All categories

3 years ago

A square tile having 50cm side is made to withstand a pressure of 200N/m^2 find out if a force of 45N can be applied on the tile

Physics

1 answer:

horrorfan [7]3 years ago

6 0

Explanation:

Given that,

Side of a square tile, a = 50 cm = 0.5 m

Pressure withstand by it, P = 200 Pa

We know that,

Pressure = force/area

$F=P\times A\\\\=200\times (0.5)^2\\\\=50\ N$

It is the force that tile withstand is 50 N. The force applied upon the time is 45 N. As 45 N can be applied on the tile because 45 N < 50 N.

You might be interested in

Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

$F = \frac{k*q_1*q_2}{d^2}$ Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)

d: distance between the charges in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

$F_{13} = \frac{k*q_1*q_3}{d_1^2}$

$F_{23} = \frac{k*q_2*q_3}{d_2^2}$

F₁₃ = F₂₃

$\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}$ We eliminate k and q₃ on both sides

$\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}$

$\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}$ We eliminate 10⁻⁶ on both sides

$(1-x)^2 = \frac{7}{5} x^2$

$1-2x+x^2=\frac{7}{5} x^2$

$5-10x+5x^2=7 x^2$

$2x^2+10x-5=0$

We solve the quadratic equation:

$x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m$

$x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m$

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in opposite way .

x = 0.458m = 45.8cm

8 0

3 years ago

A crane is used to lower weights into a lake for an underwater construction project. Determine the tension in the cable of the c

atroni [7]

Answer:

Tension in the cable = 591292.8 N or 591.3 kN

Explanation:

Tension in the cable = Weight of object in air - upthrust/bouyancy due to the liquid

Weight of object in air = mass * acceleration due to gravity, g

Since mass is not given, it is obtained from the formula, mass = density * volume

Volume of the sphere = 4/3πr³ where r, radius = 2 m; π = 22/7

volume = 4/3 * 22/7 * (2m)³

volume of sphere = 33.52 m³

Density of concrete sphere = specific gravity * density of water

where, specific gravity = 2.8, density of water = 1000 kg/m³

density of concrete = 2.8 * 1000 kg/m³

density of concrete = 2800 kg/m³

acceleration due to gravity, g = 9.8 m/s²

Thus, weight of concrete = 2800 kg/m³ * 33.52 m³ * 9.8 m/s² = 919788.8 N

Upthrust = density of water * volume of sphere * g

Upthrust = 1000 kg/m³ * 33.52 m³ * 9.8 m/s² = 328496 N

The tension in the cable is the calculated as below;

Tension in the cable = Weight of object in air - upthrust/bouyancy due to the liquid

Tension in cable = 919788.8 N - 328.496 N = 591292.8 N

Therefore, tension in the cable is 591292.8 N or 591.3 kN

5 0

3 years ago

4. Describe how the velocity of an object changes if it undergoes uniformly acceleration motion. Can its direction change?

valentinak56 [21]

Answer:

n the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.

In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.

Explanation:

Velocity is a vector therefore it has magnitude and direction, a change in either of the two is the consequence of an acceleration on the system.

In the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.

$a_{t}$ = (v₂-v₁)/Δt

In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.

$a_{c}$ = v2/R

In the general case, both the module and the address change

a = Ra ( a_{t}^2 + a_{c}^2)

4 0

3 years ago

Discuss how does a Van de Graaff generator works?

Svetllana [295]

Answer:

How does a Van de Graaff generator works?

A Van de Graaff generator pulls electrons from the Earth, moves them along a belt and stores them on the large sphere. These electrons repel each other and try to get as far away from each other as possible, spreading out on the surface of the sphere. ... It provides a convenient path for electrons to move to the ground.

source-

scienceworld.ca

5 0

3 years ago

An electric train moving at 5m/s accelerates to a speed of 8m/s in 20 seconds. Fine the distance travelled in meters during the

adelina 88 [10]

<h3>Answer: 130 meters</h3>

===================================================

Explanation:

vi = 5 and vf = 8 are the initial and final velocities respectively. The change in time is t = 20 seconds.

So,

x = 0.5*(vi + vf)*t

x = 0.5*(5+8)*20

x = 130 meters

represents the distance traveled. The first equation shown above is one of the four kinematics equations.

3 0

3 years ago

Read 2 more answers