Answer:
-4×-2y=14 (1)
-10×+7y=-25 (2)
multiplying eq 1 by 7 and eq 2 by 2 and add eq. 1 and 2
-28×-14y=98
-20×+14y=-50
___________
-28×=48
×=48/-28
×=-12/7
now
-4×-2y=14
-4*-12/7-2y=14
48/7-2y=14
-2y=14-48/7
-2y=(98-48)/7
-2y=50/7
y=-50/14
y=-25/7
Answer:
weight at height = 100 N .
Explanation:
The problem relates to variation of weight due to change in height .
Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .
At the surface :
Applying Newton's law of gravitation
mg₀ = G Mm / R²
At height h from centre
mg₁ = G Mm /h²
Given mg₀ = 400 N
400 = G Mm / R²
400 = G Mm / (6400 x 10³ )²
G Mm = 400 x (6400 x 10³ )²
At height h from centre
mg₁ = 400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²
= 400 / 4
= 100 N .
weight at height = 100 N
When you double capacitance and inductance, the new resonance frequency becomes f/2.
The resonance frequency of RLC series circuit, is the frequency at which the capacity reactance is equal to inductive reactance.
It can also be defined as the natural frequency of an object where it tends to vibrate at a higher amplitude.
Xc = Xl
which gives the value for resonance frequency:

where;
f is the resonance frequency
L is the inductance
C is the capacitance
When you double capacitance and inductance, the new resonance frequency becomes;




Thus from above,
When you double capacitance and inductance, the new resonance frequency becomes f/2.
Learn more about resonance frequency here:
<u>brainly.com/question/13040523</u>
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maximum static friction acting on the object will be

plug in all values

So here it means that if applied force is less than or equal to 58.8 N then the object will remain stationary as friction can balance the external force upto this limit of external force
So here it is given that applied force is 20 N
so here object will not move due to this force and it will remain at rest always
due to this applied force