us:

7s = 5x

So,  s = (5/7)x

Next, we differentiate both sides of this equation with respect to time (t), noting that both s and x are functions of time:

ds/dt = (5/7)dx/dt

Given that the rate at which the person walks (dx/dt) is 4 feet per second, we find:

ds/dt = (5/7) * 4 = 20/7 ≈ 2.86 feet per second

Thus, the length of the person's shadow is growing at a rate of approximately 2.86 feet per second.

Answer: A. Wavelength

Explanation: USAtestprep

The average angular speed can be calculated by dividing the total rotation of the planet (2π radians) by the time taken for one rotation (converted to seconds). This gives an average angular speed of approximately 8.11x10-5 radians per second.

The first step in tackling this problem is understanding what angular speed is. Angular speed is the rate at which an object moves through an angle. It is measured in radians per second. In your case, you want to find the angular speed of the planet about its own axis of rotation.

To do this, we need to recall that one complete rotation is 2π radians. Since one day on this distant planet lasts 21.5 hours, we convert this to seconds (1 hour = 60 minutes = 3600 seconds). So, 21.5 hours is 21.5 x 3600 = 77400 seconds.

The angular speed (ω) is therefore calculated by dividing the total rotation (2π radians) by the time (t) taken for one rotation. That is ω = 2π/t. Substituting for t in this formula, we get ω = 2π/77400 = 8.11x10-5 radians per second. Note that this answer is an approximation, and actual planetary motion can be influenced by a number of factors.

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Answer:

Kinetic Energy (K.E) of the car is 630000 J

Explanation:

Kinetic energy (K.E) of a body is the energy of the body in motion and it is given by the product of half its mass (m) and the square of its velocity (v). It is expressed in Joules.

Mathematically written as;

K.E = [tex]\frac{1}{2}[/tex] x m x [tex]v^{2}[/tex]

According to the question,

The mass m of the car is 1400kg

=> m = 1400kg

The speed (velocity) of the car is 30m/s

=> v = 30m/s

Substituting these values into the equation above gives;

K.E = [tex]\frac{1}{2}[/tex] x 1400 x [tex]30^{2}[/tex]

K.E = 630000 J

Therefore the kinetic energy of the car is 630000Joules

The kinetic energy of a 1400 kg car traveling at 30 m/s is calculated using the kinetic energy formula, resulting in a total energy of 630,000 Joules or 630 kJ.

The answer is 630,000 Joules or 630 kJ.

Kinetic energy is a fundamental concept in physics, representing the energy of an object in motion. It depends on two factors: the object's mass (m) and its velocity (v). The formula for kinetic energy is KE = 0.5 * m * v^2. Essentially, the greater an object's mass or speed, the more kinetic energy it possesses. This energy is vital in understanding various natural phenomena, such as the motion of vehicles, the flight of birds, and the behavior of particles in atomic and subatomic physics. It is also crucial in engineering and everyday applications.

The kinetic energy of an object can be calculated using the formula: Kinetic Energy = 0.5 * mass * speed². In this case, the mass of the car is 1400 kg and the speed is 30 m/s. To calculate the kinetic energy, you would substitute these values into the formula:

Kinetic Energy = 0.5 * 1400 kg * (30 m/s)²

Therefore, Kinetic Energy = 0.5 * 1400 * 900

So, the kinetic energy of the car is 630,000 Joules or 630 kJ.

Hence The answer is 630,000 Joules or 630 kJ.

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The horizontal and vertical components of the tension at the given point in the swing are  281.6 N and 351 N respectively.

Given data:

The length of chain is, L = 2.5 m.

The magnitude of tension on each chain is, T = 450 N.

Distance above the lowest point is, d = 55 cm = 0.55 m.

In problem, first we need to obtain the angle of inclination made by string horizontally.

So, the angle inclined by the string with horizontal is given as,

[tex]cos \theta =\dfrac{L-d}{L}\\\\cos \theta =\dfrac{2.5-0.55}{2.5}\\\\\theta = cos^{-1}(\dfrac{1.95}{2.5})\\\\\theta=38.74^{\circ}[/tex]

Now, the horizontal component of tension force acting on the string is,

[tex]T_{H}=T \times cos \theta\\T_{H}=450 \times cos 38.74\\T_{H}=281.6 \;\rm N[/tex]

And, the vertical component of tension force acting on the string is,

[tex]T_{V}=T \times sin \theta\\T_{V}=450 \times sin 38.74\\T_{V}=351 \;\rm N[/tex]

Thus, the horizontal and vertical components of the tension at this point in the swing are 281.6 N and 351 N respectively.

