calculate the magnitude and the direction of the resultant forces​

To determine theangle to which the balls wing to the opposite side, C. Both conservation of momentum and conservation of mechanical energy (option c).

Conservation of energy and momentum is one of the most important and useful principles in physics. Conservation of momentum states that if no external forces are acting on a system of bodies, the total momentum is always the same (conserved). Furthermore, if we can give a potential to all forces, whether external or internal, the total energy is also constant. This is the law of conservation of energy.

1.Conservation of energy to calculate the velocity of the first ball of clay by the time it descends:

mgh=1/2mu2

[calculate h by using h=L(1-cos45)]

2.Conservation of momentum to find out the final velocity of the balls of clay using:

m1u1+m2u2=(m1+m2)v

3.Conservation of energy to calculate the max height the balls of clay will go by using:

1/2(m1+m2)v2=mgh

Once you have h,

h=L(1-cosθ), solve for θ

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Answer:m 1= 36 g=0.036 kg - the mass of ice T1= −77 ◦C. - temperature of ice m2=589 g=0.589 kg - mass of water C1=2090 J/kg · ◦C - specific heat of ice  λ = 3.33 × 10^5 J/kg - latent heat of fusion of water T2=26◦C. - temperature of water C2= 4186 J/kg · ◦C . - - specific heat of water m3=74 g=0.074kg - mass of copper T3=26◦C. - temperature of copper C3=387 J/kg ·◦C - specific heat of copper is and of ice is T - ? - final temperature  1  1 ( 0 −  1 ) −  ℎ  0  1  −  1  2 (  − 0 ) −  ℎ   2  2 (  −  2 ) −   3  3 (  −  3 ) −  m 1 ​ C 1 ​ (0−T 1 ​ )−ice heating to 0 o C m1λ−ice melting m 1 ​ C 2 ​ (T−0)−melted water heating m 2 ​ C 2 ​ (T−T 2 ​ )−water cooling m 3 ​ C 3 ​ (T−T 3 ​ )−copper cooling  1  1 ( 0 −  1 ) +  1  +  1  2 (  − 0 ) +  2  2 (  −  2 ) +  3  3 (  −  3 ) = 0 m 1 ​ C 1 ​ (0−T 1 ​ )+m1λ+m 1 ​ C 2 ​ (T−0)+m 2 ​ C 2 ​ (T−T 2 ​ )+m 3 ​ C 3 ​ (T−T 3 ​ )=0 0.036 ⋅ 2090 ⋅ 77 + 0.036 ⋅ 3.33 ⋅ 1 0 5 + 0.036 ⋅ 4186 ⋅  + 0.589 ⋅ 2090 ⋅ (  − 26 ) + 0.074 ⋅ 387 ⋅ (  − 26 ) = 0 0.036⋅2090⋅77+0.036⋅3.33⋅10 5 +0.036⋅4186⋅T+0.589⋅2090⋅(T−26)+0.074⋅387⋅(T−26)=0 1410.344 ⋅ = 14969.368  = 10.6 1  1410.344⋅T=14969.368 T=10.61 o C Answer:  = 10.6 1  T=10.61 o C

Explanation:

Answer:

yes

Explanation:

Approximately 4300 hours is required for the decay of 2/3 of its nuclei.

Radioactive decay is the random process by which the nuclei of radioactive substances disintegrate into smaller particles. The time taken for half of the radioactive nuclei in a substance to decay is known as the half-life of the substance. In the following paragraphs, I'll explain how to calculate how much time is required for the decay of 2/3 of its nuclei if the half-life of a radioactive element is 2000 hours.The half-life of a radioactive element is the time it takes for half of its original nuclei to decay. The quantity of the substance that has decayed by half of its original quantity is called the half-life. If the half-life of a substance is T, the fraction of the original amount of the substance that remains after a time t is given by the equation:N(t) = N₀(1/2)^(t/T),Where N₀ is the initial number of radioactive nuclei and N(t) is the number of radioactive nuclei remaining after a time t. If we want to know the time it takes for two-thirds of the nuclei to decay, we must solve for t when N(t)/N₀ = 1/3. Putting this into the equation, we have:(1/3)N₀ = N₀(1/2)^(t/T).Simplifying this equation, we get:(1/2)^(t/T) = 1/3.Dividing both sides of the equation by (1/2)^(2000/T), we get:(1/3)/(1/2)^(2000/T) = (1/2)^(t/T - 2000/T).Taking the logarithm of both sides, we get:(t/T - 2000/T)log(1/2) = log(1/3).Simplifying this equation, we get:t/T = -log(1/3)/log(1/2) + 2000/T.To get an approximate solution to this equation, we can make use of the fact that the half-life of a substance is much smaller than the time it takes for most of its nuclei to decay. Therefore, the quantity 2000/T is much larger than one. This allows us to neglect the second term on the right-hand side of the equation. Then we have:t/T = 2.15.Substituting the value of T as 2000 hours, we get:t = 4300 hours.

