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Irina18 [472]
2 years ago
8

4. A crane has a weight of 85000 N and it rests on two caterpillar tracks which have a total

Physics
1 answer:
vagabundo [1.1K]2 years ago
4 0

Answer:

50kPa and if the area decreases, the pressure increases

Explanation:

P=F/A; P=pressure, F=force, A=area

85000N/1.7 m^2 = 50000 N/m^2; Pa = Pascal = N/m^2

=50 kPa

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A body is placed on a rough inclined plane. Why does the frictional force decrease with the increase of angle of inclination
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The frictional force is directly proportional to the force that is perpendicular on the surface.

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Now as the inclination of the surface increases, the gravitational force is no longer the perpendicular force of the body, its value decreases, which means only a part is used to generate frictional force. Consequently, frictional force decreases.

When the inclination reaches 90 degrees, the gravitational force does not act along the normal and accordingly, no friction force is generated.

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3 years ago
How does a substance's temperatura chanté when the average kinetic energy of its particles increases? When it decreases?
Aleks [24]
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7 0
2 years ago
A car going at v = 29.7 m/s (67 mph) rounds a curve of radius R = 50.0 m, where the road is banked at an angle of θ = 30.0°. Wha
Nikolay [14]

Answer:

μ = 0.6

Explanation:

given,

speed of car = 29.7 m/s

Radius of curve = 50 m

θ = 30.0°

minimum static friction = ?

now,

writing all the forces acting along y-direction

N cos θ - f sinθ = mg

N cos θ -μN sinθ = mg

N = \dfrac{m g}{cos\theta-\mu sin \theta}

now, writing the forces acting along x- direction

N sin θ + f cos θ = F_{net}

N cos θ + μN sinθ = F_{net}

\dfrac{m g}{cos\theta-\mu sin \theta}(cos \theta + \mu sin\theta)=F_{net}

taking cos θ from nominator and denominator

F_{net} =\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg

\dfrac{mv^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg

\dfrac{v^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}g

\mu=\dfrac{v^2 -r g tan\theta}{v^2tan\theta + r g}

now, inserting all the given values

\mu=\dfrac{29.7^2 -50 \times 9.8tan 30^0}{29.7^2\times tan 30^0 +50 \times 9.8}

μ = 0.6

7 0
3 years ago
A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km. What is his displacement from the
Readme [11.4K]

Explanation:

Given that,

A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km.

We need to find his displacement from the starting position.

We know that,

Displacement = shortest path covered

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For direction,

\theta=\tan^{-1}(\dfrac{d_y}{d_x})\\\\\theta=\tan^{-1}(\dfrac{12}{9})\\\\= $$53.13^{\circ}

Hence, this is the required solution.

5 0
2 years ago
The sphere that refers to Earth's water is called what?
WARRIOR [948]

Answer:

Its called the hydrosphere UwU!

Explanation:

6 0
3 years ago
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