#1
As we know that
now plug in all data into this
now from the formula of strain
#2
As we know that
pressure * area = Force
here we know that
now force is given as
#3
As we know that density of water will vary with the height as given below
here we know that
now density is given as
#4
as we know that pressure changes with depth as per following equation
here we know that
now we will have
here we will have
so it is 20.1 m below the surface
#5
Here net buoyancy force due to water and oil will balance the weight of the block
so here we will have
so it is 3.48 cm below the interface
Answer:
A. 19.8 cm.
Explanation:
The apparent depth of the combination is
As it mentioned that the two clear but non-mixing liquid having depth of 15 cm that placed in a glass container together
Also the refractive indices would be 1.75 and 1.33
Based on the above information
As we know that
Refractive indices = real depth ÷ apparent depth
1.33 ÷ 1.75 = 15 ÷ apparent depth
So, it would be 19.736842 cm
Now the combination of apparent depth would be
= ( 19.736842 + 15) ÷ (1.75)
= 19.8 cm
hence, the correct option is A.
Answer: 17.5N
Explanation:
Given that,
spring constant (k) = 70N/m
Original length (lo) = 40 cm
New length (ln) = 15 cm
extension of spring (e) = new length - original length
i.e e = ln - lo
e = 15 cm - 40 cm
e = -25cm (the negative sign is negligible, just the same magnitude matters)
So, convert 25cm to metres
If 100cm = 1m
25cm = 25/100 = 0.25m
(Note that how hard the spring push refers to the Force, hence, find force.)
Now, apply the Hooke law (F = ke)
Force = spring constant x extension
F = 70N/m x 0.25m
F = 17.5N
Thus, the spring push with a force of 17.5N when compressed.
Answer:
R1 + R2 = R = 12 for resistors in series - so R1 = R2 if they are identical
2 R1 = 12 and R1 = R2 = 6 ohms
1 / R = 1 / R1 + 1 / R2 for resistors in parallel
R = R1 * R2 / (R1 + R2) = 6 * 6 / (6 + 6) = 3
The equivalent resistance would be 3 ohms if connected in parallel
Answer:
Velocity of Afrom B=21m/s
Acceleration of A from B=1.68m/s°2
Explanation:
Given
Radius r=150m
Velocity of a Va= 54km/hr
Va=54*1000/3600=15m/s
Velocity of b Vb=82km/hr
VB=81*1000/3600=22.5mls
The velocity of Car A as observed from B is VBA
VB= VA+VBA
Resolving the vector into X and Y components
For X component= 15cos60=7.5m/s
Y component=22 5sin60=19.48m/s
VBA= √(X^2+Y^2)
VBA= ✓(7.5^2+19.48^2)=21m/s
For acceleration of A observed from B
A=VA^2/r= 15^2/150=1.5m/s
Resolving into Xcomponent=1.5cos60=0.75m/s
Y component=3cos60=1.5
Acceleration BA=√(0.75^2+1.5^2)
1.68m/s