Answer:
d=510.2m
t=10.2s
Explanation:
The formulas for accelerated motion are:
From them we can get 
.
We have:
And substitute:
We multiply both sides by 2a, and continue:
Being d the displacement 
, we have
For our exercise, we will write this as:
And taking upwards direction positive and imposing final velocity 0m/s (for maximum height), we have:
For the time we use:
<span>This statement simply implies that the 50 million tons of coal will produce the same energy output in one year as the 0.6 million barrels per day of oil. Assuming coal is more plentiful than oil, this is significant as 50 million tons is much smaller than 219 million barrels of oil.</span>
Answer:
Because the zinc is reluctant
Explanation:
A leclanche cell contains a conducting solution (electrolyte) of ammonium chloride, a cathode (positive terminal) of carbon, a depolarizer of manganese dioxide (oxidizer), and an anode (negative terminal) of zinc (reductant).
As the Zn2+ ions move away from the anode, leaving their electrons on its surface,
Zn → Zn2+ + 2e−
the anode becomes more negatively charged than the cathode. When the cell is connected to an external electrical circuit, the excess electrons on the zinc anode flow through the circuit to the carbon rod, the movement of electrons forming an electric current.
Answer:
Explanation:
Given:
Specific heat of gold = 0.031cal/°C
Specific heat of silver = 0.057cal/°C
To know the metals that will heat up faster, we must understand the meaning of specific heat capacity.
It is the amount of heat required to raise the temperature of 1g of a substance by 1°C.
Now,
The higher the specific heat capacity the more energy it is required to heat up the substance.
So, Gold with a specific heat capacity of 0.031cal/°C will heat up faster.
Answer:
b) 68,9 km/h a) picture
Explanation:
In this problem, since velocity is expressed in km/h and time in minutes, we have to convert either time to hours or velocity to km/min. It is easier to use hours.
Using this formula we pass time to hours:
Now we can plot speed vs time (image 1). The problem says that the driver uses constant speed, so all lines have to be horizontal.
Using the values of the speed we calculate the distance in each interval
Using these values and the fact that she was having lunch in the third one (therefore stayed in the same position), we plot position vs time, using initial position zero (image 2, distance is in km, not meters).
Finally, we compute the average speed with the distance over time:
