Salt and water is a colloid. <br> True or false?

8. Fig. 4.1 shows a heavy ball B of weight W suspended from a fixed beam by two ropes P and Q.

mart [117]

Answer:

The resultant tension of the two ropes is approximately 42.4 N

The length of the line representing the resultant tension is approximately 8.48 cm

Please find included with the answer the scale drawing created with Microsoft Word

Explanation:

The given parameters are;

The tension in rope P, $T_P$ = 30 N

The tension in rope Q, $T_Q$ = 30 N

The angle the rope, 'P', makes with the horizontal = 45°

The angle the rope, 'Q', makes with the horizontal = 45°

The scale factor of the scale diagram, S.F. = 5.0 N/cm

By the resolution of forces at equilibrium, we have;

The sum of the vertical forces, $\Sigma F_y$ = $T_P_y$ + $T_Q_y$ + W = 0

∴ W = -( $T_P_y$ + $T_Q_y$ )

W = -(30 × sin(45°) + 30 × sin(45°)) = -42.4264068712

The weight of the heavy ball, W ≈ 42.4 N acting downwards

The sum of the horizontal forces, $\Sigma F_x$ = $T_P_x$ + $T_Q_x$ = 0

The length of the resultant force, W = W/(S.F.) ≈ 42.4 N/(5.0 N/cm) = 8.48 cm

The drawing of the vectors using the scale factor of 5.0 N/cm is created using Microsoft Word is included

3 0

3 years ago

You have a 100 ohm resistor. How

sp2606 [1]

Answer:

R2 = 300 Ohms

Explanation:

Let the two resistors be R1 and R2 respectively.

RT is the total equivalent resistance.

Given the following data;

R1 = 100 Ohms

RT = 75 Ohms

To find R2;

Mathematically, the total equivalent resistance of resistors connected in parallel is given by the formula;

$RT = \frac {R1*R2}{R1 + R2}$

Substituting into the formula, we have;

$75 = \frac {100*R2}{100 + R2}$

Cross-multiplying, we have;

75 * (100 + R2) = 100R2

7500 + 75R2 = 100R2

7500 = 100R2 - 75R2

7500 = 25R2

R2 = 7500/25

R2 = 300 Ohms

4 0

2 years ago

Refrigerant-134a enters a compressor at 180 kPa as a saturated vapor with a flow rate of 0.35 m3/min and leaves at 900 kPa. The

Alexus [3.1K]

Answer:

52.5°C

Explanation:

The final enthalpy is determined from energy balance where initial enthalpy and specific volume are obtained from A-12 for the given pressure and state

mh1 + W = mh2

h2 = h1 + W/m

h1 + Wα1/V1

242.9 kJ/kg + 2.35.0.11049kJ/ 0.35/60kg

=287.4 kJ/kg

From the final enthalpy and pressure the final temperature is obtained A-13 using interpolation

i.e T2 = T1 + T2 -T1/h2 -h1(h2 - h1)

= 50°C + 60 - 50/295.15 - 284.79

(287.4 - 284.79)°C

= 52.5°C

7 0

3 years ago

Identify the positive and negative about trade blocs.

ohaa [14]

Answer:

if this is on odyssey ware then i can edit my answer to help you

Explanation:

5 0

3 years ago

A boat is traveling at an initial velocity of 2.7 meters per second in the positive direction. It accelerates at a rate of 0.15

cupoosta [38]

Answer:

$\boxed {\boxed {\sf 4.5 \ m/s \ in \ the \ positive \ direction}}$

Explanation:

We are asked to find the final velocity of the boat.

We are given the initial velocity, acceleration, and time. Therefore, we will use the following kinematic equation.

$v_f= v_i + at$

The initial velocity is 2.7 meters per second. The acceleration is 0.15 meters per second squared. The time is 12 seconds.

$v_i$ = 2.7 m/s
a= 0.15 m/s²
t= 12 s

Substitute the values into the formula.

$v_f = 2.7 \ m/s + (0.15 \ m/s^2)(12 \ s)$

Multiply the numbers in parentheses.

$v_f= 2.7 \ m/s + (0.15 \ m/s/s * 12 \ s)$

$v_f = 2.7 \ m/s + (0.15 \ m/s *12)$

$\v_f=2.7 \ m/s + (1.8 \ m/s)$ $v_f=2.7 \ m/s + (1.8 \ m/s)$

Add.

$v_f=4.5 \ m/s$

The final velocity of the boat is <u>4.5 meters per second in the positive direction.</u>

5 0

3 years ago

Salt and water is a colloid. True or false?