A) The acceleration is due to gravity at any given point if you look at it vertically, so
.
b)
, so
. We use
and then the final speed must be 0 because it stops at the highest point. So
. Solve for
and you get
c)
, and then we plug the values:
and we already have the time from "b)", so
![Y_m_a_x = [(32sin(25))*(32sin(25)/10)] - 5(32sin(25)/10)^2](https://tex.z-dn.net/?f=Y_m_a_x%20%3D%20%5B%2832sin%2825%29%29%2A%2832sin%2825%29%2F10%29%5D%20-%205%2832sin%2825%29%2F10%29%5E2)
; then we just rearrange it
![Y_m_a_x = 10[(32sin(25))^2/100] - 5 [(32sin(25))^2/100]](https://tex.z-dn.net/?f=Y_m_a_x%20%3D%2010%5B%2832sin%2825%29%29%5E2%2F100%5D%20-%205%20%5B%2832sin%2825%29%29%5E2%2F100%5D%20)
and finally
![Y_m_a_x = 5[(32sin(25))^2/100] = (32sin(25))^2/20](https://tex.z-dn.net/?f=Y_m_a_x%20%3D%205%5B%2832sin%2825%29%29%5E2%2F100%5D%20%3D%20%2832sin%2825%29%29%5E2%2F20)
The values with (1/2)(9.8 m/s2) (4 - X)^2 = (1.7 × 10^8 m/s) (X) —> We discover with the calculator X = 4.40 × 10^-7. The depth of the ice is then calculated by multiplying by the speed 74.8 m. Through ice, radio waves move at a speed of 1.7 x 108 m/s.
What is radio waves?
A radio wave pulse that is delivered into the Antarctic ice returns after reflecting off the bottom rock. A radio wave pulse that is transmitted into the Antarctic ice bounces off the subsurface rock and resurfaces. ice in Antarctica Through ice, radio waves move at a speed of 1.7 108 m/s. The rock reflects the radio wave pulse that was transmitted into the Antarctic ice. The amount of time that passed between the signal being sent and being received by the earth station.
To learn more about Antarctic from given link
brainly.com/question/3200132
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By Newton's second law, we have
So, in order to give a 0.15kg body an acceleration of 40m/s^2, you need a force of
Answer:
Explanation:
Let's use projectile motion equations. First of all we need to find the travel time. So we are going to use the next equation:
(1)
Where:
In this case:
Because the dog jumps horizontally
Let's asume the gravity constant as:
Because when the dog reach the base the height is 0
Now let's replace the data in (1)
Isolating t:
Finally let's find the horizontal displacement using this equation:
Replacing the data:
Answer:
Explanation:
The process is modelled after the First Law of Thermodynamics:
![(6000\,g) \cdot \left(0.385\,\frac{J}{g\cdot ^{\textdegree}C} \right)\cdot (T -0^{\textdegree}C ) = (2000\,g)\cdot \left(2.108\,\frac{J}{g\cdot ^{\textdegree}C} \right)\cdot [0^{\textdegree}C -(-10^{\textdegree}C)] + (1500\,g)\cdot \left(334\,\frac{J}{g} \right)](https://tex.z-dn.net/?f=%286000%5C%2Cg%29%20%5Ccdot%20%5Cleft%280.385%5C%2C%5Cfrac%7BJ%7D%7Bg%5Ccdot%20%5E%7B%5Ctextdegree%7DC%7D%20%5Cright%29%5Ccdot%20%28T%20-0%5E%7B%5Ctextdegree%7DC%20%29%20%3D%20%282000%5C%2Cg%29%5Ccdot%20%5Cleft%282.108%5C%2C%5Cfrac%7BJ%7D%7Bg%5Ccdot%20%5E%7B%5Ctextdegree%7DC%7D%20%5Cright%29%5Ccdot%20%5B0%5E%7B%5Ctextdegree%7DC%20-%28-10%5E%7B%5Ctextdegree%7DC%29%5D%20%2B%20%281500%5C%2Cg%29%5Ccdot%20%5Cleft%28334%5C%2C%5Cfrac%7BJ%7D%7Bg%7D%20%5Cright%29)
The initial temperature of the piece of solid copper is:
