Three spheres (water, iron and ice) of the exact same volume are submerged in a tub of water. After the spheres are lined up, they are released. The spheres are made of plastic with the same density as water, ice, and iron.

Required:
a. Compare the weights of the three spheres.
b. Compare the buoyant forces on the three spheres.
c. What direction does the net force push on each of the spheres?
d. What happens to each sphere after it is released?

Answers

Answer 1

Answer:

(a) Iron > plastic > ice

(b) Same on all

(c) Iron downwards, plastic net force zero, ice upwards.

(d) Iron sphere sinks, plastic sphere is in equilibrium and ice sphere will floats.

Explanation:

Three spheres have same volume , plastic, ice and iron.

(a) The weight is given by

Weight = mass x gravity = volume x density x gravity

As the density of iron is maximum and the density of ice is least so the order of the weight is

Weight of iron > weight of plastic > weight of ice

(b) Buoyant force is given by

Buoyant force = Volume immersed x density of fluid x g

As they have same volume, density of fluid is same so the buoyant force is same on all the spheres.

(c) Net force is

F = weight - buoyant force  

So, the net force on the iron sphere is downwards

On plastic sphere is zero as the density of plastic sphere is same as water. On ice sphere it is upwards.

(d) Iron sphere sinks, plastic sphere is in equilibrium and ice sphere will floats.  


Related Questions

the rate of cooling determines ....... and ......​

Answers

Answer:

freezing point and melting point

A kind of variable that a researcher purposely changes in investigation is

Answers

Answer:

independent variable

Explanation:

A motor is designed to operate on 117 V and draws a current of 17.7 A when it first starts up. At its normal operating speed, the motor draws a current of 2.78 A. Obtain (a) the resistance of the armature coil, (b) the back emf developed at normal speed, and (c) the current drawn by the motor at one-third normal speed.

Answers

Answer:

Resistance of the armature coil = 6.61 ohms

Back emf developed at normal speed = 98.62 V (Approx.)

Current drawn by the motor at one-third normal speed = 12.73 A

Explanation:

Given:

Potential difference V = 117 V

Current = 17.7 A

Motor drawn current = 2.78 A

Find:

Resistance of the armature coil

Back emf developed at normal speed

Current drawn by the motor at one-third normal speed

Computation:

A] Resistance of the armature coil R = V/ I

Resistance of the armature coil = 117 / 17.7

Resistance of the armature coil = 6.61 ohms

B] Back emf developed at normal speed  = V- IR

Back emf developed at normal speed = 117 V - (2.78 A)(6.61 ohms)

Back emf developed at normal speed = 117 V - 18.37

Back emf developed at normal speed = 98.62 V (Approx.)

C] Current drawn by the motor at one-third normal speed = 17.7 A - (98.62/3)/(6.61 ohms)

Current drawn by the motor at one-third normal speed = 17.7 - 4.97

Current drawn by the motor at one-third normal speed = 12.73 A

What is the maximum speed at which a car can round a curve of 25-m radius on a level road if the coefficient of static friction between the tires and road is 0.80? ​

Answers

I assume the curve is flat and not banked. A car making a turn on the curve has 3 forces acting on it:

• its weight, mg, pulling it downward

• the normal force from contact with the road, n, pushing upward

• static friction, f = µn, directed toward the center of the curve (where µ is the coefficient of static friction)

By Newton's second law, the net forces on the car in either the vertical or horizontal directions are

F (vertical) = n - mg = 0

F (horizontal) = f = ma

where a is the car's centripetal acceleration, given by

a = v ²/r

and where v is the maximum speed you want to find and r = 25 m.

