Question: How is a triple beam balance used to find mass?1. Observe: The riders have masses of 10 grams (top), 100 grams (middle), and 1 gram (bottom). Drag the 100-gram rider to 300. At this position it balances a 300-gram mass.What happens to the pointer? ________________________________________________2. Compare: Place each object on the measurement tray, one at a time. Which objects have a mass greater than 300 grams? ________________________________________________ How do you know? _________________________________________________________

Answer:

Explanation:

The reaction time will be equal to the time the ruler takes to fall by 6.4 cm .

the ruler is falling with acceleration due to gravity which is equal to

9.8 m /s²

S = ut + 1/2gt²

s is distance of fall , u is initial velocity , t is time and g is acceleration due to gravity .

6.4 = 0 + 1/2 x 9.8 t²

t = 1.14 s .

So reaction time is 1.14 s .

Answer:

0.114s

Explanation:

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The reaction time is simply how long it takes the ruler to fall 6.4 cm.

For this problem, all you gotta do is convert your units and then plug them into a kinematic equation for t (time).

{*0.01 the conversion from cm to m}

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Units:

Δx (change in x) or d = 6.4 cm = 0.064m

Vi = 0 m/s

a = 9.81 m/s^2

t = ?

--------------------------------------------------------------------------------------------------------------

[tex]d = Vi * t + \frac{1}{2} *a*t^{2}[/tex]

(0.064) = 0 * t + 1/2 * (9.81) * t^2

↓

0.064 = 1/2 * (9.81) * t^2

↓

0.064 = 4.905 * t^2

↓

0.01304791 = t^2

↓

t = 0.114s

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Your reaction time is 0.114s

Answer:

I think its false...........

Answer: hope this helps

Explanation:

Explanation:

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Given parameters:

Distance 1 = 319 miles  

Time taken 1 = 6 hours

Distance 2 = 221 miles

Time taken 2 = 3 hours

Unknown:

Average speed  = ?

To find the average speed we need find the total distance traveled and divided by the total time taken for the travel.

Speed is a scalar quantity that expresses the distance covered per unit of time.

Mathematically;

   Average speed  = [tex]\frac{Total distance }{Total time taken }[/tex]

                                = [tex]\frac{distance 1 + distance 2}{time taken 1 + time taken 2}[/tex]

                               = [tex]\frac{319 + 221}{6 + 3}[/tex]

                               = 60miles/hr

The average speed of the bus is 60miles/hr

Answer:

The answer is #4: there is a period of time when the deep space network won't be gathering data.

Explanation:

I think this is the answer. I hope it helps.

Answer:

ΔE = 7.559 eV ,     λ = 1,645 10⁻⁷ m

Explanation:

For this exercise we can use the Bohr model for ionized atom with only one free electron,

         r_n = n² a₀ / Z

         E_n = -13,606 Z² / n²

Where a₀ is the Bohr radius of the hydrogen atom (a₀ = 0.0529 nm), Z is the atomic number of the atom under study and 13.606 eV is the energy of the ground state of Hydrogen.

In our case the Helium atom has two protons Z = 2

let's calculate the quantum number and the energy of each orbit

r_n = 0.30 nm

          n₁ = √ (r_n Z / a₀)

          n₁ = √ (0.30 2 / 0.0529)

Note that we do not have to reduce the radius since they are all in nanometers

          n₁ = 3.3

since n is an integer we approximate it to

         n₁ = 3

r_n = 0.20 nm

          n₂ = √ (0.2 2 / 0.0529)

          n₂ = 2.7

To approximate this value we must assume that there could be some error in the medicinal radio,

          n₂ = 2

having the quantum numbers of the two radius we can calculate their energy

        E₃ = - 13,606 2²/3²

        E₃ = - 6.047 eV

        E₂ = -13.606 2²/2²

         E₂ = -13.606 eV

the energy of the emitted photon is

          ΔE = E₃ - E₂

          ΔE = -6.047 + 13.606

          ΔE = 7.559 eV

You do not indicate in the exercise if you want the energy or the wavelength of the photon,

to find the wavelength We use the Planck relation

          E = h f

          c = λ f

          E = h c /λ

          λ = h c / E

we must reduce the energy to the SI system

          E = 7.559 ev (1.6 10⁻¹⁹ J / 1eV) = 12.09 10⁻¹⁹ J

          λ = 6.63 10⁻³⁴ 3 10⁸ / 12.09 10⁻¹⁹

          λ = 1,645 10⁻⁷ m

Answer:

