A traffic light weighing 200N hangs from a vertical cable tied to two other cables that are fastened to to a support ,as shown . The upper cables make angles 41° and 63° with the horizontal . Calculate the tension in of the three cables

Answers

Answer 1

Answer:

Answer:

93.6 N in the 41° cable
155.6 N in the 63° cable
200 N in the vertical cable

Explanation:

Let T and U represent the tensions in the 41° and 63° cables, respectively. In order for the system to be stationary, the horizontal components of these tensions must balance, and the vertical components of these tensions must total 200 N.

Tcos(41°) =Ucos(63°) . . . . . balance of horizontal components

U = Tcos(41°)/cos(63°) . . . . write an expression for U

The vertical components must total 200 N, so we have ....

Tsin(41°) +Usin(63°) = 200

Tsin(41°) +Tcos(41°)sin(63°)/cos(63°) = 200

T(sin(41°)cos(63°) +cos(41°)sin(63°))/cos(63°) = 200

T = 200cos(63°)/sin(41° +63°) ≈ 93.6 . . . newtons

U = 200cos(41°)/sin(41° +63°) ≈ 155.6 . . . newtons

The vertical cable must have sufficient tension to balance the weight of the traffic light, so its tension is 200 N.

Then the tensions in the 3 cables are ...

41°: 93.6 N

63°: 155.6 N

90°: 200 N

Answer 2

Answer:

The tension in each of the three cables are 94.29, 155.56 and 200 Newton respectively.

Given the following data:

Force = 200 Newton.

Angle 1 = 41°

Angle 2 = 63°

How to calculate the tension.

First of all, we would determine the third tension force based on the vertical component as follows:

$\sum F_y = 0\n\nT_3 - F_g =0\n\nT_3 - F_g=200\;N$

Next, we would apply Lami's theorem to resolve the forces acting on the traffic light at equilibrium:

For the horizontal component:

$\sum F_x = -T_1cos41+T_2cos 63=0\n\n0.7547T_1=0.4540T_2\n\nT_1=(0.4540T_2)/(0.7547)\n\nT_1 = 0.6016T_2$ ....equation 1.

For the vertical component:

$\sum F_y = T_1sin41+T_2sin 63-T_3=0\n\n\sum F_y = T_1sin41+T_2sin 63-200=0\n\n0.6561T_1+0.8910T_2 =200$ ...equation 2.

Substituting eqn. 1 into eqn. 2, we have:

$0.6561 * (0.6016T_2)+0.8910T_2 =200\n\n0.3947T_2+0.8910T_2 =200\n\n1.2857T_2 =200\n\nT_2 = (200)/(1.2857) \n\nT_2 = 155.56\;Newton$

For the first tension:

$T_1 = 0.6061T_2\n\nT_1 = 0.6061 * 155.56\n\nT_1 = 94.29\;Newton$

Read more on tension here: brainly.com/question/4080400

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Two charged particles are separated by 10 cm. suppose the charge on each particle is doubled. By what factor does the electric force between the particles change?

Answers

Answer:

The electric force increases by a factor of 4.

Explanation:

The electric force between two charges $q_1$ and $q_2$ separated a distance d can be calculated using Coulomb's Law:

$F=(kq_1q_2)/(d^2)$

where $k=9*10^9Nm^2/C^2$ is the Coulomb constant.

If the value of each charge is doubled, then we will have a force between them which is:

$F'=(k(2q_1)(2q_2))/(d^2)=4(kq_1q_2)/(d^2)=4F$

So the new force is 4 times larger than the original force.

Final answer:

Doubling the charge on each particle increases the electric force between them by a factor of 4.

Explanation:

The force between two charged particles is given by Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. So, if we denote the electric force as F, the charges as q1 and q2, and the distance as r, we can write Coulomb's law as F = k* q1*q2/r^2, where k is a constant.

Now if you double the charges (q1 and q2 become 2q1 and 2q2), and use these values in the formula, we get Fnew = k*(2q1) *(2q2)/r^2 = 4 * k*q1*q2/r^2 = 4F.

