Answer:
The magnitude of the net force is √2F.
Explanation:
Since the two particles have the same charge Q, they exert the same force on the test charge; both attractive or repulsive. So, the angle between the two forces is 90° in any case. Now, as we know the magnitude of these forces and that they form a 90° angle, we can use the Pythagorean Theorem to calculate the magnitude of the resultant net force:
Then, it means that the net force acting on the test charge has a magnitude of √2F.
The emerging velocity of the bullet is <u>71 m/s.</u>
The bullet of mass <em>m</em> moving with a velocity <em>u</em> has kinetic energy. When it pierces the block of wood, the block exerts a force of friction on the bullet. As the bullet passes through the block, work is done against the resistive forces exerted on the bullet by the block. This results in the reduction of the bullet's kinetic energy. The bullet has a speed <em>v</em> when it emerges from the block.
If the block exerts a resistive force <em>F</em> on the bullet and the thickness of the block is <em>x</em> then, the work done by the resistive force is given by,
This is equal to the change in the bullet's kinetic energy.
If the thickness of the block is reduced by one-half, the bullet emerges out with a velocity v<em>₁.</em>
Assuming the same resistive forces to act on the bullet,
Divide equation (2) by equation (1) and simplify for v<em>₁.</em>
Thus the speed of the bullet is 71 m/s
Answer:
I₁ = (7.78 i ^ - 6.71 j ^) 10⁻³ J s , I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s , I₃ = (19.1i ^ + 18.6 j ^) 10⁻³ J s and I₄ = (-9.14i ^ + 7.24 j ^) 10⁻³ J s
Explanation:
The impulse is equal to the variation of the moment, to apply this relationship to our case, we will assume that initially the mouse was at rest
I = Δp = m -m v₀
I = m ( -v₀)
Bold indicates vector quantities, let's calculate the momentum of each mouse in for the x and y axes
We recommend bringing all units to the SI system
Mouse 1.
It has a mass of 22.3 g = 22.3 10⁻³ kg, a final velocity of (v = 0.349 i ^ - 0.301 j ^) m / s with an initial velocity of zero
Iₓ = m ( - v₀ₓ)
Iₓ = 22.3 10⁻³ (0.349 -0)
Iₓ = 7.78 10⁻³ J s
= m (
-
)
= 22.3 10⁻³ (-0.301)
= -6.71 10⁻³ J s
I₁ = (7.78 i ^ - 6.71 j ^) 10⁻³ J s
Mouse 2
Mass 17.9 g = 17.9 10⁻³ kg
Speed (-0.699 i ^ - 0.815 j ^) m / s
Iₓ = m ( - v₀ₓ)
Iₓ = 17.9 10⁻³ (-0.699 -0)
Iₓ = -12.5 10⁻³ J s
= 17.9 10⁻³ (-0.815 - 0)
= -14.6 10⁻³ J s
I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s
Mouse 3
Mass 19.1 g = 19.1 10⁻³ kg
Speed (0.745i ^ + 0.975 j ^) m / s
Iₓ = 19.1 10⁻³ (0.745 -0)
Iₓ = 14.2 10⁻³ J s
= 19.1 10⁻³(0.975 -0)
= 18.6 10⁻³ J s
I₃ = (19.1i ^ + 18.6 j ^) 10⁻³ J s
Mouse 4
Mass 10.1 g = 10.1 10⁻³ kg
Speed (-0.905i ^ + 0.717j ^) m / s
Iₓ = 10.1 10⁻³ (-0.905 -0)
Iₓ = -9.14 10⁻³ J s
= 10.1 10⁻³ (0.717 -0)
= 7.24 10⁻³ J s
I₄ = (-9.14i ^ + 7.24 j ^) 10⁻³ J s