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aleksandr82 [10.1K]
3 years ago
9

Pressure is defined as

Physics
1 answer:
Gnesinka [82]3 years ago
4 0
Pressure is defined as force per area.

It is normally further advantageous to apply pressure preferably than force to explain the importance toward liquid reaction.
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A boy pulls a 28.0-kg box with a 230-N force at 35° above a horizontal surface. If the coefficient of kinetic friction between t
ddd [48]

Answer:

1977.696 J

Explanation:

Given;

Weight of the box = 28.0 kg

Force applied by the boy = 230 N

angle between the horizontal and the force = 35°

Therefore,

the horizontal component of the force = 230 × cosθ

= 230 × cos 35°

= 188.405 N

Coefficient of kinetic friction, μ = 0.24

Force by friction, f = μN

here,

N = Normal force = Mass × acceleration due to gravity

or

N = 28 × 9.81 = 274.68 N

therefore,

f = 0.24 × 274.68

or

f = 65.9232 N

Now,

work done by the boy, W₁ = 188.405 N × Displacement  

= 188.405 N × 30

= 5652.15 J

and,

the

work done by the friction, W₂ = - 65.9232 N × Displacement  

= - 65.9232 N × 30 m

= - 1977.696 J

[ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]

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3 years ago
show the trajectory of a body projected with an initial velocity vat an angle of departure thita is a paraboler​
Jobisdone [24]

Answer:

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Explanation:

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5 0
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A 2kg book is moved from a shelf that is 2m off the ground to a shelf that is 1.5m off the ground, what is it’s change in gravit
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The gravitational energy is going up subtracting the energy that was on the ground
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A soccer player extends her lower leg in a kicking motion by exerting a force with the muscle above the knee in the front of her
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Answer:

10.09 N

Explanation:

Analogously to Newton's second law, torque can be defined as:

\tau=I\alpha

Here, I is the moment of inertia and \alpha is the angular acceleration. We have:

\tau=(0.65kg*m^2)(29.5\frac{rad}{s^2})\\\tau=19.18N*m

Torque is the vector product of the position vector of the point at which the force is applied by the force vector:

\vec{\tau}=\vec{r}\times \vec{F}

Since the effective lever arm is perpendicular to the force, the angle between them is 90^\circ. The magnitud of this vector product is defined as:

\tau=rFsen\theta.

Solving for F and replacing the known values:

F=\frac{\tau}{rsen\theta}\\F=\frac{19.18N*m}{1.9m(sen90^\circ)}\\F=10.09N

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