Omar observes that many buildings in his city were built using limestone. He has read that acid rain can damage limestone. He also knows that limestone reacts with acids and that chemical reactions are affected by temperature. With this in mind, he conducts an investigation to see how the amount of damage to the limestone is affected by the amount of acid on the stone.Which statement describes the correct plan for his procedure?Rhonda watched a video taken by a camera that was lifted into the upper atmosphere by a weather balloon. She saw the balloon pop when it reached a certain height. Afterward, Rhonda wondered what effect the air pressure at high altitudes has on the volume of gas in balloons.

What scientific practice is Rhonda performing

Answers

Answer 1

Answer:

Answer:he conducts the investigation to see the effect acidic water has on limestone

Explanation:

Related Questions

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Monochromatic coherent light shines through a pair of slits. If the distance between these slits is decreased, which of the following statements are true of the resulting interference pattern?A) The distance between the maxima stays the same.B) The distance between the maxima decreases.C) The distance between the minima stays the same.D) The distance between the minima increases.E) The distance between the maxima increases.
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. Using your knowledge of circular (centripetal) motion, derive an equation for the radius r of the circular path that electrons follow in terms of the magnetic field B, the electrons' velocity v, charge e, and mass m. You may assume that the electrons move at right angles to the magnetic field.2. Recall from electrostatics, that an electron obtains kinetic energy when accelerated across a potential difference V. Since we can directly measure the accelerating voltage V in this expierment, but not the electrons' velocity v, replace velocity in your previous equation with an expression containing voltage. The electron starts at rest. Now solve this equation for e/m.

You should obtain e/m = 2V/(B^2)(r^2)

3. The magnetic field on the axis of a circular current loop a distance z away is given by

B = mu I R^2 / 2(R^2 + z^2)^ (3/2)

where R is the radius of the loops and I is the current. Using this result , calculate the magnetic field at the midpoint along the axis between the centers of the two current loops that make up the Helmholtz coils, in terms of their number of turns N, current I, and raidus R.Helmholtz coils are separated by a distance equal to their raidus R. You should obtain:

|B| = (4/5)^(3/2) *mu *NI/R = 9.0 x 10^-7 NI/R

where B is magnetic field in tesla, I is in current in amps, N is number of turns in each coil, and R is the radius of the coils in meters

Answers

Answer:

Explanation:

Magnetic field creates a force perpendicular to a moving charge in its field which is equal to Bev where B is magnetic field , e is amount of charge on the moving charge and v is the velocity of charge particle .

This force provides centripetal force for creation of circular motion. If r be the radius of the circular path

Bev = mv² / r

r = mv / Be

2 ) If an electron is accelerated by an electric field created by potential difference V then electric field

= V / d where d is distance between two points having potential difference v .

force on charged particle

electric field x charge

= V /d x e

work done by field

= force x distance

= V /d x e x d

V e

This is equal to kinetic energy created

V e = 1/2 mv²

= 1/2 m (r²B²e² / m² )

V = r²B²e/ 2 m

e / m = 2 V/ r²B²

3 )

B = $(\mu* I* R^2)/(2(R^2+Z^2)^(3)/(2) )$

In Helmholtz coils , distance between coil is equal to R so Z = R/2

B = $(\mu* I* R^2)/(2(R^2+(R^2)/(4) )^(3)/(2) )$

For N turns of coil and total field due to two coils

B = $(\mu* I* N)/(R*((5)/(4))^(3)/(2) )$

= $(\mu* I* N)/(R)* ((4)/(5))^(3)/(2)$

= 9.0 x 10^-7 NI/R

A positive magnification means the image is inverted compared to the object

Answers

False

Explanation:

A positive magnification means the image is erect compared to the object. Magnifications with values greater than one represent images that are smaller than the object. A magnification of 1 (plus or minus) means that the image is the same size as the object. If m has a magnitude greater than 1 the image is larger than the object, and an m with a magnitude less than 1 means the image is smaller than the object. If the magnification is positive, the image is upright compared to the object; if m is negative, the image is inverted compared to the object.

The surface tension of a liquid is to be measured using a liquid film suspended on a U-shaped wire frame with an 12-cm-long movable side. If the force needed to move the wire is 0.096 N, determine the surface tension of this liquid in air.

Answers

Answer:

σ = 0.8 N/m

Explanation:

Given that

L = 12 cm

We know that 1 m = 100 cm

L = 0.12 m

The force ,F= 0.096 N

Lets take surface tension = σ

We know that surface tension is given as

$\sigma =(F)/(L)\n\sigma =(F)/(L)\nNow\ by\ putting\ the\ values\n\sigma =(0.096)/(0.12)\ N/m\n\sigma=0.8\ N/m$

Therefore the surface tension σ will be 0.8 N/m .

σ = 0.8 N/m

Final answer:

The surface tension of the liquid in air is 0.8 N/m.

