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2 years ago

Applied sciences refers to the study of scientific principles. True False

Physics

2 answers:

Darina [25.2K]2 years ago

7 0

This i believe is true , brainlest ?

Hoochie [10]2 years ago

3 0

True, sorry if wrong

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Una columna de mármol, cuya área de sección transversal es de 2.0 m2 sostiene una masa de 25.000 kg. Encontrar: (3 pto )a) El es

bazaltina [42]

Responder:

122,500 Pa; 2.45 × 10 ^ -6; 2.94 × 10 ^ -5m

Explicación:

Dado lo siguiente:

Área de sección transversal (A) = 2m ^ 2

Masa (m) = 25000 kg

Módulo de Young = 50 x 10 ^ 9 N / m2

(1) estrés en la columna:

Estrés = Fuerza / Área

F = masa * aceleración debido a la gravedad

F = 25000kg * 9.8m / s ^ 2 = 245,000J

Estrés = 245,000J / 2m ^ 2

Estrés = 122,500 Pa

2) Deformación de la unidad (deformación):

Usando la relación:

Módulo de Young = Estrés / tensión

50 × 10 ^ 9 = 122,500 / CEPA

Cepa = 122500 / (50 × 10^9)

Cepa = 0.00000245

C) Si la altura es de 12 m, ¿cuánto se acorta la columna?

Cepa = extensión / longitud

0.00000245 = extensión / 12

0,00000245 * 12

0,0000294 m

7 0

2 years ago

Suppose that you purchased a water bed with the dimensions 2.55 m Ã 2.53 dm Ã 245 cm. what mass of water does this bed contain

Nadya [2.5K]

dimensions of the bed is given as

$length = 2.55 m$

$thickness = 2.53 dm = 0.253 m$

$width = 245 cm = 2.45 m$

now the volume of the bed is given as

$V = 2.55 * 0.253 * 2.45 m^3$

$V = 1.581m^3$

now the mass of water in it is given as

$mass = density * volume$

$mass = 1000* 1.581$

$mass = 1581 kg$

<em>so it will contain 1581 kg mass in it</em>

6 0

2 years ago

A 2600-m-high mountain is located on the equator. how much faster does a climber on top of the mountain move than a surfer at a

puteri [66]

The climber move 0.19 m/s faster than surfer on the nearby beach.

Since both the person are on the earth, and moves with the constant angular velocity of earth, however there linear velocity is different.

Number of seconds in a day, t=24*60*60=86400 sec

The linear speed on the beach is calculated as

V1= $\frac{2πr}{t}$

Here, t is the time

Plugging the values in the above equation

V1= $\frac{2π*6.4*10^6}{86400}$ =465.421 m/s

Velocity on the mountain is calculated as

V2= $\frac{2π(r+h)}{t}$

Plugging the values in the above equation

V2= $\frac{2π(6.4*10^6+2600}{86400}$ =465.61 m/s

Therefore person on the mountain moves faster than the person on the beach by 465.61-465.421=0.19 m/s

5 0

2 years ago

Terminal velocity. A rider on a bike with the combined mass of 100kg attains a terminal speed of 15m/s on a 12% slope. Assuming

Firlakuza [10]

Answer:

0.9378

Explanation:

Weight (W) of the rider = 100 kg;

since 1 kg = 9.8067 N

100 kg will be = 980.67 N

W = 980.67 N

At the slope of 12%, the angle θ is calculated as:

$tan \ \theta = \dfrac{12}{100} \\ \\ tan \ \theta = 0.12 \\ \\ \theta = tan^{-1}(0.12) \\\\ \theta = 6.84^0$

The drag force D = Wsinθ

$\dfrac{1}{2}C_v \rho AV^2 = W sin \theta$

where;

$\rho = 1.23 \ kg/m^3$

A = 0.9 m²

V = 15 m/s

∴

Drag coefficient $C_D = \dfrac{2 *W*sin \theta}{\rho *A *V^2}$

$C_D =\dfrac{2 *980.67*sin 6.84}{1.23 *0.9 *15^2}$

$C_D =0.9378$

8 0

2 years ago

Point charge A with a charge of +4.00 μC is located at the origin. Point charge B with a charge of +7.00 μC is located on the x

never [62]

Answer:

210.3 degrees

Explanation:

The net force exerted on charge A = 59.5 N

Use the x and y coordinates of net force to get the direction

arctan (y/x)

8 0

2 years ago

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