PLEASE HELPPPPPP <333​

Answer:

\(F=1.159\)

Explanation:

From the question we are told that:

Mass of pulley \(M=1kg\)

Radius \(r=12cm\)

Mass of block A \(M_a=2.1kg\)

Mass of block B \(m_b=4.1kg\)

Spring constant\(\mu= 358 J/m2\)

Generally the equation for Torque is mathematically given by

Since \(\sumF=ma\)

At mass A

 \(T_2-f_3=2.1a\)

At mass B

 \(4.8-T_1=4.1a\)

At  Pulley

 \(R(T_1-T_2)=\frac{1*1*R^2}{2}\frac{a}{R}\)

 \(R(T_1-T_2)=0.55a\)

Therefore the equation for total force F

At mass A+At mass B+At  Pulley

 \((T_2-f_3+4.8-T_1+R(T_1-T_2)=2.1a+4.1a+0.55a\)

 \((T_2-f_3+4.8-T_1+R(T_1-T_2)=2.7a+4.8a+0.55a\)

 \(-f_3+4.1=6.75a\)

 \(-f_3=6.75a+4.8\)

Since From above equation

\(M_{eff}=6.7kg\)

Therefore

\(T=2\pi \sqrt{{\frac{M_{eff}}{k}}\)

\(T=2\pi \sqrt{{\frac{6.75}{\mu}}\)

\(T=0.862s\)

Generally the equation for frequency is mathematically given by

\(F=\frac{1}{T} \\F=\frac{1}{0.862}\)

\(F=1.159\)

The potential difference between points A and B in the circuit is 12 V.

To find the potential difference between points A and B in the given circuit, determine the total voltage across the circuit.

From the information provided, there are four resistors labeled as 12 Ω each. Assuming these resistors are connected in series, the total resistance in the circuit is 12 Ω + 12 Ω + 12 Ω + 12 Ω = 48 Ω.

Next, determine the current flowing through the circuit. The voltage is given as 24 V.

Using Ohm's Law (V = IR), calculate the current:

I = V / R

I = 24 V / 48 Ω

I = 0.5 A

Since the resistors are connected in series, the current remains the same throughout the circuit.

Finally, to find the potential difference between points A and B, multiply the current by the resistance between those points:

V_AB = I × R_AB

V_AB = 0.5 A * 24 Ω

V_AB = 12 V

Therefore, the potential difference between points A and B in the circuit is 12 V.

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Answer:

Explanation:

1.  \(V_{x}\) = \(V_{0}\) * cos\(\alpha\) ⇒ 16*cos32 ≈ 13.6 m/s (13.56)

2. \(V_{y}\) = \(V_{0}\) * sin\(\alpha\) ⇒ 16* sin32 ≈ 9.4 m/s

3. \(y_{max}\) = \(\frac{v_{0}^2*sin^2\alpha}{2g}\)= \(\frac{16^2*sin^232}{2*9.8}\) (the g (gravity) depends on the country but i'll take the average g which is 9.2m/s^2)

\(y_{max}\) ≈ 3.6677+1.5 ≈ 5.2m

4.  \(x_{max}\) = \(\frac{v_{0}^2*sin(2\alpha)}{g}\)=\(\frac{16^2*sin(2*32)}{9.8}\) ≈ 23.5m (23.47)

5. -

answer 4 could be wrong, not certain about that one and i don't know 5

Answer:

T = 764.41 N

Explanation:

In this case the tension of the string is determined by the centripetal force. The formula to calculate the centripetal force is given by:

\(F_c=m\frac{v^2}{r}\)  (1)

m: mass object = 2.3 kg

r: radius of the circular orbit = 0.034 m

v: tangential speed of the object

However, it is necessary to calculate the velocity v first. To find v you use the formula for the kinetic energy:

\(K=\frac{1}{2}mv^2\)

You have the value of the kinetic energy (13.0 J), then, you replace the values of K and m, and solve for v^2:

\(v^2=\frac{2K}{m}=\frac{2(13.0J)}{2.3kg}=11.3\frac{m^2}{s^2}\)

you replace this value of v in the equation (1). Also, you replace the values of r and m:

\(F_c=(2.3kg)(\frac{11.3m^2/s^2}{0.034})=764.41N\)

hence, the tension in the string must be T =  Fc = 764.41 N

The given velocity is 54km/hr, which, when converted to m/s using 1km = 1000m and 1 hour = 3600 seconds, equals 15m/s.

