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Fynjy0 [20]
3 years ago
11

I need this type out answer question.

Physics
1 answer:
AveGali [126]3 years ago
8 0

Answer:

<em>The ball has 7.35 joules of energy at position B.</em>

<em>The velocity of the ball at position A is 3.13 meter/second</em>

Explanation:

<u>Kinetic and Gravitational Potential Energy</u>

Kinetic energy is the form of energy that an object or a particle has by reason of its motion. An object of mass m and speed v has kinetic energy calculated by:

\displaystyle K=\frac{1}{2}m.v^2

Gravitational Potential Energy is the form of energy that an object has by reason of its height h relative to a certain reference. It can be calculated as follows:

U=m.g.h

Where g is the acceleration of gravity.

The figure shows a pendulum with a bob (ball) of mass m=1.5 Kg. When it's pulled to point B, it has a height of h= 0.5 m and set to rest.

The potential energy at that point is:

U=1.5\ Kg\cdot 9.8 \ m/s^2\cdot 0.5\ m

U=7.35\ J

The ball has 7.35 joules of energy at position B.

When the ball is released, all of the potential energy is transformed into kinetic energy when reaches point A. Thus:

K=7.35 J

From the equation of kinetic energy, we solve for v:

\displaystyle v=\sqrt{\frac{2K}{m}}

\displaystyle v=\sqrt{\frac{2\cdot 7.35}{1.5}}

\displaystyle v=\sqrt{\frac{14.7}{1.5}}

\displaystyle v=\sqrt{9.8}

v = 3.13 m/s

The velocity of the ball at position A is 3.13 meter/second

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In an intergalactic competition, spaceship pilots compete to see who can cover the distance between two asteroids in the short-
pogonyaev

Answer:

a)  truc is C,  b) correct result is the B

Explanation:

As the speed of the competition is very high, for the judges the speed is

           v = d / t

           v = 3 109 m / 20

           v = 1.5 108 m / s

This is half the speed of light. For these high speeds we must use the relations of special relativity.

For the time          t = to γ

For distance         L = Lo / γ

                            γ = √ (1-v2 / c2)

Own time and distance (to and Lo) corresponds to the observer who is not moving the judges in this case

Let's look for the range value

                     γ = 1 / √ (1 - (1.5 / 3) 2) = 1 / 0.866 = 1.15

The time              t = 20 1.15 = 23 s

The distance       L = 3 10 9 /1.15 = 2.60 109 m

From these results we see that time increases and the distance is shorter.

Let's review the claims

A) False. It's the opposite

B) False

C) True. It is according to the result found

D) False.

In the nuclear fusion process, we will also use the special relativity that has a relationship between energy and mass

         ΔE = c² Δm

As in the process energy is released, for the law of conservation of the mass of energy to be fulfilled, the total mass of the products, He atom, must be reduced.

Therefore the correct result is the B

4 0
4 years ago
A 0.520-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of
Veseljchak [2.6K]

Answer:

Explanation:

Given that,

Mass attached to spring

M = 0.52kg

Force constant K = 8N/m

Amplitude A = 11.6 cm

a. Maximum speed?

Angular velocity is calculated using

w = √k/m

w = √8/0.52

w = √15.385

w = 3.922rad/s

Then, the relation ship between angular velocity and linear velocity is given as

v = - w•A

v = - 3.922 × 11.6

v = - 45.5 cm/s

Then, the maximum velocity is

vmax = |v|= 45.5cm/s

b. Acceleration a?

Acceleration can be determine using the formula

a = -w²• A

a = -3.922² × 11.6

a = -178.46 cm/s²

Magnitude of the acceleration is 178.46cm/s²

c. Speed when the object is at 9.6cm from equilibrium position?

Generally,

The position of the object at equilibrium is

x(t) = A•Cos(wt)

x(t) = 11.6 Cos (3.922t)

Then, when x(t) = 9.6cm

9.6 = 11.6 Cos(3.92t)

Cos(3.922t) = 9.6/11.6

Cos(3.922t) = 0.8276

3.922t = ArcCos(0.8276)

Note: the angle is in radiant

3.922t = 0.596

t = 0.596/3.922

t = 0.152 second

Then, v(t) at that time is

v(t) = x'(t) = -11.6×3.92Sin(3.922t)

v(t) = -45.5Sin(3.922t)

Now, when t =0.152

v(t) = -45.5 Sin(3.922×0.152)

v(t) = -45.5Sin(0.596)

v(t) = -25.5 cm/s

Then, it's magnitude is 25.5cm/s

d. Acceleration at same position

t = 0.152s

a(t) = v'(t) = - 45.5×3.922Cos(3.922t)

a(t) = -178.46Cos(3.92t)

a(t) = -178.46 Cos(3.92×0.152)

a(t) = -178.46 Cos(0.596)

a(t) = -147.68 cm/s²

Magnitude of the acceleration is 147.68 cm/s²

5 0
4 years ago
Determine the distance between a newly discovered planet and its single moon if the orbital period of the moon is 1.2 Earth days
Vinil7 [7]

Answer:

The distance is r = 55430496 \  m  

Explanation:

From the question we are told that

   The period of the moon T =  1.2 days = 1.2 * 24 * 3600 = 103680 \  s

    The mass of the planet is  m_p =  9.38*10^{24} kg

Generally the period of the moon is mathematically represented as

       T  =  2 *  \pi * \sqrt{ \frac{r^3 }{ G * m_p } }

Here G is the gravitational constant with value

        G  =  6.67 *10^{-11} \  N \cdot m^2/kg^2

=>   T  =  2 *  \pi * \sqrt{ \frac{r^3 }{ G * m_p } }

=>   103680   =  2 *  3.142  * \sqrt{ \frac{r^3 }{ 6.67*10^{-11} * 9.38*10^{24} } }

=>    272218492.31 = \frac{r^3}{ 6.67 *10^{-11} * 9.38*10^{24}}

=>    r = \sqrt[3]{ 1.7031241*10^{23}}j

=>   r = 55430496 \  m        

8 0
3 years ago
Pendulum mass is 4 kg. Use your equations for gravitational potential energy and kinetic energy to determine these values based
mrs_skeptik [129]

its should be 2.0 and 4.5 on it

7 0
3 years ago
A bullet of mass 0.107 kg traveling horizontally at a speed of 300 m/s embeds itself in a block of mass 3 kg that is sitting at
melisa1 [442]

Answer:

165.77J

Explanation:

M₁ = 0.107kg

u₁ = 300m/s

m₂ = 3kg

u₂ = 0

v =

m₁u₁ + m₂u₂ = (m₁ + m₂)V

(0.107*300) + 0 = (0.107 + 3)V

V = 32.1 / 3.107 = 10.33m/s

kinetic energy of the system after collision =

½m1v² + ½m2v²

K.E = ½(m1 + m2)v²

K.E = ½(0.107+3) * 10.33²

K.E = 165.77J

5 0
3 years ago
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