(1) The vertical magnitude of the ball's initial velocity is 16.075 m/s,
(2) The maximum height of the ball is 13.18 m
(1) To Calculate the vertical component of the ball's initial velocity, we use the formula below.
Formula:
- Vy = Vsin∅........................ Equation 1
Where:
- Vy = vertical component of the ball's initial velocity
- V = Initial velocity of the ball
- ∅ = angle to the horizontal.
From the question,
Given:
Substitute these values into equation 1
- Vy = 25(sin40°)
- Vy = 25×0.643
- Vy = 16.075 m/s.
(2) To calculate the maximum height reached by the ball, we use the formula below.
Formula:
- H = (Vy)²/2g............... Equation 2
Where:
- g = Acceleration due to gravity = 9.8 m/s²
Substitute the value above into equation 2
- H = (16.075)²/(2×9.8)
- H = 13.18 m
Hence, (1) The vertical magnitude of the ball's initial velocity is 16.075 m/s, (b) The maximum height of the ball is 13.18 m.
Learn more about maximum height here: brainly.com/question/13665920

1. Calculate the equivalent resistance in the circuit :
In the given diagram, the resistors are in series connection,
So equivalent resistance = sum of resistance of both resistors.

Therefore, Equivalent resistance = 840 ohms.
2. Calculate the current through the battery :

Hence, current through the battery = 0.142 A
3. How much current is passing through each resistor :
Since the resistors are joined in series connection, the current flowing through each resistor will be equal = 0.142 Amperes.
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Answer:
Star A is closer than Star B
Explanation:
As we know that in parallax method of distance measurement the angle subtended by the star when it covers a distance of one Parsec arc length, it is known as parallax angle
Here we can say

so we have

so here we have
angle subtended by Star A = 1 arc sec
angle subtended by star B = 0.75 arc sec
now we have
distance for star A is given as

distance of star B is given as

So star A is closer than star B
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)
Answer:
The answer is: letter c, density.
Explanation:
Gold is a chemical element that is <em>very rare.</em> It has a high atomic number and is represented by the symbol "Au."
The problem above is asking about the physical property of gold that will help you determine whether a shiny gold nugget is, indeed, real gold. Remember that you are hiking and not anywhere else in the world.
If you look at choice a, appearance. This is definitely incorrect because looking at the object will not be able to help you verify whether it is true gold or "fool's gold." Considering melting point (letter b) would also be hard because you will be needing the necessary equipment to melt gold. Gold melts at 1,064 degrees Celsius. So,this sounds very impractical. (thus, melting point is incorrect) Now, if you look at letter d, physical state, it goes the same way with appearance. It will only tell you that the object is solid and that's it.
Letter c, density is the answer because if you analyze the density of gold, which is 19.32 grams per cubic meter, it is far from the density of iron pyrite (fool's gold) which is 4.8-5 grams per cubic meter. This means that gold is denser or heavier than iron pyrite. You will definitely be able to tell which such big gap in their densities.