From the first equation, we have n = mg, and so f = µmg. Then in the second equation, we have

µmg = mv ²/r   ==>   v ² = µgr   ==>   v = √(µgr )

So the maximum speed at which the car can make the turn without sliding off the road is

v = √(0.80 (9.80 m/s²) (25 m)) = 14 m/s

A load of 25 kg is applied to the lower end and of a steal wire of length 25 m and thickness 3.0mm .The other end of wire is suspeded from a rigid support calculate strain and stress produced in the wire​

Answers

Answer:

the weight of the wire + 25kg

Explanation:

Partial tides _______. Question 7 options: represent various components of local tides that are resolved mathematically are predicted individually are added together to predict the height and timing of astronomical tides All of the above are correct. Only a and c are correct.

Answers

Complete Question

Partial tides __________.

Question 7 options:  

a. represent various components of local tides that are resolved mathematically  

b. are added together to predict the height and timing of astronomical tides  

c. consist of 4 components due to the influence of celestial bodies  

d. consist of up to 60 components due to astronomical and non-astronomical factors  

e. All of the above except c are correct.

Answer:

Option E

Explanation:

Generally

Partial tides represent various components of local tides that are resolved mathematically

Partial tides  are added together to predict the height and timing of astronomical tides

Partial tides consist of up to 60 components due to astronomical and non-astronomical factors

But Partial tides do not consist of 4 components due to the influence of celestial bodies

Therefore

All of the above except c are correct.

Option E

b. Projectile on cliff (range)
An object of mass 5 kg is projected at an angle of 25° to the horizontal with a speed of 22 ms-1 from the top of the cliff.
The height of the cliff is 21 m. Take g, the acceleration due to gravity, to be 9.81 ms2
How far horizontally (to 1 decimal place) from the base of the cliff does the object land?

Answers

Answer:

x = 41.28 m

Explanation:

This is a projectile launching exercise, let's find the time it takes to get to the base of the cliff.

Let's start by using trigonometry to find the initial velocity

         cos 25 = v₀ₓ / v₀

         sin 25 = Iv_{oy} / v₀

         v₀ₓ = v₀ cos 25

         v_{oy} = v₀ sin 25

         v₀ₓ = 22 cos 25 = 19.94 m / s

         v_{oy} = 22 sin 25 = 0.0192 m / s

let's use movement on the vertical axis

         y = y₀ + v_{oy} t - ½ g t²

     

when reaching the base of the cliff y = 0 and the initial height is y₀ = 21 m

         0 = 21 + 0.0192 t - ½ 9.81 t²

         4.905 t² - 0.0192 t - 21 = 0

         t² - 0.003914 t - 4.2813 =0

we solve the quadratic equation

        t = [tex]\frac{ 0.003914\ \pm \sqrt{0.003914^2 + 4 \ 4.2813 } }{2}[/tex]

        t = [tex]\frac{0.003914 \ \pm 4.13828}{2}[/tex]

        t₁ = 2.07 s

        t₂ = -2.067 s

since time must be a positive scalar quantity, the correct result is

        t = 2.07 s

now we can look up the distance traveled

         x = v₀ₓ t

         x = 19.94  2.07

         x = 41.28 m

A 13.6 kg block is tied at the top of an incline to a tree. If the incline is 35.5 degrees and the coefficient of friction between the sled and the incline is .45, What is the tension force between the block and the tree

Answers

Answer:

Explanation:

ASSUMING that block = sled AND that the rope is parallel to the slope.

The force acting parallel due to the weight is

13.6(9.81)sin35.5 = 77.475 N

The maximum friction force is

(0.45)13.6(9.81)cos35.5 = 48.877 N

If rope tension is T

77.475 - 48.877 < T < 77.475 + 48.877

            28.6 N < T < 126 N

28.6 N will occur if the block is on the verge of sliding downhill

126 N will occur if the block is on the verge of sliding uphill

Could be any value between them.

A parallel plate capacitor is constructed using two square metal sheets, each of side L = 10 cm. The plates are separated by a distance d = 2 mm and a voltage applied between the plates. The electric field strength within the plates is E = 4000 V/m. The energy stored in the capacitor is

Answers

Answer:

The energy stored is 1.4 x 10^-9 J.

