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3 years ago

6

A force in the negative direction of an x-axis is applied for 17ms to a 0.18 kg ball initially moving at 16.0 m/s in the positiv

e direction of the axis. the force varies in magnitude and results in an impulse of magnitude 4.28 ns. what is the velocity of the ball just after the force is applied? indicate the direction of the velocity by its sign.

1 answer:

Marizza181 [45]3 years ago

5 0

Impulse = Ft=mΔv => Δv = Ft/m = 4.28/0.18 = 23.78 m/s

But,
Δv = v1-v2, where v1 = initial velocity = 16 m/s, v2 = final velocity

Therefore,
v1 - v2 = 23.78 => v2 = v1 - 23.78 => v2 = 16 - 23.78 = -7.78 m/s

The velocity of ball after the force is 7.78 m/s in the direction of the force.

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3 years ago

A car starts from rest and is moving at 60.0 m/s after 7.50 s. What is the car's average acceleration?

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Answer:

${ \boxed{ \bold{ \sf{Acceleration \: ( \: a) = 8 \: m/ {s \: }^{2} }}}}$

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♨ Question :

A car starts from rest and is moving at 60.0 m/s after 7.50 s. What is the car's average acceleration ?

♨ $\underbrace{ \sf{Required \: Answer : }}$

☄ Given :

Initial velocity ( u ) = 0
Final velocity ( v ) = 60.0 m/s
Time ( t ) = 7.50 s

☄ To find :

Acceleration ( a )

✒ We know ,

$\boxed{ \underline{ \bold{ \sf{Acceleration \: ( \: a) = \frac{Final velocity ( v ) - Initial velocity ( u)}{t} }}}}$

Substitute the values and solve for a.

➛ $\sf{a = \frac{60.0 - 0}{7.50}}$

➛ $\sf{a = \frac{60.0}{7.50}}$

➛ $\boxed{ \boxed{ \sf{a = 8 \: m/ {s \: }^{2} }}}$

---------------------------------------------------------------

✑ Additional Info :

When a certain object comes in motion from rest , in the case , initial velocity ( u ) = 0
When a moving object comes in rest , in the case , final velocity ( v ) = 0
If the object is moving with uniform velocity , in the case , u = v.
If any object is thrown vertically upwards in the case , a = -g
When an object is falling from certain height , in the case , final velocity at maximum height ( v ) = 0.

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2 years ago

Suppose that the inverse market demand for silicone replacement tips for Sony earbud headphones is p = pN - 0.1Q, where p is t

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Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The effect of a change in the price of a new pair of headphones on the equilibrium price of replacement tips ( dp/dpN) is

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b

The value of Q and p at equilibruim is

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The consumer surplus is $C= 3125$

The producer surplus is $P = 375$

Explanation:

From the question we are told that

The inverse market demand is $p_d = p_N -0.1Q$

The inverse supply function is $p_s = 2+ 0.012Q$

a

The effect of change in the price is mathematically given as

$\frac{\delta p}{\delta p_N}$

Now differntiating the inverse market demand function with respect to $p_N$

We get that

$\frac{\delta p}{\delta p_N} =1$

b

We are told that $p_N =$ $30

Therefore the inverse market demand becomes

$p_d = 30 -0.1Q$

At equilibrium

$p_d = p_s$

So we have

$30 -0.1Q_e = 2+ 0.012Q_e$

Where $Q_e$ is the quantity at equilibrium

$28 = 0.112Q_e$

$Q_e = \frac{28}{0.112}$

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Substituting the value of Q into the equation for the inverse market demand function

$p_d = 30 - 0.1 (250 )$

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Looking at the equation for $p_d \ and \ p_s$ we see that

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And for Q = 250

$p_d = 5$

$p_s = 5$

Hence the consumer surplus is mathematically evaluated as

$C = \frac{1}{2} * Q_e * (30 -5)$

Substituting value

$C = \frac{1}{2} * 250 (30-5)$

$C= 3125$

And

The producer surplus is mathematically evaluated as

$P = \frac{1}{2} *250 * (5-2)$

$P = 375$

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3 years ago