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(a) The amplitude of the wave is 0.2 m.

(b) The period of the wave is  4 s.

(c) The wavelength of the wave is 100 m.

(a) The amplitude of the wave is the maximum displacement of the wave.

amplitude of the wave = 0.2 m

(b) The period of the wave is the time taken for the wave to make one complete cycle.

period of the wave = 5.5 s - 1.5 s = 4 s

(c) The wavelength of the wave is calculated as follows;

λ = v / f

where;

f = 1/t = 1 / 4s = 0.25 Hz

λ = ( 25 m/s ) / 0.25 Hz

λ = 100 m

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Answer:

Option (C)

Explanation:

Vector A has two components,

Ax = 3.5 and Ay = 5.7

Therefore, from the figure attached,

From triangle ABC,

BC represents the vertical component and AC represents the horizontal component of vector A.

For the angle of inclination,

tanθ = \(\frac{\text{Vertical component}}{\text{Horizontal component}}\)

       = \(\frac{3.5}{5.7}\)

θ = \(\text{tan}^{-1}(\frac{35}{57})\)

θ = 31.55 degrees

Therefore, Option (C) will be the correct option.

The same unbalanced force acting on a 2.0-kilogram object will accelerate this object toward the north at 6.0 m/s².

What is Newton's Second Law?
According to Newton's second law, an object's acceleration is determined by its mass and the net force that is acting on it. The body's acceleration is inversely related to its mass and directly proportional to the net force applied on it

According to Newton's second Law of motion the resultant force is directly proportional to the rate of change in momentum.

Therefore; F = ma , where F is the force, m is the mass and a is the acceleration

F = 4 × 3
 = 12 Newtons
The same force acts on a 2.0 kg object
F = ma
a = F/ m
 = 12 /2
 = 6.0 m/s²

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The description that tells two processes that scientists think move Earth's lithospheric plates is convection in the mantle and pressure of magma on the edge of the plate.

The Earth's lithosphere is the rocky outer part of Earth which is made up of the brittle crust and the top part of the upper mantle.

The Earth's lithosphere deflects the convections and as the convections churn clockwise of anticlockwise, they drag the  lithosphere with it via friction an this is what is stipulated to cause tectonic plate movements.  

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Answer: convection in the mantle and pressure of magma on the edge of the plate

Explanation: I took the unit test

The rocket's x and t coordinates are 688.32 meters and 277.44 meters, respectively, from where it will launch.

Velocity, as opposed to speed, refers to the pace & direction of such an object's movement as it moves down a path. In other respects, whereas velocity is a scalar, speed is really a scale parameter.

Speed does not necessarily have to match average velocity in magnitude. Many people mistakenly believe that cruising velocity and flow rate are merely two different labels for the same quantity. Nevertheless, average speed relies on both distance and displacement.

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The pressure of the reading of tube is given as,

The relation between the before and after pressure is,

where v1 is the velocity of the speed initially, and v2 is the velocity at the final state.

Substituting the known values,

Thus, the final pressure reading of the pilot tube is

Answer:

The initial velocity of this shark is \(8.0\; \rm m \cdot s^{-1}\). (Assuming that the unit of the acceleration in this question is \(\rm m\cdot s^{-2}\).)

Explanation:

Let \(a\) denote the acceleration of this shark.

Let \(v_0\) denote the initial velocity of this shark.

Assume that the acceleration \(a\) of this shark is constant (as it is in this question.) Over a period of time \(t\), the shark would have travelled a distance of:

\(\displaystyle x = \frac{1}{2}\, a\, t^2 + v_0\, t\).

This question states that:

This question is asking for \(v_0\), the initial velocity of this shark at the beginning of this \(0.8\)-second period. Substitute the three known values into the equation:

\(\displaystyle 11.52 = \frac{1}{2}\times 16\times (0.8)^2 + 0.8\, v_0\).

Solve for \(v_0\):

\(v_0 = \displaystyle \frac{11.52 - (1/2) \times 16 \times (0.8)^2}{0.8} = 8.0\; \rm m \cdot s^{-1}\).

In contrast to data in a database, data in a data warehouse is described as subject-oriented, which means that it focuses on a specific area.

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Answer:

5

Explanation:

give thanks pls

After considering the given data we come to the conclusion that the increase in the internal energy of the tie and spike is 16.322 J, under the condition that  a given weight of  0.76 kg spike had been jamed into a railroad tie .

