(c)
A metal, X has a work function of 2.0 eV.
Explain the underlined statement.
If X is illuminated with light of wavelength 4.5 x 10-7 m, calculate the:
cut-off wavelength;
maximum energy of the liberated electrons;
stopping potential.
[h=6.6 x 10-34 J s, c = 3.0 x 108 m s¹, 1 eV = 1.6 x 10-19 J
e = 1.6 × 10-¹⁹ C]
(ii)
(a)
(B)
(Y)

The energy transferred in Joules by the radio when it has been switched on for 2 minutes would be 6000 Joules.

Power is defined as the rate of energy transfer or the rate at which work is done, and is given by the equation:

Power = Energy transferred / Time

Rearranging the equation to solve for energy transferred, we get:

Energy transferred = Power x Time

We are given:

Power = 50 W

Time = 2 minutes = 120 seconds

Therefore, the energy transferred by the radio when it has been switched on for 2 minutes is:

Energy transferred = Power x Time = 50 W x 120 s = 6000 J

In other words, the energy transferred by the radio is 6000 Joules.

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Answer:

falseeee

Explanation:

Answer:

C. R^2

Explanation:

A cyclotron is a particle accelerator which employs the use of electric and magnetic fields for its functioning. It consists of two D shaped region called dees and the magnetic field present in the dee is responsible for making sure the charges follow the half-circle and then to a gap in between the dees.

R is denoted as the radius of the final orbit then the final particle energy is proportional to the radius of the two dees. This however translates to the energy being proportional to R^2.

D. The canister with the fastest moving particle is Z due to concentration of particle.

The speed of the particles depend on the mean distance of the particles.

The canister with the largest concentration per particle will contain particles with the greatest speed.

If the particle concentration is increasing from W to Z, then Z will have fastest moving particle.

Thus, the canister with the fastest moving particle is Z due to concentration of particle.

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Answer:

d. Z

Explanation:

If all the canisters had the same amount of particles, which canister would have the fastest moving particles?

Z

Answer:

Average Speed = 6.37 m/s

Explanation:

The average speed is simply given by the following formula:

Average Speed = Total Distance Traveled/Total Time Spent

here,

Total Time Spent = 1.1 min + 1.5 min = (2.6 min)(60 s/min) = 156 s

Now, for total distance, we have to calculate the distance traveled on tortoise and distance traveled while flying, separately. Therefore,

Distance Traveled on Tortoise = (Time spent on Tortoise)(Speed of Tortoise)

Distance Traveled on Tortoise = (1.1 min)(60 s/min)(0.06 m/s) = 3.96 m

Similarly,

Flying Distance = (Flying Time)(Flying Speed) = (1.5 min)(60 s/min)(11 m/s)

Flying Distance =  990 m

Since, total distance is the sum of both distances, therefore,

Total Distance = 3.96 m + 990 m = 993.96 m

Now, using the values in equation of average speed, we get:

Average Speed = 993.96 m/156 s

Average Speed = 6.37 m/s

Answer:

not using it too much and getting addicted

Explanation:

A uniform metal rod with of length 80cm and a mass of 3.2kg is supported horizontally by two vertical spring balances C and D. Balance C is 20cm from one end while D is 30cm from the other end would show the reading of 1.06 Kg and 2.13 kg respectively

It can be defined as the force by which a body attracts another body towards its center as the result of the gravitational pull of one body and another, The gravity varies according to the mass and size of the body for example the force of gravity on the moon is the 1/6th times of the force of gravity on the earth.

As given in the problem, A uniform metal rod of the length of 80cm and mass of 3.2kg is supported horizontally by two vertical springs balance C and D. Balance C is 20cm from one end while D is 30cm from the other end

The weight of the rod acting downward is from the center of the rod at 40 cm

Let us suppose the reading on the spring balance C and D are P and Q respectively

By using the equilibrium for the vertical force

Fv=0

P + C = 3.2

By using the equilibrium for the moment around the left corner

20×P+ 50×Q= 40 ×3.2

By solving for both P and Q from the above two equations we would get

P =1.06 and Q = 2.13

Thus, the reading on the spring balance C and D would be 1.06 Kg and 2.13 kg respectively

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Based on the image provided and the location of deposition and erosion, it seems that the longshore current is moving towards the right.

