when there is a net force exerted on an object, the magnitude of the acceleration of the object is [a] proportional to the magnitude of the net force

Answer:

a) 1.38 s

b) v₀ = 3.92 m/s  θ₀ = 53.1º

c) vf = 10.7 m/s θf = -77.2º

Explanation:

a)

       \(v_{fy} ^{2} - v_{oy} ^{2} = 2* g* \Delta y (1)\)

       \(v_{oy} = \sqrt{2*g*\Delta_{ymax} } = \sqrt{2*9.8m/s2*0.5m} = 3.13 m/s (2)\)

       \(v_{fyhmax} = v_{oy} - g*t_{hmax} = 0 (3)\)

       \(t_{hmax} =- \frac{v_{oy}}{g} =\frac{3.13m/s}{9.8m/s2} = 0.32 s (4)\)

       \(\Delta y = \frac{1}{2}* g * t^{2} = 5.5 m (5)\)

       \(t_{2} = \sqrt{\frac{2*\Delta y}{g}} = \sqrt{\frac{2*5.5m}{9.8m/s2} } = 1.06 s (6)\)

b)

      \(v_{ox} =\frac{\Delta x}{\Delta t} =\frac{3.25m}{1.38s} = 2.36 m/s (8)\)

       \(v_{oy} = 3.13 m/s (9)\)

      \(v_{o} = \sqrt{v_{ox}^{2} + v_{oy}^{2}} = \sqrt{(2.36m/s)^{2} + (3.13m/s)^{2}} = 3.92m/s (10)\)

       \(tg_{(\theta o)} = \frac{v_{oy}}{v_{ox} } = \frac{3.13}{2.36} = 1.33 (11)\)

       ⇒ θ₀ = tg⁻¹ (1.33) = 53.1º

c)

        \(v_{fy} ^{2} - v_{oy} ^{2} = 2* g* \Delta y (12)\)

       \(v_{fy} = \sqrt{2*g*\Delta_{y} } = \sqrt{2*9.8m/s2*5.5m} = -10.4 m/s (13)\)

       \(v_{f} = \sqrt{v_{fx}^{2} + v_{fy}^{2}} = \sqrt{(2.36m/s)^{2} + (-10.4m/s)^{2}} = 10.7 m/s (14)\)

       \(tg_{(\theta f)} = \frac{v_{fy}}{v_{fx} } = \frac{-10.4}{2.36} = -4.41 (15)\)

      ⇒ θf = tg⁻¹ (-4.41) = -77.2º

ANSWER:

444 N

STEP-BY-STEP EXPLANATION:

We can calculate the force, by means of Pascal's principle, which is stated as follows

We need to know F1, which would be the force exerted on the small input piston, we solve for F1:

The input force is exerted on the small input piston is 444 N

Answer:

48N

Explanation:

because as the distance is halfed. the force on the two objects are quadrupled.

Question :-

Answer :-

Explanation :-

As per the provided information in the given question, we have been given that the Velocity of the ball is 12 m/s . Time is given as 36 sec . And, we have been asked to calculate the Acceleration .

For calculating the Acceleration , we will use the Formula :-

\( \bigstar \: \: \boxed{ \sf{ \: Acceleration \: = \: \dfrac{v \: - \: u}{t} \: }} \)

Where ,

Therefore , by Substituting the given values in the above Formula :-

\( \dag \: \: \: \sf { Acceleration \: = \: \dfrac{Final \: Velocity \: - \: Initial \: Velocity}{Time} } \)

\( \longmapsto \: \: \sf { Acceleration \: = \: \dfrac{0 \: - \: 12}{36}} \)

\(\longmapsto \: \: \sf {Acceleration \: = \: \dfrac{ \: 12 \: }{36}}\)

\(\longmapsto \: \: \sf {Acceleration \: = \: \dfrac{ \: 1 \: }{3}}\)

\( \longmapsto \: \bf {Acceleration \: = \: 0.33 \: m/s^{2}} \)

Hence :-

\( \underline {\rule {212pt} {4pt}} \)

The graph of the velocity and the time  can be shown by option D.

We know that the movement of an object as it is falling under gravity would have a constant acceleration. The constant acceleration means that the velocity of the object is also held a constant.

We now have to look at the graphs as we have them here. The graph as it has been shown has the the velocity on the vertical axis and it has the time on the horizontal axis. The gradient of the slope is what we would refers to as the acceleration of the body.

We also need to recall that the acceleration has to be a constant since tje tow object would have to reach the ground at the same time if they are released from the same height at the same time.

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Answer:

The answer is D) diagonal line with varying slope, from 3 to 5 on USATestprep

Explanation:

The separation distance in B is a number times greater than in A, then the force of attraction in B is less than A.

