Answer:
d = 4 d₀o
Explanation:
We can solve this exercise using the relationship between work and the variation of kinetic energy
W = ΔK
In that case as the car stops v_f = 0
the work is
W = -fr d
we substitute
- fr d₀ = 0 - ½ m v₀²
d₀ = ½ m v₀² / fr
now they indicate that the vehicle is coming at twice the speed
v = 2 v₀
using the same expressions we find
d = ½ m (2v₀)² / fr
d = 4 (½ m v₀² / fr)
d = 4 d₀o
228 - 224 = 4
there is 4g of solute in the solution.
Answer:
Every action has an equal and opposite reaction. If the student doesn't push, nothing moves, is one student pushes, both move which is an example of newtons third law.
Explanation:
Answer:
a) L=0. b) L = 262 k ^ Kg m²/s and c) L = 1020.7 k^ kg m²/s
Explanation:
It is angular momentum given by
L = r x p
Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum
One of the easiest ways to make this vector product is with the use of determinants
![{array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]](https://tex.z-dn.net/?f=%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%26y%26z%5C%5Cpx%26py%26pz%5Cend%7Barray%7D%5Cright%5D)
Let's apply this relationship to our case
Let's start by breaking down the speed
v₀ₓ = v₀ cosn 45
voy =v₀ sin 45
v₀ₓ = 9 cos 45
voy = 9 without 45
v₀ₓ = 6.36 m / s
voy = 6.36 m / s
a) at launch point r = 0 whereby L = 0
. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero
vfy² = voy²- 2 g y
y = voy² / 2g
y = (6.36)²/2 9.8
y = 2.06 m
Let's calculate the angular momentum
L= ![\left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%26y%260%5C%5Cpx%260%260%5Cend%7Barray%7D%5Cright%5D)
L = -px y k ^
L = - (m vox) (2.06) k ^
L = - 20 6.36 2.06 k ^
L = 262 k ^ Kg m² / s
The angular momentum is on the z axis
c) At the point of impact, at this point the height is zero and the position on the x-axis is the range
R = vo² sin 2θ / g
R = 9² sin (2 45) /9.8
R = 8.26 m
L = ![\left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%260%260%5C%5Cpx%26py%260%5Cend%7Barray%7D%5Cright%5D)
L = - x py k ^
L = - x m voy
L = - 8.26 20 6.36 k ^
L = 1020.7 k^ kg m² /s
First electromagnet
Explanation:
The first electromagnet is the strongest and it is stronger than the given electromagnet above.
An electromagnet is a temporary magnet made by passing current through a wire wounded round an iron core or metallic core.
- When current is passed through, the metal becomes magnetic.
- The strength of the electromagnet depends on the number of coil round the metal core and also the intensity of current passed through it.
- The higher the number of coils wounded round the metal core, the stronger the electromagnet that will be produced.
- Also, the higher the intensity of electricity passed through the wire, the stronger it is.
learn more:
Electromagnet brainly.com/question/2191993
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