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3 years ago

Luis and Aisha conducted an experiment. They exerted different forces on four objects. Their results are shown in the table.

Physics

2 answers:

lesya692 [45]3 years ago

4 0

Answer:

Object 3 has greatest acceleration.

Explanation:

Objects Mass Force

1 10 kg 4 N

2 100 grams 20 N

3 10 grams 4 N

4 1 kg 20 N

Acceleration of object 1,

$a_1=\dfrac{F_1}{m_1}\\\\a_1=\dfrac{4}{10}\\\\a_1=0.4\ m/s^2$

Acceleration of object 2,

$a_2=\dfrac{F_2}{m_2}\\\\a_2=\dfrac{20}{0.1}\\\\a_2=200\ m/s^2$

Acceleration of object 3,

$a_3=\dfrac{F_3}{m_3}\\\\a_3=\dfrac{4}{0.01}\\\\a_3=400\ m/s^2$

Acceleration of object 4,

$a_4=\dfrac{F_4}{m_4}\\\\a_4=\dfrac{20}{1}\\\\a_3=20\ m/s^2$

It is clear that the acceleration of object 3 is $400\ m/s^2$ and it is greatest of all. So, the correct option is (3).

Tasya [4]3 years ago

3 0

Answer:

Object 3

Explanation:

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Voltmeter is the right answer

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3 years ago

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In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i

RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is $2.7 rad/s^2$

Explanation:

In a uniform circular motion, the centripetal acceleration is given by

$a_c=\omega^2 r$

where:

$\omega$ is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

$\omega=1.7 rad/s$ is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity ( $g=9.8 m/s^2$ ), so:

$a_c=3.2 g = 3.2(9.8)=31.4 m/s^2$

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

$r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m$

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

$a=4.4 g$

So, 4.4 times the acceleration due to gravity.

The total acceleration is the resultant of the centripetal acceleration ( $a_c$ ) and the tangential acceleration ( $a_t$ ):

$a=\sqrt{a_c^2+a_t^2}$

We know that:

a = 4.4g

$a_c = 3.2 g$

So, we can find the tangential acceleration:

$a_t = \sqrt{a^2-a_c^2}=\sqrt{(4.4g)^2-(3.2g)^2}=29.6 m/s^2$

The angular acceleration is related to the tangential acceleration by

$\alpha = \frac{a_t}{r}$

where r = 10.9 m is the length of the centrifuge. Substituting,

$\alpha = \frac{29.6}{10.9}=2.7 rad/s^2$

Learn more about centripetal and angular acceleration here:

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#LearnwithBrainly

8 0

3 years ago

A sample of monatomic ideal gas occupies 5.00 L at atmospheric pressure and 300 K (point A). It is warmed at constant volume to

leonid [27]

Answer:

(a) 0.203 moles

(b) 900 K

(d) 15 L

(e) A → B, W = 0, Q = Eint = 1,518.91596 J

B → C, W = Q ≈ 1668.69974 J Eint = 0 J

C → A, Q = -2,531.5266 J, W = -1,013.25 J, Eint = -1,518.91596 J

(g) ∑Q = 656.089 J, ∑W = 655.449 J, ∑Eint = 0 J

Explanation:

At point A

The volume of the gas, V₁ = 5.00 L

The pressure of the gas, P₁ = 1 atm

The temperature of the gas, T₁ = 300 K

At point B

The volume of the gas, V₂ = V₁ = 5.00 L

The pressure of the gas, P₂ = 3.00 atm

The temperature of the gas, T₂ = Not given

At point C

The volume of the gas, V₃ = Not given

The pressure of the gas, P₃ = 1 atm

The temperature of the gas, T₂ = T₃ = 300 K

(a) The ideal gas equation is given as follows;

P·V = n·R·T

Where;

P = The pressure of the gas

V = The volume of the gas

n = The number of moles present

R = The universal gas constant = 0.08205 L·atm·mol⁻¹·K⁻¹

n = PV/(R·T)

∴ The number of moles, n = 1 × 5/(0.08205 × 300) ≈ 0.203 moles

The number of moles in the sample, n ≈ 0.203 moles

(b) The process from points A to B is a constant volume process, therefore, we have, by Gay-Lussac's law;

P₁/T₁ = P₂/T₂

∴ T₂ = P₂·T₁/P₁

From which we get;

T₂ = 3.0 atm. × 300 K/(1.00 atm.) = 900 K

The temperature at point B, T₂ = 900 K

(d) For a constant temperature process, according to Boyle's law, we have;

P₂·V₂ = P₃·V₃

V₃ = P₂·V₂/P₃

∴ V₃ = 3.00 atm. × 5.00 L/(1.00 atm.) = 15 L

The volume at point C, V₃ = 15 L

(e) The process A → B, which is a constant volume process, can be carried out in a vessel with a fixed volume

The process B → C, which is a constant temperature process, can be carried out in an insulated adjustable vessel

The process C → A, which is a constant pressure process, can be carried out in an adjustable vessel with a fixed amount of force applied to the piston

(f) For A → B, W = 0,

Q = Eint = n·cv·(T₂ - T₁)

Cv for monoatomic gas = 3/2·R

∴ Q = 0.203 moles × 3/2×0.08205 L·atm·mol⁻¹·K⁻¹×(900 K - 300 K) = 1,518.91596 J

Q = Eint = 1,518.91596 J

For B → C, we have a constant temperature process

Q = n·R·T₂·㏑(V₃/V₂)

∴ Q = 0.203 moles × 0.08205 L·atm/(mol·K) × 900 K × ln(15 L/5.00 L) ≈ 1668.69974 J

Eint = 0

Q = W ≈ 1668.69974 J

For C → A, we have a constant pressure process

Q = n·Cp·(T₁ - T₃)

∴ Q = 0.203 moles × (5/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -2,531.5266 J

Q = -2,531.5266 J

W = P·(V₂ - V₁)

∴ W = 1.00 atm × (5.00 L - 15.00 L) = -1,013.25 J

W = -1,013.25 J

Eint = n·Cv·(T₁ - T₃)

Eint = 0.203 moles × (3/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -1,518.91596 J

Eint = -1,518.91596 J

(g) ∑Q = 1,518.91596 J + 1668.69974 J - 2,531.5266 J = 656.089 J

∑W = 0 + 1668.69974 J -1,013.25 J = 655.449 J

∑Eint = 1,518.91596 J + 0 -1,518.91596 J = 0 J

5 0

3 years ago

A spaceprobe has an 29.0 m length when measured at rest. What length

elixir [45]

Answer:

The observer sees the space-probe 9.055m long.

Explanation:

Let $L_0$ be the length of the space-probe when measured at rest, and $L$ be its length as observed by an observer moving at velocity $v$ , then

$(1).\: \: L = L_0\sqrt{1-\dfrac{v^2}{c^2} }$

Now, we know that $L_0 = 29.0m$ and $v = 0.95c$ , and putting these into $(1)$ we get:

$L = 29\sqrt{1-\dfrac{(0.95c)^2}{c^2} }$

$L = 29\sqrt{1-0.95^2 }$

$\boxed{L = 9.055m}$

Thus, an observer moving at 0.95c observes the space-probe to be 9.055m long.

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3 years ago

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valina [46]

Gold is a soft metal with a number of properties..It is both malleable and ductille,Also it's a heavy metal ,It is soft,Yellow....etc. Hope that helps

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