If a 1.3 kg mass stretches a spring 4 cm, how much will a 5.8 kg mass stretch the

____ can lift the 403,342 ton pioneering spirit crane vessel 10 meters in 30 seconds as if it was a cork. This about 36 GJ if wo

azamat

Answer: Trough can lift the 403,342 ton pioneering spirit crane vessel 10 meters in 30 seconds as if it was a cork. This about 36 GJ if work and 1 GW of power.

Explanation:

Why trough?

Trough is the correct answer because pioneering scale usually abide only on trough not on the other given options. A long , narrow depression between the waves or ridges is known as a trough. The lower point in the period is the trough.

Speed -: Speed is the distance per unit of time that a body moves. It's a quantity scaler that has just magnitude.
Wave energy -: The transmission and capture of energy by ocean surface waves is wave energy (or wave power). The energy collected is then used for all sorts of useful work, including the generation of electricity, water desalination, and water pumping.
Crest -: A crest point within a cycle on a wave with the highest value of upward displacement. A crest is a point on a surface wave where the medium's displacement is at its height.
Amplitude -: The maximum displacement or distance measured from its equilibrium position, moved by a point on a vibrating body or wave, is called amplitude. It is equal to half of the vibration path's length.
Period-: The duration T is the time needed to pass a given point for one complete cycle of vibration. The wave length decreases as the frequency of a wave increases.
Wavelength-: The distance between two successive crests or troughs of a wave can be described as the wavelength. The frequency is inversely proportional to the wavelength. This implies that the longer the wavelength, the smaller the frequency. Similarly, the shorter the wavelength, the higher the frequency would be.
Frequency -: Frequency defines the number of waves in a given amount of time that travel through a fixed location. In the Hertz unit, frequency is normally measured.
Information -: A piece of data is a basic fact about the identity or properties of an object, i.e. a portion of its example.
Milli -: Milli is known as a merged form meaning 'thousand' (millipede) used in the metric system for unit names equal to one thousandth of the base unit (millimeter) given.

Hence , the answer is TROUGH.

7 0

3 years ago

In a simple RC circuit, at t=0 the switch is closed with the capacitor uncharged. If C=30µF, =50V and R=10k, what is the poten

sergij07 [2.7K]

Answer:

Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W

Explanation:

As we know that the current in the circuit at given instant of time is

i = 2.0 mA

R = 10 k ohm

now we know by ohm's law

$V = iR$

$V = (2 mA)(10 kohm)$

$V = 20 volts$

so voltage across the capacitor + voltage across resistor = V

$V_c + 20 = 50$

$V_c = 30 V$

Now we know that

$U = \frac{q^2}{2C}$

here rate of change in energy of the capacitor is given as

$\frac{dU}{dt} = \frac{q}{C} \frac{dq}{dt}$

$\frac{dU}{dt} = (30)(2 mA)$

$\frac{dU}{dt} = 0.06 W$

3 0

3 years ago

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The

Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}$

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

$F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, mm is mass of the object, and $g$ is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass ${m_1}$

Substitute for in the expression of maximum static friction ${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute for , [/tex]{m_1}[/tex] for in the equation .

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation ${F_{\rm{f}}} = {m_1}a$

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute for in the above equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0

3 years ago

Investigar la función de los espejos esféricos y como funciona, en el microscopio.

NikAS [45]

Answer:

yea

Explanation:

4 0

3 years ago

You’re squeezing a springy rubber ball in your hand. If you push inward on it with a force of 1 N, it dents inward 2 mm. How far

creativ13 [48]

Answer:

10mm

Explanation:

According to Hooke's law which states that "the extension of an elastic material is directly proportional to the applied force provided the elastic limit is not exceeded. Direct proportionality there means, increase/decrease in the force leads to increase/decrease in extension.

Mathematically, F = ke where;

F is the applied force

k is the elastic constant

e is the extension

from the formula k = F/e

k = F1/e1 = F2/e2

Given force of 1N indents the spring inwards by 2mm, this means force of 1N generates extension of 2mm

Let F1 = 1N e1 = 2mm

The extension that will be produced If force of 5N is applied to the string is what we are looking for. Therefore F2 = 5N; e2= ?

Substituting this values in the formula above we have

1/2=5/e2

Cross multiplying;

e2 = 10mm

This shows that we must have dent it by 10mm before it pushes outwards by a 5N force

8 0

4 years ago