Explanation:
- The applications are, hydraulic lift- to transmit equal pressure throughout a fluid.
- Hydraulic jack- used in the braking system of cars.
- use of a straw- to suck fluids, which goes because of air pressure.
<h3>The question simply asks, where pressure can be applied. There are many others, such as
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R(parallel) = product/ sum
50×30/50+30
1500/80
18,75 ohms
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^
