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3 years ago

1. Calculate the force of gravity in newtons if your weight is 110 lbs?

Physics

2 answers:

lianna [129]3 years ago

6 0

Given that,

mass m = 110 lbs

Force of gravity in Newton, F =?

Since,

1 lbs = 0.45 kg

110 lbs = 110* 0.45 = 49.5 kg

Since, force of gravity equals the weight of the body.

F = W = mg = 49.5 * 9.81

F = 485.595 N

Force of gravity will be 485.595 N.

In order for motion to occur net force is greater than 485.595 N.

Tom [10]3 years ago

3 0

First of all you need to know that the force of gravity =m×g. In which, m is the mass and g is the gravitational constant, which is 9.81N/Kg. Since the constant is in Newtons per Kilograms, you need to have the mass in Kg. So your 110 lbs is approximately 49.89 Kg.

Now, just plug in everything in the equation.

F= m×g

F= 49.89kg×9.81N/kg

F= 489.4N

Hope this helps!

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A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo

jeka57 [31]

Answer:

a. A = 0.735 m

b. T = 0.73 s

c. ΔE = 120 J decrease

d. The missing energy has turned into interned energy in the completely inelastic collision

Explanation:

4 kg * 10 m /s + 6 kg * 0 m/s = 10 kg* vmax

vmax = 4.0 m/s

¹/₂ * m * v²max = ¹/₂ * k * A²

m * v² = k * A² ⇒ 10 kg * 4 m/s = 100 N/m * A²

A = √1.6 m ² = 1.26 m

At = 2.0 m - 1.26 m = 0.735 m

T = 2π * √m / k ⇒ T = 2π * √4.0 kg / 100 N/m = 1.26 s

T = 2π *√ 10 / 100 *s² = 1.99 s

T = 1.99 s -1.26 s = 0.73 s

E = ¹/₂ * m * v²max =

E₁ = ¹/₂ * 4.0 kg * 10² m/s = 200 J

E₂ = ¹/₂ * 10 * 4² = 80 J

200 J - 80 J = 120 J decrease

The missing energy has turned into interned energy in the completely inelastic collision

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Mnenie [13.5K]

Answer:

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Explanation:

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Answer:

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(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength λ ? (b) W

Kitty [74]

Answer:

The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

Wavelength = λ

For D to be small,

We need to calculate the minimum width

Using formula of minimum width

$D\sin\theta=n\lambda$

$D=\dfrac{n\lambda}{\sin\theta}$

Where, D = width of slit

$\lambda$ = wavelength

Put the value into the formula

$D=\dfrac{n\lambda}{\sin\theta}$

Here, $\sin\theta$ should be maximum.

So. maximum value of $\sin\theta$ is 1

Put the value into the formula

$D=\dfrac{1\times\lambda}{1}$

$D=\lambda$

(b). If the minimum number is 50

Then, the width is

$D=\dfrac{50\times\lambda}{1}$

$D=50\lambda$

(c). If the minimum number is 1000

Then, the width is

$D=\dfrac{1000\times\lambda}{1}$

$D=1000\lambda$

Hence, The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

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anygoal [31]

Answer:

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Explanation:

15 minutes is 1/4 of an hour.

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3 years ago