Only a fraction of the electric energy supplied to a tungsten lightbulb is converted to visible light. The rest of the energy shows up as infrared radiation (i.e., heat). A 25−W lightbulb converts 15.0 percent of the energy supplied to it into visible light (assume the wavelength to be 505 nm). How many photons are emitted by the light bulb per second (1 W = 1 J / s)

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 3.8 × 2.5

We have the final answer as

Hope this helps you

The net force on particle q₂, located between particles q₁ and q₃, is approximately 189000 N. The force exerted by particle q₁ on q₂ is positive and equals 252000 N, while the force exerted by particle q₃ on q₂ is negative and equals -63000 N.

To find the net force on particle q₂, we need to calculate the individual forces exerted on q₂ by particles q₁ and q₃ and then determine their sum.

The force between two charged particles can be calculated using Coulomb's law:

F = k * |q₁ * q₂| / r²

Where F is the force between the particles, k is the electrostatic constant (k ≈ 9.0 x \(10^9\) Nm²/C²), q₁ and q₂ are the charges of the particles, and r is the distance between them.

First, let's calculate the force exerted on q₂ by q₁:

F₁₂ = k * |q₁ * q₂| / r₁₂²

F₁₂ = (9.0 x \(10^9\) Nm²/C²) * |(8.0 μC) * (3.5 μC)| / (0.10 m)²

F₁₂ ≈ 252000 N

The force is positive because q₁ and q₂ have opposite charges.

Next, let's calculate the force exerted on q₂ by q₃:

F₂₃ = k * |q₂ * q₃| / r₂₃²

F₂₃ = (9.0 x \(10^9\)Nm²/C²) * |(3.5 μC) * (-2.5 μC)| / (0.15 m)²

F₂₃ ≈ -63000 N

The force is negative because q₂ and q₃ have the same charge.

Finally, we can find the net force on q₂ by summing the individual forces:

Net force = F₁₂ + F₂₃

Net force = 252000 N + (-63000 N)

Net force ≈ 189000 N

The net force on particle q₂ is approximately 189000 N.

For more such information on: net force

https://brainly.com/question/14361879

#SPJ8

Answer:

The horizontal range of the cannonball is 9180.4 meters.

Explanation:

The horizontal range of a cannonball can be calculated using the following formula:

range = (speed * speed * sin (2 * angle)) / acceleration due to gravity

In this case, we are given that the cannon is fired at an angle of 45 degrees and that the speed of the ball is 300 m/s. We can plug these values into the formula to calculate the horizontal range of the ball:

range = (300 * 300 * sin (2 * 45)) / 9.8

= (90000 * sin (90)) / 9.8

= (90000 * 1) / 9.8

= 9180.4 m

Answer:

3.456 × 10^(8) m

Explanation:

We are given;

Mass of earth; m_e = 5.98 × 10^(24) kg

Mass of moon; m_m = 7.36 × 10^(22) kg

Distance from earth to moon; L = 3.84 × 10^(8) m.

We are told that the Moon’s gravitational pull is stronger than that of Earth’s.

Thus;

F_EA ≤ F_MA

Formula for force due to gravity is;

F = GMm/r²

Applying to this question, we have;

(Gm_e•m_a)/x² ≤ (Gm_m•m_a)/(l - x)²

Where x is his distance from the center of the Earth

Now, G and m_a will cancel out from both sides and we plug on other values to get;

(5.98 × 10^(24))/x² ≤ (7.36 × 10^(22))/(3.84 × 10^(8) - x)²

Taking square root of both sides gives;

2445403852127.4966/x ≤ 271293199325.0107/(3.84 × 10^(8) - x)

Rearranging gives;

2445403852127.4966(3.84 × 10^(8) - x) ≤ 271293199325.0107x

Simplifying this gives;

9.0139(3.84 × 10^(8) - x) ≤ x

(34.613 × 10^(8)) - 9.0139x ≤ x

(34.613 × 10^(8)) ≤ x + 9.0139x

(34.613 × 10^(8)) ≤ 10.0139x

x ≥ (34.613 × 10^(8))/10.0139

x ≥ 3.456 × 10^(8)

