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3 years ago

The motion of a parachuists isn't considered free fall? give reason

Physics

1 answer:

lidiya [134]3 years ago

5 0

Answer:

cuz he use parachute

Explanation:

free fall is when you are falling and accelerating with G, so as parachuist using parachute and decelerating it's not free fall.

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What kind of food service is the most expensive and requires the largest staff of skilled servers?

MrMuchimi

Your answer is french

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3 years ago

A 500 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exert

liberstina [14]

Answer:

x = 7.62 m

Explanation:

First we need to calculate the weight of the rocket:

W = mg

we will use the gravity as 9.8 m/s². We have the mass (500 g or 0.5 kg) so the weight is:

W = 0.5 * 9.8 = 4.9 N

We know that the rocket exerts a force of 8 N. And from that force, we also know that the Weight is exerting a force of 4.9. From here, we can calculate the acceleration of the rocket:

F - W = m*a

a = F - W/m

Solving for a:

a = (8 - 4.9) / 0.5

a = 6.2 m/s²

As the rocket is accelerating in an upward direction, we can calculate the distance it reached, assuming that the innitial speed of the rocket is 0. so, using the following expression we will calculate the time which the rocket took to blast off:

y = vo*t + 1/2 at²

y = 1/2at²

Solving for t:

t = √2y/a

t = √2 * 20 / 6.2

t = √6.45 = 2.54 s

Now that we have the time, we can calculate the horizontal distance:

x = V*t

Solving for x:

x = 3 * 2.54 = 7.62 m

4 0

4 years ago

How do you know that light travels faster than

Dahasolnce [82]

Answer:

Explanation:

Answer

How about a thunder storm. I don't know if you live in a city or out in the sticks as I do. Lighting is very obvious and it is not a good idea to be out when experiencing a thunderstorm, especially in an open field. You might be the only thing around that will cause the lightning to be connected to the ground.# Seconds later, you will hear the thunder which is quite harmless. If you live in the city, observing this is not quite so easy. There are all kinds of buildings around and some of them may block your view.

#The great golfer, Lee Travino, was "hit" by lightening storm twice while playing in tournaments.

8 0

2 years ago

Amoving object is in equilibrium. Which best describes the motion of the object if no forces change?

kaheart [24]

I believe the answer is C. It will maintain its state of motion

4 0

4 years ago

Read 2 more answers

Resistors 1 and 2− R1 = 50 Ω , R2 = 90 Ω − are connected in series to a 6.0-V battery. Part APart complete What is the potential

kondor19780726 [428]

Answer:

Part A: The voltage across resistor R1 is approximately $\rm 2.1 \; V$ .

Part B: When the value of resistor R1 decreases, the current in this circuit will increase.

Part C: When the value of resistor R1 decreases, the voltage across resistor R1 will decrease.

Explanation:

Resistor R1 and and R2 are connected in series. That's equivalent to a single resistor of $R_1 + R_2 = 50 + 90 = 140\; \Omega$ . The voltage across the two resistor, combined, is equal to $\rm 6\; V$ . Hence by Ohm's Law, the current through the circuit will be equal to $\rm \dfrac{6\; V}{140\; \Omega} = \dfrac{3}{70}\; A$ .

These two resistors are connected in series. The voltage across each of them might differ. However, the current through each of them should both be equal to the current through the circuit. In this case, the current through both R1 and R2 should be equal to $\rm \dfrac{3}{70}\; A$ . Apply Ohm's Law (again) to find the voltage across R1:

$V = I \cdot R = \dfrac{3}{70} \times 50 \approx \rm 2.1\; V$ .

Since the equivalent resistance is equal to $R_1 + R_2$ , when the value of $R_1$ decreases, the equivalent resistance will also decrease. By Ohm's Law, $I = \dfrac{V}{R}$ . When the value of the denominator ( decreases, the value of the quotient, $I$ the current through the circuit, will increase.

Keep in mind that if two resistors are connected in series,

$I(R_1) = I(\text{Circuit}) = I(R_2)$ .

The resistance of R1 decreases, while the current through it increases. Applying Ohm's Law on R1 won't give much useful information. However, since the resistance of R2 stays the same, the voltage across it will increase when its current increases (again by Ohm's Law.)

Again, since the two resistors are connected in series,

$V(R_1) + V(R_2) = V(\text{Circuit}) = \rm 6 \; V$ ,

when the voltage across R2 increases, the voltage across R1 will decrease.

4 0

3 years ago