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2 years ago

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A motion sensor is used to create the graph of a student’s horizontal velocity as a function of time as the student moves toward

and away from the sensor, as shown above. The positive direction is defined as the direction away from the sensor. Which of the following describes the student’s final position xf in relation to the starting position x0 and the student’s average horizontal acceleration ax between 0.0 s and 3.0s? A.)Position xf is farther away from the sensor than x0, and ax is positive. B.)Position xf is farther away from the sensor than x0, and ax is negative. C.)Position xf is closer to the sensor than x0, and ax is positive. D.)Position xf is closer to the sensor than x0, and ax is negative.

1 answer:

Marrrta [24]2 years ago

3 0

Answer:

Position xf is farther away from the sensor than x0, and ax is negative

Explanation:

Area of trapezoidal are=

= $\frac{1}{2} *(1.5+0.75)+\frac{1}{2} (1+0.75)(-0.5)$

= $\frac{1}{2} *(2.25-1.75*0.5)$

=0.6875 m

As the area is positive therefore displacement from xo is positive

ax=(change in velocity)/(Time)

ax= $\frac{-0.5-0}{3} =-\frac{1}{6} ms2$

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Answer:

(a). The width of the river is 90.5 m.

The current speed of the river is 3.96 m.

(b). The shortest time is 15.0 sec and we would end 59.4 m east of our starting point.

Explanation:

Given that,

Constant speed = 6.00 m/s

Time = 20.1 sec

Speed = 9.00 m/s

Time = 11.2 sec

We need to write a equation for to travel due north across the river,

Using equation for north

$v^2-c^2=\dfrac{w^2}{t^2}$

Put the value in the equation

$6.00^2-c^2=\dfrac{w^2}{(20.1)^2}$

$36-c^2=\dfrac{w^2}{404.01}$ ....(I)

We need to write a equation for to travel due south across the river,

Using equation for south

$v^2-c^2=\dfrac{w^2}{t^2}$

Put the value in the equation

$9.00^2-c^2=\dfrac{w^2}{(11.2)^2}$

$81-c^2=\dfrac{w^2}{125.44}$ ....(II)

(a). We need to calculate the wide of the river

Using equation (I) and (II)

$45=\dfrac{w^2}{125.44}-\dfrac{w^2}{404.01}$

$45=w^2(0.00549)$

$w^2=\dfrac{45}{0.00549}$

$w=\sqrt{\dfrac{45}{0.00549}}$

$w=90.5$

We need to calculate the current speed

Using equation (I)

$36-c^2=\dfrac{(90.5)^2}{(20.1)^2}$

$36-c^2=20.27$

$c^2=20.27-36$

$c=\sqrt{15.73}$

$c=3.96\ m/s$

(b). We need to calculate the shortest time

Using formula of time

$t=\dfrac{d}{v}$

$t=\dfrac{90.5}{6}$

$t=15.0\ sec$

We need to calculate the distance

Using formula of distance

$d=vt$

$d=3.96\times15.0$

$d=59.4\ m$

Hence, (a). The width of the river is 90.5 m.

The current speed of the river is 3.96 m.

(b). The shortest time is 15.0 sec and we would end 59.4 m east of our starting point.

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An electron is moving northward in a magnetic field. The magnetic force on the electron is toward the northeast. What is the dir

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a bobsled has a momentum of 1500 kg*m/s to the south. Air resistance reduces its momentum to 750 kg*m/s to the south. What impul

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The impulse is exactly the change in momentum,
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The impulse imparted to the bobsled by the air
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2 years ago

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Lightning bolts can carry currents up to approximately 20 kA. We can model such a current as the equivalent of a very long, stra

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Given Information:

Current due to Lightning bolt = I₁ = 20 kA

Current of household = I₂ = 10 A

Distance = r₁ = 4.9 m

Distance = r₂ = 4.9 cm = 0.049 m

Required Information:

a) Magnetic field due Lightning bolt = B₁ = ?

b) Magnetic field due to household current = B₂ = ?

Answer:

a) Magnetic field due Lightning bolt = B₁ = 8.16×10⁻⁴ T

b) Magnetic field due to household current = B₂ = 4.08×10⁻⁵ T

B₁ = 20B₂

Explanation:

The magnetic field produced in a long straight wire carrying a current (I) at distance r is given by

B = μ₀I/2πr

Where μ₀ is the permeability of free space and its value is 4π×10⁻⁷

a) The magnetic field produced due to the lightning bolt current is

B₁ = μ₀I₁/2πr₁

B₁ = (4π×10⁻⁷*20,000)/2π*4.9

B₁ = 0.000816

B₁ = 8.16×10⁻⁴ T

Therefore, the strength of magnetic field due to the lightning bolt current is 8.16×10⁻⁴ T

b) The magnetic field produced due to the household current is

B₂ = μ₀I₂/2πr₂

B₂ = (4π×10⁻⁷*10)/2π*0.049

B₂ = 0.00004081

B₂ = 4.08×10⁻⁵ T

Therefore, the strength of magnetic field due to the household current is 4.08×10⁻⁵ T

The ratio of magnetic field produced by the lightning bolt current to the magnetic field produced by the household current is

B₁/B₂ = 8.16×10⁻⁴/4.08×10⁻⁵

B₁/B₂ = 20

B₁ = 20B₂

Which means that the magnetic field produced by the lightning bolt current is 20 time greater than the magnetic field produced by the household current.

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