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4 years ago

You weigh 716 newtons on Earth. You

Physics

1 answer:

Otrada [13]4 years ago

8 0

Answer:

537 N

Explanation:

The force due to gravity of a planet is:

F = GMm / r²

where G is the universal gravitational constant

M is the mass of the planet

m is the mass of the object

and r is the distance between the object and the center of the planet

On Earth, you weigh 716 N, so:

716 N = GMm / r²

On planet X:

F = G (3M) m / (2r)²

F = 3/4 GMm / r²

F = 3/4 (716 N)

F = 537 N

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3 years ago

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The rate of rotation of the disk is gradually increased. The coefficient of static friction between the coin and the disk is 0.5

MrRissso [65]

Question is not complete and the missing part is;

A coin of mass 0.0050 kg is placed on a horizontal disk at a distance of 0.14 m from the center. The disk rotates at a constant rate in a counterclockwise direction. The coin does not slip, and the time it takes for the coin to make a complete revolution is 1.5 s.

Answer:

0.828 m/s

Explanation:

Resolving vertically, we have;

Fn and Fg act vertically. Thus,

Fn - Fg = 0 - - - - eq(1)

Resolving horizontally, we have;

Ff = ma - - - - eq(2)

Now, Fn and Fg are both mg and both will cancel out in eq 1.

Leaving us with eq 2.

So, Ff = ma

Now, Frictional force: Ff = μmg where μ is coefficient of friction.

Also, a = v²/r

Where v is linear speed or velocity

Thus,

μmg = mv²/r

m will cancel out,

Thus, μg = v²/r

Making v the subject;

rμg = v²

v = √rμg

Plugging in the relevant values,

v = √0.14 x 0.5 x 9.8

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v = 0.828 m/s

3 0

3 years ago

The breaks on a 15,680 N car exert a stopping force of 640 N.

BartSMP [9]

The brakes on a 15,680 N car exert a stopping force of 640 N. The car's velocity changes from 20.0 m/s to 0 m/s.

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3 years ago

A computational model predicts the speed of a roller coaster car at different heights given its speed at the lowest height of it

s2008m [1.1K]

Answer:

the answers, the correct one is A v= 22.5 m/s

Explanation:

We can solve this problem using the concepts of conservation of mechanical energy

Starting point. Lowest point of the trajectory

Em₀ = K = ½ m v₀²

Final point. When it is at a height of y = 20 m

$Em_{f}$ = K + U = ½ m v² + mg y

how energy is conserved

Emo = Emf

½ m v₀² = ½ m v² + m g y

v² = v₀² + 2 g y

let's calculate

v²2 = 30 2 + 2 9.8 20

v² = 508

v²= ra 508

v² = 22.54 m / s

When checking the answers, the correct one is A

4 0

3 years ago

using hooke's law F spring = k Δ x find the force needed to stretch a spring 2 cm if it has elastic constant of 3 N/cm. A. 2/3 N

Assoli18 [71]

Data:
Hooke represented mathematically his theory with the equation:

F = K * x

On what:
F (elastic force) = ?
K (elastic constant) = 3 N/cm
x (deformation or elongation of the elastic medium) = 2cm

Solving:

 $F = K * x$
$F = 3 \frac{N}{\diagup\!\!\!\!\!\!cm} * 2\diagup\!\!\!\!\!\!cm$
$\boxed{\boxed{F = 6N}}$

Answer:
C. 6 N

4 0

4 years ago