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2 years ago

An object is thrown straight up into the air at 10 m/s. how fast is the object traveling after 2 seconds?

Physics

1 answer:

ziro4ka [17]2 years ago

7 0

First write down all the known variables:

vi = 10m/s

t = 1

a = -9.8m/s^2

vf = ?

Now choose the kinematic equation to relate these variables:

vf = vi + at

vf = (10) + (-9.8)(1)

vf = 0.2 m/s

The speed of the object travelling after a second would be 0.2m/s in reality, however, it is closest to the choice A, which would suffice as the answer for your question.

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pochemuha

Answer:

تكون دائمًا متعامدة مع سرعة الجسم وتكون دائمًا في اتجاه مركز انحناء المسار

Explanation:

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2 years ago

Read 2 more answers

A machine part has the shape of a solid uniform sphere of mass 250 g and a diameter of 4.30 cm. It is spinning about a frictionl

zysi [14]

Answer: $\alpha =9.302\ rad/s^2$

Explanation:

Given

mass of sphere $m=250\ gm$

diameter of sphere $d=4.30\ cm$

radius $r=\frac{4.30}{2}\ cm$

$f=0.0200\ N$

friction will provide resisting torque so

$f\times r=I\times \alpha$

where $I=\text{moment of Inertia}$

$f=\text{friction force}$

$\alpha =\text{angular acceleration}$

$I=\frac{2}{5}mr^2$

$0.02\times r=\frac{2}{5}mr^2\times \alpha$

$\alpha =\frac{5}{2r}\times f$

$\alpha =\frac{5}{2}\times \frac{2}{4.3\times 10^{-2}}\times 0.02$

$\alpha =9.302\ rad/s^2$

(b)time taken to decrease its rotational speed by $21\ rad/s$

$t=\dfrac{\Delta \omega }{\alpha }$

$t=\dfrac{21}{9.302}$

$t=2.25\ s$

6 0

2 years ago

Select the correct answer. The motion of a car on a position-time graph is represented with a horizontal line. What does this in

enyata [817]

If the car's motion appears as a horizontal line on a position-time graph, it shows that as time changes, the car's position doesn't change.

This is just a complicated way to say that the car is not moving. (A)

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2 years ago

A particle's trajectory is described by x = (0.5t^3-2t^2) meters and y = (0.5t^2-2t), where time is in seconds. What is the part

mestny [16]

Differentiate the components of position to get the corresponding components of velocity :

$v_x = \dfrac{\mathrm dx}{\mathrm dt} = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) t^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)t$

$v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}$

At t = 5.0 s, the particle has velocity

$v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}$

$v_y = \left(1\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)-2\dfrac{\rm m}{\rm s} = 3.0\dfrac{\rm m}{\rm s}$

The speed at this time is the magnitude of the velocity :

$\sqrt{{v_x}^2 + {v_y}^2} \approx \boxed{17.8\dfrac{\rm m}{\rm s}}$

The direction of motion at this time is the angle $\theta$ that the velocity vector makes with the positive x-axis, such that

$\tan(\theta) = \dfrac{3.0\frac{\rm m}{\rm s}}{17.5\frac{\rm m}{\rm s}} \implies \theta = \tan^{-1}\left(\dfrac{3.0}{17.5}\right) \approx \boxed{9.73^\circ}$

4 0

2 years ago

During a chemical reaction, Bromine (Br) would be expected to

Andrew [12]

Answer:

During a chemical reaction, Bromine (Br) would be expected to gain 1 valence electron to have a full octet.

Explanation:

In the periodic table the elements are ordered so that those with similar chemical properties are located close to each other.

The elements are arranged in horizontal rows, called periods, which coincide with the last electronic layer of the element. That is, an element with five electronic shells will be in the fifth period.

The columns of the table are called groups. The elements that make up each group coincide in their electronic configuration of valence electrons, that is, they have the same number of electrons in their last.

The elements tend to resemble the closest noble gases in terms of their electronic configuration of the last layer, that is, having eight electrons in the last layer to be stable.

Bromine belongs to group 17 (VII A), which indicates that it has 7 electrons in its last shell. So bromine requires more energy to lose all 7 electrons and generate stability, than it does to gain 1 electron and fill in 8 electrons to be stable. So:

During a chemical reaction, Bromine (Br) would be expected to gain 1 valence electron to have a full octet.

6 0

2 years ago