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Anuta_ua [19.1K]
3 years ago
8

Which forces include attractive and repulsive interactions? A. electric and magnetic forces B. static friction and air resistanc

e C.rolling friction and fluid friction D. gravitational and spring forces
Physics
2 answers:
Ivahew [28]3 years ago
5 0

Answer: A. electric and magnetic forces

Explanation:

  • Magnetic force acts between two magnetic objects. A magnet has two poles- North pole and South pole. Like poles repel each other and unlike poles attract each other.
  • Electric force acts between two charged objects. Two similarly charged bodies have repulsive force between them and attractive force exists between two dissimilar charged objects.

Thus, an electric force and magnetic force are repulsive and attractive in nature.

saul85 [17]3 years ago
3 0
The answer is : A



i hoped this helped and have a wonderful day!!


:)
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Does wavelength affect the energy of a wave?
Galina-37 [17]

Answer:

Not quite

Explanation:

The frequency of a wave is inversely proportional to its wavelength. That means that waves with a high frequency have a short wavelength, while waves with a low frequency have a longer wavelength

What determines the strength of a wave?

Wave height is affected by wind speed, wind duration (or how long the wind blows), and fetch, which is the distance over water that the wind blows in a single direction. If wind speed is slow, only small waves result, regardless of wind duration or fetch.

So,

As Wavelength increases, The energy of the wave spreads and it decreases

4 0
2 years ago
Water pressurized to 3.5 x 105 Pa is flowing at 5.0 m/s in a horizontal pipe which contracts to 1/3 its former area. What are th
masha68 [24]

Answer:

the pressure after contraction is 2×10^5 Pa

the speed after contraction is 15m/s

Explanation:

We were given Pressure P to be 3.5 x 10^5 that is Flowing with speed of 5.0 m/s,

For us to calculate pressure we need to calculate the area first as;

Let initial Area = A₁

And Final area A₂

We were told that in a horizontal pipe it contracts to 1/3 its former area. Which means

A₂= A₁/3.................

V₁ is the speed

the pressure and speed of the water after the contraction can be calculated using equation of continuity below

A₂V₂ = A₁V₁

But

If we substitute given value in the expresion we have

V₂ = (3A *5)/A

V₂ = 15m/s

Therefore, the speed after contraction is 15m/s

Now we can calculate the pressure using

Bernoulli's equation

p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂

But we know that the pipe is horizontal, then "h" terms cancel out then

p₁ + ½ρv₁² = p₂ + ½ρv₂²

Making P₂ subject of formula we have

p₂ = 0.5ρ( V ₁² - v₂² ) + P₁

P₂=. 0.5 × 1000 (5² -15² ) + 3*10^5

=2×10^5 Pa

Therefore, the pressure after contraction is 2×10^5 Pa

7 0
3 years ago
Which statement describes a property of a proton?
lesantik [10]

Answer:

They have no charge and are present in the nucleus of an atom. They have a negative charge and travel around the nucleus of an atom. They have a positive charge and travel around the nucleus of an atom

Explanation:

7 0
3 years ago
Block m1 of mass 2 m and velocity v0 is traveling to the right (+x) and makes an elastic head-on collision with block m2 of mass
OLga [1]
Ness awaits you sir, please be theoretical and positive eeeeeeeee
7 0
4 years ago
Two spherical asteroids have the same radius R. Asteroid 1 has mass M and asteroid 2 has mass 1.97·M. The two asteroids are rele
nekit [7.7K]

Answer:

0.536\sqrt{\frac{GM}{R}}

Explanation:

We are given that

Mass of one  asteroid 1,m_1=M

Mass of asteroid 2,m_2=1.97 M

Initial distance between their centers,d=13.63 R

Radius of each asteroid=R

d'=R+R=2R

Initial velocity of both asteroids

u=0

We have to find the speed of second asteroid just before they collide.

According to law of conservation of momentum

(m_1+m_2)u=m_1v_1+m_2v_2

(M+1.97 M)\times 0=Mv_1+1.97Mv_2

Mv_1=-1.97 Mv_2

v_1=-1.97v_2

According to law of conservation of energy

Gm_1m_2(\frac{1}{d'}-\frac{1}{d})=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2

GM(1.97M)(\frac{1}{2R}-\frac{1}{13.63R})=\frac{1}{2}M(-1.97v_2)^2+\frac{1}{2}(1.97M)v^2_2

1.97M^2G(\frac{13.63-2}{27.26R})=\frac{1}{2}Mv^2_2(3.8809+1.97)

1.97MG(\frac{11.63}{27.26 R})=\frac{1}{2}(5.8509)v^2_2

v^2_2=\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}

v_2=\sqrt{\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}}

v_2=0.536\sqrt{\frac{GM}{R}}

Hence, the speed of second asteroid =0.536\sqrt{\frac{GM}{R}}

8 0
3 years ago
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