Answer:
A) 35 ft
B) 5 ft
C) Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick
Explanation:
A) Total distance covered by the dog = 20 + 15
= 35 ft
B) Since the other distance covered by the dog before chewing the stick, after the retrieval, was in an opposite direction to the initial direction, then;
total displacement of the dog = 20 - 15
= 5 ft
C) Net displacement = distance covered by the dog to retrieve the stick + distance covered before the dog starts chewing the stick
But, displacement involves a specified direction. The distance covered before the dog starts chewing the stick was in an opposite direction to the initial direction.
Thus,
Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick
Answer:
1.843 x 10^-5 C
Explanation:
<u><em>Givens:
</em></u>
It is given that the air starts ionizing when the electric field in the air exceeds a magnitude of 3 x 10^6 N/C, which means that the max electric field can stand without forming a spark is 3 x 10^6 N/C.
Also it is given that the radius of the disk is 50 cm, it is required to find out the max amount of charge that the disk can hold without forming spark, which means the charge that would produce the max magnitude of the electric field that air can stand without forming spark, and since we know that the electric field in between 2 disk "Capacitor" is given by the following equation
E = (Q/A)/∈o (1)
Where,
Q: total charge on the disk.
A: the area of the disk.
<u><em>Calculations: </em></u>
We want to find the quantity of charge on the disk that would produce an electric field of 3 x 10^6 N/C, knowing the radius of the disk we can find the cross-section of the disk, thus substituting in equation (1) we find the maximum quantity of charge the disk can hold
Q = EA∈o
= (3 x 10^6) x (π*0.50) x (8.85 x 10^-12)
= 1.843 x 10^-5 C
note:
calculations maybe wrong but method is correct
The friction force between the box and the incline if the box does not slide down the incline will be 0.577
The force preventing sliding against one another of solid surfaces, fluid layers, and material components is known as friction. There are several kinds of friction: Two solid surfaces in touch are opposed to one another's relative lateral motion by dry friction.
Given the box resting on the inclined plane above has a mass of 20kg and the The incline sits at a 30 degree angle
We have to find the friction force between the box and the incline if the box does not slide down the incline
Since the frictional force F₁ must equal or exceed gravitational force F₂ down the incline:
F₁ = F₂
μmgcosΘ = mgsinΘ
μ = (mgsinΘ)/(mgcosΘ)
μ = tanΘ
μ = 0.577
Hence the friction force between the box and the incline if the box does not slide down the incline will be 0.577
Learn more about friction force here:
brainly.com/question/24386803
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Answer:
towards west
Explanation:
As we know that the speed of the blue car as appear to the bicycle rider is given as
towards west
also it is given that bicycle is moving at speed of 10 km/h towards East
so here we have

so we have

towards west
now speed of the red car is given as 15 km/h towards west
so here the relative speed of blue car with respect to red car is given as

towards west
30 grams of radioactive isotope have passed.