How would you calculate the number of moles of oxygen you had?

Answer:

magnification is 4

Explanation:

m= image height / object height

m= 12/3

m= 4

Answer:

F = ⅔ F₀

Explanation:

For this exercise we use Coulomb's law

         F = k q₁q₂ / r²

let's use the subscript "o" for the initial conditions

          F₀ = k q² / r²

now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r

we substitute

          F = k 4q² / 9 r²

          F = k q² r² 4/9

          F = ⅔ F₀

Answer:

The second one.

Explanation:

It caused both to change speed because they have both the same mass.

Answer:

Did you try searching it up

Explanation:

Answer:

so 9/3=3 current is 3 amperes

Explanation:

The fomula to calculate resistance is:

voltage/cutrent

9 V/3 A= 3 ohms

Answer:

Explanation:

Im going to be using the rules for significant digits properly so I hope you're quite familiar with them. The equation we need for this is

F - f = ma where F is the applied force (our unknown), f is the frictional force, m is the mass, and a is the acceleration. Filling in:

F - 18.2 = 2.25(1.50) and

F = 2.25(1.50) + 18.2  Do the multiplication first and round to get

F = 3.38 + 18.2   The addition rules tell us that we will be rounding to the tenths place after we add to get

F = 21.6 N

Answer:

This plasma mostly consists of electrons

Answer:

67.5

Explanation:

The plane accelerates at 2.7m/s,^2

Time is 25 seconds

The velocity can be calculated as follows

= 25×2.7

= 67.5

Hence the speed f the plane is 67.5

Answer:

Explanation:

F = ma is a linear equation. This means that the Force change as the accleration changes. And vice versa. If the Force is cut in thirds, then the acceleration is also cut in thirds. Let's do some math on this just to prove it, shall we?

We know that at first, the F = 15. Let's give this object a mass of 5kg. That means that

15 = 5a so

a = 3

Then the F is cut into thirds, so

5 = 5a so

a = 1

The second acceleration is one-third of the first one, where the Force is 3 times greater.

Answer:

[tex]W=17085KJ[/tex]

Explanation:

From the question we are told that:

Height [tex]H=16m[/tex]

Radius [tex]R=3[/tex]

Height of water [tex]H_w=9m[/tex]

Gravity [tex]g=9.8m/s[/tex]

Density of water [tex]\rho=1000kg/m^3[/tex]

Generally the equation for Volume of water is mathematically given by

 [tex]dv=\pi*r^2dy[/tex]

 [tex]dv=\frac{\piR^2}{H^2}(H-y)^2dy[/tex]

Where

   y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

 [tex]dw=(pdv)g (H-y)[/tex]

Substituting dv

 [tex]dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)[/tex]

 [tex]dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy[/tex]

Therefore

 [tex]W=\int dw[/tex]

 [tex]W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy[/tex]

 [tex]W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)[/tex]

 [tex]W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0[/tex]

 [tex]W=3420.84*0.25[2401-65536][/tex]

 [tex]W=17084965.5J[/tex]

 [tex]W=17085KJ[/tex]

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Answer:

P = V * Q      potential energy = potential * charge

V = =3.09 J / 6.93 * 10E-4 C = 4460 Joules / Coulomb

The electric potential, V at the point given the data from the question is –4458.87 V

The electric potential or electromotive force (EMF) is defined as the energy supplied by a battery per unit charge. Mathematically, it can be expressed as:

Electromotive force (EMF) = Work (W) / charge (Q)

V = EMF = W / Q

V = W / Q

V = –3.09 / 6.93×10⁻⁴

V = –4458.87 V

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Answer: 400 m

Explanation:

Vf= 20 + (4*10)

Vf= 60 [m/s]

x= (60^2 - 20^2) / (2*4)

x= 400 m

Answer:

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

Explanation:

The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:

[tex]I = m\cdot (\vec{v}_{2} - \vec{v_{1}})[/tex] (1)

Where:

[tex]I[/tex] - Impulse, in kilogram-meters per second.