Now substitute the initial speed u = 15m/s values.

v = 0m/s as the final velocity

Time is 6 seconds.

In the a= -3 m/ s2 equation,

s= ut + 1/2at2 s= 15 6+ 1/2(-3)(6)2 s= 90-45 s = 45m

As a result, after using the brakes, the car will halt for up to 45 meters.

The inverse of acceleration is deceleration. The deceleration will be calculated by dividing the final velocity minus the initial velocity by the time required for the velocity drop.

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The minimum coefficient of static friction with the floor is 0.3846.

To find the minimum coefficient of static friction with the floor, we need to consider the forces acting on the couch. In this case, the force of gravity is pulling the couch downward with a magnitude of mg, where m is the mass of the couch (40 kg) and g is the acceleration due to gravity (9.8 m/s²).

Since the couch does not move, the force of static friction between the couch and the floor must be equal in magnitude but opposite in direction to the horizontal pushing force of 150 N.

Therefore, we have the equation F_friction = F_push, where F_friction is the force of static friction.

The force of static friction can be calculated using the formula F_friction = μ_s * N, where μ_s is the coefficient of static friction and N is the normal force.

Since the couch is on a level floor and is not accelerating vertically, the normal force N is equal in magnitude but opposite in direction to the force of gravity, which is mg.

Substituting the values into the equation, we have μs * mg = 150 N.
Solving for μs, we get μs = 150 N / (mg).
Substituting the given values, we have μ_s = 150 N / (40 kg * 9.8 m/s²).
Simplifying, we find that μs = 0.3846.

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The breaking rate refers to the rate at which an object slows down due to braking or deceleration. In other words, it is the rate of change of velocity in the opposite direction of the object's motion.

Looking at the data provided, we can see that the velocity decreases from 50 m/s to 0 m/s over a period of 10 seconds, which means the object is decelerating at a constant rate. To calculate the breaking rate, we can use the formula:

breaking rate = (final velocity - initial velocity) / time taken

In this case, the breaking rate is:

breaking rate = (0 - 50) / 10 = -5 m/s^2

So, the object is decelerating at a rate of 5 m/s^2.

To compare this to the acceleration, we need to know the acceleration of the object before it starts breaking. If we assume that the object was accelerating at a constant rate of 5 m/s^2 before it started breaking, then the acceleration and breaking rates are equal in magnitude but opposite in direction. In other words, the acceleration and breaking rates are both 5 m/s^2, but the acceleration is positive while the breaking rate is negative.

It's worth noting that the breaking rate can vary depending on various factors such as the mass of the object, the friction between the object and the surface it is moving on, and the force applied to the brakes.

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Answer:

the same question I want to know

a. Spheres A and B carry the same charge \(\rm \( Q \)\), the force can be written as:

\(\rm \[ F = \frac{k Q^2}{R^2} \]\), b. The net force on sphere A now is the force due to sphere B, which is: \(\rm \[ F = \frac{k \frac{Q}{2} \cdot Q}{R^2} = \frac{k Q^2}{2R^2} \]\), c. The magnitude of the force on sphere A in this third case is \(\rm \( \frac{3}{8} \)\) of the original force.

a) The magnitude of the force (F) sphere B exerts on sphere A can be calculated using Coulomb's law:

\(\rm \[ F = \frac{k Q_1 Q_2}{R^2} \]\)

where:

k is Coulomb's constant \(\rm \( k = \frac{1}{4\pi\epsilon_0} \)\) where \(\rm \( \epsilon_0 \)\) is the vacuum permittivity constant),

\(\rm \( Q_1 \)\) and \(\rm \( Q_2 \)\) are the charges on spheres A and B respectively, and

\(\rm \( R \)\) is the distance between the two spheres.

Since both spheres A and B carry the same charge \(\rm \( Q \)\), the force can be written as:

\(\rm \[ F = \frac{k Q^2}{R^2} \]\)

b) When an identical sphere C makes contact with sphere B, they share the charge equally. Sphere B now carries \(\rm \( \frac{Q}{2} \)\) charge, and sphere C carries \(\rm \( \frac{Q}{2} \)\) charge.

When sphere C is moved far away, it exerts no force on sphere A. So, the net force on sphere A now is the force due to sphere B, which is:

\(\rm \[ F = \frac{k \frac{Q}{2} \cdot Q}{R^2} = \frac{k Q^2}{2R^2} \]\)

c) When sphere C makes contact with sphere A, they both share the charge equally. Each sphere now carries \(\rm \( \frac{Q}{2} \)\) charge.

When sphere C is moved far away, the net force on sphere A now is the force due to the charge on sphere A itself, which is:

\(\rm \[ F = \frac{k \frac{Q}{2} \cdot \frac{Q}{2}}{R^2} \\\\\rm = \frac{3}{8} \frac{k Q^2}{R^2} \]\)

So, the magnitude of the force on sphere A in this third case is \(\rm \( \frac{3}{8} \)\) of the original force.

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Answer:

(a) the mechanical energy of the system, U = 0.1078 J

(b) the maximum speed of the object, Vmax = 0.657 m/s

(c) the maximum acceleration of the object, a_max = 15.4 m/s²

Explanation:

Given;

Amplitude of the spring, A = 2.8 cm = 0.028 m

Spring constant, K = 275 N/m

Mass of object, m = 0.5 kg

(a) the mechanical energy of the system

This is the potential energy of the system, U = ¹/₂KA²

U = ¹/₂ (275)(0.028)²

U = 0.1078 J

(b) the maximum speed of the object

\(V_{max} =\omega*A= \sqrt{\frac{K}{M} } *A\\\\V_{max} = \sqrt{\frac{275}{0.5} } *0.028\\\\V_{max} = 0.657 \ m/s\)

(c) the maximum acceleration of the object

\(a_{max} = \frac{KA}{M} \\\\a_{max} = \frac{275*0.028}{0.5}\\\\a_{max} = 15.4 \ m/s^2\)

The characteristics of the simple harmonic motion allows to find the results for the questions of the oscillating mass are:

     a) The total energy is: Em = 0.1078 J

     b) The maximum speed is: v = 0.657 m / s

    c) the maximum acceleration is: a = 15.4 m / s²

Given parameters

To find

    a) Mechanical energy

    b) Maximum speed

    c) Maximum acceleration

the simple harmonic movement is an oscillatory movement where the restoring force is proportional to the displacement, it is described by the expression:

         x = A cos (wt + Ф)

         w² = k / m

Where x is the displacement, A the amplitude w the angular velocity, t the time, Ф a phase constant, k the spring constant and m the mass.

A) The mechanical energy is

        Em = ½ k A²

Let's calculate.

       Em = ½ 275 (2.8 10⁻²) ²

       Em = 0.1078 J

b) Velocity is defined as the change of position with respect to time.

       v = \(\frac{dx}{dt}\)  = - Aw sin ( wt + fi)

To obtain the maximum velocity, the sine function must be ±1

        \(v_{max}\) = w A

Let's calculate

        w = \(\sqrt{\frac{275}{0.5} }\)  

        w = 23.45 rad / s

        \(v_{max}\) = 23.45 2.8 10⁻²

        \(v_{max}\) = 0.657 m / s

c) maximum acceleration.

Acceleration is defined as the change in velocity withtrspect to time.

       a = \(\frac{dv}{dt}\)  = - A w² cos (wt + fi)

To have the maximum value, the cosine function must be maximum, that is ±1

      a = A w²

let's calculate

       a = 2.8 10⁻²  23.45²

       a = 15.4 m / s²

In conclusion using the characteristics of the simple harmonic motion we can find the results for the questions of the oscillating mass are:

     a) The total energy is:  Em = 0.1078 J

     b) The maximum speed is:  v = 0.657 m / s

     c) the maximum acceleration is: a = 15.4 m / s²

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The worker must apply 600N of force to lift the sound system. The mechanical advantage of 2.5 means that the worker only needs to apply 600N of force to lift the sound system.

