Explanation:
Given Data:
mass of dog = 12 Kg
dog's center of mass = 0.20m
length of dog = 0.50m
height of dog's jump = ?
Solution:
Work done of gravitational force = Gain in Potential energy
2.1 × mgΔh = mg (h - 0.1)
2.1 × (0.3 - 0.1) = (h - 0.1)
h = 0.52 m
Answer:
Total distance = 400+700+1200= 2300km
Explanation:
the resultant of d 1st right angle triangle + 1200
= 806.2 + 1200 = 2006.2km
The correct answer is option C. <span>This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger.
</span><span>
Keisha follows the instructions for a demonstration on gas laws.
1. Place a small marshmallow in a large plastic syringe.
2. Cap the syringe tightly.
3. Pull the plunger back to double the volume of gas in the syringe.
Now, this activity is being done at the same temperature, because there is no mention of the temperature change. Thus, when the plunger is pulled back, the volume doubles, so pressure will decrease. Therefore, </span>This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger.
Answer:
-2.3 × 10^-9 Coulombs(C).
Explanation:
So, we are given the following data or information or parameters that is going to help us to solve the problem effectively and efficiently;
=> " the shuttle's potential is typically changed by -1.4 V during one revolution. "
=> " Assuming the shuttle is a conducting sphere of radius 15 m".
So, in order to estimate the value for the charge we will be making use of the equation below:
Charge, C =( radius × voltage or potential difference) ÷ Coulomb's law constant.
Note that the value of Coulomb's law constant = 9 x 10^9 Nm^2 / C^2.
So, charge = { 15 × (- 1.4)} / 9 x 10^9 Nm^2 / C^2.
= -2.3 × 10^-9 Coulombs(C).
Answer:
434 Hz
Explanation:
According to the Doppler effect, when a source of a wave is moving towards an observer at rest, then the observer will observe an apparent frequency which is higher than the original frequency of the source.
In this situation, Tina is driving towards Rita. Tina is the source of the sound wave (the horn), while RIta is the observer. Since the original frequency of the sound is 400 Hz, Rita will hear a sound with a frequency higher than this value.
The only choice which is higher than 400 Hz is 434 Hz, so this is the frequency that Rita will hear.
