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topjm [15]
3 years ago
11

why does the velocity in the horizontal direction remain constant while the velocity in the vertical direction changes?

Physics
1 answer:
Bond [772]3 years ago
8 0
The horizontal velocity<span> of a projectile is </span>constant<span> (a never </span>changing<span> in value), There is a </span>vertical<span>acceleration caused by gravity; its value is 9.8 m/s/s, down, The </span>vertical velocity<span> of a projectile </span>changes<span> by 9.8 m/s each second, The </span>horizontal<span> motion of a projectile is independent of its </span>vertical<span> motion.</span>
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Answer:

yes

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Sandy is on a road trip. She leaves at 8:00 AM. It takes her 2 hours to drive 200 kilometers. She stops at a rest stop for half
lutik1710 [3]
The average velocity of Sandy is given by the total distance covered S divided by the total time taken t:
v= \frac{S}{t}

The total distance covered is
S=200 km+0+100 km=300 km
while the total time taken is 2 hours + half an hour (for the rest) + 1 hour and half, so
t=2h+0.5h+1.5 h=4 h
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0 0
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
Lina20 [59]

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
The box leaves position x=0x=0 with speed v0v0. The box is slowed by a constant frictional force until it comes to rest at posit
const2013 [10]

Answer:

fr = ½ m v₀²/x

Explanation:

This exercise the body must be on a ramp so that a component of the weight is counteracted by the friction force.

The best way to solve this exercise is to use the energy work theorem

            W = ΔK

Where work is defined as the product of force by distance

           W = fr x cos 180

The angle is because the friction force opposes the movement

          Δk =K_{f} –K₀

          ΔK = 0 - ½ m v₀²

We substitute

         - fr x = - ½ m v₀²      

           fr = ½ m v₀²/x

8 0
3 years ago
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