Answer:
1. The image of the person is 1.41 m, virtual and formed at the back of the surface of the globe.
2. The person's image is 3.38 m tall.
Explanation:
From the given question, object distance, u = 0.75 m, object height = 1.8 m, radius of curvature of the reflecting globe, r = 8 cm = 0.08 m.
f = 
= 
= 0.04 m
1. The image distance, v, can be determined by applying mirror formula:
= 
+ 
= 
+
- 
=
= 
= - 
⇒ v = -
= - 1.41 m
The image of the person is 1.41 m, virtual and formed at the back of the surface of the globe.
2. 
= 
= 
v = 
= 3.384
v = 3.38 m
The person's image is 3.38 m tall.
Answer:
Explanation:
Due to heat energy , metal expands . Formula for linear expansion is as follows .
L = l ( 1 + α Δt )
where L is expanded length , l is original length , α is coefficient of linear expansion and Δt is increase in temperature .
To pass the sphere through the ring , the diameter of both ring and sphere should be same after heating . Let after increase of temperature Δt , their diameter becomes same as L . The linear coefficient of brass and steel are
20 x 10⁻⁶ and 12 x 10⁻⁶ respectively .
For steel sphere ,
L = 25 ( 1 + 12 x 10⁻⁶ Δt )
For brass ring
L = 24.9 ( 1 + 20 x 10⁻⁶ Δt )
25 ( 1 + 12 x 10⁻⁶ Δt ) = 24.9 ( 1 + 20 x 10⁻⁶ Δt )
1.004( 1 + 12 x 10⁻⁶ Δt ) = ( 1 + 20 x 10⁻⁶ Δt )
1.004 + 12.0482 x 10⁻⁶ Δt = 1 + 20 x 10⁻⁶ Δt
.004 = 7.9518 x 10⁻⁶ Δt
Δt = 4000 / 7.9518
= 503⁰C.
final temp = 503 + 15 = 518⁰C .
Answer:
The gravitational force is 130.
Explanation:
During this problem you have to multiply the 65 and the 0.6.
Answer:
Ro = 133 [kg/m³]
Explanation:
In order to solve this problem, we must apply the definition of density, which is defined as the relationship between mass and volume.
where:
m = mass [kg]
V = volume [m³]
We will convert the units of length to meters and the mass to kilograms.
L = 15 [cm] = 0.15 [m]
t = 2 [mm] = 0.002 [m]
w = 10 [cm] = 0.1 [m]
Now we can find the volume.
![V = 0.15*0.002*0.1\\V = 0.00003 [m^{3} ]](https://tex.z-dn.net/?f=V%20%3D%200.15%2A0.002%2A0.1%5C%5CV%20%3D%200.00003%20%5Bm%5E%7B3%7D%20%5D)
And the mass m = 4 [gramm] = 0.004 [kg]
![Ro = 0.004/0.00003\\Ro = 133 [kg/m^{3}]](https://tex.z-dn.net/?f=Ro%20%3D%200.004%2F0.00003%5C%5CRo%20%3D%20133%20%5Bkg%2Fm%5E%7B3%7D%5D)
Answer:
C.Supersaturated
Explanation:
There are three types of solution:
<u>SATURATED SOLUTION</u>:
It is the solution that contains maximum amount of solute dissolved in a solution in normal conditions.
<u>UNSATURATED SOLUTION</u>:
It is the solution that contains less than the maximum amount of solute dissolved in a solution in normal conditions. It has space for more solute to be dissolved in it.
<u>SUPERSATURATED SOLUTION:</u>
It contains more than the maximum amount of solute dissolved in it. Such a solution has no capacity to dissolve any more solute under any conditions.
Since the sugar is no more dissolving in the tea and has settled down. Therefore, the solution is:
<u>C.Supersaturated</u>
