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3 years ago

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At what speed will a car round a 52-m-radius curve, banked at a 45???? angle, if no friction is required between the road and ti

res to prevent the car from slipping? (g = 9.8 m/s2)a. 27 m/s
b. 17 m/s
c. 23 m/s
d. 35 m/s

1 answer:

Serhud [2]3 years ago

4 0

Answer:

c. 23 m/s

Explanation:

$\theta$ = Angle of banking = 45°

r = Radius of turn = 52 m

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity of car

As the car is tilted we have the relation

$tan\theta=\dfrac{v^2}{rg}\\\Rightarrow v=\sqrt{tan\theta rg}\\\Rightarrow v=\sqrt{tan45\times 52\times 9.8}\\\Rightarrow v=22.57432\ m/s\approx 23\ m/s$

The velocity of the car is 23 m/s if the car is not slipping

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A 1990 kg car moving at 20.0 m/s collides and locks together with a 1540 kg car at rest at a stop sign. Show that momentum is co

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Answer:

The momentum before collision and momentum after collision is equal in a frame of reference moving at 10.0 m/s in the direction of the moving car.

Explanation:

$m_1$ = Mass of moving car = 1990 kg

$u_1$ = Velocity of moving car in stationary frame = 20 m/s

$m_2$ = Mass of stationary car = 1540 kg

$u_2$ = Velocity of stationary car in stationary frame = 0 m/s

v = Combined velocity in stationary frame

Momentum conservation for stationary frame

$m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{1990\times 20+0}{1990+1540}\\\Rightarrow v=\dfrac{3980}{353}\ \text{m/s}$

In frame moving at 10 m/s the velocities change in the following ways

$u_1=20-10\\\Rightarrow u_1=10\ \text{m/s}$

$u_2=0-10\\\Rightarrow u_1=-10\ \text{m/s}$

$v=\dfrac{3980}{353}-10\\\Rightarrow v=\dfrac{450}{353}\ \text{m/s}$

Momentum before collision

$m_1u_1+m_2u_2=1990\times 10+1540\times -10\\\Rightarrow m_1u_1+m_2u_2=4500\ \text{kg m/s}$

Momentum after collision

$(m_1+m_2)v=(1990+1540)\times \dfrac{450}{353}\\\Rightarrow (m_1+m_2)v=4500\ \text{kg m/s}$

The momentum before collision and momentum after collision is equal. So, the momentum is conserved in a reference frame moving at 10.0 m/s in the direction of the moving car.

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3 years ago

Mary spins a ball in a horizontal circle on a massless string. Karen has a similar ball but

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Answer: d

Explanation:

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3 years ago

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Answer:

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A period of a pendulum is how long it takes to swing back to it's original position, and this time (2 seconds) is given in the question as well

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An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 49.1 N,

balu736 [363]

When the spring is stretched by 15.2 cm = 0.152 m, the spring exerts a restorative force with magnitude (due to Hooke's law)

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where $k$ is the spring constant. Solve for $k$ .

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The amount of work required to stretch or compress a spring by $x\,\mathrm m$ from equilibrium length is

$W = \dfrac12 kx^2$

Then the work needed to stretch the spring by 15.2 cm is

$W_1 = \dfrac12 \left(343\dfrac{\rm N}{\rm m}\right) (0.152\,\mathrm m)^2 \approx 3.73\,\mathrm J$

and by 15.2 + 13.7 = 28.9 cm is

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so the work needed to stretch from 15.2 cm to 28.9 cm from equilibrium is

$\Delta W = W_2 - W_1 \approx \boxed{9.76\,\mathrm J}$

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2 years ago

Two isolated copper plates, each of area 0.40 m2, carry opposite charges of magnitude 7.08 × 10-10 C. They are placed opposite e

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Answer:

The potential difference between the plates is 8 V.

Explanation:

Given that,

Area of plates = 0.40 m²

Charge $q=7.08\times10^{-10}\ C$

Distance = 4.0 cm

We need to calculate the electric field

Using for formula of electric field

$E=\dfrac{2q}{2\epsilon_{0}A}$

Where, q = charge

A = area

Put the value into the formula

$E=\dfrac{7.08\times10^{-10}}{8.85\times10^{-12}\times0.40}$

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Using formula of potential difference

$V=E\times d$

Where, E = electric field

d = distance

Put the value into the formula

$V=200\times0.04$

$V=8\ V$

Hence, The potential difference between the plates is 8 V.

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4 years ago