Where are you likely to find the oldest oceanic crust?

Physics

1 answer:

Stells [14]3 years ago

4 0

Answer:

it would be letter E. near oceanic ridges

Explanation:

new ocean crust is formed at the mid ocean ridges

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Dean is hunting in the Northwest Territories at a location where earths magnetic field is 7.0x10^-5T. He shoots by mistake at a

velikii [3]

The magnetic force acting on an object of charge q is
$F=qvB \sin \theta$
where
q is the charge
v is the speed of the object
B is the magnetic field intensity
$\theta$ is the angle between the directions of v and B.

In our problem, the direction of the bullet is perpendicular to the magnetic field, so $\theta=90^{\circ}$ and $\sin \theta=1$ , so we can ignore it in the formula.

Therefore, we can use the data of the problem to calculate the magnetic force on the bullet:
$F=qvB=(2.0 \cdot 10^{-12}C)(300 m/s)(7.0 \cdot 10^{-5} T)=4.2 \cdot 10^{-14} N$

6 0

3 years ago

A car with a mass of 500 kg, gets off road and goes straight into a cliff of 30 m with an

mars1129 [50]

Answer:

The velocity of the car when it hits the ground is approximately 55.574 meters per second.

Explanation:

According to this expression, the car goes straight into the cliff and goes down due to gravity, whose situation is described by the Principle of Energy Conservation and supposing that all non-conservative forces are negligible:

$U_{g,1}+K_{1} = U_{g,2}+K_{2}$ (1)

Where:

$U_{g,1}$ , $U_{g,2}$ - Gravitational potential energies of the car at the top and the bottom, measured in joules.

$K_{1}$ , $K_{2}$ - Translational kinetic energies of the car at the top and the bottom, measured in joules.

By definitions of gravitational potential and translational kinetic energies, we expand and simplify the equation above:

$\frac{1}{2}\cdot m\cdot v_{2}^{2} = \frac{1}{2}\cdot m \cdot v_{1}^{2}+m\cdot g \cdot (z_{1}-z_{2})$ (2)

$v_{2}^{2} = v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})$

$v_{2} = \sqrt{v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})}$

Where:

$m$ - Mass, measured in kilograms.

$v_{1}$ , $v_{2}$ - Velocity of the car at the top and the bottom, measured in meters per second.

$g$ - Gravitaitional acceleration, measured in meters per square second.

$z_{1}$ , $z_{2}$ - Height of the car at the top and at the bottom, measured in meters.

If we know that $v_{1} = 50\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ , $z_{1} = 30\,m$ and $z_{2} = 0\,m$ , then the velocity of the car when it hits the ground is: