What is the average velocity in m/y of a tectonic plate that has traveled 9000 km to the south in 60 million years

An athlete executing a long jump leaves the ground at a 28.0º angle and travels 7.80 m. (A) What was the takeoff speed?

Naddika [18.5K]

Answer:

9.6 m/s

Explanation:

Angle of projection, θ = 28°

Horizontal distance, R = 7.8 m

Let the velocity of projection is given by u.

The formula used to find the velocity of projection is given by

$R=\frac{u^{2}Sin2\theta }{g}$

$7.8=\frac{u^{2}Sin\left (2\times 28 \right ) }{9.8}$

$7.8=\frac{u^{2}\times 0.829}{9.8}$

u = 9.6 m/s

Thus, the velocity of projection is 9.6 m/s.

7 0

3 years ago

A diverging lens has a focal length of magnitude 22.8 cm. (a) Locate the images for each of the following object distances. 45.6

Nostrana [21]

(a) -15.2 cm

We can solve the problem by using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f = -22.8 cm is the focal length of the lens (negative because it is a diverging lens)

p = 45.6 cm is the distance of the object from the lens

q is the distance of the image from the lens

Solving the equation for q, we find the position of the image:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-22.8 cm}-\frac{1}{45.6 cm}=-0.066 cm^{-1}$

$q=\frac{1}{-0.066 cm^{-1}}=-15.2 cm$

and the negative sign means that the image is virtual.

(b) -11.4 cm

In this case, the distance of the object from the lens is

p = 22.8 cm

Substituting into the lens equation, we can find the new image distance, q:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-22.8 cm}-\frac{1}{22.8 cm}=-\frac{2}{22.8 cm}$

$q=\frac{-22.8 cm}{2}=-11.4 cm$

and the negative sign means that the image is virtual.

(c) -7.6 cm

In this case, the distance of the object from the lens is

p = 11.4 cm

Substituting into the lens equation, we can find the new image distance, q:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-22.8 cm}-\frac{1}{11.4 cm}=-0.132 cm^{-1}$

$q=\frac{1}{0.132 cm^{-1}}=-7.6 cm$

and again, the negative sign means that the image is virtual.

3 0

3 years ago

In the simple Bohr model of the eighth excited state of the hydrogen atom, the electron travels in a circular orbit around a fix

Ksju [112]

Answer:

Net force on electron will be $12.568\times 10^{-12}N$

Explanation:

We have given that the orbit $r=4.28\times 10^{-9}m$

Speed of the electron $v=2.43\times 10^5m/sec$

Mass of the electron is given as $m=9.11\times 10^{-31}kg$

Centripetal acceleration is given by $a_c=\frac{v^2}{r}=\frac{(2.43\times 10^5)^2}{4.28\times 10^{-9}}=1.3796\times 10^{19}m/sec^2$

Force is given by F = ma

So force $F=9.11\times 10^{-31}\times 1.3796\times 10^{19}=12.568\times 10^{-12}N$

8 0

3 years ago

ANSWER QUICK 30 POINTS

labwork [276]

Answer:

Gravity controls the movement of the planets around the sun, holds together stars grouped in galaxies, and galaxies grouped in clusters. The Universal Law of Gravitation depends on two things. First it depends on mass of each object and the second factor is the distance between two objects. If the mass of one object is Larger, the gravitational pull towards it will be larger and the smaller distance, the larger the gravitational pull will be between the objects. Therefore the Larger planets have more moon and the inner planets have less.

Explanation:

6 0

3 years ago

Init. A

Gekata [30.6K]

Answer:

v = 1/3 m / s = 0.333 m / s

in the direction of the truck

Explanation:

The average speed is defined by the variation of the position between the time spent

v = Δx / Δt

since the position is a vector we must add using vectors, we will assume that the displacement to the right is positive, the total displacement is

Δx = 20 - 15 +20

Δx = 25 m

therefore we calculate

v = 25/75

v = 1/3 m / s = 0.333 m / s

in the direction of the truck

7 0

3 years ago