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3 years ago

5

A net force of 12.0 N causes an object to increase its speed from 15.0 m/s to 18.0 m/s during a 2.0 second interval. What is the

mass of the object?

1 answer:

Nat2105 [25]3 years ago

8 0

Answer:

Mass of object = 8kg

Explanation:

<em>Since</em><em> </em><em>a</em><em>=</em><em> </em><em> </em><em>change</em><em> </em><em>in</em><em> </em><em>v</em><em>/</em><em> </em><em>change</em><em> </em><em>in</em><em> </em><em>t</em><em>,</em><em> </em><em>a</em><em>=</em><em> </em><em>(</em><em>1</em><em>8</em><em>-16</em><em>)</em><em>/</em><em>2</em>

<em>=</em><em> </em><em>1.5</em>

<em>And</em><em> </em><em>F</em><em>=</em><em>ma</em><em> </em>
<em>Force</em><em> </em><em>=</em><em>1</em><em>2</em>
<em>a</em><em>=</em><em> </em><em>1.5</em>
<em>m</em><em>=</em><em>?</em>

<em>You</em><em> </em><em>can</em><em> </em><em>now</em><em> </em><em>solve</em><em> </em><em>for</em><em> </em><em>your</em><em> </em><em>unknown</em><em> </em><em>value</em><em> </em><em>of</em><em> </em><em>mass</em><em>.</em>

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2 years ago

a race car accelerates uniformly from 18.5 m/s to 46.1m/s in 2.47 seconds detrrmine the acceleration of the car and distance tra

LenaWriter [7]

Because acceleration is constant, the acceleration of the car at any time is the same as its average acceleration over the duration. So

$a=\dfrac{\Delta v}{\Delta t}=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.47\,\mathrm s}=11.2\,\dfrac{\mathrm m}{\mathrm s^2}$

Now, we have that

${v_f}^2-{v_0}^2=2a\Delta x$

so we end up with a distance traveled of

$\left(46.1\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(11.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)\Delta x$

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3 years ago

An equipoterntial surface that surrounds a + 3.0 pC point charge has a radius of 2.0 cm. What is the potential of this surface?

mestny [16]

Answer:

Electric potential = 0.00054 V

Explanation:

We are given;

Charge; q = 3 pC = 3 × 10^(-12) C

Radius; r = 2 cm = 0.02 m

Formula for the electric potential of this surface will be;

V = kqr

Where;

K is a constant = 9 × 10^(9) N⋅m²/C².

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8 0

3 years ago

How to calculate displacement?

Blizzard [7]

-- Take a straight ruler.

-- Lay it down with the 'zero' mark at the start point.

-- Rotate it around the start point until the end point is also touching the edge of the ruler.

-- From the marks on the ruler, read the straight-line distance from the start point to the end point.

-- Without moving the ruler, observe and write down the DIRECTION from the start point to the end point.

-- The Displacement is the straight-line distance and direction from the start point to the end point.

4 0

3 years ago

Whats the answer???????????

Leviafan [203]

Answer: Less than 4 ohms

Explanation:

We have three resistors with the following resistance:

$R_{1}=4\Omega$

$R_{2}=6\Omega$

$R_{3}=8\Omega$

Now, when the resistors are connected in parallel, the total resistance $R$ is calculated as follows:

$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$

Isolating $R$ :

$R=\frac{R_{1}R_{2}R_{3}}{R_{3}(R_{1}+R_{2})+R_{1}R_{2}}$

Rewriting with th known values:

$R=\frac{(4\Omega)(6\Omega)(8\Omega)}{8\Omega(4\Omega+6\Omega)+(4\Omega)(6\Omega)}$

Finally:

$R=1.84 \Omega$

Hence, the correct option is less than 4 ohms.

4 0

3 years ago