<span>If the refrigerator weights 1365 and you are not exerting any vertical force on it, then the normal force is also 1365N. so Fn=1365
Fsf = Static frictional force = (coefficient of static friction) * (Normal force)
So the least for you could exert to move it is equal to the Fsf.
Fsf = (0.49)(1365N)</span><span>
</span>
Answer:
magnitude of the magnetic field 0.692 T
Explanation:
given data
rectangular dimensions = 2.80 cm by 3.20 cm
angle of 30.0°
produce a flux Ф = 3.10 × 
Wb
solution
we take here rectangular side a and b as a = 2.80 cm and b = 3.20 cm
and here angle between magnitude field and area will be ∅ = 90 - 30
∅ = 60°
and flux is express as
flux Ф = 
.................1
and Ф = BA cos∅ ............2
so B = 
and we know
A = ab
so
B = 
..............3
put here value
B = 
solve we get
B = 0.692 T
Answer:
im sure your already past this but it's E.
Explanation:
This is because in this case potential energy is linear to height, which means that the higher the more potential energy.
Answer:
30.0625 W
Explanation:
325 g/h x (1h x 1kg)/(3600s x 1000g) x 3,33 x 10^5 J/Kg = 30.0625 J/Kg = 30.0625 W
Answer:
- The emf of the generator is 6V
- The internal resistance of the generator is 1 Ω
Explanation:
Given;
terminal voltage, V = 5.7 V, when the current, I = 0.3 A
terminal voltage, V = 5.1 V, when the current, I = 0.9 A
The emf of the generator is calculated as;
E = V + Ir
where;
E is the emf of the generator
r is the internal resistance
First case:
E = 5.7 + 0.3r -------- (1)
Second case:
E = 5.1 + 0.9r -------- (2)
Since the emf E, is constant in both equations, we will have the following;
5.1 + 0.9r = 5.7 + 0.3r
collect similar terms together;
0.9r - 0.3r = 5.7 - 5.1
0.6r = 0.6
r = 0.6/0.6
r = 1 Ω
Now, determine the emf of the generator;
E = V + Ir
E = 5.1 + 0.9x1
E = 5.1 + 0.9
E = 6 V