The force of attraction between two charged object is determined by applying Coulomb's law.

Coulomb's law states that the force of attraction or repulsion between two charged objects is directly proportional to the product of the charges and inversely proportional to the square of the distance between the charges.

Mathematically, this law is written as;

F = Kq₁q₂/r²

where;

From the formula given above, as the distance of separation increases, the magnitude of the force of attraction between the charges decreases. Also, as the distance of separation between the charges decreases, the magnitude of the force of attraction between the charges increases.

Thus, the magnitude of the force of attraction between charged objects is a function of the magnitude of the charges and distance of separation between the charges.

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Answer:

C

Explanation:

Answer is C.

Answer:

The end that faces the north is called the north-seeking pole, or north pole, of the magnet. The other end is called the south pole. When two magnets are brought together, the opposite poles will attract one another, but the like poles will repel one another. This is similar to electric charges.

Explanation:

Answer:

North pole, south pole

Explanation:

The end of the pole that faces the north is the north pole, and the other end is called the south pole.

Opposite poles attract (North-South), but like poles repel (South-South, North-South).

The picture shown in the figure represents the Milky Way Galaxy. The Galaxy in which the entire solar system is present.

The million and trillion of stars in the universe form Galaxy. The galaxy in which the entire solar system is present is called Milky Way Galaxy. The Milky Way Galaxy is spiral in shape. This Galaxy has four major arms. The major arms have both old and young stars and the minor arms have the gas and star formation activity. This galaxy also has a black hole at its center. Galileo Galilei was the first to see the Galaxy.

The Milky Way Galaxy is made up of a dense cloud of gas that stretches across the sky as seen from the Earth. The age of the Milky Way Galaxy is 13.61 billion years ago and the Andromeda Galaxy is the nearest galaxy to the Milky Way Galaxy.

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In general, option c - warming it up on the stove - is often an effective method to increase the reaction rate.

Increasing the temperature of a reaction generally leads to faster reaction rates. This is because higher temperatures provide more thermal energy to the reactant particles, causing them to move faster and collide more frequently. The increased collision frequency and energy lead to more successful collisions and a higher likelihood of effective molecular interactions, which speeds up the reaction. On the other hand, options a and b - putting it in the fridge where it is cold or covering it with a blanket to make it dark - are unlikely to have a significant effect on the reaction rate. While temperature can influence reaction rates, cooling the reaction or making it dark typically reduces the kinetic energy of the particles, resulting in slower reaction rates. Option d - there is nothing you can do to change the speed of the reaction - is not accurate. The reaction rate can be influenced by various factors such as temperature, concentration, catalysts, and surface area, among others. By manipulating these factors, it is often possible to control and change the speed of a reaction. Hence option c, is correct

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An observer on the spacecraft would see the light beam moving away from the ship at the speed of light, while an observer on Mars would also see the light beam moving away from the ship at the speed of light. This is because the speed of light is always constant, regardless of the motion of the observer or the light source.

When the ball is hit from the deck of the yacht at 100 mph, it is moving at a speed relative to the yacht.

Since the yacht is moving at 20 mph, the ball would appear to be moving away from the yacht at 100 mph.

This is because the speed of the ball relative to the yacht is 100 mph, while the speed of the yacht relative to the ground is 20 mph.

Therefore, the total speed of the ball relative to the ground would be the sum of the speed of the yacht and the speed of the ball relative to the yacht, which is 120 mph (20 mph + 100 mph).

Now let's consider the Hermes spacecraft traveling at 0.1c past Mars and shining a laser from the front of the ship.

According to the theory of relativity, the speed of light is always the same for all observers, regardless of their motion or the motion of the light source.

So, regardless of the speed of the spacecraft, the light beam would travel away from the ship at the speed of light, c.

An observer on the spacecraft would see the light beam moving away from the ship at the speed of light, while an observer on Mars would also see the light beam moving away from the ship at the speed of light.

This is because the speed of light is always constant, regardless of the motion of the observer or the light source.

In summary, the theory of relativity tells us that the speed of light is always the same for all observers, regardless of their motion or the motion of the light source.

This means that the speed of light is a fundamental constant of the universe, and it plays a crucial role in our understanding of the laws of physics.

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Protons, neutrons, and electrons are the three main subatomic particles that make up atoms.

An oxygen atom has 8 protons, 8 electrons, and its number of neutrons may vary depending on the isotope of oxygen. The more frequently encountered isotope of oxygen is oxygen-16, with 8 neutrons.

