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AURORKA [14]
3 years ago
10

A 140-kg box is being lowered at constant velocity. A rope is attached to the top of the box, and Jimmy is slightly supporting t

he box from the bottom. If Jimmy is applying 50 N of force to the box, how much force is the rope applying?
Physics
1 answer:
Zinaida [17]3 years ago
4 0

Answer:

the anwser is 90

Explanation:

because 140-50=90

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An object moves in uniform circular motion at 25 m/s and takes 1.0 second to go a quarter circle. What is the radius of the circ
FromTheMoon [43]
Object Motion: 25 m/s

Circumference of Circle: 
1/4 Circumference of Circle in 1 second = 25 meters
25 meters times 4 = Circumference of Circle 
Circumference = 100 meters

Formula to Find Circumference of Circle: (work opposite)
C = 2<span>πr

100 = </span>2πr          divided
100/2π = r         simplify
50/π = r         (exact radius)

Answer:
50/π meters = r         (exact radius)

4 0
3 years ago
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What is the correct equation kinetic enery
Ivahew [28]

Kinetic energy is directly proportional to the mass of the object and to the square of its velocity: K.E. = 1/2 m v2.

4 0
4 years ago
The nonreflective coating on a camera lens with an index of refraction of 1.21 is designed to minimize the reflection of 570-nm
lord [1]

Answer: 117.8 nm

Explanation:

Given,

Nonreflective coating refractive index : n = 1.21

Index of refraction: n_0 = 1.52

Wave length of light = λ = 570 nm = 570\times10^{-9}\ m

\text{ Thickness}=\dfrac{\lambda}{4n}

=\dfrac{570\times10^{-9}\ m}{4\times1.21}\\\\\approx\dfrac{117.8\times 10^{-9}\ m}{1}\\\\=117.8\text{ nm}

Hence, the minimum thickness of the coating that will accomplish= 117.8 nm

5 0
4 years ago
You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land.
Alika [10]

Answer:

d = 2021.6 km

Explanation:

We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them

Airplane 1

Height   y₁ = 800m

Angle θ = 25°

           cos 25 = x / r

           sin 25 = z / r

           x₁ = r cos 20

           z₁ = r sin 25

          x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

          z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m

2 plane

Height   y₂ = 1100 m

Angle θ = 20°

          x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

          z₂ = 20 103 without 25 = 8.452 103 m = 8452 m

The distance between the planes using the Pythagorean Theorem is

         d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Let's calculate

        d² = (18126-16314)²  + (1100-800)² + (8452-7607)²

        d² = 3,283 106 +9 104 + 7,140 105

        d² = (328.3 + 9 + 71.40) 10⁴

        d = √(408.7 10⁴)

        d = 20,216 10² m

        d = 2021.6 km

7 0
4 years ago
A car of mass 998 kilograms moving in the positive y–axis at a speed of 20 meters/second collides on ice with another car of mas
goldfiish [28.3K]
    <span> Let’s determine the initial momentum of each car.
#1 = 998 * 20 = 19,960
#2 = 1200 * 17 = 20,400

This is this is total momentum in the x direction before the collision. B is the correct answer. Since momentum is conserved in both directions, this will be total momentum is the x direction after the collision. To prove that this is true, let’s determine the magnitude and direction of the total momentum after the collision.

Since the y axis and the x axis are perpendicular to each other, use the following equation to determine the magnitude of their final momentum.

Final = √(x^2 + y^2) = √(20,400^2 + 19,960^2) = √814,561,600

This is approximately 28,541. To determine the x component, we need to determine the angle of the final momentum. Use the following equation.

Tan θ = y/x = 19,960/20,400 = 499/510
θ = tan^-1 (499/510)

The angle is approximately 43.85˚ counter clockwise from the negative x axis. To determine the x component, multiply the final momentum by the cosine of the angle.

x = √814,561,600 * cos (tan^-1 (499/510) = 20,400</span>
3 0
3 years ago
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