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2 years ago

Does a wheel barrow have mechanical advantage???

Physics

1 answer:

Rus_ich [418]2 years ago

6 0

Answer: YES.

Explanation: OK, lets say you have a basket of fruit, it has a large mass, and a large gravity, it takes more force to lift it up. now, if we try to carry it lets say, across the yard, its gonna be pretty heavy! now, when we put it in a wheel barrow, the force goes onto the wheels. it takes less force to move it.

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A crane lifts an air conditioner to the top of a building. If the building is 12 m high, and the air conditioner has a mass of 2

Pepsi [2]

Work needed = 23,520 J

<h3> Further explanation </h3>

Given

height = 12 m

mass = 200 kg

Required

work needed by the crane

Solution

Work is the transfer of energy caused by the force acting on a moving object

Work is the product of force with the displacement of objects.

Can be formulated

W = F x d

W = Work, J, Nm

F = Force, N

d = distance, m

F = m x g

Input the value :

W = mgd

W = 200 kg x 9.8 m/s²x12 m

W = 23520 J

8 0

3 years ago

The components of vector A are:

Korvikt [17]

here as it is given that x component of the vector is positive while y component of the vector is negative so we can say the vector must inclined in Fourth quadrant.

So angle must be more than 270 degree and less than 360 degree

Now in order to find the value we can say that

$tan\theta = \frac{opposite\: side}{adjacent\: side}$

$tan\theta = \frac{8.6}{6.1}$

$\theta = tan^{-1}1.41$

$\theta = 54.65^0$

so it is inclined at above angle with X axis in fourth quadrant

Now if angle is to be measured counterclockwise then its magnitude will be

$\theta = 360 - 54.65 = 305.3^0$

so the correct answer will be 305 degree

3 0

3 years ago

Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular

RUDIKE [14]

Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

Given;

initial angular velocity of the electric drill, $\omega _i$ = 5.21 rad/s

angular acceleration of the electric drill, α = 0.311 rad/s²

time of motion of the electric drill, t = 4.13 s

The angular displacement of the electric drill at the given time interval is calculated as;

$\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad$