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Dmitry_Shevchenko [17]
3 years ago
5

A(n) 30 kg boy rides a roller coaster. The acceleration of gravity is 9.8 m/s 2 . With what force does he press against the seat

when the car moving at 7.3 m/s goes over a crest whose radius of curvature is 15 m? Answer in units of N.
Physics
1 answer:
forsale [732]3 years ago
5 0

Answer:

Force, F = 187.42 N

Explanation:

It is given that,

Mass of boy, m = 30 kg

Acceleration due to gravity, a=9.8\ m/s^2

Radius of curvature of the roller coaster, r = 15 m

Speed of the car, v = 7.3 m/s

The force acting on the boy are force of gravity and the centripetal force. The net force acting on him is as follows :

F=mg-\dfrac{mv^2}{r}

F=m(g-\dfrac{v^2}{r})

F=30\times (9.8-\dfrac{(7.3)^2}{15})

F = 187.42 N

So, he press against the seat with a force is 187.42 N. Hence, this is the required solution.

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Two cars A and B are 100m apart moving towards each other with
maxonik [38]

Let car A's starting position be the origin, so that its position at time <em>t</em> is

A: <em>x</em> = (40 m/s) <em>t</em>

and car B has position at time <em>t</em> of

B: <em>x</em> = 100 m - (60 m/s) <em>t</em>

<em />

They meet when their positions are equal:

(40 m/s) <em>t</em> = 100 m - (60 m/s) <em>t</em>

(100 m/s) <em>t</em> = 100 m

<em>t</em> = (100 m) / (100 m/s) = 1 s

so the cars meet 1 second after they start moving.

They are 100 m apart when the difference in their positions is equal to 100 m:

(40 m/s) <em>t</em> - (100 m - (60 m/s) <em>t</em>) = 100 m

(subtract car B's position from car A's position because we take car A's direction to be positive)

(100 m/s) <em>t</em> = 200 m

<em>t</em> = (200 m) / (100 m/s) = 2 s

so the cars are 100 m apart after 2 seconds.

3 0
3 years ago
describe the forces acting on a car as it movies along a level highway in still air at a constant speed
Travka [436]

Answers

The car's forward motion is opposed by the friction between the road and the tires and by the resistance of the air.

3 0
2 years ago
A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite
sladkih [1.3K]

Answer:

The  kinetic energy is KE  =  7.59  *10^{10} \  J

Explanation:

From the question we are told that

       The  radius of the orbit is  r =  2.3 *10^{4} \ km  = 2.3  *10^{7} \ m

       The gravitational force is  F_g  = 6600 \ N

The kinetic energy of the satellite is mathematically represented as

       KE  =  \frac{1}{2} * mv^2

where v is the speed of the satellite which is mathematically represented as

     v  = \sqrt{\frac{G  M}{r^2} }

=>  v^2  =  \frac{GM }{r}

substituting this into the equation

      KE  =  \frac{ 1}{2} *\frac{GMm}{r}

Now the gravitational force of the planet is mathematically represented as

      F_g  = \frac{GMm}{r^2}

Where M is the mass of the planet and  m is the mass of the satellite

 Now looking at the formula for KE we see that we can represent it as

     KE  =  \frac{ 1}{2} *[\frac{GMm}{r^2}] * r

=>    KE  =  \frac{ 1}{2} *F_g * r

substituting values

       KE  =  \frac{ 1}{2} *6600 * 2.3*10^{7}

         KE  =  7.59  *10^{10} \  J

 

7 0
3 years ago
Light travels 186 000 miles per second.how many miles dose light travel in one year
Troyanec [42]

       (186,000 mi/sec) x (3,600 sec/hr) x (24 hr/da) x (365 da/yr)

  =   (186,000 x 3,600 x 24 x 365)  mi/yr

  =      5,865,696,000,000  miles per year  (rounded to the nearest million miles)
8 0
3 years ago
un movil que parte del reposo alcanza una velocidad de 75 m/s en 13 segundos ¿cual su aceleracion y el espacio que recorrio en l
Dmitriy789 [7]

Answer:

Acceleration = 5.77 m/s²

Distance cover in 13 seconds = 487.56 meter

Explanation:

Given:

Final velocity of mobile device = 75 m/s

initial velocity of mobile device = 0 m/s

Time taken = 13 seconds

Find:

Acceleration

Distance cover in 13 seconds

Computation:

v = u + at

75 = 0 + (a)(13)

13a = 75

a = 5.77

Acceleration = 5.77 m/s²

s = ut + (1/2)(a)(t²)

s = (0)(t) + (1/2)(5.77)(13²)

Distance cover in 13 seconds = 487.56 meter

8 0
3 years ago
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