In order to evaluate the increase in internal energy of the tie and spike we have to apply the formula of kinetic energy which is
1/2 × mv²
Here,
m = mass
v = velocity
Staging the values in the formula
Kinetic energy of the spike = (1/2)× (0.76)×(4.8)²= 87.2064 J
Energy absorbed by the tie and spike = 87.2064 J × 0.187 = 16.322 J

Hence, the evaluate rise in internal energy of the tie and spike is 16.322 J.
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Louis Armstrong's average velocity in the time trial was approximately 1.078 km/min.

To calculate the average velocity, divide the total displacement by the total time taken. In this case, Louis Armstrong rode his bike 55 km to the east in a time trial lasting 51 minutes.

Average Velocity = Displacement / Time

Displacement = 55 km (since he rode 55 km to the east)

Time = 51 minutes

Average Velocity = 55 km / 51 min

To express the average velocity in km/min, convert the time from minutes to minutes.

1 hour = 60 minutes

Average Velocity = 55 km / 51 min × (1 hour / 60 min)

Average Velocity = 55 km / 51 min × (1/60) hour

Simplifying the expression:

Average Velocity = 1.078 km/min

Therefore, Louis Armstrong's average velocity in the time trial was approximately 1.078 km/min.

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Answer:

It is Conductivity because it is the measure of the ease.

Answer:

There are no options, so....

Explanation:

Chlorophyll a absorbs its energy from the Violet-Blue and Reddish orange-Red wavelengths, and little from the intermediate (Green-Yellow-Orange) wavelengths.

The distance of your hometown from school is 318.64 km.

Given data

Distance S = 190 km

Speed v = 97 km / h

Now, the time taken to travel S distance is calculated as,

t = S / v

t = 180 / 95

t = 1.958 hrs

Now, the total time is given as

Total time T = 4.0 h

So, remaining time t ' = T - t = 4 hrs - 1.958 hrs = 2.042 hrs

Now, the velocity after travel 190 km is v ' = 63 km/ h

Distance travel in this velocity S ' = v ' t '

= 128.6 km

Now, the distance of your hometown from school , S " = S + S '

= 190 km + 128.6 km = 318.64 km

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The dragonfly must generate approximately 4.8 × 10^-4 Joules of energy to spin itself at a rate of 1100 rpm.

Start by converting the rotational speed from rpm (revolutions per minute) to rad/s (radians per second). Since 1 revolution is equal to 2π radians, we can use the conversion factor:

Angular speed (ω) = (1100 rpm) × (2π rad/1 min) × (1 min/60 s)

ω ≈ 115.28 rad/s

The moment of inertia (I) is given as 2.0 × 10^-7 kg⋅m².

Use the formula for rotational kinetic energy:

Rotational Kinetic Energy (KE_rot) = (1/2) I ω²

Substituting the given values:

KE_rot = (1/2) × (2.0 × 10^-7 kg⋅m²) × (115.28 rad/s)²

Calculate the value inside the parentheses:

KE_rot ≈ (1/2) × (2.0 × 10^-7 kg⋅m²) × (13274.28 rad²/s²)

KE_rot ≈ 1.331 × 10^-3 J

Round the result to the proper number of significant figures, which in this case is three, as indicated by the given moment of inertia.

KE_rot ≈ 4.8 × 10^-4 J

Therefore, the dragonfly must generate approximately 4.8 × 10^-4 Joules of energy to spin itself at a rate of 1100 rpm.

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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 9 kg explosive detonates into two fragments, each weighing 3 kg and 6 kg. The kinetic energy of mass 6 kg is 1.92 J, and the velocity of mass 3 kg is 1.6 m/s.

Therefore,

The total momentum of the bomb is unchanged before and after the explosion thanks to the conservation of linear momentum.

0 = 3 × 1.6 + 6v

v = 0.8 m/s

Kinetic energy of mass 6 kg is:

1/2 mv² = 1/2 × 6 × 0.8²

= 1.92 J

Hence, the correct option is c) 1.92J

The complete question is:

A bomb of mass 9kg explodes into 2 pieces of mass 3kg and 6kg. The velocity of mass 3kg is 1.6m/s, the kinetic energy of mass 6kg is :

A) 3.84J

B) 9.6J

C) 1.92J

D) 2.92J

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Answer:

1.29*10^6 N/C

1135.6 V

9.18 cm

Explanation:

Given that

radius of the metal, r = 19 cm

charge of the metal, q = 2.4*10^-8 C

coulomb's constant, k = 8.99*10^9

to find the electric field, we use the formula E = kq/r², where

E = electric field

k = coulomb constant

q = charge on the metal and

r = radius of the metal

E = (8.99*10^9 * 2.4*10^-8) / 0.19²

E = 215.76 / 0.0361

E = 1.29*10^6 N/C

to find the electric potential, we use this relation

V = kq/r

V = (8.99*10^9 * 2.4*10^-8) / 0.19

V = 215.76 / 0.19

V = 1135.6 V

V = kq/r,

r = kq/V

r = (8.99*10^9 * 2.4*10^-8)/ (1135.6 - 370

r = 215.76 / 765.6

r = 0.281 = 28.1 cm

distance from the sphere

28.18 - 19 = 9.18 cm

In precipitation, Rainwater can fall as either Liquid or Solid.

