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WITCHER [35]
2 years ago
8

A shopping mall has a length of 200 m and a width of 75 m. What is the area of the mall?

Physics
1 answer:
sweet [91]2 years ago
6 0
To find the area, we just need to multiply the length by the width. 200 x 75 = 15,000. The mall has an area of 15,000 sq. feet.
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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
2 years ago
an object at rest starts accelerating if it travels 30 meters to end up going 10 m/s what was it’s acceleration
vfiekz [6]

First we have to calculate the time taken to travel the distance 30 m, is

t = \frac{distance}{velocity} = \frac{30 \ m}{10 \ m/s } =  3 \ s.

Now from equation of motion,

v = u + at

Given, v = 10 \ m/s .

As object starts from rest, so  u = 0.

Substituting these values in above equation, we get

 10 \ m/s = 0 + a \times 3 \ s \\\\ a = \frac{10}{3}  = 3.33 \ m/s^2.

Thus, the acceleration is 3.33 \ m/s^2

3 0
3 years ago
75% of what number is 27? EXPLAIN
zmey [24]
75 percent (calculated percentage %) of what number equals 27? Answer: 36.
8 0
2 years ago
Read 2 more answers
Six new refrigerator prototypes are tested in the laboratory. For each refrigerator, the electrical power P needed for it to ope
Mandarinka [93]

Answer:

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s.

the rate at which they raise the temperature of the room.

2.1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

Explanation:

A refrigerator is a device that uses work to remove heat energy from a cold reservoir and deposit it into a hot reservoir. .A good refrigerator (with a large performance coefficient) will remove a large amount of heat energy from the cold reservoir for a small amount of work input

The performance coefficient  of a refrigerator is defined as the ratio of the heat energy removed from the cold reservoir  to the work  input to the refrigerator:

k=QC/W

power is defined as work per unit time

1.k=1500/750=2

2. 1200/400=3

3.2000/500=4

4.1000/250=4

5.1500/500=3

6.3000/1000=3

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s

2, Rate at which they raise the temperature of the room.

rate at which temperature rises in the inner chamber of the refrigerator is proportional to the rate of energy used to dispel heat from the refrigerator

1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

5 0
3 years ago
The curved section of a horizontal highway is a circular unbanked arc of radius 740 m. If the coefficient of static friction bet
kirill [66]
The answer would be 54 m/s as the maximum speed
8 0
3 years ago
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