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Stells [14]
3 years ago
6

a 74 kg man jumps from a window 1.6 m above a sidewalk. the acceleration of gravity is 9.81 m/s 2 . a) what is his speed just be

fore his feet strike the pavement? answer in units of m/s
Physics
1 answer:
olasank [31]3 years ago
4 0

Answer:

  5.60 m/s

Explanation:

His initial potential energy is equal to his final kinetic energy.

  mgh = 1/2mv^2

  2gh = v^2 . . . . . . . multiply by 2/m

  v = √(2gh) = √(2(9.81 m/s²)(1.6 m)) = √(31.392 m²/s²)

  v ≈ 5.60 m/s

_____

<em>Additional comment</em>

The mass is irrelevant in the equations we typically use for this purpose.

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What is it called when data are arranged in rows and columns?
Tpy6a [65]

Answer:

C. Table

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The information kept in the database table is referred to as a secondary key. This kind is a primary key, which implies it is a secondary key. A primary key can be used as an alternate key to distinguish between different records in a table and to uniquely identify a record inside a database. Multiple fields that can be either main or alternative make up a composite key.

Depending on the type of data being stored and its intended use, a database table may have a single row or several. The main key determines the number of rows, which might be 50 or more. A secondary key can be used to establish connections between various tables and to distinguish between various data types in a database. A data table also includes a secondary key, which is essential for data retrieval and is separate from the primary key.

You can have local or global database tables. The two types of tables differ in terms of names, availability, and visibility. A global table is accessible to all users and may be utilized by any other user, while a local table is only visible to the present user. It is distinct from a typical table and requires more effort to build than a straightforward one does. Its main function is to maintain a database and store data.

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8 0
2 years ago
Read 2 more answers
A man on a bicycle of total mass of 10kg has a Velocity of 2 ms.' He Paddles faster for 5 seconds and the velocity Increase to i
sesenic [268]

Answer:

the value of force, F=4.0N

Explanation:

Firstly, recall velocity-time equation

  • v=u+at
  • (4)=(2)+a(5)
  • a=0.4m/s²

Secondly, recall the Newton's 2nd Law

  • <em>F</em><em>=</em><em>ma</em>
  • <em>F</em><em>=</em><em>(</em><em>1</em><em>0</em><em>)</em><em>(</em><em>0</em><em>.</em><em>4</em><em>)</em>
  • <em>F</em><em>=</em><em>4</em><em>.</em><em>0</em><em>N</em>
7 0
2 years ago
A point charge q1q1 is held stationary at the origin. A second charge q2q2 is placed at point aa, and the electric potential ene
Yuri [45]

Explanation:

The given data is as follows.

            U_{a} = 5.4 \times 10^{-8} J

        W_{/text{a to b}} = -1.9 \times 10^{-8} J

        Electric potential energy (U_{b}) = ?

Formula to calculate electric potential energy is as follows.

            U_{b} = U_{a} - W_{/text{a to b}}

                        = 5.4 \times 10^{-8} J - (-1.9 \times 10^{-8} J)

                        = 7.3 \times 10^{-8} J

Thus, we can conclude that the electric potential energy of the pair of charges when the second charge is at point b is 7.3 \times 10^{-8} J.

6 0
4 years ago
Scenario
Anvisha [2.4K]

Answer:

1) t = 23.26 s,  x = 8527 m, 2)   t = 97.145 s,  v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

         v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

         y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

         y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

         0 = y₀ + 0 - ½ g t2

          t = \sqrt{ \frac{2y_o}{g} }

          t = √(2 2650/9.8)

          t = 23.26 s

this is the horizontal scrolling time

          x = v₀ t

          x = 366.67  23.26

          x = 8527 m

the speed at the point of arrival is

         v_y = v_{oy} - g t = 0 - gt

         v_y = - 9.8 23.26

         v_y = -227.95 m / s

Module and angle form

        v = \sqrt{v_x^2 + v_y^2}

         v = √(366.67² + 227.95²)

        v = 431.75 m / s

         θ = tan⁻¹ (v_y / vₓ)

         θ = tan⁻¹ (227.95 / 366.67)

         θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

           cos θ = v₀ₓ / v₀

           sin θ = v_{oy} / v₀

            v₀ₓ = v₀ cos θ

            v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

            y = y₀ + v_{oy} t - ½ g t²

            0 = y₀ + vo sin θ t - ½ g t²

            0 = -50 + vo sin 50 t - ½ 9.8 t²

            x = x₀ + v₀ₓ t

            0 = x₀ + vo cos θ t

            0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

          50 = 0,766 vo t – 4,9 t²

          400 =   0,643 vo t

resolved

          50 = 0,766 ( \frac{400}{0.643 \ t}) t – 4,9 t²

          50 = 476,52 t – 4,9 t²

          t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

         t = [97.25 ± \sqrt{97.25^2 - 4 \ 10.2}] / 2

         t = 97.25 ±97.04] 2

         t₁ = 97.145 s

         t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

         t = 97.145 s

let's find the initial velocity

         x = x₀ + v₀ cos 50 t

         0 = -400 + v₀ cos 50 97.145

         v₀ = 400 / 62.44

         v₀ = 6.4 m / s

5 0
3 years ago
Because of your success in physics class you are selected for an internship at a prestigious bicycle company in its research and
tiny-mole [99]

To develop the problem it is necessary to apply the equations related to the moment of inertia.

The given values can be defined as,

M = 1.0 kg

r = 0.5 m

m = 10 g

I = 0.280 kg.m^2

According to the definition of the moment of inertia applied to the exercise we can arrive at the equation that,

I = I_{rim} + n * I_{spoke}

Where n is the number of spokes necessary to construct the wheel.

I_{rim} = M*r^2 = 1.0 * 0.5^2

I_{spoke} = \frac{1}{3} * m * r^2 = \frac{1}{3}* 10 * 10^-3 * 0.5^2

Replacing the values at the general equation we have,

0.280 = 1.0 * 0.5^2 + n * (1/3 * 10 * 10^-3 * 0.5^2 )

Solving for n,

n = 36

Therefore the number of spokes necessary to construct the wheel is 36

PART B) The mass of the wheel is given by the sum of all masses and the total spokes, then

M_w= M + n*m

M_w = 1.0 + 36* 10 * 10^{-3} Kg

M_w = 1.36 Kg

Therefore the mass of the wheel must be of 1.36Kg

4 0
4 years ago
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