All categories

3 years ago

Un lote de construcción rectangular mide 100 metros por 150 metros. Calcula el área de este lote en yardas cuadradas.

Physics

1 answer:

antoniya [11.8K]3 years ago

8 0

Answer: 17939.74 yards

Explanation:

Given , A rectangular measures 100 meters by 150 meters

To find : Area of rectangle.

Formula :

Area of rectangle = Length x width

Here, let length = 100 meters and width = 150 meters

Then, Area of rectangle = 100 meters x 150 meters = 15,000 square meters

Also , 1 meter = 1.09361 yards

Then, Area of rectangle = 15,000 x 1.09361 x 1.09361 square yards

= 17939.7424815 square yards

≈ 17939.74 yards

Hence, the area of rectangle is 17939.74 yards .

You might be interested in

Anyone want text if yes ill give u my phone number

Jlenok [28]

Answer:

Explanation:OFC!!!

3 0

3 years ago

Read 2 more answers

1. How much heat must be absorbed by 375 grams of water to raise its

antiseptic1488 [7]

Answer:

39225J

Explanation:

Given parameters:

Mass of water = 375grams of water

Change in temperature = 25°C

Specific heat capacity of water = 4.184J/g°C

Unknown:

Amount of heat absorbed = ?

Solution:

To solve this problem, we use the expression below:

H = m c Ф

H is the heat absorbed

m is the mass

c is the specific heat capacity

Ф is the change in temperature

Insert the parameters and solve;

H = 375 x 4.184 x (25) = 39225J

5 0

3 years ago

A 0.5 kg block slides down a frictionless semicircular track from a height, h, and collides with a second 1.5 kg block at the bo

Ede4ka [16]

Answer:

$h' = 0.062\cdot h$

Explanation:

The speed of the 0.5 kg block just before the collision is found by the Principle of Energy Conservation:

$U_{g} = K$

$m\cdot g \cdot h = \frac{1}{2}\cdot m \cdot v^{2}$

$v = \sqrt{2\cdot g \cdot h}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot h}$

$v \approx 4.429\cdot \sqrt{h}$

Knowing that collision is inelastic, the speed just after the collision is determined with the help of the Principle of Momentum Conservation:

$(0.5\,kg) \cdot (4.429\cdot \sqrt{h}) = (2\,kg)\cdot v$

$v = 1.107\cdot \sqrt{h}$

Lastly, the height reached by the two blocks is:

$K = U_{g}$

$\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot h'$

$h' = \frac{v^{2}}{2\cdot g}$

$h' = \frac{(1.107\cdot \sqrt{h})^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$h' = 0.062\cdot h$

3 0

3 years ago

You're sitting in the back of a taxi and place your bag on the seat next to you. The taxi enters a traffic circle and travels th

svp [43]

Answer:

(d) III only

Explanation:

We have to observe the motion of the bag with respect to taxi , considering taxi as stationary or inertial frame . Since bag is not moving with respect to taxi , the inertial frame that means , net force on it is zero .So option i and ii are ruled out .

Now how to explain motion of the bag ie why it is stationary ie what are the balancing force acting on it. We know that on a body on circular path , a force called centripetal force is acting on it . So that force must be acting on it . The balancing force is the frictional force which is keeping it stationary with respect to taxi . Hence the third option is correct.

5 0

3 years ago

8.92 A 45.0 kg woman stands up in a 60.0 kg canoe 5.00 m long. She walks from a point 1.00 m from one end to a point I .00 m fro

omeli [17]

The distance that the canoe moves in this process is 1.29 meters.

We first have to find the center of mass

$X = \frac{MsXs+McXc}{Mw+Mc} \\\\$

Where

Ms = Woman's mass = 45

Mc = Canoe's mass = 60kg

Xs = position from left= 1 cm

Xc = position from left end of canoe's mass = 2.5cm

When we put these values into the equation we have:

$X=\frac{45*1+60*2.5}{45+60} \\\\= 1.857\\$

The center of gravity lies at the center of this boat. Therefore,

$Xc = \frac{L}{2} \\\\L = 5 m long\\\\\frac{5}{2} =2.5$

5.00 - 1. 00 = 4 meters

$\frac{45*4+60*2.5}{45+60} = 3.14meters\\\\$

To get the distance that is moved by this canoe

distance = 3.143-1.857

= 1.286

≈ 1.29 meters

The distance that the canoe moves in this process is 1.29 meters.

6 0

2 years ago