Answer:

W=1560 Joules

Explanation:

Use W=fd

so, W=520N(3m)= 1560J

Answer:

\(f = g \times \frac{m1 \times m2}{ {d}^{2} } \)

\(f = 6.67 \times {10}^{ - 11} \times \frac{40 \times 40}{9} \)

The electron number cannot be used instead because electrons can be removed from or added to an atom (option C)

The element of an atom is determined by its proton count, while the electron count can exhibit variability. Take, for instance, a sodium atom, which encompasses 11 protons and 11 electrons. However, it has the capacity to relinquish one electron, transforming into a sodium ion housing only 10 electrons.

This occurs due to the relatively loose binding of electrons to the nucleus, enabling their removal through the influence of an electric field or alternative mechanisms.

Learn about electron here https://brainly.com/question/13998346

#SPJ1

The value of m is 0.534 kg

f1 =0.92  Hz

f2 = 0.58  Hz

Δm = m2 - m1 =  800 g kg

Mass-spring oscillation occurs with a frequency of:

f = (1/2π)√(k/m)

k = m(2πf)²

When we have: we may conclude that k is constant for both trials.

k = k

m1(2πf1)² = m2(2πf2)²

m1 = m2(f2/f1)²

m1 = (m1+Δm)(f2/f1)²

m1 = Δm/((f1/f2)²-1)

m 1 = 0. 800 g/ (0.92/0.58)^2 – 1

= 0.534 kg or 0.53kg or 534 g

A mass m linked to a spring produces an oscillation with a period of 4 s. A spring's time period lengthens by 1 s when a 2 kg additional mass is added to it.

To know more about oscillates  visit:-

https://brainly.com/question/28994371

#SPJ4

True, Food manufacturers create fats by adding hydrogen to fats.

Hydrogenation is used in the oil industry in making trans fats or margarine by increasing the melting point through reducing the carbon-to-carbon double bonds..

During hydrogenation, unsaturated fats undergo a chemical reaction in the presence of hydrogen gas and a catalyst, typically nickel.

The process involves the addition of hydrogen atoms to the carbon double bonds in the fatty acid molecules, resulting in the conversion of some of the double bonds into single bonds.

This reduces the number of unsaturated bonds in the fat, making it more saturated.

So we can conclude that the statement is true.

Learn more about Hydrogenation here: https://brainly.com/question/13910028

#SPJ1

The hiker travelled 4.02 km North/South and 4.74 km East/West during her hike.

This question involves vector addition, the resolution of vectors, the use of bearings, and trigonometry in the calculation of the hiker's movement.

This may appear to be a difficult problem, but with some visual aid and the proper use of mathematical formulas, the issue can be addressed correctly.

Resolution of VectorThe resolution of a vector is the process of dividing it into two or more components.

The angle between the resultant and the given vector is equal to the inverse tangent of the two rectangular components.

Angles will always be expressed in degrees in the solution.

The sine, cosine, and tangent functions in trigonometry are denoted by sin, cos, and tan.

The tangent function can be calculated using the sine and cosine functions as tan x = sin x/cos x. Also, in right-angled triangles, Pythagoras’ theorem is used to find the hypotenuse or one of the legs.

Distance Travelled North/SouthThe hiker traveled North for the first part of the hike and South for the second.

The angles that the hiker traveled in the first part and second parts are 53 degrees and 17 degrees, respectively.

The angle between the two is (180 - 53 - 17) = 110 degrees.

The angle between the resultant and the Northern direction is 110 - 53 = 57 degrees.

Using sine and cosine, we can calculate the north/south distance traveled to be 5 sin 57 = 4.02 km, and the east/west distance to be 5 cos 57 = 2.93 km.

Distance Travelled East/WestThe hiker walked East for the second part of the hike.

To calculate the distance travelled East/West, we must first calculate the component of the first part that was East/West.