[tex]m[/tex] - Mass, in kilograms.

[tex]\vec{v_{1}}[/tex] - Initial velocity of the hockey park, in meters per second.

[tex]\vec{v_{2}}[/tex] - Final velocity of the hockey park, in meters per second.

If we know that [tex]m = 0.2\,kg[/tex], [tex]\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right][/tex] and [tex]\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right][/tex], then the impulse applied by the stick to the park is approximately:

[tex]I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right][/tex]

[tex]I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right][/tex]

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

Answer:

Coefficient of kinetic friction = 0.38 (Approx.)

Explanation:

Given:

Applied force = 45 N

Mass of wooden crate = 12 kg

Find:

Coefficient of kinetic friction

Computation:

Coefficient of kinetic friction = Applied force / (Mass)(Acceleration due to gravity)

Coefficient of kinetic friction = 45 / (12)(9.8)

Coefficient of kinetic friction = 45 / 117.6

Coefficient of kinetic friction = 0.3826

Coefficient of kinetic friction = 0.38 (Approx.)

Answer:

B. Current in the circuit is 5.

Ci. Voltage across 7.2 Ω (R₁) is 36 V

Cii. Voltage across 6.2 Ω (R₂) is 31 V

Ciii. Voltage across 8.6 Ω (R₃) is 43 V

Explanation:

We'll begin by calculating the number equivalent resistance in the circuit. This can be obtained as follow:

Resistor 1 (R₁) = 7.2 Ω

Resistor 2 (R₂) = 6.2 Ω

Resistor 3 (R₃) = 8.6 Ω

Equivalent Resistance (R) =?

Since the resistors are in series connection, the equivalent resistance can be obtained as follow:

R = R₁ + R₂ + R₃

R = 7.2 + 6.2 + 8.6

R = 22 Ω

B. Determination of the current.

Voltage (V) = 110 V

Resistance (R) = 22 Ω

Current (I) =?

V = IR

110 = I × 22

Divide both side by 22

I = 110 / 22

I = 5 A

Therefore, the current in the circuit is 5.

Ci. Determination of the voltage across 7.2 Ω (R₁)

Resistor 1 (R₁) = 7.2 Ω

Current (I) = 5 A

Voltage 1 (V₁) =?

V₁ = IR₁

V₁ = 5 × 7.2

V₁ = 36 V

Therefore, the voltage across 7.2 Ω (R₁) is 36 V

Bii. Determination of the voltage across 6.2 Ω (R₂)

Resistor 2 (R₂) = 6.2 Ω

Current (I) = 5 A

Voltage 2 (V₂) =?

V₂ = IR₂

V₂ = 5 × 6.2

V₂ = 31 V

Therefore, the voltage across 6.2 Ω (R₂) is 31 V

Ciii. Determination of the voltage across 8.6 Ω (R₃)

Resistor 3 (R₃) = 8.6 Ω

Current (I) = 5 A

Voltage 3 (V₃) =?

V₃ = IR₃

V₃ = 5 × 8.6

V₃ = 31 V

Therefore, the voltage across 8.6 Ω (R₃) is 43 V

Answer:

a)  3.0  10⁴ m / s, b) 3.7 10¹ m, c) 0.750 kg, d) 4 10¹² m²,  e)  75 m, f) 60 m²

g) 5.207 10³ m², e) 4.847 10⁷ s

Explanation:

The international system (SI) of measurements has as fundamental units the meter for length, the second for time and kilogram for mass.

Let's reduce the different magnitudes to the SI system

a) 30 km / h (1000m / 1 km) (1 h / 3600 s) = 3.0 10⁴ m / s

b) 37 Dm (10 m / 1 Dm) = 3.7 10¹ m

c) 750 g (1 kg / 10,000 g) = 0.750 kg

d) 4 10⁶ km² (1000 m / 1km) ² = 4 10¹² m²

e) 7500 cm (1 m / 100 cm) = 75 m

f) 600000 cm² (1m / 10² cm) ² = 60 m²

g) 520700000 mm³ (1 m / 10³ mm) ³ = 5.20700000 109/10 ^ 6

  = 5.207 10³ m²

e) 3.4 years (l65 days / 1 yr) (24 h / 1 day) (3600 s / 1h) = 4.847 10⁷ s

Answer:

Thus, the time for the first lamp is 44 minutes.

