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3 years ago

11

At a wedding reception, you notice a child who looks like their mass is about 25 kg running across the dance floor then sliding

on their knees until they stop. If the coefficient of kinetic friction between the child's pants and the floor is 0.15, what is the friction force acting on them as they slide?

2 answers:

Step2247 [10]3 years ago

7 0

Answer: i think the answer is 25.0 and divided by 0.15

Explanation:

Vitek1552 [10]3 years ago

4 0

Mass of the object m = 25 kg

Coefficient of friction Uk = 0.15

Frictional force Ff = Uk x F => Ff = Uk x m x g

Ff = 0.15 x 25 x 9.8

Frictional Force Ff = 36.75 N

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A 5.50-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical

iris [78.8K]

Answer:

17.71N/m

Explanation:

The period of the spring is expressed according to the expression;

$T = 2 \pi \sqrt{\frac{m}{k} } \\$

m is the mass of the object

k is the force constant

Given

m = 5.50kg

T = 3.50s

Substitute into the formula;

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3 years ago

A tank contains 350 liters of fluid in which 10 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu

aleksandr82 [10.1K]

Answer:

$A(t) = -340e^{-t/70} + 350$

Explanation:

Since fluid is pumping in and out at the same rate (5L/min), the total fluid volume in the tank stays constant at 350L. Only the amount of salt and its concentration changed overtime.

Let A(t) be the amount of salt (g) at time t and C(t) (g/L) be the concentration at time t

A(0) = 10 g

Brine with concentration of 1g/L is pouring in at the rate of 5L/min so the salt income rate is 5 g/min

The well-mixed solution is pouring out at the rate of 5L/min at concentration C(t) so the salt outcome rate is 5C g/min

But the concentration is total amount of salt over 350L constant volume

C = A / 350

Therefore our rate of change for salt A' is

A' = 5 - 5A/350 = 5 - A/70

This is a first-order linear ordinary differential equation and it has the form of y' = a + by. The solution of this is

$y = ce^{bt} + \frac{a}{b}$

So $A = ce^{\frac{-t}{70}} + \frac{5}{1/70} = ce^{-t/70} + 350$

with A(0) = 10

c + 350 = 10

c = 10 - 350 = -340

$A(t) = -340e^{-t/70} + 350$

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3 years ago

Why is it that an object can accelerate while

posledela

For the same reason that you can skate around a curve at constant speed but not with constant velocity.

The DIRECTION you're going is part of your velocity, but it's not part of your speed.

If the DIRECTION changes, that's a change of velocity.

The object doesn't have to change speed to have a different velocity. A change of direction is enough to do it.

And any change of velocity is called acceleration.

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2 years ago

a vector has an x-component of 19.5m and a y-component of 28.4m. Find the magnitude and direction of the vector

Delicious77 [7]

Answer:

magnitude=34.45 m

direction= $55.52\°$

Explanation:

Assuming the initial point P1 of this vector is at the origin:

P1=(X1,Y1)=(0,0)

And knowing the other point is P2=(X2,Y2)=(19.5,28.4)

We can find the magnitude and direction of this vector, taking into account a vector has a initial and a final point, with an x-component and a y-component.

For the magnitude we will use the formula to calculate the distance $d$ between two points:

$d=\sqrt{{(Y2-Y1)}^{2} +{(X2-X1)}^{2}}$ (1)

$d=\sqrt{{(28.4 m - 0 m)}^{2} +{(19.5 m - 0m)}^{2}}$ (2)

$d=\sqrt{1186.81 m^{2}}$ (3)

$d=34.45 m$ (4) This is the magnitude of the vector

For the direction, which is the measure of the angle the vector makes with a horizontal line, we will use the following formula:

$tan \theta=\frac{Y2-Y1}{X2-X1}$ (5)

$tan \theta=\frac{24.8 m - 0m}{19.5 m - 0m}$ (6)

$tan \theta=\frac{24.8}{19.5}$ (7)

Finding $\theta$ :

$\theta= tan^{-1}(\frac{24.8}{19.5})$ (8)

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3 years ago

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Answer:

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Part b)

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Explanation:

Part a)

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$I_1 = \frac{2}{5}m_1r_1^2$

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$I_1 = 6.14 \times 10^{-3} kg m^2$

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$I_2 = \frac{2}{3} m_2r_2^2$

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$I_2 = 11.3 \times 10^{-3} kg m^2$

now total inertia of the ball is given as

$I = I_1 + I_2$

$I = 17.4 \times 10^{-3} kg m^2$

Part b)

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

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3 years ago