We can employ pulley systems to provide ourselves a mechanical advantage by multiplying our input effort to apply more force to a load.

They can be used to apply tension within a system, such as in a Tensioned Line or Tyrolean, in addition to their traditional uses of hauling and lifting loads. The basic operating principles of pulley systems are explained on this page; for details on how to use them for hauling, check the topic on hauling systems.

The dynamic unit of Newtons is commonly used to express force, which is an impact that has both magnitude and direction (N). On this page, kilograms have been utilized for the sake of clarity.

Therefore, The worker must apply 600N of force to lift the sound system. The mechanical advantage of 2.5 means that the worker only needs to apply 600N of force to lift the sound system.

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The waveform of a signal is the shape of its graph as a function of time in the domains of electronics, acoustics, and allied sciences, regardless of its time and magnitude scales or any shift in time.  

Thus, Waveforms with periodic variations are those that recur consistently at set intervals.

The phrase is typically used in electronics to describe periodically changing voltages, currents, or electromagnetic fields. It is typically used in acoustics to describe constant periodic sounds caused by changes in air pressure or other media.

In these situations, the signal's frequency, amplitude, or phase shift have no bearing on the waveform, which is a characteristic. Additionally, non-periodic signals like chirps and pulses can be referred to by this name.

Thus, The waveform of a signal is the shape of its graph as a function of time in the domains of electronics, acoustics, and allied sciences, regardless of its time and magnitude scales or any shift in time.  

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Answer: Brainliest?

Explanation:

Isobars and isotherms are both types of contour lines used to represent data on weather maps, specifically for atmospheric pressure and temperature, respectively.

The similarities between isobars and isotherms are:

Both are contour lines that connect points of equal value on a map.

Both are used to depict weather patterns and conditions.

Both help to identify areas of high and low values.

The differences between isobars and isotherms are:

Isobars connect points of equal atmospheric pressure, whereas isotherms connect points of equal temperature.

Isobars are measured in units of pressure such as millibars, while isotherms are measured in units of temperature such as degrees Celsius or Fahrenheit.

Isobars are typically used to show pressure patterns associated with wind, while isotherms are used to show temperature patterns.

Isobars are often used to forecast weather conditions, including the movement and intensity of storm systems. Isotherms are used to identify areas of warm and cold air masses, which can affect local weather patterns.

In summary, both isobars and isotherms are useful tools for understanding weather patterns, but they represent different types of data and are used for different purposes.

Isobars and isotherms are both concepts used in meteorology and climatology to represent important variables that help to describe atmospheric conditions. While they share some similarities, they also have several key differences.

Isobars refer to lines of equal pressure, meaning they connect points on a map or graph where the atmospheric pressure is the same. Isobars are drawn on weather maps to indicate areas of high and low pressure, and to show the general movement of air masses. When isobars are closely spaced, it indicates a steep pressure gradient, which can result in strong winds.

On the other hand, isotherms refer to lines of equal temperature, meaning they connect points on a map or graph where the temperature is the same. Isotherms are often drawn on weather maps to show the boundaries between warmer and cooler air masses, and to indicate areas where temperature changes rapidly.

One similarity between isobars and isotherms is that they are both used to describe atmospheric conditions in terms of spatial variation. They are also both used to infer information about the movement of air masses and the development of weather patterns.

However, there are also some key differences between isobars and isotherms. The most obvious difference is that isobars represent pressure while isotherms represent temperature. Additionally, while isobars are generally oriented parallel to each other and indicate the direction of winds, isotherms are typically oriented perpendicular to isobars and indicate the location of temperature gradients. Finally, while isobars are more commonly used to describe weather conditions associated with areas of high and low pressure, isotherms are often used to identify the location of fronts and other weather boundaries.