Explanation:

Power of first lamp, P' = 1000 W

Power of second lamp, P'' = 4400 W

time for second lamp, t'' = 10 minutes

Let the time for first lamp is t'.

As the energy is same, so,

P' x t' = P'' x t''

1000 x t' = 4400 x 10

t' = 44 minutes

Answer:

[tex] { \huge{s}} = ut + \frac{1}{2} g {t}^{2} \\ 1.5 = 3t + \frac{1}{2} \times 10\times {t}^{2} \\ 1.5 = 3t + 5 {t}^{2} \\ 5 {t}^{2} + 3t - 1.5 = 0 \\ t = 0.3 \: seconds[/tex]

Answer:

20 m/s

Explanation:

Applying,

a = (v-u)/t.................... Equation 1

Where a = acceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.

make v the subject of the equation

v = u+at.............. Equation 2

From the question,

Given: u = 0 m/s(start from rest), a = 2 m/s², t = 10 seconds

Substitute these values into equation 2

v = 0+(2×10)

v = 20 m/s

Answer:

I am not sure if this is the answer

(B) what is the height of the building if it hits the ground after those 5 seconds.

Answer:

0.893 rad/s in the clockwise direction

Explanation:

From the law of conservation of angular momentum,

angular momentum before impact = angular momentum after impact

L₁ = L₂

L₁ = angular momentum of bullet = + 9 kgm²/s (it is positive since the bullet tends to rotate in a clockwise direction from left to right)

L₂ = angular momentum of cylinder and angular momentum of bullet after collision.

L₂ = (I₁ + I₂)ω where I₁ = rotational inertia of cylinder = 1/2MR² where M = mass of cylinder = 5 kg and R = radius of cylinder = 2 m, I₂ = rotational inertia of bullet about axis of cylinder after collision = mR² where m = mass of bullet = 0.02 kg and R = radius of cylinder = 2m and ω = angular velocity of system after collision

So,

L₁ = L₂

L₁ = (I₁ + I₂)ω

ω = L₁/(I₁ + I₂)

ω = L₁/(1/2MR² + mR²)

ω = L₁/(1/2M + m)R²

substituting the values of the variables into the equation, we have

ω = L₁/(1/2M + m)R²

ω = + 9 kgm²/s/(1/2 × 5 kg + 0.02 kg)(2 m)²

ω = + 9 kgm²/s/(2.5 kg + 0.02 kg)(4 m²)

ω = + 9 kgm²/s/(2.52 kg)(4 m²)

ω = +9 kgm²/s/10.08 kgm²

ω = + 0.893 rad/s

The angular velocity of the cylinder bullet system is 0.893 rad/s in the clockwise direction-since it is positive.

Answer:

Slow-moving rivers generate extensive floodplains and meanders through erosion and deposition. Stream and river deposition can result in the formation of alluvial fans and deltas. Natural levees may be formed by floodwaters. Caves and sinkholes can arise as a result of groundwater erosion and deposition.

Explanation:

s

The erosion and deposition of sediments in a river creates broad floodplains and meanders.

Deposition occurs when sediment, a combination of soil and rock produced by weathering, is eroded and transported to a new area.

Deposition is the act of depositing silt that has been transported by the wind, water, sea, or ice.

Earthen materials are worn away during erosion, a geological process in which they are moved by wind or water.

The removal of soil, rock, or dissolved material from one area on the Earth's crust and subsequent transport to another region for deposition are known as erosional processes. Erosion differs from weathering, which is a static process.

Given data ,

Let the erosion and deposition of sediments be deposited in a river

Now , Pebbles, sand, mud, and salts that have been dissolved in water can all be used to convey sediment. Afterwards, salts may be left behind by organic action.