Explanation:

To determine the surface tension of the liquid, we need to use the formula F = yL, where F is the force needed to move the wire, y is the surface tension, and L is the length of the wire. In this case, F = 0.096 N and L = 12 cm. We can rearrange the formula to solve for y: y = F / L. Plugging in the values, we get y = 0.096 N / 0.12 m = 0.8 N/m. So, the surface tension of the liquid in air is 0.8 N/m.

Learn more about Surface tension here:

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An electron enters a region of uniform electric field with an initial velocity of 50 km/s in the same direction as the electric field, which has magnitude E = 50 N/C, (a) what is the speed of the electron 1.5 ns after entering this region? (b) How far does the electron travel during the 1.5 ns interval?

Answers

Answer:

(a). The speed of the electron is $3.68*10^(4)\ m/s$

(b). The distance traveled by the electron is $4.53*10^(-5)\ m$

Explanation:

Given that,

Initial velocity = 50 km/s

Electric field = 50 N/C

Time = 1.5 ns

(a). We need to calculate the speed of the electron 1.5 n s after entering this region

Using newton's second law

$F = ma$ .....(I)

Using formula of electric force

$F = qE$ .....(II)

from equation (I) and (II)

$-qE= ma$

$a = (-qE)/(m)$

(a). We need to calculate the speed of the electron

Using equation of motion

$v = u+at$

Put the value of a in the equation of motion

$v = 50*10^(3)-(1.6*10^(-19)*50)/(9.1*10^(-31))*1.5*10^(-9)$

$v=36813.18\ m/s$

$v =3.68*10^(4)\ m/s$

(b). We need to calculate the distance traveled by the electron

Using formula of distance

$s = ut+(1)/(2)at^2$

Put the value in the equation

$s = 3.68*10^(4)*1.5*10^(-9)-(1)/(2)*(1.6*10^(-19)*50)/(9.1*10^(-31))*(1.5*10^(-9))^2$

$s=0.0000453\ m$

$s=4.53*10^(-5)\ m$

Hence, (a). The speed of the electron is $3.68*10^(4)\ m/s$

(b). The distance traveled by the electron is $4.53*10^(-5)\ m$

A small ball of mass m is held directly above a large ball of mass M with a small space between them, and the two balls are dropped simultaneously from height H. (The height is much larger than the radius of each ball, so you may neglect the radius.) The large ball bounces elastically off the floor and the small ball bounces elastically off the large ball. a) For which value of the mass m, in terms of M, does the large ball stop when it collides with the small ball? b) What final height, in terms of H, does the small ball reach?

Answers

a) The large sphere has 3 times the mass of the small sphere

b) The final height at which small ball reach y = 4H

What will be the mass of the sphere and height covered by the small ball?

We must start this problem by calculating the speed with which the spheres reach the floor

$V_f^2=V_o^2-2gy$

As the spheres are released v₀ = 0

$V_f^2=2gH$

$V_f=√(2gH)$

The two spheres arrive at the same speed to the floor.

The largest sphere clashes elastically so that with the floor it has a much higher mass, the sphere bounces with the same speed with which it arrived, the exit speed of the spheres

$V_(10)=\sqrt2gH$

The big sphere goes up and the small one down, the two collide, let's form a system that contains the two spheres, let's use moment conservation

Let's call

$V_h=\sqrt2gH$

Small sphere m₂ and $V_(20)=-\sqrt2gH=-V_h$

Large sphere m₁ and $V_(10)=√(2gH)=V_h$

Before crash

$P_o=m_1V_(10)+m_2V_(20)$

After the crash

$P_f=m_1V_(1f)+m_2V_(2f)$

$P_o=P_f$

$m_1V_(10)+m_2V_(20)=m_1V_(1f)+m_2V_(2f)$

The conservation of kinetic energy

$K_o=(1)/(2) m_1V_(10)^2+(1)/(2) m_2V_(20)^2$

$K_f=(1)/(2) m_1V_(1f^2)+(1)/(2) m_2V_(2f)^2$

$K_o=K_f$

$K_o=(1)/(2) m_1V_(10)^2+(1)/(2) m_2V_(20)^2$ $=$ $K_f=(1)/(2) m_1V_(1f^2)+(1)/(2) m_2V_(2f)^2$

Let's write the values

$-m_1V_h+m_2V_h=m_1V_(1f)+m_2V_(2f)$

$m_1V_h^2+m_2V_h^2=m_1V_(1f)^2+m_2V_(2f)^2$

The solution to this system of equations is

$m_t=m_1+m_2$

$V_(1f)=((m_1-m_2))/(m_tV_(10)) +(2m_2)/(m_tV_2)$

$V_(2f)=(2m_1)/(m_tV_(10)) +(m_2-m_1)/(m_tV_2)$

The large sphere is labeled 1, we are asked for the mass so that $V_(1f)$ = 0, let's clear the equation

