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krok68 [10]
2 years ago
7

When you run around a track which type of energy are you not demonstrating ?

Physics
2 answers:
melomori [17]2 years ago
7 0
The answer is Kinetic Energy
mina [271]2 years ago
6 0

Answer: Kinetic energy

Explanation:

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What is the reaction force of the table with a weight of 558N
Delicious77 [7]

Answer: reaction force = -558N

Explanation:

w = f = 558N

since action force and reaction force are equal in magnitude and opposite in direction,

reaction force = -(f)

reaction force = -558N

if that helps.

8 0
3 years ago
Is telekinesis real??? I really wanna start learning it!
Lesechka [4]

Answer:

if you only have to control your chakra and know how to get all your vibes to pass it to objects and it takes time to practice

4 0
3 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
3 years ago
A very long solid insulating cylinder has radius R = 0.1 m and uniform charge density rho0= 10-3 C/m3. Find the electric field a
Galina-37 [17]

Answer:

E   = (0.56 \times 10^8 ) r   \   \ N/c

Explanation:

Given that:

\rho_o = (10^{-3} ) \ c/m^3

R = (0.1) m

To find  the electric field for r < R by using Gauss Law

{\oint}E^{\to}* da^{\to} = \dfrac{Q_{enclosed}}{\varepsilon_o} --- (1)

For r < R

Q_{enclosed}=(\rho) ( \pi r^2 ) l

E*(2 \pi rl)= \dfrac{\rho ( \pi r ^2 l)}{\varepsilon_o}

E= \dfrac{\rho ( r)}{2 \varepsilon_o}

where;

\varepsilon_o = 8.85 \times 10^{-12}

E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}

E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}

E   = (0.56 \times 10^8 ) r   \   \ N/c

4 0
3 years ago
Two parallel slits are illuminated by light composed of two wavelengths. One wavelength is λA = 622nm. The other wavelength is λ
Sergio039 [100]

Answer:

\lambda_{B}=414.67 nm

Explanation:

In this question we have given

\lambda_{A}=622nm

we have to find

\lambda_{B}=?

We know that

optical path difference for bright fringe is given as=n\lambda

Here,

n is order of fringe

and optical path difference for dark fringe is given as=(n+.5)\lambda

since the light with wavelength \lambda_{A} produces its third-order bright fringe at the same place where the light with wavelength \lambda_{B} produces its fourth dark fringe  

it means

optical path difference for 3rd order bright fringe= optical path difference for forth order dark fringe

Therefore,

3\lambda_{A}=(4+.5)\lambda_{B}...............(1)

Put value of \lambda_{A} in equation (1)

3 \times 622=(4+.5)\lambda_{B}

1866=4.5\lambda_{B}

\lambda_{B}=414.67 nm

3 0
3 years ago
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