Answer:
6 days.
Explanation:
From radioactivity, The expression for half life is given as,
R/R' = 2⁽ᵃ/ᵇ)................... Equation 1
Where R = original mass of the radioactive substance, R' = Remaining mass of the radioactive substance after decay, a = Total time taken to decay, b = half life.
Given: R = 80 g, R' = 10 g, b = 2 days.
Substitute into equation 1
80/10 = 2⁽ᵃ/²⁾
8 = 2⁽ᵃ/²⁾
2³ = 2⁽ᵃ/²)
Equating the base and solving for a
3 = a/2
a = 2×3
a = 6 days.
Answer:
Volume of balloon = 1000 cm^3
Explanation:
The head of a normal person can be assumed as a sphere with radius 10 cm.
Volume of sphere
, where r is the radius.
We have approximate radius = 10 cm.
Approximate volume of head 
In the given options the closest value to the approximate volume is 1000 cm^3.
So, volume of head = Volume of balloon = 1000 cm^3
Answer:
r=P/C, where P is the amount of useful output ("product") produced per the amount C ("cost") of resources consumed.
Explanation:
Efficiency is often measured as the ratio of useful output to total input, which can be expressed with the mathematical formula r=P/C, where P is the amount of useful output ("product") produced per the amount C ("cost") of resources consumed.
Explanation:
Let's say right is positive and left is negative.
F₁ = -150 N
F₂ = 220 N
Fnet = F₁ + F₂
Fnet = -150 N + 220 N
Fnet = 70 N
The magnitude of Fnet is 70 N, and since it's positive, the direction is to the right.
And since Fnet isn't 0, the force is unbalanced and the motion is changing.
Answer:(a) With our choice for the zero level for potential energy of the car-Earth system when the car is at point B ,
U
B
=0
When the car is at point A, the potential energy of the car-Earth system is given by
U
A
=mgy
where y is the vertical height above zero level. With 135ft=41.1m, this height is found as:
y=(41.1m)sin40.0
0
=26.4m
Thus,
U
A
=(1000kg)(9.80m/s
2
)(26.4m)=2.59∗10
5
J
The change in potential energy of the car-Earth system as the car moves from A to B is
U
B
−U
A
=0−2.59∗10
5
J=−2.59∗10
5
J
(b) With our choice of the zero configuration for the potential energy of the car-Earth system when the car is at point A, we have U
A
=0. The potential energy of the system when the car is at point B is given by U
B
=mgy, where y is the vertical distance of point B below point A. In part (a), we found the magnitude of this distance to be 26.5m. Because this distance is now below the zero reference level, it is a negative number.
Thus,