Explanation:

Side of square, L = 10 cm = 0.1 m

Distance, d = 2 mm = 0.002 m

Electric field, E = 4000 V/m

The energy stored in the capacitor is

[tex]U = 0.5 C V^2[/tex]

The capacitance is given by

[tex]C = \frac{\varepsilon o A}{d}\\\\So \\\\U = 0.5\frac{\varepsilon o A}{d}\times E^2 d^2\\\\U = 0.5\times 8.85\times 10^{-12}\times 0.1\times 0.1\times 4000\times 4000\times 0.002\\\\U = 1.4\times10^{-9} J[/tex]

Ashley, a psychology major, remarks that she has become interested in the study of intelligence. In other words, Ashley is interested in?

Answers

Group of answer choices.

a) the capacity to learn from experience, solve problems, and to adapt to new situations.

b) how behavior changes as a result of experience.

c) the factors directing behavior toward a goal.

d) the ability to generate novel

Answer:

a) the capacity to understand the world, think rationally, and use resources effectively.

Explanation:

Psychology can be defined as the scientific study of both the consciousness and unconsciousness of the human mind such as feelings, emotions and thoughts, so as to understand how it functions and affect human behaviors in contextual terms.

This ultimately implies that, psychology focuses on studying behaviors and the mind that controls it.

In this scenario, Ashley who is a psychology major, stated that she's interested in the study of intelligence.

Intelligence can be defined as a measure of the ability of an individual to think, learn, proffer solutions to day-to-day life problems and effectively make informed decisions.

In other words, Ashley is interested in the capacity of humans to understand the world, think rationally, and use resources effectively to produce goods and services that meet the unending requirements, needs or wants of the people (consumers or end users) living around the world.

why the stone moves away when the string is broken rotation​

Answers

Answer:

When a stone is going around a circular path, the instantaneous velocity of stone is acting as tangent to the circle. When the string breaks, the centripetal force stops to act. Due to inertia, the stone continues to move along the tangent to circular path. So, the stone flies off tangentially to the circular path

Answer:

when the string's rotation is broken, there will be no centripetal force to keep the stone stationary. Thus, the stone will flung away when the rotation is stopped

i don't understand this, can someone help please?? ​

Answers

Explanation:

N2 + H2 --> NH3

balance them:

N2 + 3 H2 --> 2 NH3

so if 6 moles of N2 react, 12 moles of NH3 will form.

(you have to look at the big number in front, in this case its N2 and 2 NH3, therefore the amount of N2 will produce double the amount of NH3 )

which of the following can not happen when a light ray strikes a new medium

Answers

Answer:

amplification

Explanation:

reflection can happen

some amount of lighr get absorbed

something gets refracted

but amplification cant

Can someone please help!!!

Answers

Answer:

W = F • ∆x

so for work to be done, a force and displacement has to be in the same direction. (Ex: a box is being pushed forward and it's also moving forward.)

the plane of a 5.0 cm by 8.0 cm rectangular loop wire is parallel to a 0.19 t magnetic field. if the loop carries a current of 6.2 amps, what is the magnitude of the torque on the loop

Answers

here is the state ment :0.2 loop

Proper physical exercise makes bones _[blank 1]_.
People with stronger muscles and bones have better _[blank 2]_.

Which option shows the words that correctly fill in blank 1 and blank 2, in that order?


longer, flexibilitylonger, flexibility , ,

stronger, posturestronger, posture , ,

longer, posturelonger, posture , ,

stronger, flexibility

Answers

stronger, posturestronger, posture

hope that helped

How can I solve the following?

In (Figure 1), let V = 15.0 V and C1=C2=C3= 24.2 μF.

Part A: How much energy is stored in the capacitor network as shown in (Figure 1)?

Part B: How much energy would be stored in the capacitor network if the capacitors were all in series?

Part C: How much energy would be stored in the capacitor network if the capacitors were all in parallel?

