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3 years ago

5

A child in a swing oscillates with a certain frequency of oscillation (the child is sitting still. another child sits next to th

e first child. how does the swing’s frequency of oscillation change when the second child sits next to the first child?

1 answer:

Tresset [83]3 years ago

8 0

,f-u.c-k .y-e,a --m-e,n

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A Dodge Stealth is driving at 70 mph on a highway. It passes a BMW going the same direction. The BMW is moving 7 mph backward re

Tasya [4]

A Dodge Stealth is driving at 70 mph on a highway. It passes a BMW going the same direction. The BMW is moving 7 mph backward relative the Dodge. 63 mph is the velocity of the BMW

Answer: Option A

<u>Explanation:</u>

Given is the speed of the dodge stealth as 70 mph passing by the BMW moving with 7 mph relative to Dodge. Thereby, to calculate the velocity of BMW, relative velocity concept has to be employed.

As we know that the relative velocity tells us about relative velocity of mobile reference system as the difference or sum of the initial reference system.

Therefore,

velocity of the BMW = $v_{b}$

Velocity of the dodge = $v_{d}$

Velocity of the BMW with respect to dodge = $V_{b d}$

As per the formula,

$v_{b d}=v_{b}-v_{d}$

$v_{b} = v_{d} + v_{b d} = 70 + (-7) = 70 - 7 = 63 \mathrm{mph}$

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3 years ago

Friends who are your age and share the same interests are called

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The answer is C, a peer group.

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4 years ago

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Adults with Down syndrome can often find work because they have received _____.

IrinaK [193]

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4 years ago

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A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

Softa [21]

Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

$C=\frac{\epsilon_0 S}{d}$

Capacitance of capacitor after moving plates

$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

$V=\frac{Q}{C}$

Potential difference between plates after moving

$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

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4 years ago

A firecracker in a coconut blows the coconut into three pieces. Twopieces of equal mass fly off south and west, perpendicular to

loris [4]

Answer:

V = 13m/s

North-East (i.e. 45 degree from North)

Explanation:

This question deals with the idea of momentum.

Since the directions of a compass are fixed, we can take south as horizontal and west as vertical. Since both pieces of coconut are of equal mass, then the resultant of the two pieces would be in between South and West and the third piece would be opposite this direction and be in the North-East (i.e. 45 degree from North) direction

To find the speed of the third piece, we first find the speed on the two pieces of the coconut in terms of $V_{x}$ (horizontal component of velocity) and $V_{y}$ (vertical component of velocity)

We have

$V_{y}$ = Velocity in South direction = 18m/s

$m_{y}$ = Mass of piece in South direction = m

$V_{x}$ = Velocity in West direction = 18m/s

$m_{x}$ = Mass of piece in West direction = m

$V_{ry}$ = Velocity of Third Piece in North direction = unknown

$V_{rx}$ = Velocity of Third Piece in East direction = unknown

$m$ = Mass of third piece = 2m

$mV_{ry} = m_{y} V_{y} \\ 2mV_{ry} = m18\\ V_{ry} = 9$

$mV_{rx} = m_{x} V_{x} \\ 2mV_{rx} = m18\\ V_{rx} = 9$

Since we now know the horizontal and vertical component of the velocity of the third piece of coconut we find the resultant velocity. Since we know the direction is North-East, we can imagine a right angled triangle with base of 9 and height 9 and the hypotenuse equal to the resultant velocity .

We can simply apply the Pythagoras theorem and find the hypotenuse

$Hypotenuse^{2} = Height^{2} + Base^{2} \\ V_{3}^2 = 9^{2} + 9^{2} \\ V = 12.728 m/s$

V = 13m/s

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4 years ago