What is one way that plants can harm the environment?

Answer:

19

Explanation:

Avg. Vel = (28 + 10)/2

= 19

BTW, don't forget your units!

Answer:

10 m/s²

Explanation:

Given

Solving

The correct answer is 10 m/s² .

The magnitude of the net force that is acting on particle q₃ is equal to 6.2 Newton.

Given the following data:

Charge = [tex]-2.35 \times 10^{-6}[/tex] C.

Distance = 0.100 m.

Scientific data:

Coulomb's constant = [tex]8.988\times 10^9 \;Nm^2/C^2[/tex]

In this scenario, the magnitude of the net force that is acting on particle q₃ is given by:

F₃ = F₁₃ + F₂₃

Mathematically, the electrostatic force between two (2) charges is given by this formula:

[tex]F = k\frac{q_1q_2}{r^2}[/tex]

Where:

Note: d₁₃ = 2d₂₃ = 2(0.100) = 0.200 meter.

For electrostatic force (F₁₃);

[tex]F_{13} = 8.988\times 10^9 \times \frac{(-2.35 \times 10^{-6} \times [-2.35 \times 10^{-6}])}{0.200^2}\\\\F_{13} = \frac{0.0496}{0.04}[/tex]

F₁₃ = 1.24 Newton.

For electrostatic force (F₂₃);

[tex]F_{13} = 8.988\times 10^9 \times \frac{(-2.35 \times 10^{-6} \times [-2.35 \times 10^{-6}])}{0.100^2}\\\\F_{13} = \frac{0.0496}{0.01}[/tex]

F₂₃ = 4.96 Newton.

Therefore, the magnitude of the net force that is acting on particle q₃ is given by:

F₃ = 1.24 + 4.96

F₃ = 6.2 Newton.

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Answer:

6.2

Explanation:

acellus

Answer:

Take your TIme maaaaaan

Explanation:

Chillax

Here's a question for you Care to go bowling?

Answer:

By taking it Slow.

Don't go too fast

The density of the liquid is 1470.43 kg/m³. Density is an important parameter in order to understand the fluid and its properties. Its unit is kg/m³.

Density is defined as the mass per unit volume. The mass and density relation is given as;

mass = density × volume

The given data in the problem is as ;

m is the mass of a solid sphere=8.2 kg

[tex]\rm \rho_s[/tex] is the density of the metal= 2300 kg/m³

T is the tension in the cord = 23 N.

The volume of the liquid is found as;

[tex]\rm V_{sphere} = \frac{m}{\rho} \\\\ \rm V_{sphere} =\frac{8.2}{2300} \\\\ \rm V_{sphere} = 0.00356 \ m^3[/tex]

The volume of the sphere is equal to the volume displaced;

Volume displaced = 0.00356 m³

The buoyant force is found as;

[tex]\rm F_B = \rho g V \\\\[/tex]

From the given condition it is observed that the weight of the sphere is equal to the sum of buoyancy force and the tension in the cord.

[tex]\rm mg = \rho g V+T \\\\ 8.2 \times 9.81 = \rho \times 9.81 \times 0.00356+23 \\\\\ 80.442=\rho \times 0.0349 +23 \\\\ \rho = 1645.90 \ kg/m^3[/tex]

Hence the density of the liquid is 1645.90 kg/m³.

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The temperature of the transformer after the transformer has been in operation for several hours at the given resistance is 21.2 degrees Celsius.

The relationship between resistance and temperature is given as;

R = R₀( 1 + αΔθ)

From the first resistance, we will have the following equation,

R₁ = R₀( 1 + αΔθ)

50 = R₀(1 + 15α) ------ (1)

From the second resistance, we will have the following equation,

58 = R₀(1 + αT) ------ (2)

Divide (1) by (2)

[tex]\frac{50}{58} = \frac{R_0(1 + 15\alpha)}{R_0(1 + T\alpha)} \\\\\frac{50}{58} = \frac{(1 + 15\alpha)}{(1 + T\alpha)} \\\\50(1 + T\alpha) = 58(1 + 15\alpha)\\\\50 + 50T\alpha = 58 + 870\alpha \\\\50(0.0425)T = (58- 50) + 870(0.0425)\\\\2.125T = 44.975\\\\T = 21.2\ ^0C[/tex]

Thus, the temperature of the transformer after the transformer has been in operation for several hours at the given resistance is 21.2 degrees Celsius.

