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kupik [55]
3 years ago
7

What effect does the vertical acceleration have on the horizontal velocity of the projectile?

Physics
1 answer:
KengaRu [80]3 years ago
3 0
Answer:
None, if air resistance is ignored.

Explanation:
At any instant, the projectile has vertical and horizontal components of velocity.
Vertical acceleration due to gravity affects the vertical velocity by accelerating the object toward the center of the earth, and by decreasing the upward vertical velocity.. 
The horizontal component of velocity makes the object travel horizontally as long as the projectile is airborne.
Thsi discussion assumes that air resistance is ignored.
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What is the wavelength of a wave that has a speed of 350 meters/second and a frequency of 140 hertz?
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What are the only elements that exist in nature as isolated atoms
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If a reaction is exothermic, then which of the following statements is always true? If a reaction is exothermic, then which of t
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Reactants ⇒ product + heat (exothermic reaction) ....(1)

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3 years ago
Three resistors are wired in parallel with a battery. Two of the resistors have resistances of 38.7 Q/ and 89.5 Q. The current i
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Answer:

214.9 \Omega

Explanation:

The three resistors are connected in parallel: this means that the potential difference across each resistor is the same as the voltage of the battery. This can be calculated using the information about the 38.7 \Omega resistor: in fact, since we know its resistance and the current flowing through it (0.155 A), we can find the potential difference across this resistor, which is equal to the voltage of the battery:

V=IR=(0.155 A)(38.7 \Omega)=6.0 V

We also know the total current in the circuit, 0.250 A. This means that we can find the total resistance of the circuit, using Ohm's law:

R_{eq}=\frac{V}{I}=\frac{6.0 V}{0.250 A}=24 \Omega

So now we now the total resistance and the resistance of two of the 3 resistors; therefore, we can find the resistance of the 3rd resistor:

\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\\\frac{1}{R_3}=\frac{1}{R_{eq}}-\frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{24 \Omega}-\frac{1}{38.7\Omega}-\frac{1}{89.5\Omega}=0.00465 \Omega^{-1}\\R_3=\frac{1}{0.00465 \Omega^{-1}}=214.9 \Omega

4 0
3 years ago
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