Answer
correct answer is C
Head-to-tail method
Here head-to-tail method is employed to determine resultant of vectors.
Let horizontal component be 1 and let vertical component be 2
To add these two vectors. Move vector 2 until its tail touches the head of 1. The tail of resultant of these two vectors touch the tail of vector 1 and head of resultant vector will touch the head of vector 2.
Answer:
Charge = 4.9096 x 10⁻⁷ C
Explanation:
First, we find the resistance of the copper wire.
R = ρL/A
where,
R = resistance = ?
ρ = resistivity of copper = 1.69 x 10⁻⁸ Ω.m
L = Length of wire = 2.16 cm = 0.0216 m
A = Cross-sectional area of wire = πr² = π(0.00233 m)² = 1.7 x 10⁻⁵ m²
Therefore,
R = (1.69 x 10⁻⁸ Ω.m)(0.0216 m)/(1.7 x 10⁻⁵ m²)
R = 2.14 x 10⁻⁵ Ω
Now, we find the current from Ohm's Law:
V =IR
I = V/R
I = 3.27 x 10⁻⁹ V/2.14 x 10⁻⁵ Ω
I = 1.52 x 10⁻⁴ A
Now, for the charge:
I = Charge/Time
Charge = (I)(Time)
Charge = (1.52 x 10⁻⁴ A)(3.23 x 10⁻³ s)
<u>Charge = 4.9096 x 10⁻⁷ C</u>
Answer:
v₃ = 9.62[m/s]
Explanation:
To solve this type of problem we must use the principle of conservation of linear momentum, which tells us that the momentum is equal to the product of mass by velocity.
We must analyze the moment when the astronaut launches the toolkit, the before and after. In order to return to the ship, the astronaut must launch the toolkit in the opposite direction to the movement.
Let's take the leftward movement as negative, which is when the astronaut moves away from the ship, and rightward as positive, which is when he approaches the ship.
In this way, we can construct the following equation.

where:
m₁ = mass of the astronaut = 157 [kg]
m₂ = mass of the toolkit = 5 [kg]
v₁ = velocity combined of the astronaut and the toolkit before throwing the toolkit = 0.2 [m/s]
v₂ = velocity for returning back to the ship after throwing the toolkit [m/s]
v₃ = velocity at which the toolkit should be thrown [m/s]
Now replacing:
![-(157+5)*0.2=(157*0.1)-(5*v_{3})\\(5*v_{3})= 15.7+32.4\\v_{3}=9.62[m/s]](https://tex.z-dn.net/?f=-%28157%2B5%29%2A0.2%3D%28157%2A0.1%29-%285%2Av_%7B3%7D%29%5C%5C%285%2Av_%7B3%7D%29%3D%2015.7%2B32.4%5C%5Cv_%7B3%7D%3D9.62%5Bm%2Fs%5D)
An egg thrown at a concrete wall will break, but an egg thrown at a sheet hanging from the ceiling will not due to high momentum and acceleration.
<h3>Why an egg thrown at a concrete wall will break?</h3>
An egg thrown at a concrete wall will break, but an egg thrown at a sheet hanging from the ceiling will not because the momentum and acceleration increases when the egg is thrown downward due to gravity but when we throw an egg in the vertical direction, they move against gravity so the momentum and acceleration decreases.
So we can conclude that an egg thrown at a concrete wall will break, but an egg thrown at a sheet hanging from the ceiling will not due to high momentum and acceleration.
Learn more about momentum here: brainly.com/question/7538238
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Answer:

Explanation:
The range of a projectile is given by the formula

where in this case, we have
d = 100 m is the range
u is the initial speed (the magnitude of the initial velocity)
g = 9.8 m/s^2 is the acceleration of gravity
is the angle of projection
Solving for u, we find:

Now we can easily find the components of the initial velocity:

So, the initial velocity of the ball is

where i and j are the unit vector indicating the horizontal and vertical direction.