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Luba_88 [7]
3 years ago
5

How do positive and negative acceleration differ?

Physics
2 answers:
horsena [70]3 years ago
6 0

Answer:

acceleration

Explanation:

acceleration determines whether you will be adding to or subtracting from the velocity. while negative acceleration means you will subtract from the current value of the velocity and a positive acceleration means you will add

patriot [66]3 years ago
6 0

ANSWER:

✔ Positive acceleration represents an object speeding up; negative acceleration represents an object slowing down.

NOTE: Negative acceleration is also called as Retar-dation.

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A train travelling at 20 m/s has 2 000 000 J of kinetic energy. What is the mass of<br> the train?
Svetlanka [38]

Answer:

<h2>10,000 kg</h2>

Explanation:

The mass of the train can be found by using the formula

m =  \frac{2k}{ {v}^{2} }  \\

k is the kinetic energy

v is the velocity

From the question we have

m \frac{2(2000000)}{ {20}^{2} }  =  \frac{4000000}{400}  =  \frac{40000}{4}  \\

We have the final answer as

<h3>10,000 kg</h3>

Hope this helps you

3 0
3 years ago
Read 2 more answers
What impulse occurs when an average force of 7.0 N is exerted on a cart for 2.5 s ?
makkiz [27]

Answer:1.7\times 10

Explanation:

Given

Force F=7 N

time interval t=2.5 s

Impulse is given by =Force \times time\ interval\ for\ applied\ force

Impulse=7\times 2.5=17.5

For two significant Figure

Impulse=1.7\times 10

                       

7 0
3 years ago
Arm ab has a constant angular velocity of 16 rad/s counterclockwise. At the instant when theta = 60
geniusboy [140]

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second.

<h3>How to determine the angular velocity of a collar</h3>

In this question we have a system formed by three elements, the element AB experiments a <em>pure</em> rotation at <em>constant</em> velocity, the element BD has a <em>general plane</em> motion, which is a combination of rotation and traslation, and the ruff experiments a <em>pure</em> translation.

To determine the <em>linear</em> acceleration of the collar (a_{D}), in inches per square second, we need to determine first all <em>linear</em> and <em>angular</em> velocities (v_{D}, \omega_{BD}), in inches per second and radians per second, respectively, and later all <em>linear</em> and <em>angular</em> accelerations (a_{D}, \alpha_{BD}), the latter in radians per square second.

By definitions of <em>relative</em> velocity and <em>relative</em> acceleration we build the following two systems of <em>linear</em> equations:

<h3>Velocities</h3>

v_{D} + \omega_{BD}\cdot r_{BD}\cdot \sin \gamma = -\omega_{AB}\cdot r_{AB}\cdot \sin \theta   (1)

\omega_{BD}\cdot r_{BD}\cdot \cos \gamma = -\omega_{AB}\cdot r_{AB}\cdot \cos \theta   (2)

<h3>Accelerations</h3>

a_{D}+\alpha_{BD}\cdot \sin \gamma = -\omega_{AB}^{2}\cdot r_{AB}\cdot \cos \theta -\alpha_{AB}\cdot r_{AB}\cdot \sin \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \cos \gamma   (3)

-\alpha_{BD}\cdot r_{BD}\cdot \cos \gamma = - \omega_{AB}^{2}\cdot r_{AB}\cdot \sin \theta + \alpha_{AB}\cdot r_{AB}\cdot \cos \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \sin \gamma   (4)

If we know that \theta = 60^{\circ}, \gamma = 19.889^{\circ}, r_{BD} = 10\,in, \omega_{AB} = 16\,\frac{rad}{s}, r_{AB} = 3\,in and \alpha_{AB} = 0\,\frac{rad}{s^{2}}, then the solution of the systems of linear equations are, respectively:

<h3>Velocities</h3>

v_{D}+3.402\cdot \omega_{BD} = -41.569   (1)

9.404\cdot \omega_{BD} = -24   (2)

v_{D} = -32.887\,\frac{in}{s}, \omega_{BD} = -2.552\,\frac{rad}{s}

<h3>Accelerations</h3>

a_{D}+3.402\cdot \alpha_{BD} = -445.242   (3)

-9.404\cdot \alpha_{BD} = -687.264   (4)

a_{D} = -693.867\,\frac{in}{s^{2}}, \alpha_{BD} = 73.082\,\frac{rad}{s^{2}}

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second. \blacksquare

<h3>Remark</h3>

The statement is incomplete and figure is missing, complete form is introduced below:

<em>Arm AB has a constant angular velocity of 16 radians per second counterclockwise. At the instant when θ = 60°, determine the acceleration of collar D.</em>

To learn more on kinematics, we kindly invite to check this verified question: brainly.com/question/27126557

5 0
2 years ago
A 3.9 kg block is pushed along a horizontal floor by a force of magnitude 30 N at a downward angle θ = 40°. The coefficient of k
Luba_88 [7]

Answer:

The frictional force  F_{fri} = 6.446 N

The acceleration of the block a = 6.04 \frac{m}{s^{2} }

Explanation:

Mass of the block = 3.9 kg

\theta = 40°

\mu = 0.22

(a). The frictional force is given by

F_{fri} = \mu R_{N}

R_{N} = mg \cos \theta

R_{N} = 3.9 × 9.81 × \cos 40

R_{N} = 29.3 N

Therefore the frictional force

F_{fri} = 0.22 × 29.3

F_{fri} = 6.446 N

(b). Block acceleration is given by

F_{net} = F - F_{fri}

F = 30 N

F_{fri} = 6.446 N

F_{net} = 30 - 6.446

F_{net} = 23.554 N

The net force acting on the block is given by

F_{net}  = ma

23.554 = 3.9 × a

a = 6.04 \frac{m}{s^{2} }

This is the acceleration of the block.

8 0
3 years ago
An LC circuit consists of a 3.4-µF capacitor and a coil with a self-inductance 0.080 H and no appreciable resistance. At t = 0 t
alexira [117]

Answer

given,

capacitance = C = 3.4-µF

inductance = L = 0.08 H

frequency is expressed as

f = \dfrac{1}{2\pi\sqrt{LC}}

time period

T = \dfrac{1}{f}=2\pi\sqrt{LC}

after time T/4 current reach maximum

 t = \dfrac{T}{4}

 t = \dfrac{2\pi\sqrt{LC}}{4}

 t = \dfrac{2\pi\sqrt{0.08 \times 3.4 \times 10^{-6}}}{4}

        t = 8.2 x 10⁻⁴ s

        t = 0.82 ms

b) using law of conservation

  \dfrac{1}{2}CV^2=\dfrac{1}{2}LI^2

  I^2 = \dfrac{CV^2}{L}

  I^2 = \dfrac{C}{L}\dfrac{Q^2}{C^2}

  I =\sqrt{\dfrac{Q^2}{CL}}

  I =\sqrt{\dfrac{(5.4 \times 10^{-6})^2}{0.08 \times 3.4 \times 10^{-6}}}

       I = 0.010 A

       I = 10 mA

4 0
3 years ago
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