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2 years ago

A man holding 7N weight moves 7m horizontal and 5m vertical , find the work done

Physics

1 answer:

SashulF [63]2 years ago

8 0

Answer:

35 J

Explanation:

The man is holding the box: this means that he is applying a force vertically upward, to balance the weight of the box (which pushes downward).

Therefore, we can ignore the horizontal displacement of the man, because the force applied (vertically upward) is perpendicular to that displacement (horizontal), so the work done for that is zero.

So, only the vertical motion contributes to the work. The work done by the man is equal to the gain in gravitational potential energy of the box, so:

$W=(mg)\Delta h$

where

$mg=7 N$ is the weight of the box

$\Delta h=5 m$ is the vertical displacement

Substituting, we find

$W=(7N)(5 m)=35 J$

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BARSIC [14]

Answer:

Explanation:

Analysis of structure gives

a=gsinθ−μkgcosθ

Notice that all the expression are right but we want to know of we can simplify the expression further.

We want to analyse if we can still further simplify the expression,

Inspecting the Right hand side of the equation, we notice that the acceleration due to gravity is common to both side, so we can bring it out i.e.

So option a is wrong because the expression can be simplified further to

a=g(sinθ−μkcosθ)

Option b is right and the best option.

Since we are given that, g=9.8m/s²

We can as well substitute that to option a

So we will have

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Also option C is correct but it is not best inserting the values of g directly without simplifying the expression first

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a=9.8metre/second²(sinθ−μkcosθ)

So the best option is B.

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2 years ago

A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m

madam [21]

Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

= .045 x 4.23 x 2/3 x .95 cos46

= .0837 units

moment of inertia of rod = ml² / 3 , m is mass of rod and l is length

= 2.95 x .95² / 3

I₁ = .8874 units

moment of inertia of rod + putty

I₁ + mr²

m is mass of putty and r is distance where it sticks

I₂ = .8874 + .045 x (2 x .95 / 3)²

I₂ = .905

Applying conservation of angular momentum

angular momentum of putty = final angular momentum of rod+ putty

.0837 = .905 ω

ω is final angular velocity of rod + putty

ω = .092 rad /s .

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2 years ago

One end of a string is fixed. An object attached to the other end moves on a horizontal plane with uniform circular motion of ra

sveticcg [70]

Answer:

If both the radius and frequency are doubled, then the tension is increased 8 times.

Explanation:

The radial acceleration ( $a_{r}$ ), measured in meters per square second, experimented by the moving end of the string is determined by the following kinematic formula:

$a_{r} = 4\pi^{2}\cdot f^{2}\cdot R$ (1)

Where:

$f$ - Frequency, measured in hertz.

$R$ - Radius of rotation, measured in meters.

From Second Newton's Law, the centripetal acceleration is due to the existence of tension ( $T$ ), measured in newtons, through the string, then we derive the following model:

$\Sigma F = T = m\cdot a_{r}$ (2)

Where $m$ is the mass of the object, measured in kilograms.

By applying (1) in (2), we have the following formula:

$T = 4\pi^{2}\cdot m\cdot f^{2}\cdot R$ (3)

From where we conclude that tension is directly proportional to the radius and the square of frequency. Then, if radius and frequency are doubled, then the ratio between tensions is:

$\frac{T_{2}}{T_{1}} = \left(\frac{f_{2}}{f_{1}} \right)^{2}\cdot \left(\frac{R_{2}}{R_{1}} \right)$ (4)

$\frac{T_{2}}{T_{1}} = 4\cdot 2$

$\frac{T_{2}}{T_{1}} = 8$

If both the radius and frequency are doubled, then the tension is increased 8 times.

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2 years ago

A 60 kilogram astronaut weighs 96 newtons on the surface of the moon. calculate the acceleration due to gravity on the moon.

34kurt

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Weight W = 96 N

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g = 1.6 m/s²

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