Ask question

Ask question

All categories

3 years ago

15

A speed skater moving across frictionless ice at 9.2 m/s hits a 5.0 m wide patch of rough ice. She slows steadily, then continue

s on at 5.8 m/s. What is her acceleration on the rough ice?

1 answer:

enot [183]3 years ago

7 0

Answer:

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

Explanation:

The distance travelled on the rough ice is equal to the width of the rough ice.

distance d = 5.0 m

Initial speed u = 9.2 m/s

Final speed v = 5.8 m/s

The time taken to move through the rough ice can be calculated using the equation of motion;

d = 0.5(u+v)t

time t = 2d/(u+v)

Substituting the given values;

t = 2(5)/(9.2+5.8)

t = 2/3 = 0.66667 second

The acceleration is the change in velocity per unit time;

acceleration a = ∆v/t

a = (v-u)/t

Substituting the values;

a = (5.8-9.2)/0.66667

a = -5.099974500127

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

You might be interested in

Which of the following formulas is used to solve for current flowing through an electrical circuit? A. current = voltage ÷ resis

Alex787 [66]

The answer to your question is a

8 0

3 years ago

Read 2 more answers

How many quantum numbers are used to describe the energy state of an electron in an atom

matrenka [14]

We have Four (4). quantum number used in description of the energy state of an electron.

6 0

2 years ago

What happens when multiple forces act upon an object?

Bogdan [553]

The object will move if the forces are unbalanced.

Newtons second tells you that when a net force (the unbalanced force) is applied to and object it will produce an acceleration (movement) in direct proportion to the force and in inverse proportion to the mass of the object.

7 0

2 years ago

Read 2 more answers

2. A particular planet has a moment of inertia of 9.74 × 1037 kg•m2 and a mass of 5.98 × 1024 kg. Based on these values, what is

mash [69]

Answer:

$6.38\cdot 10^6 m$

Explanation:

The planet can be thought as a solid sphere rotating around its axis. The moment of inertia of a solid sphere rotating arount the axis is

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

where

M is the mass

R is the radius

For the planet in the problem, we have

$M=5.98\cdot 10^{24} kg$

$I=9.74\cdot 10^{37} kg\cdot m^2$

Solving the equation for R, we find the radius of the planet:

$R=\sqrt{\frac{5I}{2M}}=\sqrt{\frac{5(9.74\cdot 10^{37}}{2(5.98\cdot 10^{24}}}=6.38\cdot 10^6 m$

3 0

3 years ago

Two charges, X and Y, are placed along the x-axis. Charge X is +18 nC and is placed at x = 0. Charge Y is placed at a location o

Helen [10]

Answer:

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m.

Explanation:

The Coulomb force between two charges, $Q_1$ and $Q_2$ , separated by a distance, $d$ , is given

$F = k\dfrac{Q_1Q_2}{r^2}$

<em>k</em> is a constant.

For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.

It is to be placed along the <em>x</em>-axis. Hence, it is on the same line as charges X and Y.

Let the charge on Z be <em>Q</em>. It is positive.

Let the distance from charge X be <em>x m.</em> Then the distance from charge Y will be (0.60 - <em>x</em>) m.

Force due to charge X

$F_X = k\dfrac{18Q}{x^2}$

Force due to charge Y

$F_Y = k\dfrac{-27Q}{(0.60-x)^2}$

Since both forces are equal and opposite,

$F_X = -F_Y$

$k\dfrac{18Q}{x^2} = -k\dfrac{-27Q}{(0.60-x)^2}$

$\dfrac{2}{x^2} = \dfrac{3}{(0.60-x)^2}$

$2(0.60-x)^2 = 3x^2$

$2(0.36-1.20x+x^2) = 3x^2$

$0.72-2.40x+2x^2 = 3x^2$

$x^2+2.40x-0.72 = 0$

Applying the quadratic formula,

$x = \dfrac{-2.40\pm\sqrt{2.40^2 - (4)(1)(-0.72)}}{2} = \dfrac{-2.40\pm\sqrt{8.64}}{2}$

$x = -2.7$ or $x = 0.27$

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m

3 0

3 years ago