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OverLord2011 [107]
3 years ago
13

Why is sun the only star that can be seen during day time​

Physics
1 answer:
hoa [83]3 years ago
7 0

Explanation:

i think the sun is near the earth that's why we see the sun during the day and the moon is behind the sun so when it is in the night the moon will remove in front the sun and there will be night I hope it will help you

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Consider three drinking glasses. All three have thesame area base, and all three are filled to the same depth withwater. Glass A
Sergio [31]

Answer:

The correct answer is C.  All three have equal non-zero pressure

Explanation:

Pressure is the relationship between the force and the area of ​​a body, when the bodies are liquid the formula that

         P = rho g h

Where rho is the density and h the height of the liquid

We see that for this expression the pressure does not depend on the shape of the container, but on its height, as the three vessels have the same height, the pressure at the bottom is the same.

The correct answer is C   All three have equal non-zero pressure

8 0
3 years ago
How high was a brick dropped from if if falls in 2.5 seconds?
jenyasd209 [6]
Using the kinematic equation d =  V_0 * t  + 1/2 * a * t^2, where d is height you can rewrite this to be d = 1/2*g*t^2 or 4.9t^2
g = a because this is a free fall 
d = 1/2 * 9.81m/s^2 * 2.5^2
d = 30.65625m
d = 30.7m
7 0
2 years ago
What are the dimensions of B
joja [24]
Well i think the answer is impossible to find because there is no picture
4 0
3 years ago
In a nuclear reactor, each atom of uranium (of atomic mass 235 u) releases about 200 MeV when it fissions. What is the change in
Ket [755]

Answer:

0.002372187708 kg

Explanation:

Each atom of Uranium 235 releases 200 MeV = 200×10⁶×1.60218×10⁻¹⁹

= 200×1.60218×10⁻¹³ Joule

Number of atoms in a 2.6 kg sample mass = (2.6/0.235)×6.02214076×10²³

⇒Number of atoms in a 2.6 kg sample mass = 66.627×10²³ atoms

Change in energy = Change in mass / (speed of light)²

ΔE = Δmc²

⇒200×1.60218×10⁻¹³×66.627×10²³ = Δm×(3×10⁸)²

⇒Δm = 200×1.60218×10⁻¹³×66.627×10²³/(3×10⁸)²

⇒Δm = 2372.187708×10⁻⁶ kg

∴Change in mass = 0.002372187708 kg

7 0
3 years ago
Two charges are located in the x-y plane. If q1=-4.55 nC and is located at x=0.00 m, y=0.680 m and the second charge has magnitu
Elden [556K]

Answer:

Ex= -23.8 N/C  Ey = 74.3 N/C

Explanation:

As the  electric force is linear, and the electric field, by definition, is just this electric force per unit charge, we can use the superposition principle to get the electric field produced by both charges at any point, as the other charge were not present.

So, we can first the field due to q1, as follows:

Due  to q₁ is negative, and located on the y axis, the field due to this charge will be pointing upward, (like the attractive force between q1 and the positive test charge that gives the direction to the field), as follows:

E₁ = k*(4.55 nC) / r₁²

If we choose the upward direction as the positive one (+y), we can find both components of E₁ as follows:

E₁ₓ = 0   E₁y = 9*10⁹*4.55*10⁻⁹ / (0.68)²m² = 88.6 N/C (1)

For the field due to q₂, we need first to get the distance along a straight line, between q2 and the origin.

It will be just the pythagorean distance between the points located at the coordinates (1.00, 0.600 m) and (0,0), as follows:

r₂² = 1²m² + (0.6)²m² = 1.36 m²

The magnitude of the electric field due to  q2 can be found as follows:

E₂ = k*q₂ / r₂² = 9*10⁹*(4.2)*10⁹ / 1.36 = 27.8 N/C (2)

Due to q2 is positive, the force on the positive test charge will be repulsive, so E₂ will point away from q2, to the left and downwards.

In order to get the x and y components of E₂, we need to get the projections of E₂ over the x and y axis, as follows:

E₂ₓ = E₂* cosθ, E₂y = E₂*sin θ

the  cosine of  θ, is just, by definition, the opposite  of x/r₂:

⇒ cos θ =- (1.00 m / √1.36 m²) =- (1.00 / 1.17) = -0.855

By the same token, sin θ can be obtained as follows:

sin θ = - (0.6 m / 1.17 m) = -0.513

⇒E₂ₓ = 27.8 N/C * (-0.855) = -23.8 N/C (pointing to the left) (3)

⇒E₂y = 27.8 N/C * (-0.513) = -14.3 N/C (pointing downward) (4)

The total x and y components due to both charges are just the sum of the components of Ex and Ey:

Ex = E₁ₓ + E₂ₓ = 0 + (-23.8 N/C) = -23.8 N/C

From (1) and (4), we can get Ey:

Ey = E₁y + E₂y =  88.6 N/C + (-14.3 N/C) =74.3 N/C

7 0
3 years ago
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