Based on the image provided, it appears that the longshore current is traveling towards the right. This can be determined by looking at where deposition and erosion are occurring. Deposition is happening on the right side of the image, where sand is building up and forming a beach. Erosion, on the other hand, is happening on the left side of the image, where the waves are breaking against the cliff face and wearing it away. Longshore currents are generated by waves hitting the shore at an angle, causing the water to move parallel to the shore. In this case, the waves are coming in from the top left of the image and hitting the shore at an angle, which creates a longshore current that moves towards the right. As the water moves along the shore, it picks up sand and sediment and carries it with it. This sand is then deposited on the right side of the image, where the water slows down and loses energy. It's important to note that longshore currents can change direction depending on the direction of the incoming waves and the shape of the shoreline. But based on the image provided and the location of deposition and erosion, it seems that the longshore current is moving towards the right.

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Answer:

The speed of the ball when it reaches equilibrium position is 3.31 m/s

Explanation:

Given;

mass of the object, m = 0.25 kg

initial displacement of the object, h₁ = 0.56 m

spring constant, k = 105 N/m

displacement at equilibrium position, h₂ = 0

initial velocity of the object, v₁ = 0

velocity of the object at equilibrium position = v₂

The change in gravitational potential energy at the equilibrium position is given as;

ΔP.E = mg(h₂ - h₁)

The change in kinetic energy of the object at the equilibrium position is given as;

ΔK.E = ¹/₂m(v₂² - v₁²)  

Apply the principle of conservation of mechanical energy;

ΔK.E  +  ΔP.E = 0

¹/₂m(v₂² - v₁²)  +  mg(h₂ - h₁) = 0

¹/₂m(v₂² - 0)  +  mg(0 - h₁) = 0

¹/₂mv₂²  -  mgh₁  =  0

¹/₂mv₂²  = mgh

¹/₂v₂² = gh

v₂² = 2gh

v₂ = √2gh

v₂ = √(2 x 9.8 x 0.56)

v₂ = 3.31 m/s

Therefore, the speed of the ball when it reaches equilibrium position is 3.31 m/s

Using a tube as a filter to reduce the amplitude of sounds in various frequency ranges relies on manipulating resonance and destructive interference to achieve the desired attenuation.

A tube can be used as a filter to reduce the amplitude of sounds in various frequency ranges through a phenomenon called resonance. When a sound wave travels through a tube, it encounters resonant frequencies at which the tube's length corresponds to an integer multiple of the sound's wavelength. At these resonant frequencies, constructive interference occurs, resulting in an amplification of the sound. To use a tube as a filter, it is necessary to exploit the destructive interference that occurs at certain frequencies. This can be achieved by designing the tube's length and shape to create resonant frequencies that coincide with the frequencies to be attenuated. When the sound wave encounters these resonant frequencies, destructive interference takes place, causing a reduction in the sound's amplitude. By carefully selecting the dimensions of the tube, it is possible to create a series of resonant frequencies that correspond to specific frequency ranges. This allows the tube to act as a filter, attenuating the amplitudes of sounds within those ranges while having less impact on sounds outside the targeted frequencies.

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Answer:

speed

and mass

Explanation:

Answer:

Mass

Explanation:

It isn´t multiple choice.

Answer:

P = 4.5 watts

Explanation:

Given that,

EMF of the circuit, E = 3 volt

The resistance  of the resistors, R = 2 ohms

We need to find the power of this circuit. The relation between power, emf and resistance is given by the formula as follows :

\(P=\dfrac{V^2}{R}\)

Substitute all the values,

\(P=\dfrac{3^2}{2}\\\\P=4.5\ W\)

So, the power of this circuit is equal to 4.5 watts.

Answer:

liquids

Explanation

Answer:

C

Explanation:

The said wave takes the shortest time to move/get transmitted

Answer:

radiation

Explanation:

Answer:

When two forces acting on an object are equal in size but act in opposite directions, we say that they are balanced forces.

Explanation:

Potatoes mostly consist of starch, a complex carbohydrate produced by plants as a means of storing energy.

As starch accounts for the majority of potatoes' calories, they are categorised as a carbohydrate vegetable rather than a lipid. Potatoes, like the majority of foods, are a balance of fat, protein, and carbohydrates.

The most prevalent biomolecules on earth are carbohydrates, which include cellulose, starch, creatine, glucose, fructose, and others. Through the process of photosynthesis, carbohydrates are created by reducing the carbon dioxide in the atmosphere with the aid of light energy.

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Mark's work decreases when he climbs the same flight of stairs again in the same amount of time without the cello.

The correct answer is option D.