The angle between the vector and the Eastern direction is 90 - 53 = 37 degrees.

Using sine and cosine, we can calculate that the distance travelled East/West for the first part of the hike is 5 cos 37 = 3.88 km.

To determine the net distance travelled East/West, we must combine this component with the distance travelled East/West in the second part of the hike.

The angle between the second vector and the Eastern direction is 17 degrees.

Using sine and cosine, we can calculate the distance traveled East/West to be 3 sin 17 = 0.86 km.

The net distance traveled East/West is 3.88 + 0.86 = 4.74 km.

Therefore, the hiker travelled 4.02 km North/South and 4.74 km East/West during her hike.

For more questions on travelled

https://brainly.com/question/750474

#SPJ8

Answer:

Diagram (3).

Explanation:

N3L states that if object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A (\(F_{A} = -F_{B}\)).

The diagram which best represents the gravitational forces, F, between a satellite, S, and Earth is; Choice (3).

The law clearly states a Force of attraction; the two objects come towards each other.

Consequently, Choice (3) best represents the gravitational forces, F, between a satellite, S, and Earth.

Read more:

https://brainly.com/question/11460810

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:

gravity

Explanation:

I don't know what the explanations would be

Based on the information, it should be noted that the resistance R is 0.5 Ω.

When S1 and S2 are open, i leads e by 30°. In this case, the circuit consists of only the inductor (L) and the capacitor (C) in series. Therefore, the impedance of the circuit can be written as:

Z = jωL - 1/(jωC)

Since i leads e by 30°, we can express the phasor relationship as:

I = k * e^(j(ωt + θ))

Z = jωL - 1/(jωC) = j(120π)L - 1/(j(120π)C)

Re(Z) = 0

By equating the real parts, we get:

0 = 0 - 1/(120πC)

Let's assume that there is a resistance (R) in series with the inductor and capacitor. The impedance equation becomes:

Z = R + jωL - 1/(jωC)

Z = R + jωL

Im(Z) = ωL > 0

Substituting the angular frequency and rearranging the inequality, we have:

120πL > 0

L > 0

This condition implies that the inductance L must be greater than zero.

When S1 and S2 are closed, i has an amplitude of 0.5 A. In this case, the impedance is:

Z = R + jωL - 1/(jωC)

Since the amplitude of i is given as 0.5 A, we can express the phasor relationship as:

I = 0.5 * e^(j(ωt + θ))

By substituting this phasor relationship into the impedance equation, we can determine the value of R. The real part of the impedance must be equal to R:

Re(Z) = R

Since the amplitude of i is 0.5 A, the real part of the impedance must be equal to 0.5 A: 0.5 = R

Therefore, the resistance R is 0.5 Ω.

Learn more about resistance on

https://brainly.com/question/17563681

#SPJ1

Answer:

Let's define t = 0s (the initial time) as the moment when Car A starts moving.

Let's find the movement equations of each car.

A:

We know that Car A accelerations with a constant acceleration of 5m/s^2

Then the acceleration equation is:

\(A_a(t) = 5m/s^2\)

To get the velocity, we integrate over time:

\(V_a(t) = (5m/s^2)*t + V_0\)

Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:

\(V_a(t) = (5m/s^2)*t\)

To get the position equation we integrate again over time:

\(P_a(t) = 0.5*(5m/s^2)*t^2 + P_0\)

Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:

\(P_a(t) = 0.5*(5m/s^2)*t^2\)

Now let's find the equations for car B.

We know that Car B does not accelerate, then it has a constant velocity given by:

\(V_b(t) =20m/s\)

To get the position equation, we can integrate:

\(P_b(t) = (20m/s)*t + P_0\)

This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:

\(P_b(t) = (20m/s)*t + 100m\)

Now we can answer this:

1) The two cars will meet when their position equations are equal, so we must have:

\(P_a(t) = P_b(t)\)

We can solve this for t.