In summary, isobars and isotherms are similar in that they both describe atmospheric conditions in terms of spatial variation, and can be used to infer information about the movement of air masses and the development of weather patterns. However, isobars represent pressure and are oriented parallel to each other, while isotherms represent temperature and are oriented perpendicular to isobars.

Option A. The angle between the force and displacement is 180°, the maximum work is considered to be done by the force.

Work is defined as the product of the force applied to an object and the displacement of the object in the direction of the force. Mathematically, work (W) is given by the equation:

W = F * d * cos(theta)

Where

F = magnitude of the force

d = magnitude of the displacement

theta = angle between the force and displacement vectors.

In order to maximize the work done by a force, we need to maximize the value of the cosine of the angle theta. The cosine function reaches its maximum value of 1 when the angle theta is 0° or 180°.

When the angle between the force and displacement is 0° (option E), the force and displacement vectors are perfectly aligned in the same direction. In this case, the work done is maximized. Therefore, the correct answer is option A.

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Answer:

a)    V = - x ( σ / 2ε₀)

c)  parallel to the flat sheet of paper

Explanation:

a) For this exercise we use the relationship between the electric field and the electric potential

          V = - ∫ E . dx        (1)

for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet

       Ф = ∫ E . dA = \(q_{int}\) /ε₀

the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product

          E A = q_{int} /ε₀

area let's use the concept of density

        σ = q_{int}/ A

       q_{int} = σ A

          E = σ /ε₀

as the leaf emits bonnet towards both sides, for only one side the field must be

          E = σ / 2ε₀

         we substitute in equation 1 and integrate

      V = - σ x / 2ε₀  

       V = - x ( σ / 2ε₀)

if the area of ​​the sheeta is 100 cm² = 10⁻² m²

      V = - x  (10⁻²/(2 8.85 10⁻¹²) = - x  ( 5.6 10⁻¹⁰)

      x = 1 cm     V = -1   V

      x = 2cm     V = -2   V

This value is relative to the loaded sheet if we combine our reference system the values ​​are inverted

       V ’= V (inf) - V

       x = 1 V = 5

       x = 2 V = 4

       x = 3 V = 3

These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.

In the attachment we can see a schematic representation of the equipotential surfaces

b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced

c) parallel to the flat sheet of paper

Answer:

5 m/s

Explanation:

We are given that 9000 kg of water flows through the pipe in 1 minute. Mass flow rate = mass/time

So, mass flow rate = 9000 kg / 1 minute = 150 kg/s

We know the cross sectional area of the pipe is 0.3 m2. From continuity equation, mass flow rate = density * area * velocity

So, 150 = 1000 * 0.3 * v (Density of water is approximately 1000 kg/m3)

Solving for v (velocity):

v = 150/(1000*0.3) = 5 m/s

Therefore, the speed of water in the pipe is 5 m/s.

The wavelength of the closed pipe, given the level the water moves up to, would be B. 5120.

The pipe, when it resonates with a sound wave, acts as a pipe closed at one end. The length of the air column in the pipe is the part of the pipe that is above the water.

Initially, the length of the air column is:

50cm - 40cm = 10cm

The initial wavelength (λi) is 4 times this length, or 40cm.

After being struck, the length of the air column changes to:

50cm - 45cm = 5cm

The new wavelength (λf) is 4 times this length, or 20cm. When converted, we get 5, 120.

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The wavelength is 5120m.

When the pipe reverberates with a sound wave, it behaves like a pipe that is closed at one end.

The portion of the pipe above the water is the length of the air column in the pipe.

The air column's initial length is:

50 cm - 40 cm = 10 cm.

This length, or 40cm, is the starting wavelength (i).

After being struck, the air column's length becomes:

50 cm - 45 cm = 5 cm

The new wavelength (f) is 20 cm, or four times this length.

In conversion, we obtain 5, 120.

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Answer:

not clear pic...but it's definitely not A)

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

1) a substance made by mixing other substances together.

2) diverse in character or content.

3) of the same kind; alike.

4) a means of solving a problem or dealing with a difficult situation.

5) a homogeneous noncrystalline substance consisting of large molecules or ultramicroscopic particles of one substance dispersed through a second substance. Colloids include gels, sols, and emulsions; the particles do not settle, and cannot be separated out by ordinary filtering or centrifuging like those in a suspension.

