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3 years ago

Glossopteris is an ancient, extinct species of found in South America, Africa, Asia, Australia, and Antarctica.

Physics

2 answers:

netineya [11]3 years ago

7 0

Answer:

Plants

Explanation:

I had this question in my science lesson

rusak2 [61]3 years ago

6 0

Answer: trees

Glossopteris is an extinct group of seed plants.

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Brendan drew a diagram to illustrate the centripetal force acting on a satellite.

harina [27]

Brendan drew a diagram to illustrate the centripetal force acting on a satellite.

How can Brendan correct his diagram?

Draw the Fg = Fc vector so it points at a 45° angle from the v vector.

Change the Fg = Fc vector label so it reads Fa = Fc.

Draw the Fg = Fc vector so it is longer than the v vector.

Change the Fg = Fc vector so it is perpendicular to the v vector.

D is the answer

4 0

4 years ago

Read 2 more answers

I NEED HELP VERY SOON PLEASE

Igoryamba

I would say d if i was you

7 0

3 years ago

if a diffraction grating produces a third-order bright spot for red light (of wavelength 650 nm ) at 68.0 ∘ from the central max

Mashutka [201]

Answer:

The angle is 25.34°.

Explanation:

Given that,

Wave length = 650 nm

Angle = 68.0°

We need to calculate the distance

For a diffraction grating

$d\sin\theta=m\lambda$

$d=\dfrac{2\times650\times10^{-9}}{\sin68.0}$

$d=2.10\times10^{-6}\ m$

We need to calculate the angle

Using formula for angle

$d\sin\theta=m\lambda$

$\sin\theta=\dfrac{m\times\lambda}{d}$

$\sin\theta=\dfrac{2\times450\times10^{-9}}{2.10\times10^{-6}}$

$\sin\theta=25.34^{\circ}$

Hence, The angle is 25.34°.

8 0

3 years ago

An ordinary workshop grindstone has a radius of 7.50 cm and rotates at 6500 rev/min. (a) Calculate the magnitude of the centripe

jonny [76]

To solve this problem it is necessary to apply the concepts related to the relationship between tangential velocity and centripetal velocity, as well as the kinematic equations of angular motion. By definition we know that the direction of centripetal acceleration is perpendicular to the direction of tangential velocity, therefore:

$a_c= \frac{v^2}{r}=r\omega^2$

Where,

V = the linear speed

r = Radius

$\omega =$ Angular speed

The angular speed is given by

$\omega =6500\frac{rev}{min}(\frac{2\pi rad}{1 rev})(\frac{1 min}{60s})$

$\omega = 680.6784 rad/s.$

Replacing at our first equation we have that the centripetal acceleration would be

$a_c = r\omega^2$

$a_c = (0.0750)(680.6784)^2$

$a_c = 34 749.23 m/s^2$

To transform it into multiples of the earth's gravity which is given as $9.8m / s$ the equivalent of 1g.

$a_c =34749.23 \frac{m}{s^2} (1\frac{g}{9.8m/s^2})$

$a_c = 3545.84g$

PART B) Now the linear speed would be subject to:

$v = \omega r$

$v= (680.6784)(0.0750)$

$v=51.05 m/s$

Therefore the linear speed of a point on its edge is 51.05m/s

8 0

3 years ago

Change of motion are caused by

AlexFokin [52]

<span>The only thing that can change the motion of an object is a net (unbalanced)</span>force acting on it. This is given by Newton's First Law of Motion, sometimes <span>also called the Law of Inertia.

I hope this helps!</span>

4 0

4 years ago