This rotation causes days. The relative position of us to the sun, and the Earth's rotation will cause the sun to appear to rise above the horizon and below, creating day and night.
Is this science because I am suck at it
Answer:
0.699 L of the fluid will overflow
Explanation:
We know that the change in volume ΔV = V₀β(T₂ - T₁) where V₀ = volume of radiator = 21.1 L, β = coefficient of volume expansion of fluid = 400 × 10⁻⁶/°C
and T₁ = initial temperature of radiator = 12.2°C and T₂ = final temperature of radiator = 95.0°C
Substituting these values into the equation, we have
ΔV = V₀β(T₂ - T₁)
= 21.1 L × 400 × 10⁻⁶/°C × (95.0°C - 12.2°C)
= 21.1 L × 400 × 10⁻⁶/°C × 82.8°C = 698832 × 10⁻⁶ L
= 0.698832 L
≅ 0.699 L = 0.7 L to the nearest tenth litre
So, 0.699 L of the fluid will overflow
Answer:
v = 2.85 m/s
Explanation:
By the law of similarity of the triangle we can say

here we know that
height of man = 2 m
let the distance of man from spotlight = x
distance of wall from spotlight = 12 m
height of image is let say "y"
so we will have


now we have

here we know

x = 4 m
now we have


Answer:
1.dr/dt=0.0096cm/s
2. dA/dt=2.19cm^2/s
Explanation:
A spherical balloon is deflating at 10 cm3/s. At what rate is the radius changing when the volume is 1000π cm3 ? What is the rate of change of surface area at this moment?
for this question, we need to analyze the parameters we know
V=volume of the spherical balloon 1000π cm3
volume of the sphere=
1000π=4/3πr^3
dividing both sides by 4
250*3=r^3
r=9.08cm, the radius of the balloon
dv/dt=dv/dr*dr/dt...................................1
dv/dr ,means
V=
dv/dr=4*pi*r^2
dv/dt=10 cm3/s
from equ 1
10=4*pi*9.08^2*dr/dt
10=1036 dr/dt
dr/dt=10/1036
dr/dt=0.0096cm/s
2. to find the rate at which the area is changing we have,
dA/dt=dA/dr*dr/dt
area of a sphere is 4πr^2
differentiate a with respect to r, radius
dA/dr=8πr
dA/dt=8πr*0.0096
dA/dt=8*pi*9.08*0.0096
dA/dt=2.19cm^2/s
is the rate of change of the surface area