$V_(1f)=(m_1-m_2)/(m_tV_(10)) +(2m_2)/(m_tV_(20))$

$0=(m_1-m_2)/(m_tV_h) +(2m_2)/(m_t(-V_h))$

$(m_1-m_2)/(m_tV_h) =(2m_2)/(m_tV_h)$

$(m_1-m_2)=2m_2$

$m_1=3m_2$

The large sphere has to complete 3 times the mass of the sphere1 because it stops after the crash.

b) Let us calculate with the other equation the speed with which the sphere comes out2 (small)

$V_(2f)=(2m_1)/(m_tV_(10)) +(m_2-m_1)/(m_tV_(20))$

$V_(2f)=(2m_1)/(m_tV_h) +(m_2-m_1)/(m_t(-V_h))$

In addition, we know that m₁ = 3 m₂

$m_t=3m_2+m_2$ mt = 3m2 + m2

$m_t=4m_2$

$V_(2f)=(2* 3m_2)/(4m_2V_h-(m_2-3m_2)4m_2V_h)$

$V_(2f)=(3)/(2) V_h+(1)/(2) V_h$

$V_(2f)=2V_h$

$V_(2f)=2\sqrt{2gh$

This is the rate of rising of sphere 2 (small. At the highest point, it zeroes velocity $V_f$ = 0

$V^2=V_(2f)^2-2gy$

$0=(2√(2gh))^2-2gy$

$y=4H$

Thus

a) The large sphere has 3 times the mass of the small sphere

b) The final height at which small ball reach y = 4H

To know more about the Laws of collisions follow

brainly.com/question/7538238

Answer:

a) the large sphere has 3 times the mass of the small sphere

b) y = 4H

Explanation:

We must start this problem by calculating the speed with which the spheres reach the floor

vf² = vo² - 2g y

As the spheres are released v₀ = 0

vf² = - 2g H

vf = √ (2g H)

The two spheres arrive with the same speed to the floor.

The largest sphere clashes elastically so that with the floor it has a much higher mass, the sphere bounces with the same speed with which it arrived, the exit speed of the spheres

V₁₀ = √2gH

The big sphere goes up and the small one down, the two collide, let's form a system that contains the two spheres, let's use moment conservation

Let's call vh = √2gH

Small sphere m₂ and v₂o = - √2gH = -vh

Large sphere m₁ and v₁o = √ 2gh = vh

Before crash

p₀ = m₁ v₁₀ + m₂ v₂₀

After the crash

pf = m₁ v₁f + m₂ v₂f

po = pf

m₁ v₁₀ + m₂ v₂₀ = m₁ v₁f + m₂ v₂f

The conservation of kinetic energy

Ko = ½ m₁ v₁₀² + ½ m₂ v₂₀²

Kf = ½ m₁ v₁f² + ½ m₂ v₂f²

Ko = KF

½ m₁ v1₁₀² + ½ m₂ v₂₀² = ½ m₁ v₁f² + ½ m₂ v₂f²

Let's write the values

-m₁ vh + m₂ vh = m₁ v₁f + m₂ v₂f

m₁ vh² + m₂ vh² = m₁ v₁f² + m₂ v₂f²

The solution to this system of equations is

mt = m₁ + m₂

v1f = (m₁-m₂) / mt v₁₀ + 2m₂ / mt v₂

v₂f = 2m₁ /mt v₁₀ + (m₂-m₁) / mt v₂

The large sphere is labeled 1, we are asked for the mass so that v1f = 0, let's clear the equation

v₁f = (m₁-m₂) / mt v₁₀ + 2m₂ / mt v₂₀

0 = (m₁-m₂) / mt vh + 2 m₂ / mt (-vh)

(m₁-m₂) / mt vh = 2 m₂ / mt vh

(m₁-m₂) = 2m₂

m₁ = 3 m₂

The large sphere has to complete 3 times the mass of the sphere1 because it stops after the crash.

b) Let us calculate with the other equation the speed with which the sphere comes out2 (small)

v₂f = 2m₁ / mt v₁₀ + (m₂-m₁) / mt v₂₀

v₂f = 2 m₁ / mt vh + (m₂-m₁) mt (-vh)

In addition, we know that m₁ = 3 m₂

mt = 3m2 + m2

mt= 4m2

v₂f = 2 3m₂ / 4m₂ vh - (m₂-3m₂) 4m₂ vh

v₂f = 3/2 vh +1/2 vh

v₂f = 2 vh

v₂f = 2 √ 2gh

This is the rate of rise of sphere 2 (small. At the highest point its zero velocity vf = 0

V² = v₂f² - 2 g Y

0 = (2√2gh)² - 2gy

2gy = 4 (2gH)

y = 4H

Let's say you have a plot for Pendulum experiment. Let's assume g for this experiment was measured (from the slope of the plot) to be 9.78 [ms^-2]. The vertical intercept, however, is 0.021 [s^2]. What might this translate to for a measurement of the length offset systematic in all the length measurements? Length Offset =[mm] .011