The mass of moon is 1/100 times and its radius 1/4 times that of earth as a result the gravitational attraction on the moon is about one sixth when compared to earth.
Answer:
A. 0.289g/mL
Explanation:
Using the equation for density which is d = m/v or density = mass/volume, we input 1.3g/4.5mL and get 0.289g/mL.
Answer:
New Resistance = 0.5556 ohm
Explanation:
Resistance = resistivity * length /area
Here since resistivity and length are constant, we only need to see how the resistance increases or decreases with change in area.
New Area = pi * (3*D)^2 / 4
Old Area = pi * D^2 / 4
The ratio of new area / old area is :
Since area increases 9 times, and it is inversely proportional to resistance:
Resistance decreases by 9 times.
So, old resistance = Voltage / Current = 10 / 2 = 5 ohm
New Resistance = 5 / 9 = 0.5556 ohm (decreases by 9 times)
Answer:
Ratio of series current to parallel
= 1 : 8
Explanation:
Total resistance Rt
For series, Rt = 2+2+2+2 = 4ohms
For parallel, 1/Rt = 1/2 + 1/2 + 1/2 + 1/2
1/Rt = 4/2, Rt = 2/4 ohms.
If we use a 1V battery, then,
I = V/Rt
I = 1/4 = 0.25 ampere for series arrangement.
I = 1/0.5 = 2 ohms.
Ratio of current of series to parallel = 0.25 : 2
= 1 : 8
Answer:
a = -0.33 m/s² k^
Direction: negative
Explanation:
From Newton's law of motion, we know that;
F = ma
Now, from magnetic fields, we know that;. F = qVB
Thus;
ma = qVB
Where;
m is mass
a is acceleration
q is charge
V is velocity
B is magnetic field
We are given;
m = 1.81 × 10^(−3) kg
q = 1.22 × 10 ^(−8) C
V = (3.00 × 10⁴ m/s) ȷ^.
B = (1.63T) ı^ + (0.980T) ȷ^
Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;
a = qVB/m
a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))
From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^
Thus;
a = -0.33 m/s² k^