Answers

Answer:

Part A - 4.084 mJ

Part B - 0.908 mJ

Part C - 8.168 mJ

Explanation:

Part A: How much energy is stored in the capacitor network as shown in (Figure 1)?

Since capacitors C₂ and C₃ are in series, their equivalent capacitance is C',

1/C' = 1/C₂ + 1/C₃      (Since C₁ = C₂ = C₃ = C)

1/C' = 1/C + 1/C

1/C' = 2/C

C' = C/2

Since C' is in parallel with C₁, the equivalent capacitance for the circuit is C" = C₁ + C' = C + C/2 = 3C/2

C" = 3C/2

The energy stored in the circuit, W = 1/2C"V² where C" = equivalent capacitance = 3C/2 and V = voltage = 15.0 V

W = 1/2C"V²

W = 1/2(3C/2)V²

W = 3CV²/4

since C = 24.2 μF = 24.2 × 10⁻⁶ F

W = 3CV²/4

W = 3 × 24.2 × 10⁻⁶ F (15.0 V)²/4

W = 3 × 24.2 × 10⁻⁶ F × 225 V²/4

W = 16335/4 × 10⁻⁶ FV²

W = 4083.75 × 10⁻⁶ J

W = 4.08375 × 10⁻³ J

W = 4.08375 mJ

W ≅ 4.084 mJ

Part B: How much energy would be stored in the capacitor network if the capacitors were all in series?

If the capacitors are connected in series, their equivalent resistance is C'

and 1/C' = 1/C₁ + 1/C₂ + 1/C₃

Since C₁ = C₂ = C₃ = C

1/C' = 1/C + 1/C + 1/C

1/C' = 3/C

C' = C/3

The energy stored in the circuit, W = 1/2C'V² where C' = equivalent capacitance = C/3 and V = voltage = 15.0 V

W = 1/2C'V²

W = 1/2(C/3)V²

W = CV²/6

since C = 24.2 μF = 24.2 × 10⁻⁶ F

W = CV²/6

W = 24.2 × 10⁻⁶ F (15.0 V)²/6

W = 24.2 × 10⁻⁶ F × 225 V²/6

W = 5445/6 × 10⁻⁶ FV²

W = 907.5 × 10⁻⁶ J

W = 0.9075 × 10⁻³ J

W = 0.9075 mJ

W ≅ 0.908 mJ

Part C: How much energy would be stored in the capacitor network if the capacitors were all in parallel?

If the capacitors are connected in parallel, their equivalent resistance is C'

and C' = C₁ + C₂ + C₃

Since C₁ = C₂ = C₃ = C

C' = C + C + C

C' = 3C

The energy stored in the capacitor network, W = 1/2C'V² where C' = equivalent capacitance = 3C and V = voltage = 15.0 V

W = 1/2C'V²

W = 1/2(3C)V²

W = 3CV²/2

since C = 24.2 μF = 24.2 × 10⁻⁶ F

W = 3CV²/2

W = 3 × 24.2 × 10⁻⁶ F (15.0 V)²/2

W = 3 × 24.2 × 10⁻⁶ F × 225 V²/2

W = 16335/2 × 10⁻⁶ FV²

W = 8167.5 × 10⁻⁶ J

W = 8.1675 × 10⁻³ J

W = 8.1675 mJ

W ≅ 8.168 mJ

Why we use semiconductor instead of metal in thermopile.

Answers

Answer:

Semiconductors are not normal materials. They have special properties which conductors/metals cannot exhibit. The main reason for the behavior of semiconductors is that they have paired charge carriers-the electron-hole pair. This is not available in metals.

4. A diver is 20 m underwater and they are startled by a shark. They are tempted to take a big breath of air, drop their gear, and swim to the surface while holding their breath. Explain why this is dangerous g

Answers

Answer:

Explanation:

The air enters their lungs at the same pressure as the water at that depth.

If they hold their breath as they rise to atmospheric pressure, the expanding volume of air (due to decreasing pressure) trapped in their lungs will hyperextend the alveoli in their lungs, likely tearing blood lines and risking death by drowning in their own blood.

