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Ipatiy [6.2K]
3 years ago
14

What ia the law of convesation of energy​

Physics
1 answer:
Free_Kalibri [48]3 years ago
8 0
In physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time. This law means that energy can neither be created nor destroyed; rather, it can only be transformed or transferred from one form to another.
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A helicopter is ascending vertically. a passenger accidentally drops her wallet out the sides of the helicopter when it is 160 m
Advocard [28]

Answer:

(E)56.0 m/s

Explanation:

Height =h=-160 m

Because the wallet moving in downward direction

Time=t=7 s

Final speed of wallet=v=0

We have to find the speed of helicopter ascending  at the moment when the passenger let go of the wallet.

v^2-u^2=2gh

Where g=9.8 m/s^2

Substitute the values

0-u^2=2(-160)\times 9.8

u^2=3136

u=\sqrt{3136}=56m/s

Option (E) is true

8 0
3 years ago
our friend is constructing a balancing display for an art project. She has one rock on the left (ms=2.25 kgms=2.25 kg) and three
Licemer1 [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The torque produced by the pile of rocks is \tau = 35.63\ N \cdot m  

b

The distance of the single for equilibrium to occur is r_s =1.62 \ m

Explanation:

From the question we are told that

     The mass of the left rock is  m_s = 2.25 \ kg

     The mass of the rock on the right m_p = 10.1 kg

    The distance from  fulcrum to the center of the pile of rocks is  r_p = 0.360 \ m

   

Generally the torque produced by the pile of rock is mathematically represented as

           \tau = m_p * g * r_p

Substituting values

         \tau = 10.1 * 9.8  * 0.360                  

          \tau = 35.63\ N \cdot m      

Generally we can mathematically evaluated the distance of the the single rock that would put the system in equilibrium as follows

   The torque due to the single rock is

           \tau = m_s  * g * r_s

At equilibrium the both torque are equal

            35.63 = m_s * r_s * g

Making r_s the subject of the formula

             r_s = \frac{35.63 }{m_s * g}

Substituting values

            r_s = \frac{35.63 }{2.25 * 9.8}

            r_s =1.62 \ m

6 0
3 years ago
The Sun delivers an average power of 1.575 W/m2 to the top of Neptune's atmosphere. Find the magnitudes of max and max for the e
FrozenT [24]

Answer:

1.1486813808\times 10^{-7}\ T

34.46 V/m

Explanation:

\mu_0 = Vacuum permeability = 4\pi \times 10^{-7}\ H/m

c = Speed of light = 3\times 10^8\ m/s

I = Intensity = 1.575 W/m²

The maximum magnetic field intensity is given by

B_m=\sqrt{\dfrac{2\mu_0I}{c}}\\\Rightarrow B_m=\sqrt{\dfrac{2\times 4\pi \times 10^{-7}\times 1.575}{3\times 10^8}}\\\Rightarrow B_m=1.1486813808\times 10^{-7}\ T

The magnetic field intensity is 1.1486813808\times 10^{-7}\ T

The maximum electric field intensity is given by

E_m=B_m\times c\\\Rightarrow E_m=1.1486813808\times 10^{-7}\times 3\times 10^8\\\Rightarrow E_m=34.46\ V/m

The  electric field intensity is 34.46 V/m

8 0
2 years ago
A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

5 0
2 years ago
The lower the__ the frequency, the __the pitch <br><br> -lower <br> -higher
rjkz [21]

Answer:

umm the lower the frequency the higher the pitch

Explanation:

6 0
2 years ago
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