The element with 13 protons is aluminum (Al). To find the mass number, we add the number of protons and neutrons in the nucleus. Therefore, the mass number of this aluminum isotope would be 13 + 14 = 27.

If an atom has 7 electrons, it must be nitrogen (N), which has an atomic number of 7.

Number of neutrons = Mass number - Atomic number

Thus, we obtain the number of neutrons by the equation: 14 - 7 = 7

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Answer:

the intellectual and practical activity encompassing the systematic study of the structure and behaviour of the physical and natural world through observation and experiment.

The displacement of a 1.5 kg mass is then determined using the formula x = F/k.  stretching a spring 2 cm from its equilibrium position need twice as much effort as stretching it a distance of x

W = 1/2kx2 = 1.96 Joules.

Actually, it requires more than twice as much labour since, as the spring extends, more power is needed to do so.

The shear strength and shear modulus of a compression spring formed of music wire with a 2mm diameter are 800 MPa and 80 GPa, respectively.

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Answer:

150   kg-m/s

Explanation:

momentum = m * v

50 * 3 = 150

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.  

A 0.842g sample of Hydrogen-3 decays to 0.0526g. Approximately 4.206 half-lives have occurred.

To determine the number of half-lives that have occurred, we can use the decay equation and the concept of exponential decay. The decay equation for radioactive decay is given by:

N(t) = N₀ * (1/2)^(t/T),\((1/2)^(^t^/^T^),\)

where N(t) is the remaining amount of the substance at time t, N₀ is the initial amount, t is the time elapsed, and T is the half-life of the substance.

In this case, we have an initial mass of 0.842g (N₀) and a remaining mass of 0.0526g (N(t)). We can set up the equation as follows:

0.0526g = 0.842g \(* (1/2)^(^t^/^1^2^.^3^2)\),

where t represents the number of half-lives that have occurred.

To solve for t, we can take the logarithm of both sides of the equation:

log(0.0526g/0.842g) = log\([(1/2)^(^t^/^1^2^.^3^2^)\)].

Using the logarithmic property log(\(a^b\)) = b*log(a), we can rewrite the equation as:

log(0.0526g/0.842g) = (t/12.32) * log(1/2).

Simplifying further:

log(0.0526g/0.842g) = (t/12.32) * (-log2),

where log2 is the logarithm base 2.

Now, we can solve for t:

t = (12.32 * log(0.0526g/0.842g)) / (-log2).

Using the given values and performing the calculation, we find:

t ≈ 4.206.

Therefore, approximately 4.206 half-lives have occurred.

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Answer:

The form of energy that can move from place to place across the universe is electromagnetic energy.

Electromagnetic energy is a form of energy that is produced by the movement of electrically charged particles. It is characterized by its wavelength or frequency and can range from gamma rays, X-rays, ultraviolet (UV) rays, visible light, infrared radiation, microwaves, to radio waves.

Notice that the K vs t graph passes through the point (0,0), which means that the initial kinetic energy was 0.

The kinetic energy increases linearly with time to reach 12J in 3s. The slope of a line is equal to the ratio of the change in the dependent variable over a change in the independent variable. In this case, the kinetic energy is the variable that depends on time. Then, the slope of the line is the ratio of a change in kinetic energy over a change in time.

The slope of a K vs t graph represents how fast the kinetic energy is transferred to the object. The magnitude that measures a change in energy over time is called Power.

Then, the slope of the graph (power delivered to the object) is:

Therefore, the correct choice is option E) The power delivered to the object is 4W.

Answer:

a. 60 m

b. 71.48 m

Explanation:

Below are the calculations:

a. The phone's height above the ground = Speed x Time

   The phone's height above the ground = 15 x 4 = 60 m

b. Speed when phone drops, u = 15 m/s

At maximum height, v = 0

Use below formula:

v² = u² -2gh

0 = 15² + 2 × 9.8 × h

h = 11.48 m

Total height = 60 + 11.48 = 71.48 m

Answer:

31 km/h in the direction which the bus is travelling

Explanation:

ball rolls at a speed of 4 km/h towards the bus back. Thus, V_bb = 4 km/h

Velocity of the bus relative to the ground; V_bg = 35 km/h

Let v_br be velocity of the ball relative to the ground.

If we define the direction to be positive, we have;

v_br = V_bg - v_bb

v_br = 35 - 4

v_br = 31 km/h in the direction which the bus is travelling

The energy stored in the circuit at any time t is given by \(U = (1/2)L*I^{2} + (1/2)Q^{2} /C = (1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C))} + (1/2)C*V_{0} ^{2} *(1 - e^{(-2t/(R*C)})).\)The units are in seconds.

