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3 years ago

How do you convert Kg's to Newtons?

1 answer:

5 0

You don't convert kilograms to newtons. By the time you've heard of these units, you know that 'kilogram' is a unit of mass, 'newton' is a unit of force or weight, and that mass and weight are different things.

Mass and force are related by Newton's second law:

Force = Mass x acceleration .

From this simple formula, you can see that in order to relate a mass to a force, you need to know an acceleration. And if the acceleration changes, then the relationship between the force and the mass also changes. So there's no direct conversion.

ON EARTH ONLY, one kilogram of mass weighs 9.8 newtons. The acceleration that connects them is the acceleration of gravity on Earth. In other places, with different gravitational accelerations, 1 kilogram weighs more or less newtons.

But they don't convert directly. That would be like asking "How do you convert miles to miles-per-hour ?"

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A bicyclist notes that the pedal sprocket has a radius of rp = 9.5 cm while the wheel sprocket has a radius of rw = 4.5 cm. The

ANTONII [103]

Answer:

Explanation:

a) ωp = 2π radians / 1.7 s = 3.7 rad/s

b) ωs = 3.7 rad/s(9.5 cm / 4.5 cm) = 7.8 rad/s

v = (ωs)R = 7.8(65) = 507 cm/s or 5.1 m/s

c) ωs = 3.5 m/s / 0.65 m = 5.38 rad/s

ωp = 5.38(4.5 cm / 9.5 cm) = 2.55 rad/s

t = θ/ω = 2π / 2.55 = 2.463... 2.5 s

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3 years ago

Answer the question below based on the periodic table entry for bromine.

Sphinxa [80]

The answer is 45 i hope i could help

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3 years ago

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Fault block mountains form when large blocks of crust are and tilted along normal faults

Katarina [22]

Fault block mountains form when large blocks of crust are uplifted and tilted along normal faults. The uplifted blocks are called block mountains. They are formed by the movement of large crustal blocks when forces in the Earth's crust pull it apart. Some parts of the Earth are pushed upward and others collapse down.

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3 years ago

. During a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from th

mixer [17]

Answer:

106.7 N

Explanation:

We can solve the problem by using the impulse theorem, which states that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object:

$F \Delta t = m (v-u)$

where

F is the average force

$\Delta t$ is the duration of the collision

m is the mass of the ball

v is the final velocity

u is the initial velocity

In this problem:

m = 0.200 kg

u = 20.0 m/s

v = -12.0 m/s

$\Delta t = 60.0 ms = 0.06 s$

Solving for F,

$F=\frac{m(v-u)}{\Delta t}=\frac{(0.200 kg) (-12.0 m/s-20.0 m/s)}{0.06 s}=-106.7 N$

And since we are interested in the magnitude only,

F = 106.7 N

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3 years ago

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Using newtons second law of motion, how fast for 100 KG object accelerates 350 N of force is applied to

fenix001 [56]

Answer:

3.5m/s^2

Explanation:

From Newton's second Law of Motion

F = ma

Where F is the applied force, m is the mass of the object and a is the acceleration.

F = 350 N

Mass = 100kg

350N = 100×a

a = 350/100

a = 3.5m/s^2

The acceleration of the object will be 3.5m/s^2

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3 years ago