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3 years ago

Explain why objects moving in fluids must have special shapes ?

Physics

1 answer:

Alexus [3.1K]3 years ago

8 0

Explanation:

Fluids exert both drag and lift forces on moving objects. Drag is the frictional force opposing motion. Lift is the force perpendicular to motion.

Some objects, like parachutes, are designed with large cross sectional areas to increase drag force. Usually though, objects are designed to minimize drag force. It's why cars, planes, and boats have sleek shapes.

Airplane wings have shapes called airfoils that generate lift. It's what makes them fly. The same shape is found in racecar spoilers. These spoilers use lift force to push down on the rear tires, increasing traction.

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Consider the two-body situation at the right. A 300kg crate rests on an inclined plane and is connected by a cable to a 100 kg m

trasher [3.6K]

Answer:

a= 0.578 m/s

T = 1037.8 N

Explanation:

Data

m₁= 300 kg

m₂= 100 kg

inclined plane, θ = 30°

μk = 0.120

Newton's second law to m₁:

We define the x-axis in the direction parallel to the movement of the 300kg (m₁) crate on the ramp and the y-axis in the direction perpendicular to it.

∑F = m₁*a Formula (1)

Forces acting on m₁

W₁: m₁ weight : In vertical direction

N : Normal force : perpendicular to the inclined plane

f : Friction force: parallel to the inclined plane

T: cable tension : parallel to inclined plane

Calculated of the W₁

W₁=m₁*g

W₁= 300kg* 9.8 m/s² = 2940 N

x-y weight components

W₁x= W₁sin θ =2940 N*sin(30)° =1470 N

W₁y= W₁cos θ =2940 N *cos(30)° =2156.4 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay ay = 0

N - W₁y = 0

N = W₁y

N = 2156.4 N

Calculated of the f

f = μk* N= (0.120)*(2156.4 N)

f = 258.77 N

Newton's second law to m₁ in direction x-axis :

∑Fx = m₁*ax ,ax =a

We assume that m₁ descends on the inclined plane and we positively take the direction of movement:

wx-f-T = m*a

wx - f - m*a =T

1470 -258.77 -300*a =T

T= 1211.23-300*a Equation (1)

Newton's second law to m₂

∑Fy = m₂*ay ,ay =a

Forces acting on m₂

W₂: m₂ weight : In vertical direction

T: cable tension:In vertical direction

Calculated of the W₂

W₂=m₂*g

W₂= 100kg* 9.8 m/s² = 980 N

∑Fy = m₂*a

Because we assume that m₁ descends on the inclined plane, then, m₂ ascends vertically, we take positive the direction of movement:

T-W₂ = m₂*a

T-980 = 100*a

T = 980 + 100*a Equation (2)

Problem development

Equation (1) = Equation (2) = T

1211.23-300*a= 980 + 100*a

1211.23- 980 = 100*a + 300*a

231.23 = 400*a

a= 231.23 / 400

a= 0.578 m/s

Because the acceleration tested positive then effectively m₁ descends on the inclined plane and m₂ ascends vertically.

We replace a= 0.578 m/s in the equatión (2)

T = 980 + 100* (0.578 )

T = 1037.8 N

5 0

3 years ago

The conventional relatively small unit for work(ignoring time) such as raising one pound one foot is the foot-pound(ft. lb.). Si

Lelechka [254]

Answer:

1 joule = 0.737 foot-pound

Joule is the unit of work.

1 J = 1 N·m

In SI units

1 J = 1 kg· m/s²

0.737 foot-pound is the amount of work to raise 0.737 pounds one foot or raising one pound to 0.737 ft.

6 0

3 years ago

our friend is constructing a balancing display for an art project. She has one rock on the left (ms=2.25 kgms=2.25 kg) and three

Licemer1 [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The torque produced by the pile of rocks is $\tau = 35.63\ N \cdot m$

The distance of the single for equilibrium to occur is $r_s =1.62 \ m$

Explanation:

From the question we are told that

The mass of the left rock is $m_s = 2.25 \ kg$

The mass of the rock on the right $m_p = 10.1 kg$

The distance from fulcrum to the center of the pile of rocks is $r_p = 0.360 \ m$

Generally the torque produced by the pile of rock is mathematically represented as

$\tau = m_p * g * r_p$

Substituting values

$\tau = 10.1 * 9.8 * 0.360$

$\tau = 35.63\ N \cdot m$

Generally we can mathematically evaluated the distance of the the single rock that would put the system in equilibrium as follows

The torque due to the single rock is

$\tau = m_s * g * r_s$

At equilibrium the both torque are equal

$35.63 = m_s * r_s * g$

Making $r_s$ the subject of the formula

$r_s = \frac{35.63 }{m_s * g}$

Substituting values

$r_s = \frac{35.63 }{2.25 * 9.8}$

$r_s =1.62 \ m$

6 0

3 years ago

Why are there so few working wave power stations?

Bond [772]

Well mostly because it's for oceans and lakes that have waves. so if you don't live near any of these, you wouldn't be getting power. <span />

3 0

4 years ago

Read 2 more answers

If a dog is mass iS 14.3 kg what is it’s weight on earth

BigorU [14]

Answer:

Weight of the dog on surface of earth is 140.14 Newton.

Given:

mass of the dog = 14.3 kg

To find:

Weight of the dog = ?

Formula used:

Weight of the dog is given by,

W = mg

Where, W = weight of the dog

m = mass of the dog

g = acceleration due to gravity

Solution:

Weight of the dog is given by,

W = mg

Where, W = weight of the dog

m = mass of the dog = 14.3 kg

g = acceleration due to gravity

W = 14.3 × 9.8

W = 140.14 Newton

Weight of the dog on surface of earth is 140.14 Newton.

8 0

3 years ago