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4 years ago

9

sophia exerts a steady 40n horizontal force on a 8kg resting on a lab bench. the box slides against a horizontal friction of 24n

. show that the box accelerates at 2m/s 2

2 answers:

astraxan [27]4 years ago

5 0

Force = 40 - 24 = 16N

Force = mass / acceleration

Acceleration = Force / mass

Acceleration = 16 / 8

Acceleration = 2 m/s^2

Fittoniya [83]4 years ago

3 0

Force = 40 - 24
= 16N.

Thus, Force = 16N.

By deducing Newton's second law of motion,

F = MA

16 = 8A

Thus, A = 16/8 = 2 m/s²

Thus, the object accelerates at 2m/s²

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nekit [7.7K]

Answer:

Explanation:

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4 0

3 years ago

A constant voltage of 12.00 V has been observed over a certain time interval across a 1.20 H inductor. The current through the i

kiruha [24]

Answer:

The time is 0.5 sec.

Explanation:

Given that,

Voltage V= 12.00 V

Inductance L= 1.20 H

Current = 3.00 A

Increases rate = 8.00 A

We need to calculate change in current

$\Delta A = 8.00-3.00= 5.00\ A$

We need to calculate the time interval

Using formula of inductor

$V=L\dfrac{\Delta A}{\Delta t}$

$\Delta t =\dfrac{L\Delta A}{V}$

Where, $\Delta A$ = change in current

V = voltage

L = inductance

Put the value into the formula

$\Delta t=\dfrac{1.20\times5.00}{12.00}$

$\Delta t=0.5\ sec$

Hence, The time is 0.5 sec.

5 0

3 years ago

PLS ANSWERR THIS QUESTION FOR ME !!

MA_775_DIABLO [31]

Explanation:

20 joule is your answer

Answer:

here

mass m =100kg

distance d=50m

acceleration due to gravity a =10m/s²

work =force×displacement

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3 0

3 years ago

Read 2 more answers

You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration

jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time $t$ the ball is in air, we can use the following equation:

$y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}$

Where:

$y=0m$ is the final height of the ball

$y_{o}=40 m$ is the initial height of the ball

$V_{o}=10 m/s$ is the initial velocity of the ball

$t$ is the time the ball is in air

$g=9.8 m/s^{2}$ is the acceleration due to gravity

$\theta=30\°$

Then:

$0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}$

$0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}$

Multiplying both sides of the equation by -1 and rearranging:

$4.9 m/s^{2}t^{2}-5m/s t-40 m=0$

At this point we have a quadratic equation of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Where:

$a=4.9$

$b=-5$

$c=-40$

Substituting the known values:

$t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}$

Solving the equation and choosing the positive result we have:

$t=3.41 s$ This is the time the ball is in air

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4 years ago

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katovenus [111]

It’s around the g force so it’s gonna be around 54 km/h

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2 years ago