All categories

2 years ago

I will mark you brainlist please help!

Physics

1 answer:

Studentka2010 [4]2 years ago

7 0

Answer:

C.) The Distance DH = 1.5 lambda

Explanation:

This statement C.) is false, because it does not count as the 1.5 wavelength, it is less than 1 wavelength.

You might be interested in

3. Twelve waves pass a dock in 3.60 If the waves are traveling at 19.5 m/s , what is the wavelength of the waves?

otez555 [7]

Answer:

65m

Explanation:

Twelve waves pass a dock in 3.60 If the waves are traveling at 19.5 m/s , what is the wavelength of the waves?

Velocity of a wave is expressed as;

V = frequency × wavelength

Given

Velocity = 19.5m/s

Frequency is the number of oscillations completed in 1sec, hence

Frequency= 3.6/12

Frequency = 0.3

Wavelength = vel/freq

Wavelength = 19.5/0.3

Wavelength = 65m

4 0

3 years ago

11. A collection of a small group of factors that are related is known as a(n)

malfutka [58]

Answer:

system

Explanation:

they are all put together as one, and they are a group

5 0

2 years ago

A carpenter is driving a 15.0-g steel nail into a board. His 1.00-kg hammer is moving at 8.50 m/s when it strikes the nail. Half

bekas [8.4K]

Answer: The increase in temperature of the nail after the three blows is 8.0636 Kelvins. The correct option is (d).

Explanation:

Kinetic energy of the hammer ,K.E.=

$\frac{1}{2}mv^2=\frac{1}{2}1.00 kg\times (8.50 m/s)^2=36.125 J$

Half of the kinetic energy of the hammer is transformed into heat in the nail.

Energy transferred to the nail in one blow =

$\frac{1}{2}K.E.=\frac{1}{2}\times 36.125 J=18.0625 J$

Total energy transferred after 3 blows,Q = $3\times 18.0625 J=54.1875 J$

Mass of the nail = 15 g = 0.015 kg

Change in temperature = $\Delta T$

Specif heat of the steel = c = 448 J/kg K

$Q=mc\Delta T$

$54.1875 J=0.015 kg\times 448 J/kg K\times \Delta T$

$\Delta T=8.0636 K\approx 8.1 K$

The increase in temperature of the nail after the three blows is 8.1 Kelvins.Hence, correct option is (d).

4 0

3 years ago

Prove that..<br>please help<br>

GaryK [48]

$\large{ \boxed{ \bf{ \color{red}{Universal \: law \: of \: gravitation}}}}$

Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The forces along the line joining the centre of the two objects.

❍ Let us consider two masses m1 and m2 line at a separation distance d. Let the force of attraction between the two objects be F.

According to universal law of gravitation,

$\large{ \longrightarrow{ \rm{F \propto m_1 m_2}}}$

Also,

$\large{ \longrightarrow{ \rm{ F \propto \dfrac{1}{ {d}^{2} } }}}$

Combining both, We will get

$\large{ \longrightarrow{ \rm{F \propto \dfrac{ m_1 m_2}{ {d}^{2}}}}}$

Or, We can write it as,

$\large{ \longrightarrow{ \rm{F \propto \: G \dfrac{ m_1 m_2}{ {d}^{2} }}}}$

Where, G is the constant of proportionality and it is called 'Universal Gravitational constant'.

☯️ Hence, derived !!

<u>━━━━━━━━━━━━━━━━━━━━</u>

8 0

3 years ago

A 0.1 kg toy contains a compressed spring. when the spring is released the toy fly 0.45 m upwards from ground level before falli

trapecia [35]

The speed of the toy when it hits the ground is 2.97 m/s.

The given parameters;

mass of the toy, m = 0.1 kg

the maximum height reached by the, h = 0.45 m

The speed of the toy before it hits the ground will be maximum. Apply the principle of conservation of mechanical energy to determine the maximum speed of the toy.

P.E = K.E

$mgh_{max} = \frac{1}{2} mv_{max}^2\\\\gh_{max} = \frac{1}{2} v_{max}^2\\\\v_{max}^2= 2gh_{max}\\\\v_{max} = \sqrt{2gh_{max}}$

Substitute the given values and solve the speed;

$v_{max} = \sqrt{2\times 9.8 \times 0.45} \\\\v_{max} = 2.97 \ m/s$

Thus, the speed of the toy when it hits the ground is 2.97 m/s.

Learn more here: brainly.com/question/7562874

7 0

2 years ago