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alexgriva [62]
3 years ago
6

What is question 22?

Physics
2 answers:
zalisa [80]3 years ago
5 0
The object moves 18 meters in 6 seconds (10 - 4).

18m/6s= 3m/s

Answer "A"
Fynjy0 [20]3 years ago
4 0

the answer is a!! its pretty simple I just read the graph.
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For any molecule, formula unit, or ion, the sum of the average atomic masses of all the atoms represented in a formula is the.
alukav5142 [94]

The mass of a substance is given in atomic mass units and is calculated by adding the average atomic masses of all the atoms in the substance's chemical formula.

<h3>What empirical formula represents the total average atomic mass of every atom?</h3>

The Method The average atomic masses of all the atoms included in a formula's representation are added to get the mass of any molecule, formula unit, or ion. It has no bearing on the number of significant figures because the number of atoms is an exact quantity. One H2O molecule weighs 18.02 amu on average.

<h3>What connection exists between the empirical formula and the molecular formula?</h3>

You can determine the number of atoms of each element in a molecule using its molecular formula. These empirical formulations provide the most basic or reduced elemental ratio of a compound. The empirical formula and the molecular formula of a substance are same if the molecular formula can no longer be decreased.

To know more about atomic mass visit:-

brainly.com/question/17067547

#SPJ4

5 0
1 year ago
Suppose a yo-yo has a center shaft that has a 0.230 cm radius and that its string is being pulled.
Fofino [41]

Answer:

Part a)

\alpha = 782.6 rad/s^2

Part B)

\omega = 587 rad/s

Part c)

a_t = 24.3 m/s^2

Explanation:

Part a)

As we know that

a = R \alpha

so we will have

a = 1.80 m/s^2

R = 0.230 cm

\alpha = \frac{a}{R}

\alpha = \frac{1.80}{0.230 \times 10^{-2}}

\alpha = 782.6 rad/s^2

Part B)

Angular speed of the yo-yo

\omega = \alpha t

so we have

\omega = 782.6 \times 0.750

\omega = 587 rad/s

Part c)

Tangential acceleration is given as

a_t = R \alpha

a_t = (3.10 \times 10^{-2})(782.6)

a_t = 24.3 m/s^2

6 0
4 years ago
If you go to a beach in the continental united states and you are standing on a long, flat strand of sand, and you look away fro
Nataly_w [17]
If you are in this kind of geographical scenery, you must be on a surfing beach in California. The Californian coastline is characterized by a numerous number of beaches that are mainly sandy and features major rivers such as the Sacramento River and the Colorado River.
3 0
3 years ago
A. What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8cm from equilibrium and the
Naddik [55]

Answer:

a) x = 8.8 cm * cos (9.52 rad/s * t)

b) x = 8.45 cm

Explanation:

This is a Simple Harmonic Motion, and most Simple Harmonic Motion equations start from the equilibrium point. In this question however, we are starting from the max displacement the equations, and thus, it ought to be different.

From the question, we are given that

A = 8.8 cm = 0.088 m

t = 0.66 s

Now, we need to find the angular speed w, such that

w = 2π/T

w = (2 * 3.142) / 0.66

w = 6.284 / 0.66

w = 9.52 rad/s

The displacement equation of Simple Harmonic Motion is usually given as

x = A*sin(w*t)

But then, the equation starts from the equilibrium point at 0 sec, i.e x = 0 m

When you have to start from the max displacement, then the equation would be

x = A*cos(w*t).

So when t = 0 the cos(0) = 1, and then x = A which is max displacement.

Thus, the equation is

x = 8.8 cm * cos (9.52 rad/s * t)

At t = 1.7 s,

x = 8.8 cos (9.52 * 1.7)

x = 8.8 cos (16.184)

x = -8.45 cm

5 0
4 years ago
Read 2 more answers
A 15.5 kg mass vibrates in simple harmonic motion with a frequency of 9.73 Hz. It has a maximum displacement from equilibrium of
algol [13]

Answer:

12.14 cm

Explanation:

mass, m = 15.5 kg

frequency, f = 9.73 Hz

maximum amplitude, A = 14.6 cm

t = 1.25 s

The equation of the simple harmonic motion

y = A Sin ωt

y =  A Sin (2 x π x f x t)

put, t = 1.25 s, A = 14.6 cm, f = 9.73 Hz

y = 14.6 Sin ( 2 x 3.14 x 9.73 x 1.25)

y = 14.6 Sin 76.38

y = 12.14 cm

Thus, the displacement of the particle from the equilibrium position is 12.14 cm.

6 0
4 years ago
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