The coefficients of friction between a race cars tyres and the track surface are

Answers

the question is about tyres of a race car, which are made of rubber and will be in contact with a race track, which is generally made from asphalt, the static coefficient of friction is in the range of (0.5–0.8), in dry conditions (Source: Friction and Friction Coefficients ).

Explanation:

please mark me as a brainlieast

The propeller on a boat motor is initially rotating at 8 revolutions per second. As the boat captain reduces the boat speed, the propeller SLOWS at a steady rate of 0.9 revolutions per second per second. After 17 revolutions, how fast is the propeller spinning in revolutions per second

Answers

Answer: [tex]5.77\ rps[/tex]

Explanation:

Given

Initial angular velocity is [tex]\omega_i=8\ rps[/tex]

rate of reduction [tex]\alpha=0.9 rev/s^2[/tex]

after 17 revolution i.e. [tex]\theta =17\ rev[/tex]

using [tex]\Rightarrow \omega_f^2-\omega_i^2=2\alpha\theta[/tex]

Insert the values

[tex]\Rightarrow \omega_f^2=8^2-2\times (0.9)\times17\\\Rightarrow \omega_f^2=33.4\\\Rightarrow \omega_f=5.77\ rps[/tex]

Consider a box with two gases separated by an impermeable membrane. The membrane can move back and forth, but the gases cannot penetrate the membrane. The left side is filled with gas A and the right side is filled with gas B. We will assume that equipartition applies to both gases, but gas A has an excluded volume due to large molecules so its entropy has a different formula.

SA=NAkln(VA+ bNA)+f(UA,NA)
SB=NBkln(VB)+f(UB,NB)

Required:
If NA= 1 moles, NB = 2 moles, the total volume of the box is 1 m3, and b= 4 × 10-4 m3/mole, then find the equilibrium value of VA by maximizing the total entropy.

Answers

Answer:

The answer is "[tex]0.3336\ m^3[/tex]"

Explanation:

Using the Promideal gas law:

[tex]P_A=P_B\\\\P_A(V_A-\eta_A b)= \eta_A RT......(1)\\\\P_B V_B=\eta_B \bar{R}T........(2)\\\\From (1) \zeta (2)\\\\[/tex]  

[tex]\frac{\eta_A}{V_A-\eta_A b}=\frac{\eta B}{V B}\\\\ \frac{V A- \eta_A b}{V B}=\frac{\eta A}{\eta B }\\\\ \frac{V A-b}{V B}=\frac{1}{2}\\\\V A+V B=1\\\\V B =1- V A\\\\\frac{V A-b}{1-V A}=\frac{1}{2}\\\\2V A-2b=1-V A\\\\3 V A=1+2b\\\\V A=\frac{1+2b}{3}\\\\[/tex]

      [tex]=\frac{1+2(4\times 10^{-4})}{3}\\\\=0.3336\ m^3[/tex]

The equilibrium value of Va is 0.3336 m³.

Ideal gas law

The equilibrium value of Va is determine by applying ideal gas law as shown below;

Pressure of gas A = Pressure of gas B

Pa = Pb

Pa(Va - nab) = naRT----(1)

PbVb = nbRT -----(2)

Solve equation (1) and (2)

[tex]\frac{P_b}{RT} = \frac{n_b}{V_b} \\\\\frac{P_b}{P_a(V_a- n_ab)/n_a} = \frac{n_b}{V_b}\\\\\frac{n_a}{V_a - n_ab} = \frac{n_b}{V_b} \\\\\frac{V_a - n_ab}{V_b} = \frac{n_a}{n_b} \\\\\frac{V_a - b}{V_b} = \frac{1}{2}[/tex]

Va + Vb = 1

Vb = 1 - Va

[tex]\frac{V_a - b}{1 - V_a} = \frac{1}{2}[/tex]

2Va - 2b = 1 - Va

3Va = 1 + 2b

[tex]V_ a = \frac{1 + 2b}{3} \\\\V_a = \frac{1 + (2 \times 4\times 10^{-4})}{3} \\\\V_a = 0.3336 \ m^3[/tex]

Thus, the equilibrium value of Va is 0.3336 m³.

