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3 years ago

A worker pushes horizontally on a large crate with force of 298 N and the crate is moved 4.3 m. How much work was done?

Physics

1 answer:

Katyanochek1 [597]3 years ago

5 0

Answer:

1281.4 J

Explanation:

F = 298 N

d = 4.3 m

the work done is defined as the product of force applied in the direction of displacement and the displacement.

W = force x displacement

Work is a scalar quantity and its SI unit is Joule.

W = F x d

W = 298 x 4.3 = 1281.4 J

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A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff

postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is 0.833.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

$F_c=\frac{mv^2}{R}$

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

$N=mg$

Therefore, frictional force, $f=\mu mg$

Now, frictional force = centripetal force

$f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}$

Plug in the given values and solve for 'μ'. This gives,

$\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833$

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

7 0

4 years ago

• What are common questions, concerns, practices, and experiences across various world religions? Include indigenous religions i

faltersainse [42]

Answer:

They are a lot of religions and faiths across the globe. 

Explanation:

There are a lot of monotheistic and polytheistic religions across the world that took origin at different points in the past.

Some of them are:

Christianity- It is a monotheistic religion based on the teachings of Jesus Christ believed to be the son of god.

Confucianism it evolved in China and is based on the teachings scholar and philosopher Confucius. It is a largely secular and humanistic.

Buddhism it is based on the life and teachings of Gautama Buddha. Buddhism advocates a way of life with meditation and instructs its followers to be compassionate, kind and wise.

Hinduism –It evolved during the period of Indus valley civilisation and is a polytheistic religion. Hinduism beliefs mention rebirths or ‘punarjanma’.

Islam- Islam is a monotheistic religion.It considers Muhammad as the messenger of only god Allah.

5 0

3 years ago

describe at least two ways in which parents are able to influence the drinking habits of their children

laiz [17]

Well parents can set rules or even punishments which can influence the drinking habits of there kids such as, they aren't allowed to go out unless they had followed the rule there parents set for the amount they needed to have drank. Parents have many ways in which there authority can influence there kids habits.

6 0

3 years ago

Read 2 more answers

What is one standard kilogramun si system

Phoenix [80]

Answer:

The kilogram (kg) is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015 ×10−34 when expressed in the unit J s, which is equal to kg m2 s−1, where the meter and the second are defined in terms of c and ∆νCs.

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3 years ago

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A 500-N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area

weqwewe [10]

Answer:

W₂= 10000 N

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:

Pressure is defined as the force (F) applied per unit area (A)

P=F/A (N/m²)

P1=P2

$\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }$

$F_{2} = \frac{F_{1}*A_{2} }{A_{1}}$ Equation (1)

Data

W₁ = weight sits on the small piston

F₁ = W₁= 500 N

A₁ = 2.0 cm²

A₂ = 40 cm²

Calculation of the weight (W₂) can the large piston support

We replace data in the equation (1)

$F_{2} = \frac{(500)*(40) }{2}$

F₂ = 10000 N

W₂= F₂= 10000 N

6 0

3 years ago