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miv72 [106K]
2 years ago
5

During an experiment a student records the net horizontal force exerted on an object moving in a straight

Physics
1 answer:
saw5 [17]2 years ago
3 0

Answer:

30 kg ⋅ m/s

Explanation:

hope it help's u :)!!

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An apple is whirled round in a horizontal circle on the end of a string which is tied to the stalk. It is whirled faster and fas
jek_recluse [69]

when the apple moves in a horizontal circle, the tension force in the string provides the necessary centripetal force to move in circle. the tension in the string is given as

T=mv²/r

where T = tension force in the string , m = mass of the apple

v = speed of apple , r = radius of circle.

clearly , tension force depends on the square of the speed. hence greater the speed, greater will be the tension force.

at some point , the speed becomes large enough that it makes the tension force in the string becomes greater than the tensile strength of the string. at that point , the string breaks

6 0
3 years ago
A swimmer bounces almost straight up from a diving board and falls vertically feet first into a pool.she starts with a speed of
garik1379 [7]

Answer:

a) 1.20227 seconds

b) 0.98674 m

c) 7.3942875 m/s

Explanation:

t = Time taken

u = Initial velocity = 4.4 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v=u+at\\\Rightarrow 0=4.4-9.81\times t\\\Rightarrow \frac{-4.4}{-9.81}=t\\\Rightarrow t=0.44852\ s

s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.4\times 0.44852+\frac{1}{2}\times -9.81\times 0.44852^2\\\Rightarrow s=0.98674\ m

b) Her highest height above the board is 0.98674 m

Total height she would fall is 0.98674+1.8 = 2.78674 m

s=ut+\frac{1}{2}at^2\\\Rightarrow 2.78674=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.78674\times 2}{9.81}}\\\Rightarrow t=0.75375\ s

a) Her feet are in the air for 0.75375+0.44852 = 1.20227 seconds

v=u+at\\\Rightarrow v=0+9.81\times 0.75375\\\Rightarrow v=7.3942875\ m/s

c) Her velocity when her feet hit the water is 7.3942875 m/s

3 0
3 years ago
What is the mass amount of a Proton and a Neutron?
zysi [14]

The masses amount of a proton and neutron are 1.0087 and 1.0073 amu respectively.

<h3>What is a Proton?</h3>

This is defined as  sub atomic particle which is positively charged and is present in the nucleus while the neutron is also a particle present in the nucleus but has a neutral charge.

Electrons on the other hand are found outside the nucleus and are negatively charged. It is the sub atomic particle which is actively involved in a chemical reaction.

The masses of neutron and proton are 1.0087 and 1.0073 amu respectively and was discovered by scientists thereby making it the most appropriate choice.

Read more about Proton and Neutron here brainly.com/question/237857

#SPJ1

6 0
1 year ago
A 2.0 kg bucket is attached to a horizontal ideal spring and rests on frictionless ice. You have a 1.0 kg mass
bogdanovich [222]

Answer:

x = A cos (w \sqrt{2y_{o}/g})

a) maximun  Ф= \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

b) minimun     Ф = \frac{\pi }{2} - \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

Explanation:

For this exercise let's use kinematics to find the time it takes for the mass to reach the floor

         y = y₀ + v₀ t - ½ g t²

   

as the mass is released from rest, its initial velocity is zero (vo = 0) and its height upon reaching the ground is zero (y = 0)

      0 = y₀ - ½ g t²

      t = \sqrt{2y_{o}/g}

The bucket-spring system has a simple harmonic motion, which is described by

     x = A cos wt

in this expression we assumed that the phase constant (Ф) is zero

let's replace the time

     x = A cos (w \sqrt{2y_{o}/g})

this is the distance where the system must be for the mass to fall into it.

a) The new system has a total mass of m ’= 3.0 kg, so its angular velocity changes

          w = \sqrt{k/m}

In the initial state

         w = \sqrt{k/2}

When the mass changes

         w ’= \sqrt{k/3}

the displacement in each case is

         x = A cos (wt)

for the new case

        x ’= A cos (w’t + Ф)

the phase constant is included to take into account possible changes due to the collision of the mass.

we see that this maximum expressions when the cosine is maximum

        cos (w´t + Ф) = 1

         w’t + Ф = 0

        Ф = -w ’t

        Ф = - \sqrt{k/3} \sqrt{2y_{o}/g}

       \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

b) the function is minimun if

        cos (w’t + fi) = 0

        w’t + Ф = π / 2

        Ф = π / 2 - w ’t

        Ф = \frac{\pi }{2} - \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

8 0
3 years ago
A passenger in the rear seat of a car moving at a steady speed is at rest relative to
Ber [7]
The person is at rest with respect to the car. So the best answer is:

c. the front seat of the car.

6 0
2 years ago
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