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4 years ago

What us the difference between accurate data and reproducible data

1 answer:

4 0

It's just in the name! Accurate data is helpful, and correct, but reproducible data is all of that, and is able to be given to other people through different sources! At least, that's what my understanding of them are. Hope it helps!

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Before being engulfed, matter that is pulled into a black hole should become very hot and emit _____.

miss Akunina [59]

A black hole is a cosmological object that is created when a massive star comes to the end of its life and collapses under its own gravity. Black holes have massive gravitational fields that even light cannot escape beyond a certain distance. Before being engulfed, matter that is pulled into a black hole should become very hot and emit electromagnetic radiation.

4 0

4 years ago

Read 2 more answers

The work done by the brakes during braking is equal to

kow [346]

Answer:

The amount of pressure delivered to them

Explanation:

6 0

2 years ago

Two boys with masses of 40 kg and 60 kg stand on a horizontal frictionless surface holding the ends of a light 10-m long rod. Th

zlopas [31]

When they meet the 40-kg boy would have moved a distance of 6 m.

<h3>Distance moved by the 40 kg boy</h3>

Apply the principle of center mass;

Take the 40 kg mass as the reference point;

X(40 kg) = (40kg x 0 + 60kg x 10 m)/(40 kg + 60 kg)

X(40 kg) = (600)/(100)

X(40 kg) = 6 m

Thus, when they meet the 40-kg boy would have moved a distance of 6 m.

Learn more about distance here: brainly.com/question/2854969

#SPJ1

4 0

2 years ago

Plz solve completely

tiny-mole [99]

Answer:

Explanation:

wouldnt it just be x=0 sorry if im

wrong

6 0

3 years ago

A 40-cm-diameter, 300 g beach ball is dropped with a 4.0 mg ant riding on the top. The ball experiences air resistance, but the

mars1129 [50]

Answer:

The normal force exerted on the ant is 0.75 N.

Explanation:

Given;

diameter of the ball, D = 40 cm = 0.4m

radius of the ball, r = 0.2m

mass of the beach ball, m₁ = 300 g = 0.3 kg

mass of the ant, m₂ = 4 x 10⁻⁶ kg

speed of the ball, v = 4 m/s

The area of the ball, assuming spherical ball is given by;

A = 4πr²

A = 4π(0.2)² = 0.5027 m²

The drag force (resistance) experienced by the spherical ball is given as;

$F_D = \frac{1}{2}C\rho Av^2$

where;

C is the drag coefficient of the spherical ball = 0.45

ρ is density of air = 1.21 kg/m³

$F_D = \frac{1}{2}C\rho Av^2\\\\F_D = \frac{1}{2}(0.45)(1.21) (0.5027)(4)^2\\\\F_D = 2.19 \ N$

The downward force of the ball due to its weight and that of the ant is given by;

$F_g = mg\\\\F_g =g(m_{ant} + m_{ball})\\\\F_g = g(4*10^{-6} \ kg\ + \ 0.3\ kg)\\\\F_g = g(0.300004 \ kg) \ \ \ (mass \ of \ the \ ant \ is \ insignificant)\\\\F_g = 9.8(0.3)\\\\F_g = 2.94 \ N$

The net downward force experienced by the ball is given by;

$F_{net} = F_g - F_D\\\\F_{net} = 2.94 \ N - 2.19 \ N\\\\F_{net} = 0.75 \ N$

This downward force experienced by the ball is equal to the normal reaction it exerts on the ant.

Thus, the normal force exerted on the ant is 0.75 N.

5 0

3 years ago