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kramer
3 years ago
12

I need help with these questions

Physics
1 answer:
Feliz [49]3 years ago
6 0
7. PE=0.5×700n/m×0.9m^2
0.9^2=0.81m
0.5×700×0.81= 283.5J

8. 2000=0.5×(x)×1.5m^2
1.5^2= 0.25
0.25×0.5=0.125
2000=0.125 (x)
2000/0.125=x
x=16000 n/m

9. 4000=0.5 (375 n/m)×(x)^2
0.5×187.5 (x)
4000/187.5=21.3333333333


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Two blocks are connected by a light string that passes over two frictionless pulleys. The block of mass m2 is attached to a spri
irina1246 [14]

(BELOW YOU CAN FIND ATTACHED THE IMAGE OF THE SITUATION)

Answer:

d=\frac{2g(m1-m2)}{k}

Explanation:

For this we're going to use conservation of mechanical energy because there are nor dissipative forces as friction. So, the change on mechanical energy (E) should be zero, that means:

E_{i}=E_{f}

K_{i}+U_{i}=K_{f}+U_{f} (1)

With K_{i} the initial kinetic energy, U_{i} the initial potential energy, K_{f} the final kinetic energy and U_{f} the final potential energy. Note that initialy the masses are at rest so K_{i} = 0, when they are released the block 2 moves downward because m2>m1 and finally when the mass 2 reaches its maximum displacement the blocks will be instantly at rest so K_{f} =0. So, equation (1) becomes:

U_{i}=U_{f} (2)

At initial moment all the potential energy is gravitational because the spring is not stretched so U_{i}=U_{gi} and at final moment we have potential gravitational energy and potential elastic energy so U_{f}=U_{gf}+U_{ef}, using this on (2)

U_{gi}=U_{gf}+U_{ef} (3)

Additional if we define the cero of potential gravitational energy as sketched on the figure below (See image attached), U_{gi}=0 and we have by (3) :

0= U_{gf}+U_{ef} (4)

Now when the block 1 moves a distance d upward the block 2 moves downward a distance d too (to maintain a constant length of the rope) and the spring stretches a distance d, so (4) is:

0=-m1gd+m2gd+\frac{kd^{2}}{2}

dividing both sides by d

0=-m1g+m2g+\frac{kd}{2}

g(m1-m2)= \frac{kd}{2}

d=\frac{2g(m1-m2)}{k}, with k the constant of the spring and g the gravitational acceleration.

7 0
3 years ago
Waste disposals cost less than recycling. true or false
Nana76 [90]

True.

Recycling programs in the United States have now become a major component  in today's waste management, unfortunately, recycling programs are not cost effective and are also considered to be one of most expensive ways of ridding waste.  According to author Harvey Black of the Environmental Health Perspectives Journal, in San Jose, California “it costs $28 per ton to landfill waste compared with $147 a ton to recycle” (Black 1006).

6 0
3 years ago
In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10
sergij07 [2.7K]

Answer:

A)\Phi=83.84\times 10^{-9}

B)\Phi=0 Wb

C)emf=5.4090\times 10^{-4}V

Explanation:

Given that:

  • no. of turns i the coil, n=200
  • area of the coil, a=13.1 \times 10^{-4}\,m^2
  • time interval of rotation, t=3.1\times 10^{-2}\,s
  • intensity of magnetic field, B=6.4\times 10^{-5}\,T

(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

\Phi=B.a\,cos \theta..................................(1)

\theta is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 0^{\circ}

\Phi=83.84\times 10^{-9} Wb

(B)

In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.

∴  \theta=90^{\circ}

From eq. (1)

\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 90^{\circ}

\Phi=0 Wb

(C)

According to the Faraday's Law we have:

emf=n\frac{B.a}{t}

emf=\frac{200\times 6.4\times 10^{-5}\times 13.1 \times 10^{-4}}{3.1\times 10^{-2}}

emf=5.4090\times 10^{-4}V

7 0
3 years ago
Two protons are aimed directly toward each other by a cyclotron accelerator with speeds of 1200 km/s , measured relative to the
Radda [10]

Answer:

The maximum electrical force is 2.512\times10^{-2}\ N.

Explanation:

Given that,

Speed of cyclotron = 1200 km/s

Initially the two protons are having kinetic energy given by

\dfrac{1}{2}mv^2=\dfrac{1}{2}mv^2

When they come to the closest distance the total kinetic energy is converts into potential energy given by

Using conservation of energy

mv^2=\dfrac{kq^2}{r}

r=\dfrac{kq^2}{mv^2}

Put the value into the formula

r=\dfrac{8.99\times10^{9}\times(1.6\times10^{-19})^2}{1.67\times10^{-27}\times(1200\times10^{3})^2}

r=9.57\times10^{-14}\ m

We need to calculate the maximum electrical force

Using formula of force

F=\dfrac{kq^2}{r^2}

F=\dfrac{8.99\times10^{9}\times(1.6\times10^{-19})^2}{(9.57\times10^{-14})^2}

F=2.512\times10^{-2}\ N

Hence, The maximum electrical force is 2.512\times10^{-2}\ N.

8 0
3 years ago
If a football player has more mass they will also have more ______ ?<br> Fill in the blank
serg [7]

If a football player has more mass, they will also have more <u>momentum</u>. This is because mass is directly proportional to momentum.

3 0
3 years ago
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