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NikAS [45]
3 years ago
5

As the shuttle bus comes to a sudden stop to avoid hitting a dog, it accelerates uniformly at -4.1 m/s^2 as it slows from 9.0 m/

s to 0.0 m/s. Find the time interval of acceleration for the bus.
Physics
1 answer:
zhuklara [117]3 years ago
3 0

Acceleration  =  (change in speed) / (time for the change)

      - 4.1 m/s²  =  (-9 m/s)  /  (time for the change)

Time for the change  =  (-9 m/s) / (-4.1 m/s²)  =   2.2 seconds
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10. A satellites is in a circular orbit around the earth at a height of 360 km above the earth’s surface. What is its time perio
Afina-wow [57]

Answer:

Orbital speed=8102.39m/s

Time period=2935.98seconds

Explanation:

For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g(R2(R+h)2)

V2R+h=g(R2(R+h)2)

V=√g(R2R+h)

V= sqrt(9.8 × (6371000)^2/(6371000+360000)

V= sqrt(9.8× (4.059×10^13/6731000)

V=sqrt(65648789.18)

V= 8102.39m/s

Time period ,T= sqrt(4× pi×R^3)/(G× Mcentral)

T= sqrt(4×3.142×(6.47×10^6)^3/(6.673×10^-11)×(5.98×10^24)

T=sqrt(3.40×10^21)/ (3.99×10^14)

T= sqrt(0.862×10^7)

T= 2935.98seconds

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2 years ago
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Read 2 more answers
An elevator (mass 4100 kg) is to be designed so that the maximum acceleration is 0.0400g. what is the maximum force the motor sh
melomori [17]

Answer:

The maximum force on the supporting cable is 80688 N.

The minimum force on the supporting cable is -164 N.

Explanation:

For maximum force movement of elevator is in upward direction. Thus, equation of motion is given by,

ma = T - mg

where m is the mass of elevator

a is acceleration of elevator

g is acceleration due to gravity

T is the maximum tension in the supporting cable

T = ma + mg

T = m (a + g)

T = 4100 ( 0.04g + 9.8)

T = 80688 N

This is the maximum force on the supporting cable.

For minimum force movement of elevator is in downward direction. Thus, equation of motion is given by,

ma = T - mg

where m is the mass of elevator

a= -0.04g is acceleration of elevator because elevator is moving downward

g is acceleration due to gravity

T is the minimum tension in the supporting cable

T = ma + mg

T = m (a + g)

T = 4100 ( 9.8 - 0.04g)

T = -164 N

This is the minimum force on the supporting cable.


7 0
3 years ago
2. A Se lanza un electrón con rapidez inicial v0 = 1.60×106 m/s hacia el interior de un campo uniforme entre las placas paralela
Vanyuwa [196]

Answer:

A)     E = 145.6 N / C , B)  y= 2,8 10-7 m with a downward direction

C) he shape of the trajectory of the two particles is to simulate a parabola,

D)     F_{e} /F_{g} = 10³⁴

Explanation:

A) For this exercise we use Newton's second law to find the acceleration of the electron, where the force is electric

           F = m a  

           - e E = m a

          a = - e E / m

with the field directed downward, the acceleration is in the vertical upward direction.

We look for how much the electron moves with kinematics, in the x direction there is no acceleration,

x axis (parallel to plates)

           x = v₀ t

           t = x / v₀

y axis (perpendicular to plates)

          y = y₀ + v_{oy} t + ½ a t²

Let's take the zero of the system in the middle of the plates y₀ = 0, also the initial vertical velocity is zero (v_{oy} = 0) the width of the plate is known

          y = ½ a t²

we substitute

         y = ½ (e E /m)  (x / v₀)²

         y = ½ e x2 /m v₀²   E

we look for the electric field

        E = 2 m y v₀² / e x²

where to use this expression the length and width of the condenser must be known, suppose that the length is x = l = 1 cm = 1 10⁻² m and the width is y = 0.5 mm = 0.5 10⁻³ m

let's calculate

         E = 2  9.1 10⁻³¹ 0.5 10⁻³ (1.6 10⁶)² / (1.6 10⁻¹⁹ (1 10⁻²)²)

         E = 145.6 N / C

B) The electron is exchanged for a proton

Let's look for the vertical displacement, in this case as the proton has a positive charge it moves towards the bottom of the plates

          y = ½ e x² / m v₀² E

          y = ½ 1.6 10⁻¹⁹ 1 10⁻⁴ / (1.67 10⁻²⁷ (1.6 10⁶)²   145.6

          y = 28.4375 10⁻⁸ m

since the distance between the plates is 0.5 10-3 m, the proton passes the condensate because its deflection is very small

In summary, its displacement is y= 2,8 10-7 m and with a downward direction (the same direction of the electric field)

C) The shape of the trajectory of the two particles is to simulate a parabola, but one for having a negative charge (electron) the force is upwards and the other for having a positive charge (proton) the trajectory is downwards

D) The force of gravity

           F_{g} = G m M / R²

electron

          Between the electron and the positive charges of the conducting plate

           F_{g}= 6.67 10⁻¹¹ 1.67 10⁻²⁷ 9.1 10⁻³¹ / (0.5 10⁻³)²

           F_{g} = 4.1 10⁻⁵¹ N

           

electric force

           F_{e} = -e E

           F_{e} = - 1.6 10⁻¹⁹ 145.6

           F_{e} = 2.3 10⁻¹⁷ N

let's look for the reason between these two forces

         F_{e} / F_{g} = 2.3 10⁻¹⁷ / 4.1 10⁻⁵¹

          F_{e} /F_{g} = 10³⁴

We see that the electric force is many orders of magnitude higher than the gravitational force.

5 0
3 years ago
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