A far-sighted person has a near-point of 80 cm. To correct their vision so that they can see objects that are as close as 10 cm
1 answer:
Answer:
Explanation:
As we know that Far sighted person has near point shifted to 80 cm distance
so he is able to see the object 80 cm
now the distance of lens from eye is 2 cm
and the person want to see the objects at distance 10 cm
so here the image distance from lens is 80 cm and the object distance from lens is 8 cm
now from lens formula we have
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Answer:
hello the diagram relating to this question is attached below
a) angular accelerations : B1 = 180 rad/sec, B2 = 1080 rad/sec
b) Force exerted on B2 at P = 39.2 N
Explanation:
Given data:
Co = 150 N-m ,
<u>a) Determine the angular accelerations of B1 and B2 when couple is applied</u>
at point P ; Co = I* ∝B2'
150 = ( (2*0.5^2) / 3 ) * ∝B2
∴ ∝B2' = 900 rad/sec
hence angular acceleration of B2 = ∝B2' + ∝B1 = 900 + 180 = 1080 rad/sec
at point 0 ; Co = Inet * ∝B1
150 = [ (2*0.5^2) / 3 + (2*0.5^2) / 3 + (2*0.5^2) ] * ∝B1
∴ ∝B1 = 180 rad/sec
hence angular acceleration of B1 = 180 rad/sec
<u>b) Determine the force exerted on B2 at P</u>
T2 = mB1g + T1 -------- ( 1 )
where ; T1 = mB2g ( at point p )
= 2 * 9.81 = 19.6 N
back to equation 1
T2 = (2 * 9.8 ) + 19.6 = 39.2 N
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Answer:
Final temperature, T2 = 31.11°C
Explanation:
Given the following data;
Mass = 4.5kg
Initial temperature = 20°C
Quantity of heat = 90000J
Specific heat capacity = 1,800. J/kg-°C
To find the final temperature;
Heat capacity is given by the formula;
Where;
Q represents the heat capacity or quantity of heat.
m represents the mass of an object.
c represents the specific heat capacity of water.
dt represents the change in temperature.
Making dt the subject of formula, we have;
Substituting into the equation, we have;
dt = 11.11°C
Now, to find the final temperature T2;
But, dt = T2 - T1
Substituting into the equation, we have;
T2 = 11.11 + 20
Final temperature, T2 = 31.11°C