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uysha [10]
3 years ago
15

The sun's core _____. A) is made of solid helium B) makes up most of the sun's size C) creates energy by fusing hydrogen into he

lium D) absorbs energy from the convection zone
Physics
2 answers:
Vikki [24]3 years ago
8 0

Answer:

the answer is C creates energy by fusing hydrogen into helium

Explanation:

i took the test

VashaNatasha [74]3 years ago
6 0
I think the correct answer is C
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Please answer will give brainliest
laila [671]

3b. No

4a. 50N


I hope this helps


8 0
3 years ago
A stone is thrown with an initial speed of 12 m/s at an angle of 30o above the horizontal from the top edge of a cliff. If it ta
kherson [118]

Answer:

d=58m

Explanation:

From the question we are told that:

Initial Speed U=12m/s

Time T=5.6s

Angle \theta=30

Generally the  Newton's equation for motion is mathematically given by

d=d'+ut+\frac{at^2}{2}

d=12cos30*5.6

d=58m

8 0
3 years ago
The magnetic field perpendicular to a single 16.7-cm-diameter circular loop of copper wire decreases uniformly from 0.750 T to z
sammy [17]

Answer:

1.24 C

Explanation:

We know that the magnitude of the induced emf, ε = -ΔΦ/Δt where Φ = magnetic flux and t = time. Now ΔΦ = Δ(AB) = AΔB where A = area of coil and change in magnetic flux = Now ΔB = 0 - 0.750 T = -0.750 T, since the magnetic field changes from 0.750 T to 0 T.

The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m

So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.

Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.

So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²

So,  4iρD/d² = 0.750πD²/4Δt.

iΔt = 0.750πD²/4 ÷ 4iρD/d²

iΔt = 0.750πD²d²/16ρ.

So the charge Q = iΔt

= 0.750π(Dd)²/16ρ

= 0.750π(16.7 × 10⁻²m 2.25 × 10⁻³ m)²/16(1.68 × 10⁻⁸ Ωm)

= 123.76 × 10⁻² C

= 1.2376 C

≅ 1.24 C

4 0
3 years ago
Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ
Irina-Kira [14]

Answer:

Magnitude of the net force on q₁-

Fn₁=1403 N

Magnitude of the net force on q₂+

Fn₂= 810 N

Magnitude of the net force on q₃+

Fn₃= 810 N

Explanation:

Look at the attached graphic:

The charges of the same sign exert forces of repulsion and the charges of   opposite sign exert forces of attraction.

Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:

F= (k*q*q)/(d)²

F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N

Magnitude of the net force on q₁-

Fn₁x= 0

Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N

Fn₁=1403 N

Magnitude of the net force on q₃+

Fn₃x= 810- 810 cos 60° = 405 N

Fn₃y= 810*sin 60° = 701.5 N

Fn_{3} = \sqrt{405^{2}+701.5^{2}  }

Fn₃ = 810 N

Magnitude of the net force on q₂+

Fn₂ = Fn₃ = 810 N

6 0
3 years ago
What would be the final size of a balloon, at -18*C, if the initial volume is 2.0 L and the initial temperature is 20*C?
Anuta_ua [19.1K]

Answer:

1.7 L

Explanation:

PV = nRT

If P, n, and R are constant:

V₁ / T₁ = V₂ / T₂

(2.0 L) / (293.15 K) = V / (255.15 K)

V = 1.7 L

7 0
3 years ago
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