Name some types of rock that have been used to construct buildings

The end of a horizontal rope is attatched to a prong of an electricity driven tuning fork that vibrates at 100hz. The other end

Darina [25.2K]

here since string is attached with a mass of 2 kg

so here tension force in the rope is given as

$T = mg$

here we will have

$T = 2(9.8) = 19.6 N$

now we will have speed of wave given as

$v = \sqrt{\frac{T}{\mu}}$

here we will have

$v = \sqrt{\frac{19.6}{0.75\times 10^{-2}}}$

$v = 16.33 m/s$

now we know that frequency is given as

F = 100 Hz

now wavelength is given as

$\lambda = \frac{v}{F}$

$\lambda = \frac{16.33}{100} = 0.16 m$

so wavelength will be 0.16 m

5 0

3 years ago

If Star A is magnitude 1.0 and Star B is magnitude 9.6 , which is brighter and by what factor?

ludmilkaskok [199]

Answer:

Star A is brighter than Star B by a factor of 2754.22

Explanation:

Lets assume,

the magnitude of star A = m₁ = 1

the magnitude of star B = m₂ = 9.6

the apparent brightness of star A and star B are b₁ and b₂ respectively

Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: $(m_{2} - m_{1}) = 2.5\log_{10}(b_{1}/b_{2})$

The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.

We need to find the factor by which star A is brighter than star B. Using the equation given above,

$(9.6 - 1) = 2.5\log_{10}(b_{1}/b_{2})$

$\frac{8.6}{2.5} = \log_{10}(b_{1}/b_{2})$

$\log_{10}(b_{1}/b_{2}) = 3.44$

Thus,

$(b_{1}/b_{2}) = 2754.22$

It means star A is 2754.22 time brighter than Star B.

3 0

3 years ago

The main advantage banking services offer is?

KATRIN_1 [288]

In short, and in general:

Advantages

Credit Unions typically pay higher dividend rates on savingsCredit Unions typically offer lower rates on loansCredit Unions typically provide better service; since they are owned and governed by their membership, they tend to prioritize the needs of their members above all elseCredit Unions operate on a not-for-profit business model, so excess earnings are returned back to the membership in form of competitive rates and lower fees, and sometimes even special dividendsMany Credit Unions offer the same products and services found at banksCredit Unions often have added-value benefits, such as free financial education, discounted theme park tickets, and special member rates for services such as home alarm systems...even discounts at online retailers like Barnes & Noble.Disadvantages

Credit Unions, and in particular smaller local credit unions, struggle to match the level of convenience (ATMs and branches) that many banks provide their customers, although many CUs are part of shared networks which enhance the breadth of delivery channels available to their membersSome Credit Unions are limited in their product offeringsOne must qualify for membership One must pay a membership fee to join. hope this helps!

7 0

3 years ago

the chef was careless and put the aluminum cookie sheet directly on the hot stove, which melted 0.05 kg of the aluminum. how muc

sergejj [24]

Heat required to melt 0.05 kg of aluminum is 28.7 kJ.

<h3>What is the energy required to melt 0.05 kg of aluminum?</h3>

The heat energy required to melt 0.05 kg of aluminum is obtained from the heat capacity of aluminum and the melting point of aluminum.

The formula to be used is given below:

Heat required = mass * heat capacity * temperature change

Assuming the aluminum sheet was at room temperature initially.;

Room temperature = 25 °C

Melting point of aluminum = 660.3 °C

Temperature difference = (660.3 - 25) = 635.3 903

Heat capacity of aluminum = 903 J/kg/903

Heat required = 0.05 * 903 * 635.3

Heat required = 28.7 kJ

In conclusion, the heat required is obtained from the heat change aluminum and the mass of the aluminum melted.

Learn more about heat capacity at: brainly.com/question/21406849

#SPJ1

3 0

2 years ago

A loop of wire is in a magnetic field such that its axis is parallel with the field direction. Which of the following would resu

ahrayia [7]

Answer:

All the given options will result in an induced emf in the loop.

Explanation:

The induced emf in a conductor is directly proportional to the rate of change of flux.

$emf = -\frac{d \phi}{dt} \\\\where;\\\\\phi \ is \ magnetic \ flux\\\\\phi = BA\ cos \theta$

where;

A is the area of the loop

B is the strength of the magnetic field

θ is the angle between the loop and the magnetic field

Considering option A, moving the loop outside the magnetic field will change the strength of the magnetic field and consequently result in an induced emf.

Considering option B, a change in diameter of the loop, will cause a change in the magnetic flux and in turn result in an induced emf.

Option C has a similar effect with option A, thus both will result in an induced emf.

Finally, considering option D, spinning the loop such that its axis does not consistently line up with the magnetic field direction will change the angle between the loop and the magnetic field. This effect will also result in an induced emf.

Therefore, all the given options will result in an induced emf in the loop.

4 0

3 years ago