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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 .

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Objects A and B, of mass M and 2M respectively, are each pushed a distance d straight up an inclined plane by a force F parallel

krek1111 [17]

Answer:

The correct answer is <u>option (A) that is KEA > KEB .</u>

Explanation:

Let us calculate -

If the object is straighten up and inclined plane , the work done is

$W=F_d- F_f_r_id-F_gh$

$W=F_d-\mu_kmgdcos\theta-mgdsin\theta$

The change in kinetic energy is ,

$\Delta K=\frac{1}{2}mv^2-\frac{1}{2}m\nu_0^2$

At the top of the inclined plane , the velocity is zero

So,

$\Delta K=\frac{1}{2} m(0)^2-\frac{1}{2}m\nu_0^2$

$\Delta KE=-\frac{1}{2}m\nu_0^2$

From the work energy theorem , we have $W=-\Delta K$ in case of friction , so

$\frac{1}{2}m\nu_0^2=Fd-\mu_kmgdcos\theta-mgdsin\theta$

$KE=Fd-\mu_kmgdcos\theta-mgdsin\theta$

For object A-

$KE_A=Fd-\mu_kmgdcos\theta-mgdsin\theta$

For object B

$KE_B= Fd -2\mu_kMgdcos\theta-2Mgdsin\theta$

$KE_B= Fd -2(\mu_kMgdcos\theta-Mgdsin\theta)$

Thus , larger mass is going to mean less total work and a lower kinetic energy .

From the above results , we get

$KE_A >KE_B$

<u>Therefore , option A is correct .</u>

6 0

2 years ago

Plz help, its an emergency, giving lots of points for it. and srry if its in the wrong subject, not really sure? I just don't un

exis [7]

Anything I can help with

4 0

3 years ago

Read 2 more answers

A Texas cockroach of mass 0.157 kg runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical a

VashaNatasha [74]

Answer:

-7.23 rad/s

Explanation:

Given that

Mass of the cockroach, m = 0.157 kg

Radius of the disk, r = 14.9 cm = 0.149 m

Rotational Inertia, I = 5.92*10^-3 kgm²

Speed of the cockroach, v = 2.92 m/s

Angular velocity of the rim, w = 3.89 rad/s

The initial angular momentum of rim is

Iw = 5.92*10^-3 * 3.89

Iw = 2.3*10^-2 kgm²/s

The initial angular momentum of cockroach about the axle of the disk is

L = -mvr

L = -0.157 * 2.92 * 0.149

L = -0.068 kgm²/s

This means that we can get the initial angular momentum of the system by summing both together

2.3*10^-2 + -0.068

L' = -0.045 kgm²/s

After the cockroach stops, the total inertia of the spinning disk is

I(f) = I + mr²

I(f) = 5.92*10^-3 + 0.157 * 0.149²

I(f) = 5.92*10^-3 + 3.49*10^-3

I(f) = 9.41*10^-3 kgm²

Final angular momentum of the disk is

L'' = I(f).w(f)

L''= 9.41*10^-3w(f)

Using the conservation of total angular momentum, we have

-0.068 = 9.41*10^-3w(f) + 0

w(f) = -0.068 / 9.41*10^-3

w(f) = -7.23 rad/s

Therefore, the speed of the lazy Susan after the cockroach stops is -7.23 and is directed in the opposite direction of the initial lazy Susan angular speed

The mechanical energy of the cockroach is not converted as it stops

8 0

3 years ago

A traveling wave on a string can be described by the equation : y = (5.26 ~\text{m}) \cdot \sin \big( (1.65 ~\frac{\text{rad}}{\

zloy xaker [14]

Answer:

t = 1.77 s

Explanation:

The equation of a traveling wave is

y = A sin [2π (x /λ -t /T)]

where A is the oscillation amplitude, λ the wavelength and T the period

the speed of the wave is constant and is given by

v = λ f

Where the frequency and period are related

f = 1 / T

we substitute

v = λ / T

let's develop the initial equation

y = A sin [(2π / λ) x - (2π / T) t +Ф]

where Ф is a phase constant given by the initial conditions

the equation given in the problem is

y = 5.26 sin (1.65 x - 4.64 t + 1.33)

if we compare the terms of the two equations

2π /λ = 1.65

λ = 2π / 1.65

λ = 3.81 m

2π / T = 4.64

T = 2π / 4.64

T = 1.35 s

we seek the speed of the wave

v = 3.81 / 1.35

v = 2.82 m / s

Since this speed is constant, we use the uniformly moving ratios

v = d / t

t = d / v

t = 5 / 2.82

t = 1.77 s

3 0

3 years ago

I NEED HELP ON 2 QUESTIONS PLEASEEE

tino4ka555 [31]

Answer:

2) c) give-way vessel

3) a) With one short blast

Explanation:

2) A vessel that is required to take early substantial action to ensure avoiding collision called Give way vessel

In overtaking, the vessel intending to overtake is the Give-Way Vessel the vessel that is going to be overtaken is the Stand-On Vessel

Therefore, the correct option is c) give-way vessel

3) When vessels use sound signals in a meeting head on situation both vessel are Give-Way vessels and both vessel pass the each other by turning to the starboard side therefore they intend to pass each other on their port side requiring one short blast

Therefore, the correct option is a) With one short blast.

4 0

3 years ago