The same car slows from 36 m/s to 15 m/s over 3 seconds. What is its average acceleration over this time interval?

Help!!! Line B touches the circle at a single point. Line A extends through the center of the circle.

Ulleksa [173]

Answer:

If I understand correctly. Line B is parallel to the circle. Also, the angle is less than 90.

The size of the circle determines.
The diameter should not be fixed either.

3 0

3 years ago

An object's starting position is at point A and its final position is at point B.

salantis [7]

B- we would need the time to fulfill the formula distance = speed x time

4 0

3 years ago

When the electron in a hydrogen atom moves from n = 6 to n = 1, light with a wavelength of ________ nm is emitted?

Oduvanchick [21]

First we find the energy level with the following formula, where a is the energy level, n1 is the final energy level, n2 is the starting energy level and r is Rydberg's constant in Joules
$a = r \times ( \frac{1}{n1} - \frac{1}{n2} )$
We insert the values

$a = 2.18 \times {10}^{ - 18} \times ( \frac{1}{ {1}^{2} } - \frac{1}{ {6}^{2} } )$
$= 2.12 \times {10}^{ - 18}$
The wavelength is found with this formula, where h is Planck's constant and c is the speed of light
$wavelength = \frac{h \times c}{a}$
Finally we insert the values
$\frac{6.626 \times {10}^{ - 34} \times 3 \times {10}^{8} }{2.12 \times {10}^{ - 18} } = 9.376 \times {10}^{ - 8}$
Which is the same as 93.8 nm

3 0

3 years ago

Which transformation of energy takes place when a slingshot launches a stone? (PE stands for potential energy.) A. elastic PE to

boyakko [2]

B. Elastic potential to kinetic energy

The elastic potential energy in the slingshot will be transferred to the stone as kinetic energy as the stone is launched.

6 0

3 years ago

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Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur

BARSIC [14]

Answer:

$v_A=7667m/s\\\\v_B=7487m/s$

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

$F_g=\frac{GMm}{R^{2} }$

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

$F_c=m\frac{v^{2}}{R}$

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

$\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}$

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So $R_A=6774km=6.774*10^6m$ and $R_B=7103km=7.103*10^6m$ (Since $R_{earth}=6371km$ ). Then, we get:

$v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s$

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0

3 years ago