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natali 33 [55]
3 years ago
5

A circus acrobat is shot out of a cannon with an initial upward speed of 34 ft/s. if the acrobat leaves the cannon 4 ft above th

e ground, how long will it take him to reach a net that is 8 ft above the ground
Physics
1 answer:
jok3333 [9.3K]3 years ago
6 0
<h2>It will take 0.125 seconds to reach the net.</h2>

Explanation:

Initial speed, u = 34 ft/s = 10.36 m/s

Acceleration, a = -9.81 m/s²

Displacement, s = Final height - Initial height = 8 - 4 = 4 ft = 1.22 m

We have equation of motion, s = ut + 0.5 at²

Substituting

              s = ut + 0.5 at²

              1.22 = 10.36 x t + 0.5 x -9.81 x t²

              4.905t² - 10.36 t + 1.22 = 0

              t = 1.99 s     or    t = 0.125 seconds

Minimum time is 0.125 seconds.

It will take 0.125 seconds to reach the net.

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Why do we need to put Airplane mode?
Serga [27]

Answer:

if we ever ride a airplane we dont mess up its signals and crash ,and its easier to ignore calls and texts

Explanation:

6 0
3 years ago
Read 2 more answers
Identify each picture as either an inelastic collision or elastic collision
Ivan

Answer:

<u>Inelastic collision:</u>

A collision in which there is a loss of Kinetic Energy due to internal friction of the bodies colliding.

<u>Characteristics of an inelastic collision:</u>

  • <em>the momentum of the system is conserved</em>
  • <em>the momentum of the system is conservedloss of kinetic energy</em><u> </u>

<em>I</em><em>n</em><em> </em><em>a perfectly elastic collision</em><em>, the two bodies </em><em>that</em><em> </em><em>collide with each other stick together.</em>

<u>Elastic </u><u>collision</u><u>:</u>

A collision in which the kinetic energy of the two bodies, before and after the collision, remains the same.

<u>Characteristic</u><u>s</u><u> </u><u>of</u><u> </u><u>elastic</u><u> </u><u>collision</u><u>:</u>

  • <em>the</em><em> </em><em>momentum</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>system</em><em> </em><em>is</em><em> </em><em>conserved</em>
  • <em>no</em><em> </em><em>loss</em><em> </em><em>o</em><em>f</em><em> </em><em>kinetic</em><em> </em><em>energy</em>

In everyday life, no collision is perfectly elastic.

__________________

ANSWER:

<u>Given examples:</u>

  • Two cars colliding with each other form an example of inelastic collision.

<u>Reason:</u>

<em>(</em><em>T</em><em>hey</em><em> </em><em>lose</em><em> </em><em>kinetic</em><em> </em><em>energy</em><em> </em><em>and</em><em> </em><em>come</em><em> </em><em>to</em><em> </em><em>a</em><em> </em><em>stop</em><em> </em><em>after</em><em> </em><em>the</em><em> </em><em>collision</em><em>.</em><em>)</em>

  • A ball bouncing after colliding with a surface is an example of elastic collision

<u>Reason:</u>

<em>(a very less amount of kinetic energy is lost)</em>

7 0
2 years ago
g A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. At t = 0, t
jok3333 [9.3K]

Answer:

(a)    f = 0.58Hz

(b)    vmax = 0.364m/s

(c)    amax = 1.32m/s^2

(d)    E = 0.1J

(e)    x(t)=0.1m*cos(2π(0.58s^{-1})t)

Explanation:

(a) The frequency of the oscillation, in a spring-mass system, is calculated by using the following formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}            (1)

k: spring constant = 20.0N/m

m: mass = 1.5kg

you replace the values of m and k for getting f:

f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz

The frequency of the oscillation is 0.58Hz

(b) The maximum speed is given by the following relation:

v_{max}=\omega A=2\pi f A      (2)

A: amplitude of the oscillations = 10.0cm = 0.10m

v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}

The maximum speed of the mass is 0.364 m/s.

The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.

(c) The maximum acceleration is given by the following formula:

a_{max}=\omega^2A=(2\pi f)^2 A

a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}

The maximum acceleration is 1.32 m/s^2

The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.

(d) The total energy of the system is calculated with the maximum potential elastic energy:

E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J

The total energy is 0.1J

(e) The displacement as a function of time is:

x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)

6 0
3 years ago
If a mass of .343 kg moves down a 5m ramp in 5.8 seconds, what is the velocity and KE developed by the moving mass
Soloha48 [4]

Velocity 0.86m/s

0.13J

Explanation:

Given parameters:

Mass = 0.343kg

distance = 5m

time taken = 5.8s

Unknown:

Velocity of mass = ?

Kinetic energy = ?

Solution:

Velocity is the rate of change of displacement with time. It is a vector quantity that shows magnitude and direction.

 Mathematically;

   Velocity = \frac{displacement}{time}

  Velocity  = \frac{5}{5.8 } = 0.86m/s

Kinetic energy is the energy due to the motion of a body. It is expressed mathematically as:

  Kinetic energy  = \frac{1}{2}  m v^{2}

m is the mass

v is the velocity

  Kinetic energy  = \frac{1}{2 }   x    0.343  x    0.86^{2} = 0.13J

learn more:

Kinetic energy brainly.com/question/6536722

#learnwithBrainly

7 0
4 years ago
A meter stick balances horizontally on a knife-edge at the 51 cm mark. With two nickels stacked over the 6.0 cm mark, the stick
Oliga [24]

Answer:

65g

Explanation:

Two main conditions for equilibrium are:

I. The resultant force must be equal to zero. That is, sum of the forces acting in one direction about a point must be equal to the sum of the forces acting in the opposite direction about the same point.

II. The resultant moment must be equal to zero. That is, sum of the moments in one direction about a point must be equal to the sum of the moments in another direction about the same point.

For the above question,

the 51cm mark is the point where the resultant weight of the meter stick lies,

the pivot or point is the 45cm mark where the stick balanced when 2 nickels ( total mass (5.0g x 2) 10g were placed at the 6cm mark.

Using the conversion factor:

1000g(1kg) = 10N, we can convert mass to weight, calculate the weight of the meter stick then reconvert to mass.

That is,

mass of 2 nickels = 10g = 10/1000 = 0.01N.

Moment = Force x distance from line of force to pivot of rotation

Applying the principle of equilibrium,

Moment of left side = Moment of right side

0.01 x (45-6) = W x (51-45)

Where W = weight of the meter stick

W x 6 = 0.01 x 39

W x 6 = 0.39

W = 0.39/6

W= 0.065N

Therefore, mass of meter stick = 0.065 x 1000 = 65g.

4 0
3 years ago
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