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valentinak56 [21]
3 years ago
12

Projectile motion is a combination of which two types of motion?

Physics
2 answers:
klio [65]3 years ago
5 0
The answer would be, Horizontal motion and Vertical motion.
Papessa [141]3 years ago
3 0

Horizontal and vertical.

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Help with these questions please , 15 PTS & brainliest
myrzilka [38]

For these two questions, first you need to know that the voltage across each branch of a parallel circuit is the same.

So, for Q5, we can first find out the voltage across R₂ by V=IR.

Voltage across R₂ = 2.5 × 8 = 20V

Since R₂ and R₃ are in parallel circuit, their voltage should be the same. Thus, voltage across R₃ is 20V.

So, by V=IR,

current of R₃ = \frac{20}{4} = 5A

Q6. voltage across R₁ = 2 × 4 = 8V

∴voltage across R₂ = 8V

current of R₂ = \frac{8}{8} = 1A

<h3><u>Alternative method</u></h3>

From these two examples, you can find out that the current of each branch of the parallel circuit is inversely proportional to the resistance of the branch.

ie. for Q5,

\frac{R2}{R3} = \frac{I3}{I2}

\frac{8}{4} = \frac{I3}{2.5}

I₃ = 5A

Q6. \frac{R1}{R2} = \frac{I2}{I1}

\frac{4}{8} = \frac{I2}{2}

I₂ = 1A

6 0
3 years ago
Which one of the statements below is true about mechanical waves?
TEA [102]
They require a medium to travel through
5 0
3 years ago
The weather conditions of an area may be expected to change over a(n) _________ period of time.
jenyasd209 [6]

Answer:

Your answer should be 2. Short

Explanation:

4 0
3 years ago
What is the relationship between area, force, and pressure? what is the relationship between area, force, and pressure? pressure
White raven [17]
Mathematically, relation between force, area and pressure is given by...
Pressure = force / area


hence, pressure is directly proportional to force but inversely proportional to area.
7 0
3 years ago
A small logo is embedded in a thick block of crown glass (n = 1.52), 4.70 cm beneath the top surface of the glass. The block is
harkovskaia [24]

The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})

Where,

d_g = Depth of glass

n_w = Refraction index of water

n_g = Refraction index of glass

n_{air} = Refraction index of air

d_w = Depth of water

I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to

d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})

d'w = (1.7cm) (\frac{1}{1.33})+(4.2cm)(\frac{1}{1.52})

d'w = 4.041cm

Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm

3 0
3 years ago
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