What is the potential energy of stretched spring, if the spring constant is 40 N/m and the elongation is 5 cm?

A Juggler is juggling a uniform rod one end of which is coated in tar and burning. He is holding the rod by the opposite end and

lyudmila [28]

The vertical speed of the rod varies inversely as the rotational speed for a given number of complete rotations

The equation that gives the vertical velocity, v₀ is $\, \underline{v_0= \dfrac{ g \cdot \pi \cdot n}{ \omega_0 }}$

Reason:

From a similar question, the given parameters appear to be correctly given as follows;

The vertical speed of the rod = v₀

Angular velocity of the = ω₀

Required;

The value of, v₀, so that the rod has made exactly, n, number of turns when it returns to his hand

Where;

n = An integer

Solution;

The time the rod spends in the air is given as follows;

$Height, \ h = u \cdot t - \dfrac{1}{2} \cdot g \cdot t^2$

Where;

g = Acceleration due to gravity

t = Time of motion

When the rod returns to his hand, we have, h = 0, therefore;

$0 = u \cdot t - \dfrac{1}{2} \cdot g \cdot t^2$

$u \cdot t = \dfrac{1}{2} \cdot g \cdot t^2$

$t^2 = \dfrac{u \cdot t }{\dfrac{1}{2} \cdot g } = \dfrac{2 \cdot u \cdot t }{g }$

$Time, \ t = \dfrac{2 \cdot u }{g }$

We have;

$Angular \ velocity, \ \omega_0 = \dfrac{ 2 \cdot \pi \cdot n}{t} \ (required)$

$\omega_0 = \dfrac{ 2 \cdot \pi \cdot n}{ \dfrac{2 \cdot v_0 }{g }} = \dfrac{ g \cdot \pi \cdot n}{ v_0 }$

Therefore;

$The \ vertical \ velocity, \, \underline{v_0= \dfrac{ g \cdot \pi \cdot n}{ \omega_0 }}$

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4 0

3 years ago

Comment l'azote entre dans l'atmosphère

Kruka [31]

Answer:

shark puppet

Explanation:

8 0

4 years ago

A system expands from a volume of 1.00 l to 2.00 l against a constant external pressure of 1.00 atm. what is the work (w) done b

stepan [7]

The work done by a system kept at constant pressure is given by:
$W=p \Delta V = p (V_f - V_i)$
where
p is the pressure
$V_f$ is the final volume
$V_i$ is the initial volume

If we plug the numbers given by the problem into this equation, we find
$W=p (V_f -V_i)=(1.00 atm)(2.00 L-1.00 L)=1.00 L\cdot atm$

And since $1 atm \cdot L = 101.3 J$ , we have that the work done is
$W= 1.00 atm \cdot L = 101.3 J$

8 0

3 years ago

Read 2 more answers

A neutron has a kinetic energy of 10 MeV. What size aperture would be necessary to observe diffraction effects?

Sedaia [141]

Answer: 9.33fm

Explanation:

Similarly as with a cut diffraction happens when part of the proliferating wave is cut off. This will occur for any object of request with a similar size as the wavelength of the diffracting wave.

Note, the object is moving at 14% of the speed of light. Be that as it may, the gamma factor is just 1% so taking care of the issue either relativistic or non relativistic conditions offers comparative responses.

Note: The size of the object has to be of the order of the wavelength of 10MeV. Therefore, size apperture that would be necessary to observe diffraction effect would be 9.33fm.

The calculations are in the attachment below:

8 0

3 years ago

1.3 kg of water, at an initial temperature of 25oC, is heated at a rate of 100 W in a well-insulated container. How much time (i

Leviafan [203]

Answer:

68.25 minutes.

Explanation:

Power = Energy/time or

Power = Heat/time

P = Q/t....................... Equation 1

Q = Pt ........................ Equation 2

Where Q = quantity of heat, P = power, t = time.

Also,

Q = cm(t₂-t₁) ................. Equation 3

Where c = specific heat capacity of water, m = mass of water, t₁ = initial temperature of water, t₂ = final temperature of water.

substitute equation 2 into equation 3

Pt = cm(t₂-t₁) ............... Equation 4

make t the subject of the equation

t = cm(t₂-t₁)/P................ Equation 5

Given: P = 100 W, m = 1.3 kg, t₁ = 25 °C, t₂ = 100 °C.

Constant: 4200 J/kg.K

Substitute into equation 5

t = 1.3(4200)(100-25)/100

t = 409500/100

t = 4095 seconds

t = (4095/60) minutes = 68.25 minutes.

Hence the time it will take the water to reach boiling point = 68.25 minutes

6 0

3 years ago