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SCORPION-xisa [38]
3 years ago
15

A curve of radius 53.1 m is banked so that a car of mass 2.9 Mg traveling with uniform speed 67 km/hr can round the curve withou

t relying on friction to keep it from slipping on the surface.
Physics
1 answer:
prisoha [69]3 years ago
6 0

Answer:

33.65°

Explanation:

radius, r = 53.1 m

m = 2.9 Mg = 2.9 x 10^6 g = 2900 kg

v = 67 km/h

convert km/h into m/s

v = 18.61 m/s

Let the angle of banking of road is θ, without friction

tan\theta =\frac{v^{2}}{rg}

tan\theta =\frac{18.61^{2}}{53.1\times 9.8}

tan  θ = 0.6655

θ = 33.65°

Thus, the angle of banking of road is 33.65°.

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One advantage is that whatever resource it is, it will never run out and you wont have to worry about not having it. A second is that there is going to be enough for everyone to use however much they want without there having to be a limit on how much you use.
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An electric motor rotating a workshop grinding wheel at 1.00 × 10² rev/min is switched off. Assume the wheel has a constant nega
ololo11 [35]

An electric engine turning a workshop sanding rotation at 1.00 × 10² rev/min is switched off. Take the wheel includes a regular negative angular acceleration of volume 2.00 rad/s². 5.25 moments long it takes the grinding rotation to control.

<h3>What is negative angular acceleration?</h3>
  • A particle that has a negative angular velocity rotates counterclockwise.
  • Negative angular acceleration () is a "push" that is hence counterclockwise.
  • The body will speed up or slow down depending on whether and have the same sign (and eventually go in reverse).
  • For instance, when an object rotating counterclockwise slows down, acceleration would be negative.
  • If a rotating body's angular speed is seen to grow in a clockwise direction and decrease in a counterclockwise direction, it is given a negative sign.
  • It is known that a change in the linear acceleration correlates to a change in the linear velocity.

Let t be the time taken to stop.

ω = 0 rad/s

Use the first equation of motion for rotational motion

ω = ωo + α t

0 = 10.5 - 2 x t

t = 5.25 second

To learn more about angular acceleration, refer to:

brainly.com/question/21278452

#SPJ4

7 0
2 years ago
the dimensions of a cuboidal block are 2.5 m * 2m * 1.2m its weight 900 n what is the minimum presseure exerted by the block on
aleksandr82 [10.1K]

Answer:

your answer is here 180

Explanation:

900/2.5 * 2 = 180

6 0
2 years ago
Read 2 more answers
A block with mass m 2.00 kg is placed against a spring on a frictionless incline with angle 30 degrees (Fig. B-43). (The block i
guajiro [1.7K]

Answer:

Explanation:

a )

The stored elastic energy of compressed spring

= 1 / 2 k X²

= .5 x 19.6 x (.20)²

= .392 J

b ) The stored potential energy will be converted into gravitational potential energy of the block earth system when the block will ascend along the incline . So change in the gravitational potential energy will be same as stored elastic potential energy of the spring that is .392 J .

c ) Let h be the distance along the incline which the block ascends.

vertical height attained ( H ) =h sin30

= .5 h

elastic potential energy = gravitational energy

.392 = mg H

.392 = 2 x 9.8 x .5 h

h = .04 m

4 cm .

=

7 0
3 years ago
A 5.00 kg mass is placed on top of a vertical spring, which compresses a distance of 3.13 cm. Calculate the force constant (in N
77julia77 [94]

Answer:

<h2>1567.09 N/m</h2>

Explanation:

Step one:

given data

mass m=5kg

compression x= 3.13cm to m= 0.0313m

<em>According to Hooke's law, provided the elastic limit of an elastic material is not exceeded the extension e is directly proportional to the applied force</em>

F=ke

where

k= spring constant in N/m

e= extension/compression in

Step two:

assume g= 9.81m/s^2

F=mg

F=5*9.81

F=49.05N

substitute in the expression F=ke

49.05=k*0.0313

k=49.05/0.0313

k=1567.09 N/m

<u>The force constant (in N/m) of the spring is 1567.09 N/m</u>

8 0
3 years ago
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