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igor_vitrenko [27]
3 years ago
15

What is the name given to this group of elements?

Physics
1 answer:
Marianna [84]3 years ago
6 0
Your answer would be A. Halogens

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2. A woman prevents a 3kg brick from falling by pressing it against a vertical wall. The coefficient of friction
Anettt [7]
<h3>Answer:</h3>

49 N

<h3>Explanation:</h3>

<u>We are given;</u>

  • Mass of the brick as 3 kg
  • The coefficient of friction as 0.6

We are required to determine the force that must be applied by the woman so the brick does not fall.

  • We need to importantly note that;
  • For the brick not to fall the, the force due to gravity is equal to the friction force acting on the brick.
  • That is; Friction force = Mg

But; Friction force = μ F

Therefore;

μ F = mg

0.6 F = 3 × 9.8

0.6 F = 29.4

      F = 49 N

Therefore, she must use a force of 49 N

6 0
3 years ago
What is one effect of the uneven heating of Earth caused by the Sun?
vladimir1956 [14]

Answer:

The uneven heating results in some of the atmosphere to be warmer than other parts and changes in volume and pressure which result in updrafts and can cause thunderstorms and other violent weather.

Explanation:

Generation of wind

8 0
3 years ago
Read 2 more answers
A plane electromagnetic wave, with wavelength 5 m, travels in vacuum in the positive x direction with its electric vector E, of
Dafna1 [17]

Answer:

Explanation:

E₀ = 229.1 V/m

E = E₀ / √2 = 229.1 / 1.414 = 162 V/m

B =  E / c  ( c is velocity of em waves )

=   162 / (3 x 10⁸)  = 54 x 10⁻⁸ T

rate of energy flow =  ( E x B )  / μ₀

= 162 x 54 x 10⁻⁸ / 4π x 10⁻⁷

= 69.65 W per m².

5 0
3 years ago
Which of following is typical stored energy used to power todays vehicles?
Gala2k [10]
Electrical energy.................
3 0
3 years ago
A locomotive accelerates a 25-car train along a level track. Every car has a mass of 7.7 ✕ 104 kg and is subject to a friction f
MaRussiya [10]

To solve the problem it is necessary to apply the concepts related to Force of Friction and Tension between the two bodies.

In this way,

The total mass of the cars would be,

m_T = 25(7.7*10^4)Kg

m_T = 1.925*10^6Kg

Therefore the friction force at 29Km / h would be,

f=250v

f= 250*29Km/h

f = 250*29*(\frac{1000m}{1km})(\frac{1h}{3600s})

f = 2013.889N

In this way the tension exerts between first car and locomotive is,

T=m_Ta+f

T=(1.925*10^6)(0.2)+2013.889

T= 3.8701*10^5N

Therefore the tension in the coupling between the car and the locomotive is 3.87*10^5N

6 0
3 years ago
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