The total energy stored in the circuit can be calculated using the formula: U = (1/2)L*I² + (1/2)Q²/C, where L is the inductance, I is the current, Q is the charge on the capacitor, and C is the capacitance.

Initially, the capacitor is uncharged, so the second term is zero.

Therefore, the initial energy stored in the circuit is U₀ = (1/2)L*I₀², where I₀ is the initial current, which is zero.

When the magnetic field is switched on, a current begins to flow in the solenoid.

This current increases until it reaches its maximum value, given by I = V/R, where V is the voltage across the solenoid and R is its resistance.

Since the solenoid is connected in series with the capacitor, the voltage across the solenoid is equal to the voltage across the capacitor, which is given by V = Q/C, where Q is the charge on the capacitor.

The charge on the capacitor is given by Q = C*V, where V is the voltage across the capacitor at any time t.

Therefore, we have I = V/R = Q/(R*C) = dQ/dt*(1/R*C), where dQ/dt is the rate of change of charge on the capacitor.

This is a first-order linear differential equation, which can be solved to give \(Q(t) = Q_{0} *(1 - e^{(-t/(R*C)}))\), where Q₀ is the maximum charge on the capacitor, given by Q₀ = C*V₀, where V₀ is the voltage across the capacitor at t=0.

The current in the solenoid is given by I(t) = \(dQ/dt*(1/R*C) = (V_{0} /R)*e^{(-t/(R*C)}).\)

The energy stored in the circuit at any time t is given by\(U = (1/2)L*I^{2} + (1/2)Q^{2} /C = (1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C))} + (1/2)C*V_{0} ^{2} *(1 - e^{(-2t/(R*C)})).\)

The time t at which the energy stored in the circuit drops to 10% of its initial value can be found by solving the equation U(t) = U₀/10, or equivalently, \((1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C)}) + (1/2)C*V_{0} /R)^{2}*(1 - e^{(-2t/(R*C)})) = (1/20)L*I_{0} /R)^{2}.\)

This equation can be solved numerically using a computer program, or graphically by plotting U(t) and U₀/10 versus t on the same axes and finding their intersection point.

The solution is t = 1.74 ms.

The units are in seconds.

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The net force acting on the particle would be and the magnitude of the force would be -12.92i + 6.49j and the magnitude of the force would be 14.46 Newtons.

Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.

The net force on the particle = mass of the particle × acceleration

                                                =3.80*(−3.40i + 1.70j)

                                                = -12.92i + 6.49j

Thus, the magnitude of the net force on the particle would be 14.46 Newtons.

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The film is positioned in the mouth perpendicular to the long axis of the tooth being radiographed, in accordance with the fundamental principles of the paralleling technique.

The parallel technique is the alignment of the central axis of the X-ray source so that it is perpendicular to both the plane of the film and the object being investigated. Unfortunately, aligning the X-ray central axis perpendicular to the film is much easier than to the object. The paralleling approach is defined by two fundamental principles: The x-ray beam is directed at right angles (perpendicular) to the film or sensor and the long axis of the teeth being radiographed in (1) and (2), respectively.

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Answer:

If the particles of the medium vibrate in a direction parallel to the direction of propagation of the wave, it is called a longitudinal wave. In longitudinal waves, the particle movement is parallel to the direction of wave propagation.

Explanation:

Answer:

56.25J

Explanation:

Kinetic energy = 1/2 x 4.50 x (5.00)^2= 1/2 x 4.50 x (5.00 x 5.00)= 1/2 x 4.50 x 25Multiply through = 112.5/2= 56.25J

The frictional force involved when a penny sinks to the bottom of a wishing well is primarily due to viscous drag or fluid friction. As the penny moves through the water, it experiences resistance from the surrounding fluid. This resistance is caused by the frictional forces between the water molecules and the penny's surface.

The electric field strength at a distance of 10 cm from a charge of 2 μC is 3.6 x 10⁵ N/C.

If there's a charge present in any form, an electric field is linked with every point in space. The strength and direction of the electric field are expressed by the value of E, sometimes referred to as the strength of the electrical field, electric field intensity, or simply the electric field.

The following formula determines the magnitude of the electric field generated by a point charge:

where,

E = Electrical Field Strength = ?

k = 9 x 10⁹ Nm²/C²

q = magnitude of charge

q = 2 μC = 2x 10⁻⁶ C

r = distance = 10 cm

r = 0.10 m

Therefore,

When the values are entered into the equation, we obtain:

E = 9 x 10⁹ * (2 x 10⁻⁶) / (0.1)²

E = 3.6 x 10⁵ N/C

Because of this, the electric field intensity at 10 cm from a 2 μC charge is  3.6 x 10⁵ N/C.
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