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[tex]\text{Kinetic energy,}\\\\E_k = \dfrac 12 mv^2\\\\~~~~~=\dfrac 12 \cdot 110 \times 10^{-3} \cdot 50^2\\\\~~~~~=137.5~ \text{J}[/tex]

Answer: 137.5 Joules

Explanation:

The SI units for kinetic energy is Joules, the SI units for mass is kilograms, and the SI units for velocity is meters per second. We are given the mass in grams, so we must convert it to kilograms. There are 1,000 grams in 1 kilogram:

[tex]110g*\frac{1kg}{1000g} =0.11kg[/tex]

Kinetic energy is the energy the system has due to its motion. It is defined to be:

[tex]KE=\frac{1}{2} mv^2[/tex]

Plug everything in:

[tex]KE=\frac{1}{2} (0.11kg)(50m/s)^2[/tex]

[tex]KE=137.5J[/tex]

Newton's Thired Law of Motion

R² = F1² + F2² + 2 F1.F2 cos α ⇒cosinus

F1 and F2 perpendicular, α = 90°, cos 90° = 0

[tex]\tt R=\sqrt{3^3+4^2}=5[/tex]

The kinetic energy of the merry-go-round after 5.00 s, when a constant force is applied tangentially to it, is determined as 3,069.5 J.

The kinetic energy of the merry-go-round is determined by applying the principle of conservation of angular momentum as follows;

Iα = rF

where;

I = ¹/₂mr²

I = ¹/₂ x 80 x (1.2)²

I = 57.6 kgm²

α = (rF)/I

α = ( 1.2 x 70) / (57.6)

α = 1.46 rad/s²

a = αR

a = 1.46 x 1.2

a = 1.752 m/s²

The velocity of the merry-go-round after 5 s is calculated as follows;

v = u + at

v = 0 + (1.752x 5)

v = 8.76 m/s

K.E = ¹/₂mv²

K.E = ¹/₂ x 80 x 8.76²

K.E = 3,069.5 J

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Answer:

From the SUN

Explanation:

Hope it helps!!!

Follow me if you want :)

The magnitude of the second charge given that the first is –6×10¯⁶ C and is located 0.05 m away is +3.0×10¯⁶ C

F = Kq₁q₂ / r²

Where

F = Kq₁q₂ / r²

Cross multiply

Fr² = Kq₁q₂

Divide both side by Kq₁

q₂ = Fr² / Kq₁

q₂ = (65 × 0.05²) / (9×10⁹ × 6×10¯⁶)

q₂ = +3.0×10¯⁶ C (since the force is attractive)

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Answer:

The formula for work is

W= fd

work= force X distance

so, this is

200 X 5 =1,000

and the unit for work is Joules

so.. 1,000 joules

Explanation:

hope that helps!

Answer:

Explanation:

Multiple choices. 1)when two groups are playing tug of war the forces of the two groups are:- A)same direction B)opposite direction C)different direction but not possible D)A and B

Explanation:

Sooooo, first,

Constant Speed means that the body isn't accelerating.

a= 0

Let the force that causes it to move up the incline be P and Q be the force that causes the body to move down the incline. Let Fr be the frictional force.

The force that pulls the body downwards is given by,

F = mgsin40°

F= 18.898N

P -(F+Fr) = ma, but, a= 0

P = F + Fr

Fr = 26-18.898

= 7.102N

If the block must move down the incline, Q and Fr will act in the same direction.

Q + Fr = F

Q = 11.796N

The difference between the pressure at the top surfaces of the condor's wings and the pressure at the bottom surfaces is 101,204 Pa.

