The parts of the body that are related to the movement of blood

A bulldozer attempts to drag a log weighing 500 N along the rough horizontal ground. The cable attached to the log makes an angl

Gemiola [76]

Answer:

T= 224.01 N

Explanation:

in imminent motion we have to :

The frictional force reaches its maximum value
The system is in balance of forces

Data

W= 500 N : weight of the log

μs = 0.5

μk = 0.35

α = 30°above the ground : angle of the cable attached to the log

Newton's first law to the log:

∑F =0 Formula (1)

∑F : algebraic sum of the forces in Newton (N)

Forces acting on the log

T: cable tension for impending movement

N: normal force

W : weight

f: frictional force , f= μsN

We apply the formula (1)

∑Fx=0

Tx-f = 0

Tcosα-μsN=0

Tcos30°-0.5N=0 Equation (1)

∑Fy=0

N+Ty-W=0

N+Tsin30°-500=0

N= 500-Tsin30° Equation (2)

We replace the value of N of the Equation (2) in the equation (1)

Tcos30°-0.5(500-Tsin30°) = 0

Tcos30°+0.5Tsin30° = 0.5*500

T( cos30°+0.5*sin30°) = 250

(1.116) T = 250

T= 250/1.116

T= 224.01 N

6 0

3 years ago

Displacement vector points due east and has a magnitude of 3.35 km. Displacement vector points due north and has a magnitude of

tekilochka [14]

Answer:

(a) 6.56 km

(b) $59.68^\circ$ north of west

Explanation:

Given:

$\vec{d}_1= 3.35\ km\ east = 3.35\ km\ \hat{i}\\\vec{d}_2= 9.31\ km\ north = 9.31\ km\ \hat{j}\\\vec{d}_3= 6.66\ km\ west = -6.61\ km\ \hat{i}\\\vec{d}_4= 3.65\ km\ south = -3.65\ km\ \hat{j}\\$

Let the resultant displacement vector be $\vec{D}$ .

As the resultant is the vector sum of all the vectors.

$\therefore \vec{D}=\vec{d}_1+\vec{d}_2+\vec{d}_3+\vec{d}_4\\\Rightarrow \vec{D} =(3.35\ km\ \hat{i})+(9.31\ km\ \hat{j})+(-6.61\ km\ \hat{i})+(-3.65\ km\ \hat{j})\\\Rightarrow \vec{D} =-3.31\ km\ \hat{i}+5.66\ km\ \hat{j}\\\textrm{Magnitude of the resultant displacement vector} = \sqrt{(-3.31)^2+(5.66)^2}\ km= 6.56\ km\\$

$\textrm{Angle with the positive west} = \theta = \tan^{-1}(\dfrac{5.66}{3.31})= 59.68^\circ$

7 0

4 years ago

Two different charges, q1 and q2, are placed at two different locations, one charge at each location. The locations have the sam

Mars2501 [29]

Answer:

No

Explanation:

Electric potential is the work done to bring a unit of charge (1 C) from infinity to a point inside an electric field.

Electric potential energy of a charge q is the energy required to keep it in an electric potential V. Electric potential energy is given by,

U = qV

Hence even if the two charges are on an equipotential surface (surface where the potential is the same at all points), the potenial energy will be different if the magnitude or nature of the charges are different.

5 0

3 years ago

Please help asap its for anatomy

Brilliant_brown [7]

The smooth muscle in the wall of the bladder when stretched triggers the micturition reflex (urination).

<h3>What is a Bladder?</h3>

This is defined as a lined layers of muscle tissue that stretch to hold urine in organisms.

In older people the elasticity of the bladder is reduced which is why it makes it harder for them to hold urine for a long time.

Read more about Micturition here brainly.com/question/26493943

6 0

2 years ago

what equastion do you use to solve Riders in a carnival ride stand with their backs against the wall of a circular room of diame

Hitman42 [59]

Answer:

μsmín = 0.1

Explanation:

There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
This friction force has a maximum value, that can be written as follows:

$F_{frmax} = \mu_{s} *F_{n} (1)$

where μs is the coefficient of static friction, and Fn is the normal force,

perpendicular to the wall and aiming to the center of rotation.

This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
This force has the following general expression:

$F_{c} = m* \omega^{2} * r (2)$