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timurjin [86]
3 years ago
11

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

Physics
1 answer:
lana [24]3 years ago
7 0
The correct answer would be

=-7m/s^2
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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.

  1. The size of the circle determines.
  2. The diameter should not be fixed either.
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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
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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
Read 2 more answers
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
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