To determine the mass of gold dissolved in oceans, we must first know the volume of the ocean in liters. To determine its volume, we multiply the area and the depth taking note of the consistency of units. The volume would be 1.38 x 10^18 cubic meters or 1.38x10^21 liters. Then,multiplying this to the concentration, the answer would be 8 x 10^12 grams or 8 billion kg.
Let original length be L. The new length is therefore 4L.
Let original cross sectional surface area of the wire be equal to πr^2.
This means original volume was L x πr^2 = Lπr^2
The volume is the same but the length is different so 4L x new surface area must be equal to Lπr^2. Let new surface area be equal to Y.
4L x Y = Lπr^2
=> Y = (πr^2 )/ 4
Using the resistivity formula,
R = pL/A. p which is resistivity is a constant so it stays the same
But this time, instead of L we have 4L and instead of πr^2 we have (πr^2)/4.
so the new resistance
= (4Lp)/ {(πr^2)/4}
= 16 (pL)/(πr^2)
= 16 (pL)/A. because πr^2 is A
since pL/A is equal to R from the formula, this is equal to
16 R.
R was 10 ohms
therefore new resistance is 16 x 10 = 160 ohms
The component of the total velocity in the x - direction is 6.96 m/s.
The component of the total velocity in the y - direction is 2.95 m/s.
<h3>
Component of the velocity in x direction </h3>
The component of the total velocity in the x - direction is calculated as follows;
v(x) = vtot cosθ
where;
- vtot is total velocity
- v(x) is velocity in x direction
v(x) = 7.56 x cos(23)
v(x) = 6.96 m/s
<h3>
Component of the velocity in y - direction</h3>
v(y) = vtot sinθ
v(y) = 7.56 x sin(23)
v(y) = 2.95 m/s
Learn more about component velocity here: brainly.com/question/24681896
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The greatest difference between high and low tide is around New Moon and Full Moon. During these Moon phases, the solar tide coincides with the lunar tide because the Sun and the Moon are aligned with Earth, and their gravitational forces combine to pull the ocean's water in the same direction.
Your answer is B Full moon
Answer:
I = 1.06886 N s
Explanation:
The expression for momentum is
I = F t = Δp
therefore the momentum is a vector quantity, for which we define a reference system parallel to the floor
Let's find the components of the initial velocity
sin 28.2 = v_y / v
cos 28.2= vₓ / v
v_y = v sin 282
vₓ = v cos 28.2
v_y = 42.8 sin 28.2 = 20.225 m / s
vₓ = 42.8 cos 28.2 = 37.72 m / s
since the ball is heading to the ground, the vertical velocity is negative and the horizontal velocity is positive, it can also be calculated by making
θ = -28.2
v_y = -20.55 m / s
v_x = 37.72 m / s
X axis
Iₓ = Δpₓ = 
since the ball moves in the x-axis without changing the velocity, the change in moment must be zero
Δpₓ = m 
- m v₀ₓ = 0
v_{fx} = v₀ₓ
therefore
Iₓ = 0
Y axis
I_y = Δp_y = p_{fy} -p_{oy}
when the ball reaches the floor its vertical speed is downwards and when it leaves the floor its speed has the same modulus but the direction is upwards
v_{fy} = - v_{oy}
Δp_y = 2 m v_{oy}
Δp_y = 2 0.0260 (20.55)
= 1.0686 N s
the total impulse is
I = Iₓ i ^ + I_y j ^
I = 1.06886 j^ N s