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Rachel has good distant vision but has a touch of presbyopia. Her near point is 0.60 m. Part A When she wears 2.0 D reading glasses, what is her near point

Answers

Answer:

The right answer is "0.273 m".

Explanation:

Given:

Power (P),

[tex]\frac{1}{f} = 2D[/tex]

Near point,

u = 0.6 m

As we know,

⇒ [tex]\frac{1}{v} -\frac{1}{u}=\frac{1}{f} = 2[/tex]

By substituting the values, we get

⇒ [tex]\frac{1}{v} -\frac{1}{0.6} =2[/tex]

            [tex]\frac{1}{v}=2+\frac{1}{0.6}[/tex]

            [tex]\frac{1}{v} =\frac{1.2+1}{0.6}[/tex]

            [tex]\frac{1}{v}=\frac{2.2}{0.6}[/tex]    

By applying cross-multiplication, we get

          [tex]0.6=2.2 \ v[/tex]

            [tex]v = \frac{0.6}{2.2}[/tex]

      [tex]S_{near} = 0.273 \ m[/tex]

The pressure of sea water increases by 1.0atm for each 10m increase in the depth, by what percentage is the density of water increased in the deepest ocean of water of 12km. Compressibility is 5.0×10^-5 atm

Answers

The percentage by which the water density increased is 4.1[tex]\mathbf{\overline 6}[/tex] %

The known values are;

The increase in pressure per 10 meter increase in depth = 1.0 atm

The depth of the deepest ocean = 12 km = 12,000 m

The compressibility of the ocean = 5.0 × 10⁻⁵ 1/atm

The unknown

The percentage the density of water increased in the deepest ocean

Strategy;

Find the pressure at the deepest point of the deepest ocean and apply the compressibility

We have;

[tex]\mathbf{Compressibility = \dfrac{1}{V} \times \dfrac{\partial V}{\partial p}}[/tex]

The change in pressure, [tex]\partial p[/tex] = (12,000 m/(10 m)) × 1.0 atm = 1,200 atm

Therefore, we have for one cubic meter of water

[tex]\mathbf{5.0 \times 10^{-5} \ atm^{-1} = \dfrac{1}{1 \, m^3} \times \dfrac{\partial V}{1,200 \, atm}}[/tex]

Therefore;

[tex]\mathbf{\partial}[/tex]V = 5.0 × 10⁻⁵ atm⁻¹ × 1 m³ × 1,200 atm = 0.06 m³

The new volume = V - [tex]\mathbf{\partial}[/tex]V

∴ The new volume = 1 m³ - 0.06 m³ = 0.94 m³

The initial density = mass/(1 m³)

The new density = mass/(0.96 m³)

The percentage increase in density, [tex]\partial[/tex]ρ%, is given as follows;

[tex]\mathbf{\partial p \% = \dfrac{ \dfrac{Mass}{0.96 \ m^3} - \dfrac{Mass}{1 \ m^3} }{ \dfrac{Mass}{1 \ m^3}} \times 100 = \dfrac{25}{6} \% = 4.1 \overline 6 \%}[/tex]

∴  [tex]\mathbf{\partial}[/tex]ρ% =  4.1[tex]\mathbf {\overline 6}[/tex] %

The percentage by which the water density increased, [tex]\partial[/tex]ρ% = 4.1[tex]\mathbf{\overline 6}[/tex] %

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If a negatively charged particle is placed inside a uniform electric field the electric force that will act on that particle points in what direction in reference to the electric field lines?

Answers

Answer:

opposite direction

Explanation:

An electric field is defined as a physical field which surrounds the electrically charged particles that exerts force on the other particles on the field.