The amount of work Mark does depends on the weight of the cello, as well as the distance he climbs and the time it takes. Work is calculated using the formula :

Work = Force × Distance.
In the given scenario, Mark is carrying a full-sized cello while climbing the stairs. The weight of the cello adds to the force he exerts. So, the total force Mark exerts is the weight of the cello plus his own weight (375 N).

When Mark climbs the stairs with the cello, he is doing work against the force of gravity.

The work done is equal to the force exerted multiplied by the distance climbed (375 N + weight of cello) × 4 m.

Now, if Mark were to climb the same flight of stairs again in the same amount of time (3 seconds), but this time without the cello, the amount of work he does would decrease. This is because without the cello, the force exerted would only be Mark's weight (375 N), which is less than the total force exerted with the cello.

Therefore, mark's work decreases.

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When a ball is dropped from the same height and at the same time as the ball that was shot horizontally, both balls will hit the ground at the same time. This is because both balls experience the same acceleration due to gravity, which is a constant 9.81 m/s² near the surface of the Earth.

The horizontal velocity of the second ball has no effect on its vertical motion. This is because the two motions (horizontal and vertical) are independent of each other, and do not interfere with each other.
The vertical motion of the second ball is determined solely by the force of gravity acting on it, which causes it to accelerate downwards at a constant rate. The time it takes for the ball to hit the ground is determined by its initial height and the acceleration due to gravity.

The first ball that is dropped vertically, travels only vertically. It doesn't have any horizontal velocity. Hence it comes down straight, whereas the other ball comes down in a parabolic path. However, both the balls will hit the ground at the same time as the horizontal motion doesn't affect the vertical motion.

Therefore, it can be concluded that if a ball is dropped from the same height and at the same time as the ball that was shot horizontally, both balls will hit the ground at the same time. This is because the time taken by both balls is the same as their motion is independent of each other.

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The initial height H of the boy can be determined by adding the height of the slide h and the horizontal distance d the boy travels after leaving the track: H = h + d.

To determine the initial height H of the boy in terms of h and d, we can use the principle of conservation of energy. The total mechanical energy of the system remains constant throughout the motion.

At the top of the slide, the boy has gravitational potential energy given by mgh, where m is the mass of the boy, g is the acceleration due to gravity, and h is the height of the slide above the ground.

As the boy slides down the slide, there is no friction or other dissipative forces, so there is no change in mechanical energy. At the bottom of the track, the gravitational potential energy is converted entirely into kinetic energy.

Therefore, we can equate the initial potential energy to the final kinetic energy:

mgh = 1/2 m\(v^{2}\),

where v is the horizontal velocity of the boy when he leaves the track.

Since the boy leaves the track horizontally, the vertical component of his velocity is zero. Therefore, we can use the relationship between horizontal distance d and horizontal velocity v:

d = vt.

Solving these equations, we can express the initial height H in terms of h and d:

H = h + d.

So the initial height H of the boy can be determined by adding the height of the slide h and the horizontal distance d the boy travels after leaving the track.G

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Answer:

Explanation:

a ) At constant pressure , work done = P x Δ V

= 200 x 10³ x ( .1 - .04 )

= 12 x 10³ J .

b )

At constant temperature work done

= n RT ln v₂ / v₁

= PV ln v₂ / v₁

= 200 x 10³ x .04 ln .1 / .04

8 x 10³ x .916

= 7.33 x 10³ J .

Explanation:

Power = change in work/change in time

P = 600 joules/ 4 seconds

P= 150 watts

hope this helps :)

Answer:

   h = 2 R (1 +μ)

Explanation:

This exercise must be solved in parts, first let us know how fast you must reach the curl to stay in the

let's use the mechanical energy conservation agreement

starting point. Lower, just at the curl

       Em₀ = K = ½ m v₁²

final point. Highest point of the curl

        \(Em_{f}\) = U = m g y

Find the height y = 2R

      Em₀ = Em_{f}

      ½ m v₁² = m g 2R

       v₁ = √ 4 gR

Any speed greater than this the body remains in the loop.