Models,Mathematics

am not sure pliz mark brainliest

Explanation:

Answer:

Kepler's three laws of planetary motion can be described as follows: ... Kepler's first law is rather simple - all planets orbit the sun in a path that resembles ... Newton was able to combine the law of universal gravitation with circular motion ... The net centripetal force acting upon this orbiting planet is given by the relationship ...

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Answer:velocity =-.65 speed=.65

Explanation:

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social impact, focused business strategy, economic viability

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Just like the journalism industry, the factors of sustainability are ever-evolving. Community Wealth Partners recommends that nonprofit organizations consider how they are performing across five key drivers of sustainability which is  social impact, focused business strategy, economic viability

Answer:

The vacuum of space

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You can't "make" things move but you can apply enough force on to an object for it to move, pls brainliest!

Answer:

1. The magnitude of the vector is 26.9 m

2. The magnitude of direction of the vector is 21.8°.

Explanation:

From the question given, we obtained the following data:

Displacement along x (Bₓ) = 25 m

Displacement along y (Bᵧ) = 10 m

Magnitude of displacement (B) =?

Magnitude of direction (θ) =.?

1. Determination of the magnitude of displacement of the vector (B)

Displacement along x (Bₓ) = 25 m

Displacement along y (Bᵧ) = 10 m

Magnitude of displacement (B) =?

B² = Bₓ² + Bᵧ²

B² = 25² + 10²

B² = 625 + 100

B² = 725

Take the square root of both side

B = √725

B = 26.9 m

Therefore, the magnitude of the vector is 26.9 m

2. Determination of the magnitude of the direction of the vector.

Displacement along x (Bₓ) = 25 m

Displacement along y (Bᵧ) = 10 m

Magnitude of direction (θ) =.?

The magnitude of direction of the vector can be obtained by using Tan ratio as illustrated below:

Adjacent = Bₓ = 25 m

Opposite = Bᵧ = 10 m

Magnitude of direction (θ) =.?

Tan θ = Opposite /Adjacent

Tan θ = Bᵧ / Bₓ

Tan θ = 10 /25

Tan θ = 0.4

Take the inverse of Tan

θ = Tan¯¹ (0.4)

θ = 21.8°

Therefore, the magnitude of direction of the vector is 21.8°.

Answer:

(a)  The distance of the chicken from the finish line is 62.5 m

(b) The stationary time of the wildebeest is 675 s

Explanation:

Given;

total distance traveled by wildebeest and chicken, d = 2 km = 2000 m

speed of the wildebeest, [tex]v_w[/tex] = 16 m/s

speed of the chicken, [tex]v_c[/tex] = 2.5 m/s

Time for wildebeest to finish the race without stopping, 2000 / 16 = 125 s

Time for chicken to finish the race without stopping, 2000/2.5 = 800 s

(b) for how long in time (in s) was the wildebeest stationary?

t(stationary) = t(chicken) - t(wildebeest)

t = 800s - 125 s

t = 675 s

(a) how far (in m) is the chicken from the Finish Line when the wildebeest resume the race?

The time taken for the wildebeest to run 1.6 km (1600 m) is given by;

t = 1600 / 16 = 100 s

The total time spent by the wildebeest before it resumed the race = stationary time + 100s

t (total) = 675 s + 100 s = 775 s

Distnace traveled by the chicken when the wildebeest resumed the race = 2.5m/s x 775s = 1937.5 m

Thus, the distance of the chicken from the finish line = 2000 m -  1937.5 m

the distance of the chicken from the finish line = 62.5 m

According to the question,

                       [tex]S_1 = 1.6 \ km \ or \ 1600 \ m[/tex]

                    [tex]V_c = 2.5 \ m/s[/tex]

                    [tex]= \frac{1600}{16}[/tex]

                    [tex]= 100 \ s[/tex]

                 [tex]t_2 = \frac{400}{16}[/tex]

                     [tex]= 25 \ s[/tex]

Now,

→ [tex]T= \frac{S}{V_c}[/tex]

      [tex]= \frac{2000}{2.5}[/tex]

      [tex]= 800 \ s[/tex]

(a)

Distance will be:

→ [tex]S_c' = V_c\times t_2[/tex]

By substituting the values,

      [tex]= 2.5\times 25[/tex]

      [tex]= 62.5 \ m[/tex]

(b)

Time will be:

→ [tex]T_{static} = T-(t_1+t_2)[/tex]

           [tex]= 800-125[/tex]

           [tex]= 675 \ s[/tex]

Thus the responses above are correct.