Precipitation is a part of the water cycle. Precipitation falls to the floor as snow and rain. It subsequently evaporates and rises back into the surroundings as a gas. In clouds, it turns back into a liquid or stable water, and it falls to Earth again

The water cycle shows the non-stop motion of water within the Earth and its surroundings. it's miles a complex system that consists of many one-of-a-kind techniques. Liquid water evaporates into water vapor and condenses.

There are four fundamental ranges in the water cycle. they are evaporation, condensation, precipitation, and collection. permit's examined every one of these ranges. Evaporation: that is whilst warm temperature from the sun causes water from oceans, lakes, streams, ice, and soils to upward push into the air and grow to be water vapor.

Disclaimer: your question is incomplete, please see below for the complete question.

Liquid

Steam cloud

2nd half

Steam

Solid

Cloud

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Answer:

3201.304 J

Explanation:

Use ideal gas equation to initial stage:

PV=nRT

P * 0.034 = 1.8 * 8.314 * 250

P = 110038.2353 Pa

Use ideal gas equation to final stage:

PV=nRT

P * 0.08 = 1.8 * 8.314 * 250

P = 46766.25 Pa

Process is isothermal (constant temperature )

Therefore,

Work= C ln (V2/V1)

(P1V1=P2V2=C)

(Above equation is taken by integration of P.dv)

Work = P1V1 ln (V2/V1) = P2V2 ln (V2/V1)

By substituting above data to the equation:

Work = (110038.2353 * 0.034) * ln (0.08/0.034)

Work = 3201.304 J

The kinetic energy of the projectile that is shot is higher than the kinetic energy of the cannon that is recoiling.

In our situation, the bowling pin exerts an equivalent force in the opposite direction on the bowling ball when the bowling ball applies a force to it.

While lighter things have less momentum, massive ones have more momentum for a given speed. Because of this, stopping a loaded vehicle requires more force than stopping an empty one. Likewise, items that move quicker have greater momentum than those that move slower.

According to Newton's third law, every force applied to an item will result in an equal and opposite force being applied to the object applying the force.

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Wundt studies consciousness using introspection because his subjects examined and shared their own feelings and thoughts. Wundt was the first psychologist.

Introspection refers to the examination of one's own mental and/or emotional mechanisms and processes.

Wilhelm Wundt was a distinguished psychologist who demonstrated that introspection is a highly practiced mechanism of self-examination.

W. Wundt developed the theory of conscious thought by indicating that conscious mental states can be scientifically (empirically) studied by using introspection.

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Answer: B His subjects examined and shared their own feelings and thoughts.

Explanation: Just took the test

Answer:

30 m/s

Explanation:

Applying,

v = u+at................ Equation 1

Where v = final speed of the ball, u = initial speed of the ball, a = acceleration, t = time.

From the question,

Given: u = 0 m/s (stationary), a = 600 m/s², t = 0.05 s

Substitute these values into equation 5

v = 0+(600×0.05)

v = 30 m/s

Hence the speed at which the ball leaves the player's boot is 30 m/s

Answer:

Average acceleration = 6m/s²

Explanation:

Given the following data;

Initial velocity, u = 4m/s

Final velocity, v = 22m/s

Time, t = 3 seconds

To find the average acceleration;

Acceleration = (v - u)/t

Substituting into the equation, we have;

Acceleration = (22 - 4)/3

Acceleration = 18/3

Acceleration = 6m/s²

Answer:

Explanation:

To find the current flowing in the circuit, we can use Ohm's Law and Kirchhoff's circuit laws.

Ohm's Law states that the current (I) flowing through a circuit is equal to the voltage (V) divided by the resistance (R):

I = V / R

In this case, the voltage (V) is the electromotive force (EMF) of the current source, which is 14 V. The total resistance (R) in the circuit is the sum of the internal resistance (r) and the resistances of the two resistors (R1 and R2):

R = r + R1 + R2

Given that the internal resistance (r) is 1Ω and each resistor (R1 and R2) has a resistance of 3Ω, we can substitute these values into the equation:

R = 1Ω + 3Ω + 3Ω = 7Ω

Now we can calculate the current (I):

I = V / R = 14 V / 7Ω = 2 A

Therefore, the current flowing in the circuit is 2 Amperes.