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Ulleksa [173]
3 years ago
14

a student drew the following model: volcano cooling crust motion plates tension which landform should the student put next in th

e sequence? a. plain b. cinder cone c. rift valley d. plateau
Physics
2 answers:
Sonbull [250]3 years ago
4 0

Answer:C.rift valley (apex)

Explanation:

Andrei [34K]3 years ago
3 0

Answer:

C) rift valley

Explanation:

A rift valley is a lowland region formed by the interaction of Earth's tectonic plates. This small rift valley has a typical formation—long, narrow, and deep. It was formed by the Thingvellir rift, where the North American and Eurasian tectonic plates are tearing, or rifting, apart over a hotspot on the island of Iceland.

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A force of 40 N stretches a spring 0.20 m what is the spring
mart [117]
It’s B lovey !!! Glad to help
8 0
3 years ago
Planet X is in a stable circular orbit around a star, as shown in the figure. Which of the following graphs best predicts the an
damaskus [11]

Answer:

C

Explanation:

Angular momentum is the product of moment of inertia and angular velocity.

L = I × ω

Since the planet follows a stable circular orbit, I and ω are constant and non-zero.  Therefore, the angular momentum is constant and non-zero.

5 0
3 years ago
James m = 85.0 kg and ramon m = 59.0 kg are 20.0 m apart on a frozen pond. midway between them is a mug of their favorite bevera
Vlad [161]
You can use conservation of momentum, Ramon has a momentum of m * v = 59.0 * 1.20 = 70.8. James must have that same momentum so m * v = p -->
p / m = v --> 70.8 / 85.0 = 0.83 m/s for James
8 0
4 years ago
. A proton, which moves perpendicular to a magnetic field of 1.2 T in a circular path of radius 0.080 m, has what speed? (qp = 1
almond37 [142]

Answer:

9.198\times 10^6 m/s

Explanation:

We are given that

Magnetic field, B=1.2 T

Radius of circular path, r=0.080 m

q_p=1.6\times 10^{-19} C

m_p=1.67\times 10^{-27} kg

\theta=90^{\circ}

We have to find the speed of proton.

We know that

Magnetic force, F=qvBsin\theta

According to question

Magnetic force=Centripetal force

q_pvBsin90^{\circ}=\frac{m_pv^2}{r}

1.6\times 10^{-19}\times 1.2=\frac{1.67\times 10^{-27}v}{0.08}

v=\frac{1.6\times 10^{-19}\times 1.2\times 0.08}{1.67\times 10^{-27}}

v=9.198\times 10^6 m/s

5 0
4 years ago
how large can the kinetic energy of an electron be that is localized within a distance (change in) x = .1 nmapproximately the di
elena-14-01-66 [18.8K]

Answer:

The kinetic energy of an electron is 1.54\times10^{-15}\ J

Explanation:

Given that,

Distance = 0.1 nm

We need to calculate the momentum

Using uncertainty principle

\Delta x\Delta p\geq\dfrac{h}{4\pi}

\Delta p\geq\dfrac{h}{\Delta x\times 4\pi}

Where, \Delta p = change in momentum

\Delta x = change in position

Put the value into the formula

\Delta p=\dfrac{6.6\times10^{-34}}{4\pi\times10^{-10}}

\Delta p=5.3\times10^{-23}

We need to calculate the kinetic energy for an electron

K.E=\dfrac{p^2}{2m}

Where, P = momentum

m = mass of electron

Put the value into the formula

K.E=\dfrac{(5.3\times10^{-23})^2}{2\times9.1\times10^{-31}}

K.E=1.54\times10^{-15}\ J

Hence, The kinetic energy of an electron is 1.54\times10^{-15}\ J

4 0
3 years ago
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