So, by doubling the charge on each particle, the electric force between them is multiplied by the factor of 4. So, the force increases fourfold.

Learn more about Coulomb's Law here:

brainly.com/question/32002600

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The left ventricle of a resting adult's heart pumps blood at a flow rate of 85.0 cm3/s, increasing its pressure by 110 mm Hg, its velocity from zero to 25.0 cm/s, and its height by 5.00 cm. (All numbers are averaged over the entire heartbeat.) Calculate the total power output (in W) of the left ventricle. Note that most of the power is used to increase blood pressure.

Answers

Answer:

P = 1.29625 W

Explanation:

Given

Q = 85.0 cm ³ / s, p₁ = 110 mmHg, u₁ = 25.0 cm / s, h = 5.0 cm

Also knowing the density of the blood is

ρₐ = 1.05 x 10 ³ kg / m³

Δp₁ = 110 mmHg * 133.322 Pa / 1 mmHg

Q = 85.0 cm³ / s = 85.0 x 10 ⁻⁶ m³ / s

To calculated the power

P = H * Q

H = Δp₁ + ¹/₂ * ρₐ * ( u₁² - v₂²) + ρₐ * g *Δh

H = 14.666 x 10 ³ Pa + 0.5 * 1.05 x 10 ³ kg / m³ * ( 25 x 10 ² m /s )² + 1.05 x 10 ³ kg / m³ * 9.8 m /s² * 0.05 m

H = 15.25 x 10 ³ Pa

P = 15.25 x 10 ³ Pa * 85.0 x 10 ⁻⁶ m³ / s

P = 1.29625 W

3.what does this stand for ??

Answers

Answer:

see below

Explanation:

The triangle stands for the change in

We would change the change in x

Answer:

ΔThis is the symbol of Delta which means Change

and x is length/distance/position.

Thus,Δxstands for Change in length/distance/position.

-TheUnknownScientist

An electron enters a region of uniform electric field with an initial velocity of 50 km/s in the same direction as the electric field, which has magnitude E = 50 N/C, (a) what is the speed of the electron 1.5 ns after entering this region? (b) How far does the electron travel during the 1.5 ns interval?

Answers

Answer:

(a). The speed of the electron is $3.68*10^(4)\ m/s$

(b). The distance traveled by the electron is $4.53*10^(-5)\ m$

Explanation:

Given that,

Initial velocity = 50 km/s

Electric field = 50 N/C

Time = 1.5 ns

(a). We need to calculate the speed of the electron 1.5 n s after entering this region

Using newton's second law

$F = ma$ .....(I)

Using formula of electric force

$F = qE$ .....(II)

from equation (I) and (II)

$-qE= ma$

$a = (-qE)/(m)$

(a). We need to calculate the speed of the electron

Using equation of motion

$v = u+at$

Put the value of a in the equation of motion

$v = 50*10^(3)-(1.6*10^(-19)*50)/(9.1*10^(-31))*1.5*10^(-9)$

$v=36813.18\ m/s$

$v =3.68*10^(4)\ m/s$

(b). We need to calculate the distance traveled by the electron

Using formula of distance

$s = ut+(1)/(2)at^2$

Put the value in the equation

$s = 3.68*10^(4)*1.5*10^(-9)-(1)/(2)*(1.6*10^(-19)*50)/(9.1*10^(-31))*(1.5*10^(-9))^2$

$s=0.0000453\ m$

$s=4.53*10^(-5)\ m$

Hence, (a). The speed of the electron is $3.68*10^(4)\ m/s$

(b). The distance traveled by the electron is $4.53*10^(-5)\ m$

An insulated rigid tank contains 3 kg of H2O in the form of a saturated mixture of liquid and vapor at a pressure of 150 kPa and a quality of 0.25. An electric heater inside the tank is turned on to heat this H2O until the pressure increases to 200 kPa. Please determine the change in total entropy of water during this process. Hint: See if you can find the electrical work consumed during this process.