Now , floodplain is caused when erosion happens

And , A floodplain is a broad, level or nearly level area of land where the stream flows.

Meandering streams that wander from side to side broaden the plain by eroding it during the formation of the plain. Flooding can occur when stream flows overflow from their channel due to very excessive rainfall or quick snow-melt.

Hence , floodplain and meandering occurs due to erosion and deposition

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Answer:

C. She can increase the number of wire loops.

Explanation:

The more wire loops the more energy.

For a coil of wire, the magnetic field strength is increased by increasing the number of coils around the nail.

An electromagnet is a soft metal core shaped into a magnet by the passing the electric current through a coil surrounding it.

The end of the wire leading to the S is attached to the positive terminal of a battery. The end of the wire leading to the N is attached to the negative terminal of the battery. The current begins to flow. Current cant be changed to increase magnetic field strength, but the no of coils will definitely increase it.

Thus, To increase the strength of the electromagnet, Lisa can increase the number of wire loops.

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Answer:

B energy can't be created or destroyed

Answer:

Electromotive force or EMF is equal to the terminal potential difference when no current flows. EMF (ϵ) is the amount of energy (E) provided by the battery to each coulomb of charge (Q) passing through.

Answer:

7. (D) uniformly accelerated vertical motion

8. (A) zero

9. (A) zero

10. (C) parabolic

Answer:

7.Uniformly accelerated vertical motion

8.0m/s²

9.9.8m/s

10.parabolic

11.vertical component.

Answer:

The given potential energy of the spring is expressed as follows;

ΔUsp = -(WB + WW)

Where;

WB = Th work done by the spring on the block to which it is attached

WW = The work done by the spring on the wall

We recall that work done, W = Force applied × Distance moved in the direction of the force

The work done by the spring on the block, WB = The spring force × The distance the block moves

The work done by the spring on the wall, WW = The spring force × The distance the wall moves

However, given that the wall does not move, we have;

The distance the wall moves = 0

∴ The work done by the spring on the wall, WW = The spring force × 0 = 0 J

Therefore, WW = 0 J, and the spring does not do work on the wall, and WW can be ignored in the next subsequent) steps

Explanation:

The complete question is as follows: A student is subjected to a reaction force of 10 N northward from a 5 kg block while pushing the block over a smooth, level surface. Ignoring friction, what is the acceleration of the block?

Answer: The acceleration of the block is [tex]2 m/s^{2}[/tex].

Explanation:

Given: Force = 10 N

Mass = 5 kg

It is known that force applied on an object is the product of mass and acceleration.

Mathematically, [tex]F = m \times a[/tex]

where,

F = force

m = mass

a = acceleration

Substitute the values into above formula as follows.

[tex]F = m \times a\\10 N = 5 kg \times a\\a = \frac{10}{5}\\= 2 m/s^{2}[/tex]

Thus, we can conclude that the acceleration of block is [tex]2 m/s^{2}[/tex].

Answer:

a. Acceleration, a = 0.28 m/s²

b. Distance, S = 156 meters

Explanation:

Given the following data;

Initial velocity = 30 km/h

Final velocity = 45 km/h

Time = 15 seconds

a. To find the acceleration;

Conversion:

30 km/h to m/s = 30*1000/3600 = 8.33 m/s

45 km/h to m/s = 45*1000/3600 = 12.5 m/s

Mathematically, acceleration is given by the equation;

[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]

Substituting into the equation;

[tex]a = \frac{12.5 - 8.3}{15}[/tex]

[tex]a = \frac{4.2}{15}[/tex]

Acceleration, a = 0.28 m/s²

b. To find the distance travelled, we would use the second equation of motion given by the formula;

[tex] S = ut + \frac {1}{2}at^{2}[/tex]

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

[tex] S = 8.3*15 + \frac {1}{2}*(0.28)*15^{2}[/tex]

[tex] S = 124.5 + 0.14*225[/tex]

[tex] S = 124.5 + 31.5 [/tex]

S = 156 meters