The following are heterogeneous mixture: MILK THAT IS SOLD AT THE STORE, A CAKE MIX AND MIXED NUT.An heterogeneous mixture is one which contained many parts which can be separated out.A smoothie is an homogeneous mixture because the juice of different fruits that are use can not be separated out.
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Answer:
The final temperature is 21.531°c
Explanation:
Heat gained by the water = heat lost by the metal
Heat gained = (m)(c)(∆tita)
Where m = mass
C = specific heat capacity
∆tita = temperature change
X = equilibrium temperature
So ....
Heat gained by water
= 175*4.185*(x-20)
= 732.2x - 14644
Heat lost by metal
= 44.5*0.321*(100-x)
=1428.45 -14.2845x
So....
1428.45 - 14.2845x = 732.2x - 14644
1428.45+14644= 732.2x + 14.2845x
16072.45= 746.4845x
16072.45/746.4845= x
21.531 = x
The final temperature is 21.531°c
Answer:
I don't have the answer but...
Explanation:
A variable is a quantity that may change within the context of a mathematical problem or experiment. Typically, we use a single letter to represent a variable. The letters x, y, and z are common generic symbols used for variables.
Answer:
3.2 m/s²
Explanation:
Acceleration can be calculated as:
v = u + at (where v is final velocity, u is initial velocity, a is acceleration and t is time)
25 m/s = 9 m/s + a(5 s) (a is unknown)
16 m/s = a(5 s)
a = 3.2 m/s²
We assume that this is a uniform acceleration (meaning that the velocity increases at an equal rate for those 5 seconds).
Answer:
<em>The second car was moving at 2.1 m/s to the east</em>
Explanation:
<u>Conservation Of Linear Momentum</u>
The total momentum of a system of bodies is constant if no external force is applied to it. The momentum of a body with mass m and velocity v is p=mv. When two objects collide and join together afterward, the total mass is the sum of the individual masses, and the final speed is common to both.
Let's say
![\displaystyle m_1,\ m_2\ ,\ v_1,\ v_2\ ,\ p_1,\ p_2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20m_1%2C%5C%20m_2%5C%20%2C%5C%20v_1%2C%5C%20v_2%5C%20%2C%5C%20p_1%2C%5C%20p_2)
are the masses of the objects 1 and 2, their speeds, and their linear momentums respectively before the collision, and
![\displaystyle v_1',\ v_2'\ ,\ p_1',\ p_2'](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_1%27%2C%5C%20v_2%27%5C%20%2C%5C%20p_1%27%2C%5C%20p_2%27)
are the speeds of each object and their linear momentums after the collision.
The principle of conservation of linear momentum states that
![\displaystyle p_1+p_2=p_1'+p_2'](https://tex.z-dn.net/?f=%5Cdisplaystyle%20p_1%2Bp_2%3Dp_1%27%2Bp_2%27)
This means that
![\displaystyle m_1\ v_1+m_2\ v_2=m_1\ v_1'+m_2\ v_2'](https://tex.z-dn.net/?f=%5Cdisplaystyle%20m_1%5C%20v_1%2Bm_2%5C%20v_2%3Dm_1%5C%20v_1%27%2Bm_2%5C%20v_2%27)
Since both cars remain together after the collision,
![\displaystyle v_1'=v_2'=v'](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_1%27%3Dv_2%27%3Dv%27)
The relation becomes
![\displaystyle m_1\ v_1+m_2\ v_2=(m_1+m_2)\ v'](https://tex.z-dn.net/?f=%5Cdisplaystyle%20m_1%5C%20v_1%2Bm_2%5C%20v_2%3D%28m_1%2Bm_2%29%5C%20v%27)
Solving for ![v_2](https://tex.z-dn.net/?f=v_2)
![\displaystyle v_2=\frac{(m_1+m_2)v'-m_1\ v_1}{m_2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_2%3D%5Cfrac%7B%28m_1%2Bm_2%29v%27-m_1%5C%20v_1%7D%7Bm_2%7D)
The given data is
![\displaystyle m_1=2150\ kg\ ,\ m_2=3250\ kg\ ,\ v_y=10\ m/s,\ v'=5.22\ m/s](https://tex.z-dn.net/?f=%5Cdisplaystyle%20m_1%3D2150%5C%20kg%5C%20%2C%5C%20m_2%3D3250%5C%20kg%5C%20%2C%5C%20v_y%3D10%5C%20m%2Fs%2C%5C%20v%27%3D5.22%5C%20m%2Fs)
Let's assume all the speeds are positive towards East
Replacing those values
![\displaystyle v_2=\frac{(2150+3250)5.22-2150(10)}{3250}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_2%3D%5Cfrac%7B%282150%2B3250%295.22-2150%2810%29%7D%7B3250%7D)
![\displaystyle v_2=\frac{6688}{3250}=2.06\ m/s](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_2%3D%5Cfrac%7B6688%7D%7B3250%7D%3D2.06%5C%20m%2Fs)
The second car was moving at 2.1 m/s (to the nearest tenth) to the east