\(0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0\)

This is a quadratic equation, the solutions are given by the Bhaskara's formula:

\(t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)} }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}\)

We only care for the positive solution, which is:

\(t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s\)

Car A reaches Car B after 11.48 seconds.

2) How far does car A travel before the two cars meet?

Here we only need to evaluate the position equation for Car A in t = 11.48s:

\(P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m\)

3) What is the velocity of car B when the two cars meet?

Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s

4)  What is the velocity of car A when the two cars meet?

Here we need to evaluate the velocity equation for Car A at t = 11.48s

\(V_a(t) = (5m/s^2)*11.48s = 57.4 m/s\)

Answer:

1- 66mL

2-16mL

3-7.8mL

Explanation:

they are numbered from left to right, and have the number of the values. Dont forget the measurements, because sometimes if they are forgotten, they are counted wrong.

A) Before the collision, the total momentum is

(1250 kg) (0 m/s) + (3550 kg) (8.33 m/s) ≈ 29,600 kg•m/s

B) Momentum is conserved, so after the collision it is still approximately 29,600 kg•m/s.

C) If v is the speed of the locked car-truck system, then

(1250 kg) (0 m/s) + (3550 kg) (8.33 m/s) = (1250 kg + 3550 kg) v

Solve for v :

29,571.5 kg•m/s = (4800 kg) v

v ≈ 6.16 m/s

Answer:

35.28m/s

Explanation:

Given

Time t = 3.6secs

Required

Final speed v

Using the equation of motion;

v = u + gt

Substitute;

v = 0 + 9.8(3.6)

v = 0 + 35.28

v = 35.28m/s

Hence the speed of the marble is 35.28m/s

Answer:

true

Explanation:

Answer:

true

Explanation:

Answer:

  a = - e E / m

a = - 1,758 10¹¹ E

Explanation:

For this exercise we can use Newton's second law

        F = m a

where the force is electric

 the forces given by the product of the charge by the electric field

         F = q E

in this case tell us that the charge is the charge of the electron

         q = -e = - 1.6 10⁻¹⁹ C

we substitute

        - e E = m a

          a = - e E / m

we calculate

           a = - 1.6 10⁻¹⁹ / 9.1 10⁻³¹ E

           a = - 1,758 10¹¹ E

The negative sign indicates that the acceleration is in the opposite direction to the electric field

The instantaneous current at the same moment in the RL circuit can be expressed as I = Imaxsin(150), where Imax represents the maximum current.

1. Given that the instantaneous voltage at a specific moment in the RL circuit is V = Vmaxsin(150).

2. We can express the current at the same moment using Ohm's Law, which states that V = IR, where V is voltage, I is current, and R is resistance.

3. In an RL circuit, the resistance is represented by the symbol R, and it is typically associated with the resistance of the wire or any resistors in the circuit.

4. However, the given equation does not explicitly mention resistance.

5. Since we are considering an RL circuit, it suggests the presence of inductance (L) along with resistance (R).

6. In an RL circuit, the voltage across the inductor (VL) can be expressed as VL = L(di/dt), where L is the inductance and di/dt represents the rate of change of current.

7. At any given instant, the total voltage across the circuit (V) can be expressed as the sum of the voltage across the resistor (VR) and the voltage across the inductor (VL).

8. Therefore, V = VR + VL.

9. Since the given equation represents the instantaneous voltage (V), we can deduce that V = VR.

10. By comparing V = VR with Ohm's Law (V = IR), we can conclude that I = Imaxsin(150), where Imax represents the maximum current.

The specific values of Vmax, Imax, and the phase angle have not been provided in the question, so we are working with the general expression.

For more such questions on current, click on:

https://brainly.com/question/1100341

#SPJ8

Answer:  Solar Photovoltaic (PV) cells generate electricity by absorbing sunlight and using that light energy to create an electrical current. There are many PV cells within a single solar panel, and the current created by all of the cells together adds up to enough electricity to help power your school, home and businesses.

