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Softa [21]
3 years ago
15

11. What are the units for acceleration?​

Physics
1 answer:
Monica [59]3 years ago
5 0

Answer:

Acceleration (a) is defined as the rate of change of velocity. Velocity is a vector quantity, and therefore acceleration is also a vector quantity. The SI unit of acceleration is metres/second2 (m/s2).

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A rod of very small diameter with a mass 2m and length 3L is placed along the xaxis with one end at the origin. An identical rod
rewona [7]

Answer:

coordinates of the center of mass for these two rods

(x_{cm}, y_{cm})= (\frac{3L}{4},  \frac{3L}{4})cm

Explanation:

given

mass of a rod = 2m

length of the rod = 3L

mass of two rods = 2(2m) = 4m

radius = diameter/2 = \frac{3L}{2}

attached is the diagram and solution to the question

5 0
3 years ago
What amount of energy is needed for an electron to jump from n = 1 to n = 4?
liberstina [14]

Answer:

E=2.04\times 10^{-18}\ J

Explanation:

We need to find the energy for an electron to jump from n = 1 to n = 4.

The energy in transition from 1 state to another is given by :

E=\dfrac{-2.18\times 10^{-18}}{n^2}\ J

The difference in energy for n = 1 to n = 4 is:

E=-2.18\times 10^{-18}\times (\dfrac{1}{4^2}-1)\\\\E=2.04\times 10^{-18}\ J

So, the required energy is equal to 2.04\times 10^{-18}\ J.

4 0
3 years ago
The pages of a book are numbered 1 to 200 and each
never [62]

Answer:

10.4mm

Explanation:

2 pages = 1 leaf

200 pages = 100 leaves

100 × 0.10 = 10 mm thickness

Total thickness = 2(0.20) +10 = 0.4+10 = 10.4mm

6 0
3 years ago
Electromagnet Fluctation
Tems11 [23]

Answer:

answer choice B

Explanation:

6 0
3 years ago
A child, hunting for his favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The ang
olga55 [171]

Answer:

9.22 s

Explanation:

One-quarter of a turn away is 1/4 of 2π, or π/2 which is approximately 1.57 rad

Let t (seconds) be the time it takes for the child to catch up with the horse. We would have the following equation of motion for the child and the horse:

For the child: s_c = \omega_ct = 0.233t

For the horse: s_h = s_0 + a_ht^2/2 = 1.57 + 0.0136t^2/2 = 1.57 + 0.0068t^2

For the child to catch up with the horse, they must cover the same angular distance within the same time t:

s_c = s_h

0.233t = 1.57 + 0.0068t^2

0.0068t^2 - 0.233t + 1.57 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{0.233\pm \sqrt{(-0.233)^2 - 4*(0.0068)*(1.57)}}{2*(0.0068)}

t= \frac{0.233\pm0.11}{0.0136}

t = 25.05 or t = 9.22

Since we are looking for the shortest time we will pick t = 9.22 s

6 0
3 years ago
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