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fredd [130]
4 years ago
12

Neglecting air resistance, which of the following statements is true regarding an object in freefall?Lector inmersivo A. An obje

ct’s acceleration increases at a constant rate. B. An object maintains a constant velocity. C. An object falls an equal distance in between each second. D. An object’s velocity and acceleration both increase at a constant rate. E. An object’s velocity changes at a constant rate, and its acceleration remains constant.
Physics
1 answer:
Dafna1 [17]4 years ago
3 0

Answer:

E. An object’s velocity changes at a constant rate, and its acceleration remains constant.

Explanation:

When an object is in freefall, it implies that the object is falling freely under gravity. If it falls towards the earth surface, the fall is in the direction of the Earth's gravitational force.

At the point of release of the object, its initial velocity is zero because it is at rest. But when released, its velocity increases at a constant rate until it is acted upon by an external force. But its acceleration remains constant, acceleration due to gravity.

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Rank the deformations of the following rods in terms of the magnitude of the average normal strain: (a) The length of a 1-m-long
kipiarov [429]

To solve this problem we will consider the concepts related to the normal deformation on a surface, generated when the change in length is taken per unit of established length, that is, the division between the longitudinal fraction gained or lost, over the initial length. In general mode this normal deformation can be defined as

\epsilon = \frac{\delta}{l} = \frac{l_0-l}{l}

Here,

\delta= Change in final length (l_0) and the initial length l

PART A)

\epsilon = \frac{\delta_1}{l}

\epsilon = \frac{l_0-l}{l_0}

\epsilon = \frac{1.02-1}{1}

\epsilon = 0.01961

PART B)

\epsilon = \frac{\delta_1}{l}

\epsilon = \frac{l_0-l}{l_0}

\epsilon = \frac{2-1.05}{2}

\epsilon = 0.475

PART C)

\epsilon = \frac{\delta_1}{l}

\epsilon = \frac{l_0-l}{l_0}

\epsilon = \frac{3.07-3}{3}

\epsilon = 0.0233

Therefore the rank of this deformation would be  B>C>A

7 0
3 years ago
Una persona de 80 kilogramos de masa se encuentra parada sobre una silla de 3 kilogramos de masa Cuál es la presión que ejerce e
MariettaO [177]

Answer:

de 20 miligramos cada pata dividiendo 80÷4=20

7 0
3 years ago
A 4 kg particle moves at a constant speed of 2.5 m/s around acircle of radius 2 m. What is its angular momentum about the center
blagie [28]

Explanation:

Given that,

Mass of the particle, m = 4 kg

Speed of the particle, v = 2.5 m/s

The radius of the circle, r = 2 m

We need to find the angular momentum about the center of the circle. The formula for the angular momentum is given by :

L=mvr

Substitute all the values,

L=4\times 2.5\times 2\\\\L=20\ kg{\cdot}m^2s

So, the angular momentum of the particle is 20 kg-m² s.

6 0
3 years ago
A box that is sliding across the floor experiences a net force of 10.0 N. If the box has a mass of 1.50 kg, what is the resultin
Maksim231197 [3]

Answer:

a = 6.67 m/s²

Explanation:

F = 10.0 N

m = 1.50 kg

a = ?

F = ma

10.0 = (1.50)a

6.67 = a

7 0
3 years ago
What is the mass of a car traveling at a velocity of 16 meters per second with a kinetic energy of 130, 048 joules
a_sh-v [17]

Explanation:

KE = ½ mv²

130,048 J = ½ m (16 m/s)²

m = 1016 kg

3 0
3 years ago
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