The density would decrease because the mass of an object deals with the amount of atoms in the object and since none of the object was reduced "a" wouldn't be the answer. Depending on the amount and period of time that the heat is applied the liquid could change into a gas so "d" wouldn't be correct. Density is the mass÷ volume, and when you add heat to an object it could take up different amounts of space because of its particles gaining energy and spreading apart. So the density would decrease because of the volume increasing. So I believe that "c" is the answer.
Answer:
80 meters high
Explanation:
The velocity of the balloon would be g*t (I won't calculate, but will us this later)
We know that the kinetic energy at the bottom equals the potential at the top.
KE = PE
1/2 * m * v^2 = m * g * h
1/2 * m * (g * t)^2 = m * g * h (substitution)
1/2 * m * g^2 * t^2 = m * g * h
1/2 * g * t^2 = h (simplification by dividing the commons between both sides)
h = 1/2 * 9.81 * 4^2
h = 78.48 m (roughly 80 m)
Answer:
ΔP = 14.5 Ns
I = 14.5 Ns
ΔF = 5.8 x 10³ N = 5.8 KN
Explanation:
The mass of the ball is given as 0.145 kg in the complete question. So, the change in momentum will be:
ΔP = mv₂ - mv₁
ΔP = m(v₂ - v₁)
where,
ΔP = Change in Momentum = ?
m = mass of ball = 0.145 kg
v₂ = velocity of batted ball = 55.5 m/s
v₁ = velocity of pitched ball = - 44.5 m/s (due to opposite direction)
Therefore,
ΔP = (0.145 kg)(55.5 m/s + 44.5 m/s)
<u>ΔP = 14.5 Ns</u>
The impulse applied to a body is equal to the change in its momentum. Therefore,
Impulse = I = ΔP
<u>I = 14.5 Ns</u>
the average force can be found as:
I = ΔF*t
ΔF = I/t
where,
ΔF = Average Force = ?
t = time of contact = 2.5 ms = 2.5 x 10⁻³ s
Therefore,
ΔF = 14.5 N.s/(2.5 x 10⁻³ s)
<u>ΔF = 5.8 x 10³ N = 5.8 KN</u>
Answer:
Explanation:
A function f(x) is a Probability Density Function if it satisfies the following conditions:
Given the function:
(1)p(x) is greater than zero since the range of exponents of the Euler's number will lie in 
(2)
![\int_{0}^{\infty} p(x)=\int_{0}^{\infty} \dfrac{1}{r}e^{-x/r}\\=\dfrac{1}{r} \int_{0}^{\infty} e^{-x/r}\\=-\dfrac{r}{r}\left[e^{-x/r}\right]_{0}^{\infty}\\=-\left[e^{-\infty/r}-e^{-0/r}\right]\\=-e^{-\infty}+e^{-0}\\SInce \: e^{-\infty} \rightarrow 0\\e^{-0}=1\\\int_{0}^{\infty} p(x)=1](https://tex.z-dn.net/?f=%5Cint_%7B0%7D%5E%7B%5Cinfty%7D%20p%28x%29%3D%5Cint_%7B0%7D%5E%7B%5Cinfty%7D%20%5Cdfrac%7B1%7D%7Br%7De%5E%7B-x%2Fr%7D%5C%5C%3D%5Cdfrac%7B1%7D%7Br%7D%20%5Cint_%7B0%7D%5E%7B%5Cinfty%7D%20e%5E%7B-x%2Fr%7D%5C%5C%3D-%5Cdfrac%7Br%7D%7Br%7D%5Cleft%5Be%5E%7B-x%2Fr%7D%5Cright%5D_%7B0%7D%5E%7B%5Cinfty%7D%5C%5C%3D-%5Cleft%5Be%5E%7B-%5Cinfty%2Fr%7D-e%5E%7B-0%2Fr%7D%5Cright%5D%5C%5C%3D-e%5E%7B-%5Cinfty%7D%2Be%5E%7B-0%7D%5C%5CSInce%20%5C%3A%20e%5E%7B-%5Cinfty%7D%20%5Crightarrow%200%5C%5Ce%5E%7B-0%7D%3D1%5C%5C%5Cint_%7B0%7D%5E%7B%5Cinfty%7D%20p%28x%29%3D1)
The function p(x) satisfies the conditions for a probability density function.
"Pluto was the first dwarf planet to be discovered" is the one statement among the following choices given in the question that is true <span>about dwarf planets. The correct option among all the options that are given in the question is the first option or option "a". Pluto was classified as a planet at first but in the year 1930 it was classified as a dwarf planet.</span>
