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vampirchik [111]
3 years ago
12

Which explanation BEST supports Newton's second law of motion which states that force = mass x acceleration?

Physics
1 answer:
GaryK [48]3 years ago
8 0
The answer is C. F=ma basically says that force is a function of mass multiplied by acceleration. The first two answers don’t make sense because there’s no necessary relationship between mass and acceleration. And for the last two, the higher the mass, the higher the force needed, therefore C is the correct answer.
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A water wave has a frequency of 2 Hertz and a wavelength of 5 cm. Calculate it speed
Llana [10]

Answer: 0.1 m/s

Explanation:

Use formula,

v = f * w where, v is speed, f is frequency and w is wavelength.

Now,

v = 2 * 5 * 10 ^ -2 ( Remember to convert all the units to SI units. Here 5 cm becomes 5 * 10 ^ -2 m. )

v = 0.1 m/s.

8 0
3 years ago
A body falls from the top of the tower and during the last second of its fall it fall through 23mvfind height of tower.
DanielleElmas [232]

Answer:

39.7 m

Explanation:

First, we conside only the last second of fall of the body. We can apply the following suvat equation:

s=ut+\frac{1}{2}at^2

where, taking downward as positive direction:

s = 23 m is the displacement of the body

t = 1 s is the time interval considered

a=g=9.8 m/s^2 is the acceleration

u is the velocity of the body at the beginning of that second

Solving for u, we find:

ut=s-\frac{1}{2}at^2\\u=\frac{s}{t}-\frac{1}{2}at=\frac{23}{1}-\frac{1}{2}(9.8)(1)=18.1 m/s

Now we can call this velocity that we found v,

v = 18 m/s

And we can now consider the first part of the fall, where we can apply the following suvat equation:

v^2-u^2 = 2as'

where

v = 18 m/s

u = 0 (the body falls from rest)

s' is the displacement of the body before the last second

Solving for s',

s'=\frac{v^2-u^2}{2a}=\frac{18.1^2-0}{2(9.8)}=16.7 m

Therefore, the total heigth of the building is the sum of s and s':

h = s + s' = 23 m + 16.7 m = 39.7 m

7 0
3 years ago
What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC?
Nastasia [14]

Answer:

F=1.26*10^{-3}N

Explanation:

Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:

F=\frac{kq_1q_2}{d^2}

We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges(q_1,q_2) and inversely proportional to the square of the distance(d) that separates them.

Replacing the given values, where k is the Coulomb constant:

F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-30*10^{-9}C)(-30*10^{-9}C)}{(8*10^{-2}m)^2}\\F=1.26*10^{-3}N

8 0
3 years ago
You throw a baseball straight up into the air with a speed of 24.5 m/s. How long does it take the baseball to reach its highest
galben [10]
Time=speed/acceleration
Gravitaional Acceleration=9.8 m/s^2
Speed=24.5 m/s
Time=24.5/9.8=2.5 s
3 0
3 years ago
Read 2 more answers
What can you say about the magnitudes of the forces that the balloons exert on each other?
maxonik [38]

Answer:

F_G=G. \frac{m_1.m_2}{R^2} gravitational force

F=\frac{1}{4\pi \epsilon_0} \times \frac{q_1.q_2}{R^2} electrostatic force

Explanation:

The forces that balloons may exert on each other can be gravitational pull due to the mass of the balloon membrane and the mass of the gas contained in each. This force is inversely proportional to the square of the radial distance between their center of masses.

The Mutual force of gravitational pull that they exert on each other can be given as:

F_G=G. \frac{m_1.m_2}{R^2}

where:

G= gravitational constant  =6.67\times 10^{-11} m^3.kg^{-1}.s^{-2}

m_1\ \&\ m_2 are the masses of individual balloons

R= the radial distance between the  center of masses of the balloons.

But when  there are charges on the balloons, the electrostatic force comes into act which is governed by Coulomb's law.

Given as:

F=\frac{1}{4\pi \epsilon_0} \times \frac{q_1.q_2}{R^2}

where:

\rm \epsilon_0= permittivity\ of\ free\ space

q_1\ \&\ q_2 are the charges on the individual balloons

R = radial distance between the charges.

3 0
3 years ago
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