Answer:
the time it takes for one complete back and forth swing
Explanation:
the Mark's is showing you the time it swings back and forth
Answer:
1. <u>F = ma</u> <em>F = 0.2kg * 20m/s² = 4Kg * m/s² =</em> 4N
2. <u>F = ma</u> <em>F - 18Kg * 3m/s² = 54Kg * m/s² =</em> 54N
3. <u>F = ma</u> <em>F = 0.025Kg * 5m/s² =</em> 0.125N
4. <u>F = ma</u> <em>F = 50Kg * 4m/s² =</em> 200N
5. <u>F = ma</u> <em>F = 70Kg * 4m/s² =</em> 280N
6. <u>F = ma</u> <em>F = 9Kg * 9.8m/s² =</em> 88.2N
Explanation:
Hope this helps ! ^^
Answer:
Impulse = Average force x time of contact
Explanation:
Impulsive force is a force which is very large but applied on a body for a very small duration of time.
Impulse is given by the change in momentum of the body.
Impulse = Average force x small time interval
When padding is there, the time interval of contact is large and thus, the force exerted by the body is small.
So, when a person falls on the tile floor, there is no compression and thus, the time of contact is very small and thus the impulsive force is very large, due to which the body may damage.
So, when a person falls on the carpeted floor, there is a compression and thus, the time of contact is comparatively large and thus the impulsive force is small, due to which the body may safe.
It’s true all the way. It’s true
Answer:
The correct answer is B
Explanation:
Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity
Φ
= ∫ E. dA = 
/ ε₀
For this case we create a Gaussian surface that is a sphere. We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product
∫ E dA = / ε₀
The area of a sphere is
A = 4π r²
E 4π r² = / ε₀
E = (1 /4πε₀
) q / r²
Having the solution of the problem let's analyze the points:
A ) r = 3R / 4 = 0.75 R.
In this case there is no charge inside the Gaussian surface therefore the electric field is zero
E = 0
B) r = 5R / 4 = 1.25R
In this case the entire charge is inside the Gaussian surface, the field is
E = (1 /4πε₀
) Q / (1.25R)²
E = (1 /4πε₀
) Q / R2 1 / 1.56²
E₀ = (1 /4π ε₀
) Q / R²
= Eo /1.56
²
= 0.41 Eo
C) r = 2R
All charge inside is inside the Gaussian surface
=(1 /4π ε₀
) Q 1/(2R)²
= (1 /4π ε₀
) q/R² 1/4
= Eo 1/4
= 0.25 Eo
D) False the field changes with distance
The correct answer is B
