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Olegator [25]
3 years ago
6

A skydiver reaches terminal velocity. Then he opens his parachute.

Physics
2 answers:
schepotkina [342]3 years ago
4 0

Answer:

it is 0.9999.10896

Explanation:

MatroZZZ [7]3 years ago
4 0

Answer:

Explanation:

Remark

He's pulled upwards by the parachute. The formula is given below. Details are found   searching for terminal velocity.

V_t=\sqrt\frac{2mg}{\rho A C_d}

Whatever that calculation turns out to be, a person has to be going a whole lot slower before he can safely hit the ground. The parachute is designed so that the person will be slowed very quickly to a safe velocity so when he hits the ground the impulse will not be as great as it could be. The force that the parachute delivers acts upward and opposes the force of gravity which is a downward force.

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A certain electric circuit obeys Ohm's law. If the resistance of the circuit is doubled, what will happen to the current through
netineya [11]

Answer:

currrent will be halved

Explanation:

v = ir

v/r = i      multiply both sides by  1/ 2

v / (2r) = 1/2 i  

4 0
3 years ago
To calculate the heat needed to melt a block of ice at its melting point what do you need to know?
AnnyKZ [126]

Answer:

Explanation:

You are not raising the temperature of the block of ice. You are making it turn phase.

H = m * hf

That's the mass of the block and the latent heat of fusion.

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3 0
3 years ago
Read 2 more answers
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, exer
ivanzaharov [21]

Answer:

The rocket has to be launched 8 m from the hoop

Explanation:

Let's analyze this problem, the rocket is on a car that moves horizontally, so the rocket also has the same speed as the car; The initial horizontal rocket speed is (v₀ₓ = 3.0 m/s).

On the other hand, when starting the engines we have a vertical force, which creates an acceleration in the vertical axis, let's use Newton's second law to find this vertical acceleration

    F -W = m a

    a = (F-mg) / m

    a = F/m  -g

    a = 7.0/0.500  - 9.8

    a = 4.2 m/s²

We see that we have a positive acceleration and that is what we are going to use in the parabolic motion equations

Let's look for the time it takes for the rocket to reach the height (y = 15m) of the hoop, when the rocket fires its initial vertical velocity is zero (I'm going = 0)

    y = v_{oy} t + ½ a t²

    y = 0 + ½ a t²

    t = √ 2y/a

    t = √( 2 15 / 4.2)

    t = 2.67 s

This time is also the one that takes in the horizontal movement, let's calculate how far it travels

    x = v₀ₓ t

    x = 3 2.67

    x = 8 m

The rocket has to be launched 8 m from the hoop

8 0
3 years ago
With no effort at all, you'll have deep and lasting relationships.<br> True<br> False
Firdavs [7]
The correct answer is false
7 0
3 years ago
Determine the automobile’s braking distance from 90 km/h when it is going up a 5° incline. (Round the final value to one decimal
NeX [460]

Answer:

366 m

Explanation:

u = 90 km/h = 25 m/s,

theta = 5 degree

acceleration, a = g Sin theta = 9.8 x Sin 5 = 0.854 m/s^2

The final velocity os zero and let the braking distance be s.

Use third equation of motion

v^2 = u^2 - 2 a s

0 = 25 x 25 - 2 x 0.854 x s

s = 366 m

8 0
3 years ago
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