Now when an electron or a negatively charged particle enters a uniform electric field, the electric forces acts on the negatively charged particles and it forces the particle to move in the direction which is opposite to the direction of the field. In an uniform electric field, the field lines are parallel.

Answer:

Explanation:

negatively charged particle is placed inside uniform electric field

The force on the charge due to the electric field is

F = q E

when the charge is negative so the force on the charge is opposite to the direction of electric field.

The electric field is opposite to the force.

A 0.3 kg mass attached to a 1.5 m long string is whirled in a horizontal circle at a speed of 6.0 m/s. What is the tension in the string? (neglect gfavity)

Answers

Answer:

Hi I hope this is correct!

Explanation:

You can use this formula to solve this question T = mv^2/R

m = 0.3 kg , v = 6.0 m/s , R = 1.5 m

T = (0.3 kg)(6.0 m/s)^2 / 1.5 m  

  = 7.2 Newtons

Hope this helps! Best of luck <3

Determine the acceleration of a pendulum bob as it passes through an angle of 15 degrees to the right of the equilibrium point.

Answers

Answer:

Explanation:

Since energy is conserved:

2

mu  

2

 

=  

2

mv  

2

 

+mgh

⇒u  

2

=v  

2

+2gh

⇒(3)  

2

=v  

2

+2(9.8)(0.5−0.5cos60)

⇒v=2m/s

Acceleration of the simple pendulum is 2.62 m/s².

What is meant by a simple pendulum ?

When a point mass is suspended from a fixed support by a light, non-extensible string, the instrument is said to be a simple pendulum.

Here,

Let the mass of the bob be m. The simple pendulum is attached to the fixed support with a string having length l. The pendulum makes an angle of 15° with the vertical from the equilibrium point.

Let T be the tension acting on the string.

As, the bob passes through the angle,

The weight of the bob becomes equal to the vertical component of the tension.

mg = T cos15°

Also, the horizontal component of the tension,

T sin15° = ma

By solving these two equations, we get that,

Acceleration of the simple pendulum,

a = g tan15°

a = 9.8 x 0.267

a = 2.62 m/s²

Hence,

Acceleration of the simple pendulum is 2.62 m/s².

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A block with a mass of 0.26 kg is attached to a horizontal spring. The block is pulled back from its equilibrium position until the spring exerts a force of 1.2 N on the block. When the block is released, it oscillates with a frequency of 1.4 Hz. How far was the block pulled back before being released?

Answers

Answer:

2

Explanation:

pulling force because of it force

Answer:

5.9 cm

Explanation:

f: frequency of oscillation

frequency of oscillationk: spring constant

frequency of oscillationk: spring constantm: the mass

[tex]f = \frac{1}{2\pi} \sqrt{ \frac{k}{m} } [/tex]

in this problem we know,

F= 1.4 Hz

m= 0.26 kg

By re-arranging the formula we get

[tex]k = {(2\pi \: f )}^{2} m = {(2\pi(1.4hz))}^{2} 0.26kg = 20.1 \frac{n}{m} [/tex]

The restoring force of the spring is:

F= kx

where

F= 1.2 N

k= 20.1 N/m

x: the displacement of the block

[tex]x = \frac{f}{k} = \frac{1.2 \: n}{20.1 \frac{n}{m} } = 0.059m \: = 5.9 \: cm[/tex]

What is a measure of how hard it is to stop a moving object?
25.
A. gravity
B. weight
C.
inertia
D. momentum

Answers

Answer:

C. inertia

Explanation:

inertia describes an object’s resistance to change in motion (or to get in motion due to a lack of motion), and momentum describes how much motion it has.

both are connected, as inertia depends on the object's momentum, but the answer here is inertia.

MULTIPLE CHOICE
What is the role of the Sun in a forest ecosystem?
What is the Role of the sun in a forest ecosystem?

Answers

Answer:

It is to produce sunlight in the forest plants

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