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Elements are arranged on the periodic table in terms of valence electrons based on their atomic number and electron configuration.

1. Elements are arranged on the periodic table in terms of valence electrons based on their atomic number and electron configuration. The valence electrons are the outermost electrons in an atom's electron shell, and they are crucial in determining the chemical properties and reactivity of elements.

2. Evidence from data tables can be shown by examining the electron configuration and the group and period numbers of various elements on the periodic table. Here is a simplified example:

Element   | Electron Configuration | Group | Period |

--------------------------------------------

Hydrogen  | 1s^1                              | 1     | 1      |

Lithium       | [He] 2s^1                     | 1     | 2      |

Carbon       | [He] 2s^2 2p^2          | 14    | 2      |

Oxygen      | [He] 2s^2 2p^4          | 16    | 2      |

Neon          | [He] 2s^2 2p^6          | 18    | 2      |

--------------------------------------------

3. The evidence from the data table supports the claim that the arrangement of elements on the periodic table is based on valence electrons.

- Group Trend: Elements within the same group (vertical columns) share the same number of valence electrons. In the example table, Hydrogen, Lithium, and Neon are all in Group 1, indicating they have 1 valence electron.

- Period Trend: Elements within the same period (horizontal rows) have the same number of electron shells. In the example table, Hydrogen and Lithium are in Period 1, indicating they have their valence electron in the first energy level. Carbon, Oxygen, and Neon are in Period 2, indicating they have their valence electrons in the second energy level.

By examining the electron configurations, group numbers, and period numbers, we can clearly see the trends and patterns in the number of valence electrons for both groups and periods on the periodic table. This evidence supports the claim that the arrangement of elements on the periodic table is based on their valence electrons, which play a crucial role in determining their chemical behavior and properties.

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Answer:

a)    #_pastes = N / 8 ,  b) 5 tablespoons oil

Explanation:

a) For this exercise we can use the inverse proportions rule. So if each pasta has more volume in the same box size, tame pasta will fit.

Let's find the volume of the new pasta

         V = 2L 2W 2H

         V = 8 L W H

          V = 8 V₀

where V₀ is the volume of the usual pasta and we assume that there is N quantity in the box. therefore for the new pasta

          #_pastas = N V₀ / 8V₀

          #_pastes = N / 8

unfortunately you do not give the number of usual pastes to finish the calculation.

b) Let's analyze the situation, if the size of the box is the same, I initially need 5 tablespoons for charring and then I have the same box I will need the same amount of oil since the number of pastes is less

Answer:

9.8 seconds

Explanation:

Let's start by changing the speeds of both animals into meters per second:

107 km/h = 107000 m/3600s = 29.72 m/s

71.8 km/h = 71800 m/3600s = 19.94 m/s

The distance covered equation for the cheetah is:

Dc = 29.72 t

The distance covered equation for the gazelle considering as origin the cheetah's starting location, is:

Dg = 95.6 + 19.94 t

We make both equations equal at the time (t) the cheetah catches with the gazelle:

Dc =Dg

29.72 t = 95.6 + 19.94 t

solve for "t":

29.72 t - 19.94 t = 95.6

9.78 t = 95.6

t = 95.6 / 9.78

t = 9.775 seconds

We can approximate it to 9.8 seconds

Explanation:

Initial time, t₁ = 2:30 pm

Final time, t₂ = 2:30:45

We need to find the motion of students in terms of time. Final time is 45 seconds more than the initial time.

Change in time,

[tex]\Delta t=t_2-t_1\\\\\Delta t=2:30:45-2.30\\\\\Delta t=45\ s[/tex]

Hence, this is the required solution.