The difference in pressure between the top and bottom of the wingspan is calculated as follows;

ΔP = P(top) - P(bottom)

A = bh

A = 2.7 m x 0.27 m

A = 0.729 m²

W = mg

W = 9 x 9.8

W = 88.2 N

The pressure at the top surface of condor's wings is due to atmospheric pressure

P(top) = 14.7 Psi = 101,325 Pa

The pressure at the bottom surface is due to weight of andean condor.

P = W/A

P(bottom) = 88.2/0.729

P(bottom) = 120.99 Pa

The difference between the pressure at the top surfaces of the condor's wings and the pressure at the bottom surfaces is calculated as;

ΔP = P(top) - P(bottom)

ΔP = 101,325 Pa - 120.99 Pa

ΔP = 101,204 Pa

The complete question is below;

An Andean condor with a wingspan of 270 cm and a mass of 9.00 kg soars along a horizontal path. Model its wings as a rectangle with a width of 27.0 cm.

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Answer:

Yes, It does mean that there was a Red Giant in the middle of the galaxy.

Explanation:

The star at the middle of the galaxy will forever stay at the middle from the gravity it has on itself and it implodes and turns into a black hole with a mass equal to at least 10 suns

Answer: Matter is anything that is made of atoms and that has mass (A)

Explanation: All everyday objects that can be touched are ultimately composed of atoms. Even you are made of atoms. Anything that you can touch has moving particles unless the object is at absolute zero, which is not possible.

Hope this helps!

Answer:

It's true that organisms get energy from food.

Answer:

Explanation:

Organisms ingest large molecules, like carbohydrates, proteins, and fats, and convert them into smaller molecules like carbon dioxide and water. This process is called cellular respiration, a form of catabolism, and makes energy available for the cell to use

Answer:

b)    pressure increases with depth because of the weight of the fluid  above

P = ρ g H        density * gravity * fluid depth

(Note - force from the fluid below is an example of Newton's Third Law - for every action there is an equal and opposite reaction)

The time taken for the capacitor voltage to be 45 % of the battery voltage is 5,750.3 s.

The rate at which the capacitor voltage discharges is given by the following formula;

[tex]V = V_0(1 - e^{-t/RC})[/tex]

Where;

The final voltage = 0.45 x 5 V = 2.25 V

[tex]2.25 = 5(1 - e^{-t/RC})\\\\(1 - e^{-t/RC}) = \frac{2.25}{5} \\\\(1 - e^{-t/RC}) = 0.45\\\\e^{-t/RC} = 1- 0.45\\\\e^{-t/RC} = 0.55\\\\\frac{-t}{RC} = ln(0.55)\\\\\frac{-t}{RC} = -0.597\\\\t = 0.597 (RC)\\\\t = 0.597(3440 \times 2.8)\\\\t = 5,750.3 \ s[/tex]

Thus, the time taken for the capacitor voltage to be 45 % of the battery voltage is 5,750.3 s.

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Based on the latitude and time of the place, the solar angle for the day at 10am and 2pm is 51.6° and 73.0°.

The solar hour angle of a point on the earth’s surface is the angle through which the earth would turn to bring the meridian of the point directly under the sun.

The rotation of the earth on its axis is 15° per hour where before noon is negative and after noon is positive. For example, at 10:00 a.m. local apparent time the hour angle is −30°.

Therefore, based on the latitude and time, the solar angle for the day at 10am and 2pm is 51.6° and 73.0°.

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Answer:

Explanation:

This depends on the height.

Using the formula mgh + 1/2mv^2 = total mechanical energy, we can determine the amount of kinetic and potential energy. I'm assuming your talking about if the marble goes all the down, then all of the energy is converted.

Explanation:

Increasing the separation distance between objects decreases the force of attraction or repulsion between the objects. And decreasing the separation distance between objects increases the force of attraction or repulsion between the objects.

Answer:

A. an area perpendicular to the direction of propagation.

Explanation:

The property of sound called intensity is proportional

to the rate at which energy flows through an area perpendicular to the direction of propagation.