In the second part we look for the speed that must have when arriving at the part with friction, we use Newton's second law

X axis

    -fr = m a                      (1)

Y Axis  

      N - W = 0

      N = mg

the friction force has the formula

     fr = μ  N

     fr = μ m g

    we substitute 1

    - μ mg = m a

     a = - μ g

having the acceleration, we can use the kinematic relations

    v² = v₀² - 2 a x

    v₀² = v² + 2 a x

the length of this zone is x = 2R

    let's calculate

     v₀ = √ (4 gR + 2 μ g 2R)

     v₀ = √4gR( 1 + μ)

this is the speed so you must reach the area with fricticon

finally have the third part we use energy conservation

starting point. Highest on the ramp without rubbing

     Em₀ = U = m g h

final point. Just before reaching the area with rubbing

     \(Em_{f}\) = K = ½ m v₀²

      Em₀ = Em_{f}

     mgh = ½ m 4gR(1 + μ)

       h = ½ 4R (1+ μ)

       h = 2 R (1 +μ)

Answer:

A medium in which speed of light is more is known as optically rarer medium and a medium in which speed of light is less is said to be optically denser medium. For example in air and water, air is raer and water is a denser medium.

Explanation:

Answer:Rarer medium is less dense

Explanation:

Rarer medium is less dense than dense medium because rarer medium doesn't have a very high amount of particles packing it together. It's easier to see through a rarer medium, so you're able to get a clearer picture of what's on the other side of it. Denser medium is, simply put, more dense.

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.

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For the two animals involved (bird and its prey), the momentum of both animals is conserved.

The principle of conservation of momentum states that if two objects collide, then the total momentum before and after the collision will be the same if there is no external force acting on the colliding objects.

Initial momentum = final momentum

Thus, for the two animals involved (bird and its prey), the momentum of both animals is conserved.

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Answer:

a) the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b) the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

c) the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

Explanation:

Given that;

height of the ramp h1 = 0.40 m

foot of the ramp above the floor h2 = 1.50 m

assuming R = 15 mm = 0.015 m

density of steel = 7.8 g/cm³

density of aluminum =  2.7 g/cm³

a) distance that the solid steel sphere sliding down the ramp without friction;

we know that

distance = speed × time

d = vt --------let this be equ 1

according to the law of conservation of energy

mgh₁ = \(\frac{1}{2}\) mv²

v² = 2gh₁  

v = √(2gh₁)

from the second equation; s = ut +  \(\frac{1}{2}\) at²

that is; t = √(2h₂/g)

so we substitute for equations into equation 1

d = √(2gh₁) × √(2h₂/g)

d = √(2gh₁) × √(2h₂/g)

d = 2√( h₁h₂ )    

we plug in our values

d = 2√( 0.40 × 1.5 )

d = 1.55 m

Therefore, the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b)

distance that a solid steel sphere rolling down the ramp without slipping;

we know that;

mgh₁ = \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) \(I_{}\)ω²

mgh₁ = \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) (\(\frac{2}{5}\)mR²) ω²

v = √( \(\frac{10}{7}\)gh₁  )

so we substitute √( \(\frac{10}{7}\)gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( \(\frac{10}{7}\)gh₁  ) × √(2h₂/g)  

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )  

d = 1.31 m

Therefore, the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

c)

distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping;

we know that;

mgh₁ = \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) \(I_{}\)ω²

mgh₁ = \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) (\(\frac{2}{3}\)mR²) ω²

v = √( \(\frac{6}{5}\)gh₁ )

so we substitute √( \(\frac{6}{5}\)gh₁ ) for v and t = √(2h₂/g) in equation 1 again

d = vt

d = √( \(\frac{6}{5}\)gh₁ ) × √(2h₂/g)

d = 1.549√( h₁h₂ )

d = 1.549√( 0.4 × 1.5 )

d = 1.2 m

Therefore, the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) distance that a solid aluminum sphere rolling down the ramp without slipping.

we know that;

mgh₁ = \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) \(I_{}\)ω²

mgh₁ = \(\frac{1}{2}\) mv² + \(\frac{1}{2}\) (\(\frac{2}{5}\)mR²) ω²

v = √( \(\frac{10}{7}\)gh₁  )

so we substitute √( \(\frac{10}{7}\)gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( \(\frac{10}{7}\)gh₁  ) × √(2h₂/g)  

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )  

d = 1.31 m

Therefore, the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

We have that for the Question it can be said that

From the question we are told

Suppose the height of the ramp is h1= 0.40m, and the foot of the ramp is horizontal, and is h2= 1.5m above the floor. What will be the horizontal distance traveled by the following four objects before they hit the floor? Assume that R= 15mm in each case; assume that the density of steel is 7.8 g/cm3; and assume that the density of aluminum is 2.7 g/cm3.