The students need 45 seconds to reach out at the edge of the school property with the final interval of 2:30:45 s. Hence, option (D) is correct.

The difference between the moment of starting an event and the completion of the same is known as the time interval.

Given data -

The exact time to start the day at school is, t1 = 2:30 pm.

The time given to reach the edge of the school property is, t2 = 2:30:45 s.

In order to describe the motion, we need to obtain the change in the time interval. Then the expression for the change in the time interval is,

Δt = t2 - t1

Solving as,

Δt = 2:30:45 - 2:30

Δt = 45 s

Thus, we can conclude that the students need 45 seconds to reach out at the edge of the school property with the final interval of 2:30:45 s.

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Answer:

Approximately [tex]1125\; \rm m[/tex].

Explanation:

Calculate the distance that sound would travel in [tex]1.5\; \rm s[/tex] at [tex]1500\; \rm m \cdot s^{-1}[/tex]:

[tex]s = v\cdot t = 1500\; \rm m\cdot s^{-1}\times 1.5\; \rm s = 2250\; \rm m[/tex].

Sound waves in this question here needs to:

Hence, the total distance that sound travelled in that [tex]1.5\; \rm s[/tex] would be the sum of:

Therefore, the distance [tex]2250\; \rm m[/tex] here is supposed to be approximately twice the depth of the sea at that place. The depth of the sea would thus be approximately:

[tex]\displaystyle \frac{1}{2}\times 2250\; \rm m = 1125\; \rm m[/tex].

Answer:

You need to be patient with them. Also, a positive attitude is the single most important quality for anyone who works with children with special needs. I’ve seen highly trained specialists unable to interact with Louie because of their negative attitude and assumptions

Explanation:

This problem is solved by determining how long it takes the arrow to fall to the ground in exactly the same manner as if the arrow was dropped with no horizontal velocity. The time required for the arrow to fall to the ground will be the same time that the arrow flies horizontally at 60.0 m/s, so

d=1/2at^2

solved for t=√2d/a

t=√(2)(2.00m)/9.80m/s^2 =0.639s

The time required for the fall is multiplied by the horizontal velocity to get the horizontal distance.

d(horizontal)=v(horizontal)×time=60.0m×0.639s =38.3m

The arrow will fly 37.8 meters in the horizontal direction before it strikes the ground.

Projectile motion is a form of motion experienced by an object or particle that is projected near Earth's surface and moves along a curved path under the action of gravity only.

Given is an arrow fired perfectly along the horizontal direction with velocity of 60 m/s. The arrow was 2 m above the ground when it was released.

initial horizontal velocity = u[H] = 60 m/s

initial vertical velocity = u[V] = 0 m/s

Horizontal acceleration = a[H] = 0 m/s²

Vertical acceleration = a[V] = 9.8 m/s²

Vertical distance = S[V] = 2 m

Using second equation of motion for vertical direction to calculate time taken to reach the ground.

S[V] = u[V] × t + 1/2 a[V] × t²

S[V] =  1/2 × a[V] × t²

2 = 1/2 x 9.8 x  t²

4/9.8 =  t²

t² = 0.4081

t = 0.63 seconds

Horizontal distance covered = S[H] = u[H] x t

S[H] = 60 X 0.63 = 37.8 meters

Therefore, the arrow will fly 37.8 meters in the horizontal direction before it strikes the ground.

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Answer: Well you didn't give any answers so my guess would be it depends how much force you put into it and where you throw it.

Explanation:

Answer:

a = 2.22 [m/s^2]

Explanation:

First we have to convert from kilometers per hour to meters per second

[tex]40 [\frac{km}{h}]*[\frac{1h}{3600s}]*[\frac{1000m}{1km}] = 11.11 [m/s][/tex]

We have to use the following kinematics equation:

[tex]v_{f}= v_{i}+(a*t)[/tex]

where:

Vf = final velocity = 11.11 [m/s]

Vi = initial velocity = 0

a = acceleration [m/s^2]

t = time = 5 [s]

The initial speed is taken as zero, as the car starts from zero.

11.11 = 0 + (a*5)

a = 2.22 [m/s^2]

Refer to the attachment.