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Rainbow [258]
3 years ago
12

Suppose the Earth's magnetic field at the equator has magnitude 0.50×10−4T and a northerly direction at all points. Part A Estim

ate the speed a singly ionized uranium ion (m=238u,we) would need to circle the Earth 5.0 km above the equator. Express your answer using two significant figures. vv = nothing m/s
Physics
1 answer:
notsponge [240]3 years ago
4 0

Answer:

v = 1.3 10⁸  m / s

Explanation:

The magnetic force is given by

       F = q v x B

Where blacks indicate vectors

The magnetic field has a direction to the north, the fixed uranium atom on Earth has a direction to the east, so the field and velocity are perpendicular, the expression of the force remains

         F = q v B

If we use Newton's second law

       F = ma

The acceleration is centripetal

      F = m v² / r

The distance is the radius of the earth plus the distance from the surface

      r = Re + 50.103

      r = 6.37 10⁶ + 5 10³

      r = 6.3705 10⁶ m

      m v² / r = q v B

      v = q/m   B r

uranium mass

     m = 238 (1.66 10⁻²⁷) kg = 395 10⁻²⁷ kg

Let's calculate

      v = 1.6 10⁻¹⁹ 0.5 10⁻⁴ 6.3705 10⁶ /395 10⁻²⁷

      v = 1.29 10⁸  m / s

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A psychologist is interested in exploring the effect tutorial support on students academic performance and assign students in to
NikAS [45]

Answer:

The dependent variable is academic performance

The independent variable is the presence/absence of tutorial support

The control group are students who did not get the tutorial support.

The experimental group were students that got the tutorial support

Explanation:

In every experiment, there is a dependent and independent variable as well as an experimental and a control group.

The experimental group receive the treatment while the control group do not receive the treatment. The independent variable is manipulated and its impact on the dependent variable is evaluated.

The control group are students who did not receive the tutorial support while the experimental group are students that received the tutorial support.

The dependent variable in this case is academic performance. Its outcome depends on the presence or absence of tutorial support (independent variable).

5 0
3 years ago
Assertion – Reason
Norma-Jean [14]

u have not displayed the question fella but i can guess the answer

Explanation:

the answer is (i)

because the current in a circuit is always directly proportional to its voltage thus making it to make a straight line in its graph

3 0
3 years ago
Every object in the universe exerts a force on every other object. This force is called
MrMuchimi

Gravity is the correct answer.

6 0
3 years ago
Bodies weighing 1 kilogram and 5 kilograms lie on a smooth horizontal surface. If a traction force of 0.6 N acts on another 5 kg
natima [27]

0.6/5,1,5

so calculate it

not so sure though

6 0
3 years ago
Consult interactive solution 2.22 before beginning this problem. a car is traveling along a straight road at a velocity of +30.0
Inessa05 [86]

Let a_1 be the average acceleration over the first 2.46 seconds, and a_2 the average acceleration over the next 6.79 seconds.

At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let v be the velocity of the car after the first 2.46 seconds.

By definition of average acceleration, we have

a_1=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

a_2=\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}

and we're also told that

\dfrac{a_1}{a_2}=1.66

(or possibly the other way around; I'll consider that case later). We can solve for a_1 in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:

1.66a_2=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

\implies1.66\left(\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}\right)=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

Now we can solve for v. We find that

v=20.8\,\dfrac{\mathrm m}{\mathrm s}

In the case that the ratio of accelerations is actually

\dfrac{a_2}{a_1}=1.66

we would instead have

\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}=1.66\left(\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}\right)

in which case we would get a velocity of

v=24.4\,\dfrac{\mathrm m}{\mathrm s}

6 0
3 years ago
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