Answer:
The battery can supply 130 W for 11.75 h
Explanation:
In order to discover the time in wich the battery can supply this energy we need to find how much current is being drawn from it, we do that by using the equation for real power that is P = V*I, since we have V and P we can solve for I as seen bellow:
I = P/V = 130/12 = 10.834 A
We can use this value to find how many hours the power can supply said current. We do that by dividing the current capacity of the battery by the current drawn:
t = 141/12 = 11.75 h
Answer:
twice
Explanation:
From magnification = height of image / height of object
Distance of image/ distance of object = magnification
If the distance and height of the object represents the initial light distance and the exposed surface respectively.
And similarly the distance and height of the image represents the final light distance and the exposed surface respectively.
Hence the new image exposure would be twice as large.
If we use the formula our point of investigation is Height of image,
H2= D2/D1× H1
H2 = 2D2/D1 × H1
H2 = 2H1
Answer:
Formation of a Precipitate
Unexpected Color Change
Formation of a Gas
Explanation:
Answer:

Explanation:
Given:
Gravity of Mars = 0.38 times the gravity at Earth
Gravity of Earth is, 
Radius of Mars (R) = 3400 km
Mass of mars (M) = ?
We know that, the acceleration due to gravity of a planet of mass 'M' and radius 'R' is given as:

Now, as per question:

Plug in 9.8 for
and solve for
. This gives,

Now, plug in this value in the above equation and solve for 'M'. This gives,

Therefore, the mass of Mars is
.
Answer:
The answer is + 4.8V
Explanation:
The electric potential is
V = k ∑ 
We apply this expression to our case with three charges
V = k (q₁ / r₁ + q₂ / r₂ + q₃ / r₃)
The distance is
R = √((x-x₀)² + (y-y₀)²+ (z-z₀)²)
Let's find the distances
q₁ = 50 10⁻⁹ C
r₁ = √ ((8-0)² + (0-6)²)
r₁ = 10m
q₂ = -80 10⁻⁹ C
r₂ = √ ((8 + 4)²
r₂ = 12m
q₃ = 70 10⁻⁹ C
r₃ = √ ((8-0)²+ (0 + 6)²)
r₃ = 10 m
We calculate the potential
V = 8.99 10⁸ 10⁻⁹ (50/10 -80/12 +70/10)
V = 4.795 V
The answer is + 4.8V