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Likurg_2 [28]
4 years ago
14

A 2-kg bowling ball is 1 meter off the ground on a post when it falls just before it reaches ground it is travelling 4.4 m/s ass

uming there is no air resistance which statement is true
mechanical energy is not conserved
initial potential energy is greater than final kinetic energy
mechanical energy is conserved
initial potential is less than the final kinetic
Physics
2 answers:
vagabundo [1.1K]4 years ago
6 0

Answer : (C) "mechanical energy is conserved".

Explanation :

It is given that :

Mass of the ball, m = 2 kg

Velocity, v = 4.4 m/s

Height, h = 1 m

Mechanical energy is the sum of kinetic energy and the potential energy. It is related with the position and the motion of the body.

So, Total\ energy = Kinetic \ energy +potential \ energy

In this case, the mechanical energy or the total energy of the ball remains conserved. This is termed as conservation of Mechanical energy.

TE=\dfrac{1}{2}mv^2+mgh

TE=\dfrac{1}{2}\times 2\ kg\times (4.4)^2\ m/s+2\ kg\times 9.8\ m/s^2\times 1\ m

TE=38.96\ J

So, the correct option is (C) "mechanical energy is conserved and it is equal to 38.96 J".

MissTica4 years ago
3 0
<span>mechanical energy is E=mgh+ 1/2 mV²=2*9.8*1+1/2*2*4.4²=38.96J, 
the only force is P, that is a conservative force so the answer is
</span>mechanical energy is conserved and its value is <span>38.96J, </span>
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algol13
Erosion:
A.) Is the same as weathering
3 0
4 years ago
A 12,000 kg railroad car is traveling at 2 m/s when it strikes another 10,000 kg railroad car that is at rest. If the cars lock
Degger [83]

Answer:

The final speed of the railroad car

V= 1.14 \frac{m}{s}

Explanation:

v_{1}=2.1\frac{m}{s} \\m_{1}=12000kg\\v_{2}=0\frac{m}{s} \\m_{2}=10000kg \\v_{t}=?

m_{1}*v_{1}+m_{2}*v_{2}= (m_{1}+m_{2})*v_{t}\\v_{t}=\frac{m_{1}*v_{1}}{(m_{1}+m_{2})} \\v_{t}=\frac{12000kg*2.1\frac{m}{s} }{(12000+10000)kg} \\v_{t}=1.14 \frac{m}{s}

That's the final speed of the both railroad car

4 0
3 years ago
Suppose you designed a spacecraft to work by photon pressure. The sail was a completely absorbing fabric of area 1.0 km2 and you
Alekssandra [29.7K]

Answer:

(a) F = 6.14 *10⁻⁴ N

(b) P = 6.14* 10⁻¹⁰ Pa

(c) t = 27.2 min

Explanation:

Area of sail A = 1.0 km² = 1.0 * 10⁶m²

Wavelength of light  λ = 650 nm = 650 * 10⁻⁹ m

Rate of impact of photons R = 1 mol/s = 6.022 * 10²³ photons/s

(a)

Momentum of each photon is Ρ = h/λ = 6.625 * 10⁻³⁴ / 650 * 10⁻⁹

      = 1.0192 * 10⁻²⁷ kg.m/s

Since the photons are absorbed completely, in every collision the above momentum is transferred to the sail.  

Momentum transferred to the sail per second is product of rate of impact of photons and momentum transferred by each photon.

dp/dt = R * h/ λ

This is the force acting on the sail.

F = R * h/λ = 6.022 * 10²³ * 1.0192 * 10⁻²⁷ = 6.14 *10⁻⁴ N

F = 6.14 *10⁻⁴ N

b)

Pressure exerted by the radiation on the sail = Force acting on the sail / Area of the sail

P = F/A =  6.14 * 10⁻⁴ / 1.0 * 10⁶ =  6.14* 10⁻¹⁰ Pa

P = 6.14* 10⁻¹⁰ Pa

c)  

Acceleration of spacecraft a = F/m = 6.14 * 10⁻⁴ /1.0 = 6.14 * 10⁻⁴m/s²

As the spacecraft starts from rest, initial speed u=0,m/s ,

final speed is u = 1.0 m/s after time t  

v = u+at  

t = 1.0 - 0/ 6.14 * 10⁻⁴ =  1629s = 27.2 min

t = 27.2 min

4 0
4 years ago
A block with mass of 10 kg is on a frictionless surface. One hand on the left side of the block is pushing it to the right. A se
igomit [66]

Answer:

W_2=-12J

Explanation:

The work of force 2 will be given by the vectorial equation W_2=F_2.d. We know the value of F_1 and have information about its movement, which relates to the net force F=F_1+F_2.

About this movement we can obtain the acceleration using the equation v_f^2=v_i^2+2ad. Since it departs from rest we have a=\frac{v_f^2}{2d}.

And then using Newton's 2dn Law we can obtain the net force F=ma, thus we will have F_2=F-F_1=ma-F1=\frac{mv_f^2}{2d}-F_1

And we had the work done by force 2 as:

W_2=F_2.d=\frac{mv_f^2}{2}-F_1d

(The sign will be given algebraically since we take positive the direction to the right.)

With our values:

W_2=\frac{(10kg)(2m/s)^2}{2}-(8N)(4m)=-12J

<em>Another (shorter but maybe less intuitive way for someone who is learning) way of doing this would have been to say that the work done by both forces would be equal to the variation of kinetic energy:</em>

<em>W_1+W_2=K_f-K_i=K_f=\frac{mv_f^2}{2}</em>

<em>Which leads us to the previous equation straightforwardly.</em>

6 0
3 years ago
An infinite long straight wire is uniformly charged, the charge density is a. Use Coulomb's law to calculate the electric field
bixtya [17]

Answer:

\vec{E} = \frac{a}{2\pi \epsilon_0 R}\^R

Explanation:

Since the wire is infinitely long, we will use Gauss' Law:

\int\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}

We will draw an imaginary cylindrical surface with height h around the wire. The electric flux through the imaginary surface will be equal to the net charge inside the surface.

In that case, the net charge inside the imaginary surface will be the portion of wire with height h. Then the charge of that portion will be equal to

Q_{enc} = ah

The left-hand side of the Gauss' Law is the flux through the imaginary surface. Since we choose our surface as a cylinder, of which we know the area, we do not have to take the surface integral.

\int\vec{E}d\vec{a} = E2\pi R h

where R is the radius of the imaginary cylinder.

Finally, Gauss' Law gives

E2\pi Rh = \frac{ah}{\epsilon_0}\\E = \frac{a}{2\pi \epsilon_0 R}

The vector expression is

\vec{E} = \frac{a}{2\pi \epsilon_0 R}\^R

As you can see, the electric field is independent from the height h, since that is merely an imaginary cylinder to apply Gauss' Law. In the end, what matters is the charge density of the wire and the distance from the wire.

4 0
4 years ago
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