Answer:
Equilibrium temperature will be 
Explanation:
We have given weight of the lead m = 2.61 gram
Let the final temperature is T
Specific heat of the lead c = 0.128
Initial temperature of the lead = 11°C
So heat gain by the lead = 2.61×0.128×(T-11°C)
Mass of the water m = 7.67 gram
Specific heat = 4.184
Temperature of the water = 52.6°C
So heat lost by water = 7.67×4.184×(T-52.6)
We know that heat lost = heat gained
So 


Answer:
I think it's a true but I'm not the smartest
Answer:
Position =
behind the mirror
Nature = Virtual and Erect
Size =
: Diminished
Explanation:
Sign convention-Distance measured to the left of pole is negative and to the right of pole is positive.
Object distance = u = -20 cm
Focal length = f = Radius of curvature/2 = 30/2 = 15 cm
We have to use mirror formula to find image distance.

Since the image distance is positive, it is formed behind the mirror or a virtual image is formed.
Magnification = 

Height of the object = 5 cm
Height of the image = 
Since the height of the image is positive and less than the size of object,it is erect and diminished.
<span>Pollination is when the male gamete or pollen reaches the female flower through different modes...wind,air,animals,...
Fertilization </span><span>is when the male gamete "join" with the female gamete to become an embryo.
I hope I helped.:) </span>
Answer
given,
vertical speed of stone,v = 12 m/s
height of the cliff = 70 m
a) time taken by the stone to reach at the bottom of the cliff
We know that,
S = u t + 1/2 a t²
- 70 = 12 t - 0.5 x 9.8 t²
4.9 t² - 12 t - 70 = 0
solving the equation
t = 5.2 s (neglecting the negative value)
b) again using equation of motion
v = u + a t
v = 12 - 9.8 x 5.2
v = -38.96 m/s
ignoring the negative sign
magnitude of velocity is equal to 38.96 m/s
c) total distance travel by the stone
vertical distance covered by the stone
v² = u² + 2 g h
0 = 12² - 2 x 9.8 x h
h = 7.34 m
to reach the stone to the same level distance travel be doubled.
Total distance travel by the stone
H = h + h + 70
H = 7.34 x 2 + 70
H = 84.7 m.