                              --Cited from Solar Schools

Answer:

3.16 × 10‐⁶

Explanation:

through coulomb's law

the force of attraction or repulsion of charges is given by the relation

F = k.q1.q2/r²

where your k = 9 × 10⁹

Answer:

Charge 3.16 mkC  microcoulomb

Explanation:

 Given:

q₁ = q₂ = q - Electrical charges

r = 60.00 sm = 0.60 m - Distance between charges

F = 0.250 N - Repulsive force between charges

___________________

q - ?

F = k·q₁·q₂ / r² = k·q·q / r² = k·q² / r²

Charge:

q = √ (F·r² / k )

q = r·√ (F / k )

q = 0.60· √ (0.250 / 9·10⁹) ≈  3.16·10⁻⁶ C or:

q = 3.16 mkC  (microcoulomb)

Answer:

A. reduce the amount of soil by one-half and apply the same amount of force

Explanation:

Answer:

Plz mark brainliest

Explanation:

As with kinetic energy, potential energy can change. When two objects interact, each one exerts a force on the other that can cause energy to be transferred to or from the other object.

The force acting on the system of two blocks connected by a rope passing over a pulley is -13.26 N.

The system of two blocks connected by a rope passing over a pulley are M1 and M2, where M1 rests on the triangle edge to the left of the pulley, which makes an angle of theta subscript 1 with the base of the triangle. The coefficient of friction between box M1 and the surface is mu subscript 1. M2 rests on the triangle edge to the right of the pulley, which makes an angle of theta subscript 2 with the base of the triangle.

The coefficient of friction between box M2 and the surface is mu subscript 2. The system sits atop a scalene triangle whose long edge forms the base. The pulley is attached to the apex of the triangle.M1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. M2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105.The free-body diagram of M1 shows that the weight of M1 acts straight downwards (vertically) and the normal force acts perpendicular to the slope.

The force of friction opposes the motion and acts opposite to the direction of motion.M1 = 2.25 kgTheta subscript 1 = 42.5 degreesMu subscript 1 = 0.205g = 9.81 m/s²In the free-body diagram of M2, the normal force acts perpendicular to the incline of the slope, the weight of the object acts vertically downwards and parallel to the incline, and the force of friction opposes the motion and acts opposite to the direction of motion.M2 = 5.55 kgTheta subscript 2 = 33.5 degreesMu subscript 2 = 0.105g = 9.81 m/s²The tension in the string is the same throughout the rope. Since the masses are being pulled by the same rope, the acceleration of the objects is the same as the acceleration of the rope.

The tension in the string is directly proportional to the acceleration of the objects and the rope.A system of two blocks connected by a rope passing over a pulley has a total mass of M. The acceleration of the system is given by the formula below:a = [(m1-m2)gsin(θ1) - μ1(m1+m2)gcos(θ1)] / (m1 + m2)Where, μ1 = 0.205 is the coefficient of friction of block M1θ1 = 42.5 degrees is the angle of the incline of block M1M1 = 2.25 kg is the mass of block M1M2 = 5.55 kg is the mass of block M2g = 9.81 m/s² is the acceleration due to gravitysinθ1 = sin 42.5 = 0.67cosθ1 = cos 42.5 = 0.75The acceleration of the system is:a = [(2.25-5.55)(9.81)(0.67) - (0.205)(2.25+5.55)(9.81)(0.75)] / (2.25 + 5.55)a = -1.7 m/s² (the negative sign indicates that the system is accelerating in the opposite direction).

The force acting on the system is given by:F = MaWhere M is the total mass of the system and a is the acceleration of the system. The total mass of the system is:M = m1 + m2M = 2.25 + 5.55M = 7.8 kgThe force acting on the system is:F = 7.8(-1.7)F = -13.26 N (the negative sign indicates that the force is acting in the opposite direction).

for more question on force

https://brainly.com/question/12785175

#SPJ8

ΔU is the change in the internal energy of the system.

Q is the net heat transferred into the system.

W is the net work done by the system.

ΔU = Q - W

All are part of the first law of Thermodynamics.