Explanation:

ya

Answer:

19,224 N

Explanation:

The given parameters are;

The mass limit of the elevator = 2,400 kg

The maximum ascent speed = 18.0 m/s

The maximum descent speed = 10.0 m/s

The maximum acceleration = 1.80 m/s²

Given that the acceleration due to gravity, g ≈ 9.81 m/s²

The minimum upward force that the elevator cable exert on the elevator car, \(F_{min}\) , is given in the downward motion as follows;

\(F_{min}\) = m·g - m·a

∴ \(F_{min}\) = 2,400 kg × 9.81 m/s² - 2,400 kg × 1.80 m/s² = 19,224 N

The minimum upward force that the elevator cable exert on the elevator car, \(F_{min}\) = 19,224 N

Accelerating objects are changing their velocity. Velocity is often thought of as an object's speed with a direction. Thus, objects which are accelerating are either changing their speed or changing their direction. They are either speeding up, slowing down or changing directions. Changing the velocity in any one of these three ways would be an example of an accelerated motion.

The arrows are drawn in the figure which shows gravitational forces on each person on earth.

Gravitational force is force of attraction between two masses. Gravitational force(F) between two bodies is directly proportion to the product of masses(m₁,m₂) of two bodies and inversely proportional to square of distance(r) between them. mathematically it is written as,

F∝ m₁.m₂

F ∝ 1/r²

F = G m₁,m₂÷r²

where G is gravitational constant, whose value is 6.6743 × 10⁻¹¹ m³ kg-1s⁻².

Force is expressed in Newton N in SI unit. its dimensions are [M¹L¹T⁻²].

This is analogous with coulomb's law which gives force between two charges.

To know more about force :

https://brainly.com/question/13191643

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The mass of Jupiter in solar mass units with the help of Kepler's Third Law is found to be 0.000935 Solar mass Unit.

According to the Kepler's Third law, "the squares of the orbital periods of the planets are directly proportional to the cubes of the semi-major axes of their orbits".

Mathematically, p² = a³M

where, p= years

           a= AU

           M= Solar Masses

In the given question,

p= 16.7 days = 0.0457 years

a= 1.88 x 10⁶ km = 0.0125 AU

M= a³/p²=(0.0125 AU)³/(0.0457)²

M= 0.000935 Solar mass Unit

M= 1.87 x 10²⁷ kg

Hence, the mass of Jupiter in solar mass units is found to be 0.000935 Solar mass Unit.

To know more about Kepler's laws, refer:

https://brainly.com/question/33230757?referrer=searchResults

Answer:

if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value. If the charges come 10 times closer, the size of the force increases by a factor of 100. The size of the force is proportional to the value of each charge.

When charges are 3 centimeters apart if the distance between the charges is doubled, the force between them is reduced by a factor of four.

Coulomb's law can be stated as the product of the charges and the square of the distance between them determine the force of attraction or repulsion acting in a straight line between two electric charges.

F= k Q₁Q₂/r²

Where F is the force between two charges

As given in the problem, the two charges are 3 centimeters apart, then we have to find out what will happen to this force if the distance between the charges is doubled.

As per Coulomb's law, the force is inversely proportional to the square of the distance between charges.

Thus, the force between them is reduced by a factor of four.

Learn more about Coulomb's law from here, refer to the link;

brainly.com/question/506926

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Answer:

here's the answer

Explanation:

given:

h= 20.0m

M= 12.0kg

g=10m/s²

solution:

potential energy = mgh = (12.0kg)(10m/s²)(20.0m)

= 2400

potential energy = 2400J

hope this help if it does help I'd be happy if you follow and give me the brainliest answer

Answer:

1.8 Kg 17.6N

Explanation:

I don't know the explanation hahaha

Answer:

The gravitational force on an object increases as the object’s mass increases.

Explanation:

This is the answer on Edmentum. :)

Momentum is conserved in a car accident whereas a part of kinetic energy is converted into heat due to friction.

Thus, the statement is false.