Answer:
Correct answer: m₂ = 1,950 g
Explanation:
When mixing the liquid, in this case water, at different temperatures, the warmer liquid will cool and the cooler will warm up until their temperatures equalize.
The amount of heat that the second fluid (water) releases is equal to the amount of heat that the first fluid (water) receives in this spontaneous process.
Q₁ = Q₂
Q₁ = m₁ c (t - t₁) and Q₂ = m₂ c (t₂ - t)
where are they: m₁ = 750 g, t₁ = 10°C, t₂ = 100°C, t = 75°C, m₂ = ?
m₁ c (t - t₁) = m₂ c (t₂ - t) \ : c
Since water is the liquid in question, it has the same specific heat capacity. Therefore, we will divide both sides of the equation by c and get:
m₁ (t - t₁) = m₂ (t₂ - t) => m₂ = m₁ (t - t₁) / (t₂ - t) =>
m₂ = 750 · (75 - 10) / (100 - 75) => m₂ = 750 · 65 / 25 = 1,950 g
m₂ = 1,950 g
God is with you!!!
Answer:
A. 5 m/s
Explanation:
From the graph, for the first 2 seconds, the graph is a straight line meaning that the slope is a constant.
Average speed of an object is the rate of change of position. Here, the position of the object changes from 0 m to 10 m for a time interval of 2 seconds.
The change in position (
) and time interval (
) are given as:

Therefore, the average speed (
) is given as the ratio of the total change in position and the time interval for the change.

Hence, the average speed is 5 m/s.
Answer: I don't do adopt me but I do have a Ro.blox username its Nvm i cant but my user name
Explanation:
Answer:
The time taken to travel is, t = 12 minutes
Explanation:
Given data,
The speed of the car, v = 60 km/h
The distance of travel, d = 12 km
The time taken for the travel is t = ?
The speed is defined as the distance divided by the time taken to travel. The formula for speed is,
v = d/t
∴ t = d/v
t = 12 km / 60 km/h
t = 0.2 h
t = 12 minutes
Hence, the time taken to travel is, t = 12 minutes.
Answer:
When have passed 3.9[s], since James threw the ball.
Explanation:
First, we analyze the ball thrown by James and we will find the final height and velocity by the time two seconds have passed.
We'll use the kinematics equations to find these two unknowns.
![y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\](https://tex.z-dn.net/?f=y%3Dy_%7B0%7D%20%2Bv_%7B0%7D%20%2At%2B%5Cfrac%7B1%7D%7B2%7D%20%2Ag%2At%5E%7B2%7D%20%5C%5Cwhere%3A%5C%5Cy%3D%20elevation%20%5Bm%5D%5C%5Cy_%7B0%7D%3Dinitial%20height%20%5Bm%5D%5C%5Cv_%7B0%7D%3D%20initial%20velocity%20%5Bm%2Fs%5D%20%3D41.67%5Bm%2Fs%5D%5C%5Ct%20%3D%20time%20passed%20%5Bs%5D%5C%5Cg%3D%20gravity%20%5Bm%2Fs%5E2%5D%3D9.81%5Bm%2Fs%5E2%5D%5C%5CNow%20replacing%3A%5C%5Cy%3D0%2B41.67%20%2A%282%29-%5Cfrac%7B1%7D%7B2%7D%20%2A%289.81%29%2A%282%29%5E%7B2%7D%20%5C%5C%5C%5Cy%3D63.72%5Bm%5D%5C%5C)
Note: The sign for the gravity is minus because it is acting against the movement.
Now we can find the velocity after 2 seconds.
![v_{f} =v_{o} +g*t\\replacing:\\v_{f} =41.67-(9.81)*(2)\\\\v_{f}=22.05[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3Dv_%7Bo%7D%20%2Bg%2At%5C%5Creplacing%3A%5C%5Cv_%7Bf%7D%20%3D41.67-%289.81%29%2A%282%29%5C%5C%5C%5Cv_%7Bf%7D%3D22.05%5Bm%2Fs%5D)
Note: The sign for the gravity is minus because it is acting against the movement.
Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.

As we can see the equation is based on Time (t).
Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t
![y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]](https://tex.z-dn.net/?f=y%3Dy_%7Bo%7D%20%2Bv_%7Bo%7D%20%2At%2B%5Cfrac%7B1%7D%7B2%7D%20%2Ag%2At%5E%7B2%7D%20%5C%5Cwhere%3A%5C%5Cv_%7Bo%7D%20%3D55.56%5Bm%2Fs%5D%20%3D%20initial%20velocity%5C%5Cy_%7Bo%7D%20%3D0%5Bm%5D%5C%5Cnow%20replacing%5C%5C63.72%20%2B22.05%20%2At-%284.905%29%2At%5E%7B2%7D%20%3D0%20%2B55.56%20%2At-%284.905%29%2At%5E%7B2%7D%20%5C%5C63.72%20%2B22.05%20%2At%20%3D0%20%2B55.56%20%2At%5C%5C63.72%20%3D%2033.51%2At%5C%5Ct%3D1.9%5Bs%5D)
Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.
Time = 2 + 1.9 = 3.9[s]