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3 years ago

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Coasting due west on your bicycle at 8.4 m/s, you encounter a sandy patch of road 7.2 m across. When you leave the sandy patch,

your speed has been reduced to 6.4 m/s. What is the bicycles acceleration in the sandy patch? assume that the acceleration is constant and that the direction of travel is the positive direction.

1 answer:

Nady [450]3 years ago

5 0

Answer:

-2.1 m/s²

Explanation:

Given:

v₀ = 8.4 m/s

v = 6.4 m/s

Δx = 7.2 m

Find: a

v² = v₀² + 2aΔx

(6.4 m/s)² = (8.4 m/s²) + 2a (7.2 m)

a = -2.1 m/s²

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A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$

Put the value into the formula

$r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}$

$r=0.0876\ m$

We need to calculate the magnitude of the charge q₃

Using formula of net force

$F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})$

Put the value into the formula

$14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})$

$(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}$

$\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}$

$q_{3}=0.0210909\times(0.0438)^2$

$q_{3}=40.46\times10^{-6}\ C$

$q_{3}=40.46\ \mu C$

Hence, The value of charge q₃ is 40.46 μC.

5 0

3 years ago

Given the distance between the crest of one wave and the crest of the next wave, you can determine the?

Sonja [21]

Answer: wavelength !!
hope this helped :)

3 0

3 years ago

What type of plate boundary do most volcanoes occur along

sertanlavr [38]

Answer:

Subducting convergent boundary

Explanation:

Generally, volcanoes occurs in both divergent and convergent boundaries. But the convergent boundary it occurs is usually associated with subduction.

Divergent boundary, plates move away from each other creating a new crust in the process. The diverging plates creates the space for magma to be squeezed through cracks and fissures. The magma's erupt to form volcanoes. In the Atlantic ocean the spreading of the plates causes an upwelling of magma through the crest of the Atlantic ridges. New oceanic crust are formed through this process. Sometimes the magma eruption forms volcanoes that are higher than the sea level.

Convergent boundary , plates collides with each other . But in the case of volcanoes existence , the collision should be between a denser plate(oceanic plates) and a less dense plates(continental plates) so that subduction can take place. The subducted plates (oceanic plates) creates trenches and get expose to high temperature and pressure as it sinks toward the mantle. The upper mantle rocks melts and migrate to the earth surface forming volcanoes . Over 75% of the volcanoes occur along the pacific basin where convergent boundary is dominant. Pacific ring of fire has one of the most number of volcanoes.

6 0

3 years ago

An electron is accelerated within a particle accelerator using a 100 MV electric potential. The 100 MeV electron moves along an

Delicious77 [7]

Answer:

The length of the tube is 3.92 m.

Explanation:

Given that,

Electric potential = 100 MV

Length = 4 m

Energy = 100 MeV

We need to calculate the value of $\gamma$

Using formula of relativistic energy

$E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)$

Put the value into the formula

$1.6\times10^{-15}= 9.1\times`10^{-31}\times9\times10^{16}(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)$

$(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)=\dfrac{1.6\times10^{-15}}{9.1\times10^{-31}\times9\times10^{16}}$

Here, $\gamma-1=(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)$

$\gamma-1=0.01953$

$\gamma=0.01953+1$

$\gamma=1.01953$

We need to calculate the length

Using formula of length

$L'=\dfrac{L}{\gamma}$

Put the value into the formula

$L'=\dfrac{4}{1.01953}$

$L'=3.92\ m$

Hence, The length of the tube is 3.92 m.

8 0

3 years ago

A -turn rectangular coil with length and width is in a region with its axis initially aligned to a horizontally directed uniform

madam [21]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The maximum emf is $\epsilon_{max}= 26.8 V$

The emf induced at t = 1.00 s is $\epsilon = 24.1V$

The maximum rate of change of magnetic flux is $\frac{d \o}{dt}|_{max} =26.8V$

Explanation:

From the question we are told that

The number of turns is N = 44 turns

The length of the coil is $l = 15.0 cm = \frac{15}{100} = 0.15m$

The width of the coil is $w = 8.50 cm =\frac{8.50}{100} =0.085 m$

The magnetic field is $B = 745 \ mT$

The angular speed is $w = 64.0 rad/s$

Generally the induced emf is mathematically represented as

$\epsilon = \epsilon_{max} sin (wt)$

Where $\epsilon_{max}$ is the maximum induced emf and this is mathematically represented as

$\epsilon_{max} = N\ B\ A\ w$

Where $\o$ is the magnetic flux

N is the number of turns

A is the area of the coil which is mathematically evaluated as

$A = l *w$

Substituting values

$A = 0.15 * 0.085$

$= 0.01275m^2$

substituting values into the equation for maximum induced emf

$\epsilon_{max} = 44* 745 *10^{-3} * 0.01275 * 64.0$

$\epsilon_{max}= 26.8 V$

given that the time t = 1.0sec

substituting values into the equation for induced emf $\epsilon = \epsilon_{max} sin (wt)$

$\epsilon = 26.8 sin (64 * 1)$

$\epsilon = 24.1V$

The maximum induced emf can also be represented mathematically as

$\epsilon_{max} = \frac{d \o}{dt}|_{max}$

Where $\o$ is the magnetic flux and $\frac{d \o}{dt}|_{max}$ is the maximum rate at which magnetic flux changes the value of the maximum rate of change of magnetic flux is

$\frac{d \o}{dt}|_{max} =26.8V$

8 0

3 years ago