Generally the equation for sliding without friction is mathematically given as

\(V = \sqrt{4Hh}\)

the equation for sliding without slipping is mathematically given as

\(X = \sqrt{\frac{4Hh}{1+I/mR^2}}\)

A) A solid steel sphere sliding down the ramp without friction.\(V = \sqrt{4*1.5*0.4}\\\\= 1.55m\)

B) .A solid steel sphere rollingdown the ramp without slipping.\(I = 2/5 mR^2\\\\X = \sqrt{\frac{4*1.5*0.4}{1+2/5}}\\\\= 1.309m\)

C) A spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping.

\(I = 2/3 mR^2\\\\X = \sqrt{\frac{4*1.5*0.4}{1+2/3}}\\\\= 1.2m\)

D) A solid aluminum sphere rolling down the ramp without slipping.

\(X = \sqrt{\frac{4*1.5*0.4}{1+2/5}}\\\\= 1.309m\)

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The slope of the graph is 0.5 m/s² and when t = 1.5 s, the predicted displacement (d) of the object is 0.75 meters.

To plot the velocity vs. time graph, we'll use the given data points:

Duration, At (s): 2.0, 4.0, 6.0, 8.0, 10.0, 12.0

Velocity, v (m/s): 6.0, 7.0, 8.0, 9.0, 10.0, 11.0

Let's plot these points on a graph:

Time (s) [x-axis] | Velocity (m/s) [y-axis]

--------------------------------------------

2.0               | 6.0

4.0               | 7.0

6.0               | 8.0

8.0               | 9.0

10.0              | 10.0

12.0              | 11.0

After plotting the points, we can connect them with a straight line to represent the motion of the object. This line represents the velocity vs. time relationship.

Now, let's calculate the slope of this line. The slope of a line represents the rate of change of the dependent variable (velocity) with respect to the independent variable (time). In this case, it gives us the acceleration of the object.

Using the formula for calculating the slope of a line:

Slope (k) = (Change in velocity) / (Change in time)

For the first two points:

Change in velocity = 7.0 - 6.0 = 1.0 m/s

Change in time = 4.0 - 2.0 = 2.0 s

Slope (k) = 1.0 m/s / 2.0 s = 0.5 m/s²

Therefore, the slope of the graph is 0.5 m/s².

Now, to answer part B, the physical significance of the slope value is that it represents the object's acceleration. In this case, the constant acceleration experienced by the object is 0.5 m/s².

Moving on to part C, we are given the equation d = kt, where d represents the displacement and t represents time. Since the object is experiencing constant acceleration, the equation can be rewritten as d = 0.5t, where 0.5 is the acceleration (k).

To predict the value of "d" when t = 1.5 s, we can substitute the value of t into the equation:

d = 0.5 * 1.5 = 0.75 meters

Therefore, when t = 1.5 s, the predicted displacement (d) of the object is 0.75 meters.

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The probable question may be:

An object is subjected to a constant acceleration along a frictionless track. A student measures its velocity (v) after specific durations (At). The student uses a graph to analyze the truck's motion.

Duration, At, (s) :- 2.0,4.0,6.0,8.0,10.0,12.0.

Velocity, v, (m/s) :- 6.0,7.0,8.0,9.0,10.0,11.0

A. Plot the velocity (in meters/sec) vs. time (seconds). The velocity is the y-axis and time is the x-axis. Use any graphing software you like or graph this data in pencil on graph paper. Excel has a nice graphing package. Calculate the slope of this graph. You will

B. What is the physical significance of the slope value computed in part A?

C. Having determined the slope of the line, you can now write d = kt. Use this equation to predict a value of "d" when t = 1.5 s.

The tension in the strings are 31.47 and 19.25 N respectively.

Mass of the block, m = 3 kg

From the figure, consider the vertical components,

T₁ sin45° + T₂ sin30° = mg

(T₁/√2) + (T₂/2) = 3 x 9.8 = 29.4

Also, consider the horizontal components,

T₁ cos45° = T₂ cos30°

T₁/√2 = T₂ x√3/2

T₁ = T₂ x √3/2 x √2

So,

T₁ = 0.612T₂

Applying in the first equation,

(T₁/√2) + (T₂/2) = 29.4

(0.612T₂/1.414) + 0.5T₂ = 29.4

0.434 T₂ + 0.5 T₂ = 29.4

0.934 T₂ = 29.4

Therefore, the tension,

T₂ = 29.4/0.934

T₂ = 31.47 N

So, the tension,

T₁ = 0.612 T₂

T₁ = 0.612 x 31.47

T₁ = 19.25 N

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