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Makovka662 [10]
2 years ago
9

If a Is close enough to a ,then it will form a nova?

Physics
1 answer:
BARSIC [14]2 years ago
4 0
Wat... are you even asking..
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The ball is displaced to the left and then oscillates backwards and forwards between the two plates. The ball touches a plate on
ELEN [110]

Answer:

Average current produced by the repeated transfer of charge is 5.6 × 10⁻⁷ ampere

Explanation:

The formula to be used here is

Q = It

where Q is the quantity of electricity and it is measured coulombs (C); 2.8 × 10⁻⁸ C or 0.000000028 C

I is current and it is measured in ampere (amps or A); unknown

t is time and it is measured in seconds (s); 0.05 s

Since, average current is what is unknown

I =Q/t

I = 0.000000028/0.05

I = 5.6 × 10⁻⁷ A

Average current produced by the repeated transfer of charge is 5.6 × 10⁻⁷ ampere

4 0
3 years ago
Whats an astral projection
Dominik [7]

Answer:Esotericism

Explanation:

it’s something that’s in intentional out of body experience

5 0
3 years ago
Read 2 more answers
A block with mass m1 = 4.50 kg and a ball with mass m2 = 7.70 kg are connected by a light string that passes over a frictionless
Allisa [31]
1.6a =  \frac{g(m_2 + m_3 - \mu km_1)}{m_1 + m_2 + m_3}  \\  \\ 1.6a(m_1 + m_2 + m_3) = g(m_2 + m_3 - \mu km_1) \\  \\ (1.6a + \mu kg)m_1 + (1.6a - g)m_2 = (g - 1.6a)m_3 \\  \\ m_3 =  \frac{1.6a +\mu kg}{g - 1.6a} m_1 - m_2 \\  \\ m_3 = 22.57 kg
4 0
3 years ago
How does the atmospheric temperature and size of inner planets compare with those of outer planets?
amid [387]
The inner planets are rocky and have diameters of less than 13,000 kilometers. The outer planets include Jupiter, Saturn, Uranus, and Neptune. The smaller, inner planets include Mercury, Venus, Earth, and Mars. Inner planet's atmosphere is thin. (Mercury has no atmosphere). Outer Planets: Outer planets' atmosphere is very thick. The four inner planets, Mercury, Venus, Earth, and Mars, are warmer than the outer gas giants. However, the temperature of the planets does not follow a linear path from the Sun.

Hope this helps!
Please give Brainliest!
6 0
2 years ago
Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m
Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

A=0.0198

From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

5 0
3 years ago
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