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3 years ago

8

A ball (with a weight of 0.25kg ) rolls horizontally across the sand and decelerates at a rate of 6 m/s2. Calculate the force th

at acted upon the ball.
please help

1 answer:

Vera_Pavlovna [14]3 years ago

3 0

Um ok so you subtract 2 on both side the plug in 6

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You are planning to make an open rectangular box from an 8-inch by 15-inch piece of cardboard by cutting congruent squares from

Phoenix [80]

Let us say that x is the cut that we will make on the sides to make a box, therefore the new dimensions are:

l = 15 – 2x

w = 8 – 2x

It is 2x since we cut on two sides.

We know that volume is:

V = l w x

V = (15 – 2x) (8 – 2x) x

V = 120x – 30x^2 – 16x^2 + 4x^3

V = 120x – 46x^2 + 4x^3

Taking the 1st derivative:

dV/dx = 120 – 92x + 12x^2

Set dV/dx = 0 to get maxima:

120 – 92x + 12x^2 = 0

Divide by 12:

x^2 – (92/12)x + 10 = 0

(x – (92/24))^2 = -10 + (92/24)^2

x - 92/24 = ±2.17

x = 1.66, 6

We cannot have x = 6 because that will make our w negative, so:

x = 1.66 inches

So the largest volume is:

V = 120x – 46x^2 + 4x^3

V = 120(1.66) – 46(1.66)^2 + 4(1.66)^3

V = 90.74 cubic inches

4 0

3 years ago

The minimum cooking temperature for eggs that will be served immediately is

solong [7]

Cooking and Serving. Cook raw shell eggs that are broken for immediate preparation and service to heat all parts of the food to a temperature of 63°C<span> (</span>145°F<span>) for 15 seconds</span>

5 0

3 years ago

Read 2 more answers

An interference pattern is produced by light with a wavelength 520 nm from a distant source incident on two identical parallel s

Goshia [24]

Answer:

1) θ = 0.00118 rad, 2) θ = 0.00236 rad , 3) I / I₀ = 0.1738, 4) I / Io = 0.216

Explanation:

In the double-slit interference phenomenon it is explained for constructive interference by the equation

d sin θ = m λ

1) the first order maximum occurs for m = 1

sin θ = λ / d

θ = sin⁻¹ λ / d

let's reduce the magnitudes to the SI system

λ = 520 nm = 520 10⁻⁹ θ = 0.00118 radm

d = 0.440 mm = 0.440 10⁻³ m ³

let's calculate

θ = sin⁻¹ (520 10⁻⁹ / 0.44 10⁻³)

θ = sin⁻¹ (1.18 10⁻³)

θ = 0.00118 rad

2) the second order maximum occurs for m = 2

θ = sin⁻¹ (m λ / d)

θ = sin⁻¹ (2 5¹20 10⁻⁹ / 0.44 10⁻³)

θ = 0.00236 rad

3) To calculate the intensity of the interference spectrum, the diffraction phenomenon must be included, so the equation remains

I = I₀ cos² (π d sin θ /λ ) sinc² (pi b sin θ /λ )

where the function sinc = sin x / x

and b is the width of the slits

we caption the values

x = π 0.310 10⁻³ sin 0.00118 / 520 10⁻⁹)

x = 2.21

I / I₀ = cos² (π 0.44 10⁻³ sin 0.00118 / 520 10⁻⁹) (sin (2.21) /2.21)²

remember angles are in radians

I / I₀ = cos² (3.0945) [0.363] 2

I / I₀ = 0.9978 0.1318

I / I₀ = 0.1738

4) the maximum second intensity is

I / I₀ = cos² (π d sinθ / λ) sinc² (πb sin θ /λ)

x =π 0.310 10⁻³ sin 0.00236 / 520 10⁻⁹)

x = 4.41

I / Io = cos² (π 0.44 10⁻³ sin 0.00236 / 520 10⁻⁹) (sin 4.41 / 4.41)²

I / Io = cos² 6.273 0.216

I / Io = 0.216

.

7 0

3 years ago

A 3.00-kg object has a velocity 1 6.00 i ^ 2 2.00 j ^2 m/s. (a) what is its kinetic energy at this moment? (b) what is the net w

tatyana61 [14]

(a) The velocity of the object on the x-axis is 6 m/s, while on the y-axis is 2 m/s, so the magnitude of its velocity is the resultant of the velocities on the two axes:
$v= \sqrt{(6.00m/s)^2+(2.00 m/s)^2}=6.32 m/s$
And so, the kinetic energy of the object is
$K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(6.32 m/s)^2=60 J$

(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is
$v= \sqrt{(8.00 m/s)^2+(4.00 m/s)^2}=8.94 m/s$
And so the new kinetic energy is
$K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(8.94 m/s)^2=120 J$

So, the work done on the object is the variation of kinetic energy of the object:
$W=\Delta K=120 J-60 J=60 J$

7 0

3 years ago

How do motion and Newton's laws apply to your everyday life?

Marina86 [1]

When you jumó ur legs put force on ground

6 0

2 years ago