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bogdanovich [222]
4 years ago
10

Elements in group 1 are known as ___________ metals. Alkaline Transition Alkali Inner Transition

Physics
1 answer:
ss7ja [257]4 years ago
8 0
Alkali

The alkali metals are six chemical elements in Group 1, the leftmost column in the periodic table. They are lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr).

Hope this is the brainliest answer.
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When using the spectroscope to view the spectra of the discharge lamps, what observations are recorded?.
natima [27]

When using the spectroscope to view the spectra of the discharge lamps, what observations are recorded by the wavelengths and colors of all bands.

<h3>Spectra of discharge lamps</h3>

The spectra of discharge lamp from spectroscope shows the different electromagnetic spectrum, they include the following;

  • Ultraviolet rays
  • Visible lights
  • Infra-red, etc

The electromagnetic spectrum consists of radiations of different wavelengths which are indicated by different color bands.

Thus, when using the spectroscope to view the spectra of the discharge lamps, what observations are recorded by the wavelengths and colors of all bands.

Learn more about spectroscope here: brainly.com/question/2114494

8 0
3 years ago
A car is driving at a velocity of 24 m/s. If its brakes can supply an acceleration of -5.0 m/s2, how much time will be required
Lady_Fox [76]
In equation form it reads: 0=24-5x
You can isolate the variable by subtracting 24 from both sides to get -24=5x.
Now all that has to be done is divide both sides by -5, which will result in -4.8=x
3 0
3 years ago
Read 2 more answers
What layer of the atmosphere contains all of the weather and thus the most water vapor?
Veronika [31]

The troposphere is the lowermost layer of the Earth's atmosphere. Most of the weather phenomena, systems, convection, turbulence and clouds occur in this layer, although some may extend into the lower portion of the stratosphere.

5 0
4 years ago
A string is stretched to a length of 376 cm and
sergey [27]

Answer: 53.09Hz

Explanation:

The fundamental frequency of an ideal taut string is:

Fn= n/2L(√T/μ)

Where:

F= frequency per second (Hz)

T= Tension of the string (cm/s sqr)

L= Length of the string (cm)

μ= Linear density or mass per unit length of the string in cm/gm

√T/μ= square root of T divided by μ

It is important to note:

Note: Typically, tension would be in newtons, length in meters and linear density in kg/m, but those units are inconvenient for calculations with strings. Here, the smaller units are used.

F1= 1/2(376cm)(0.01/1) × (√574/(0.036g/cm)(0.1kg/m÷1g/cm)

F1= 0.1329 × 399.30

= 53.09Hz

7 0
3 years ago
I have three questions. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s.
AfilCa [17]

1. 5.5 m/s

We can solve the problem by applying the law of conservation of momentum. The total momentum before the collision must be equal to the total momentum after the collision, so we have:

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where

m1 = 0.4 kg is the mass of the ball

u1 = 18 m/s is the initial velocity of the ball

m2 = 0.2 kg is the mass of the bottle

u2 = 0 is the initial velocity of the bottle (which is initially at rest)

v1 = ? is the final velocity of the ball

v2 = 25 m/s is the final velocity of the bottle

Substituting and re-arranging the equation, we can find the final velocity of the ball:

v_1 = \frac{m_1 u_1 - m_2 v_2}{m_1}=\frac{(0.4 kg)(18m/s)-(0.2 kg)(25 m/s)}{0.4 kg}=5.5 m/s


2. 22.2 m/s

We can solve the problem again by using the law of conservation of momentum; the only difference in this case is that the bullet and the block, after the collision, travel together at the same speed v. So we can write:

m_1 u_1 + m_2 u_2 = (m_1 +m_2) v

where

m1 = 0.04 kg is the mass of the bullet

u1 = 300 m/s is the initial velocity of the bullet

m2 = 0.5 kg is the mass of the block

u2 = 0 is the initial velocity of the block (which is initially at rest)

v = ? is the final velocity of the bullet+block, which stick and travel together

Substituting and re-arranging the equation, we can find the final velocity of bullet+block:

\frac{m_1 u_1}{m_1 +m_2}=\frac{(0.04 kg)(300 m/s)}{0.04 kg+0.5 kg}=22.2 m/s


3. 6560 N

The impulse exerted on the ball is equal to its change in momentum:

I=\Delta p (1)

The impulse can be rewritten as product between force and time of collision:

I=F \Delta t

while the change in momentum of the ball is equal to the product between its mass and the change in velocity:

\Delta p = m\Delta v = m(v_f -v_i)

So, eq.(1) becomes

F \Delta t = m(v_f -v_i)

where:

F = ? is the unknown force

\Delta t = 0.002 s is the duration of the impact

m = 0.16 kg is the mass of the ball

v_f = 44 m/s is the final velocity of the ball

v_i = -38 m/s is its initial velocity (we must add a negative sign, since it is in opposite direction to the final velocity)

So, by using the equation, we can find the force:

F=\frac{m (v_f -v_i)}{\Delta t}=\frac{(0.16 kg)(44 m/s-(-38 m/s))}{0.002 s}=6560 N

7 0
4 years ago
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