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The discus, after completing one revolution in 0.90 seconds starting from rest, will be released at a speed of approximately 6.98 rad/s.

This question is related to the concept of rotational motion in physics. As it is stated that the discus thrower takes 0.90s to complete one revolution, and the discus is starting from rest, the rotational speed or the angular velocity (ω) can be calculated using the formula ω = 2π/T, where T is the period of rotation which is the time to complete one revolution. Substituting the given values into the formula gives us ω = 2π/0.90 s which is approximately 6.98 rad/s.

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To determine the speed of a discus at release, given the time for one revolution is 0.90 seconds, we first find the angular velocity to be approximately 6.98 rad/s. Then, using the radius of the circle, we can calculate the linear speed. For a radius of 1 meter, the speed is approximately 6.98 m/s.

When calculating the speed of the discus at release, we first need to determine the angular velocity. Given that the thrower completes one revolution in 0.90 seconds, we can use the formula for angular velocity:

ω = 2π / T

where ω is the angular velocity and T is the period of one revolution. Substituting the given values, we get:

ω = 2π / 0.90 s ≈ ( 2 x 3.14 ) / 0.90s  ≈ 6.98 rad/s

Next, to find the linear speed at release, we use the relationship between linear speed (v), angular velocity (ω), and radius (r):

v = ωr

Assuming we know the radius of the circle in which the discus is being rotated, we can substitute r. If r is, for example, 1 meter, then:

v ≈ 6.98 rad/s * 1 m ≈ 6.98 m/s

Therefore, the speed of the discus at release would be approximately 6.98 meters per second.

To find the rate at which oil is flowing through a constricted pipe with different diameters and pressures, apply the principle of continuity.

A horizontal pipe of diameter 0.985 m has a smooth constriction to a section of diameter 0.591 m. The pressure in the pipe is 8100 N/m2, and in the constricted section, it is 6075 N/m2. The density of oil flowing in the pipe is 821 kg/m3.

To find the rate at which oil is flowing, we can apply the principle of continuity, which states that the product of the cross-sectional area and the fluid velocity is constant in a pipe with steady flow.

By applying the principle of continuity, you can calculate the rate at which oil is flowing through the pipe.

Answer: a = (2 v²)/d = 1.9 m/s²

Explanation:

In circular motion, the acceleration is given by:

a = v²/r = v²/(d/2) = (2 v²)/d

where v is the velocity and r is the radius of the circular path in which the vehicle is moving. d is the diameter of the circular path.

It is given that:

v = 28.5 m/s

r = d/2 = 0.85 km /2 = 0.425 km = 425 m

⇒ a = (28.5 m/s)²/425 m = 1.9 m/s²

An expression for the magnitude of the acceleration (a) of the car in terms of the given parameters is: [tex]A_c = \frac{2V^2}{D}[/tex]

Given the following data:

To write an expression for the magnitude of the acceleration (a) of the car in terms of the given parameters:

The acceleration of an object along a circular track is referred to as centripetal acceleration.

Mathematically, the centripetal acceleration of an object is given by the formula:

[tex]A_c = \frac{V^2}{r}[/tex]   .....equation 1

Where:

But, [tex]Radius, \;r = \frac{D}{2}[/tex]  .....equation 2

Substituting the eqn 2 into eqn 1, we have:

[tex]A_c = \frac{V^2}{\frac{D}{2}}[/tex]

Simplifying further, we have:

[tex]A_c = \frac{2V^2}{D}[/tex]

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Answer:

B broad leaves

Explanation:

Plants in a tropical rainforest usually have large as well as broad leaves. The correct option is B.

Tropical rainforests are rainforests that occur in tropical rainforest climate areas where there is no dry season and all months have an average precipitation of at least 60 mm. They are also known as lowland equatorial evergreen rainforest.

Tropical rainforests are dominated by broad-leaved trees that form a dense upper canopy and contain a diverse array of vegetation and other life. They are one of Earth's largest biomes (major life zones).

Large leaves also allow tropical plants to capture more sunlight energy, which, when combined with a ready supply of water, allows for rapid growth.

Thus, the correct option is B.

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