Can someone please explain to me why the direction at the final answer is said to be left.​

In order to start the lawn mower engine, a constant tension of 80 N is applied to a cord wrapped around a hub with a radius of 7.0 cm. The hub makes three revolutions before the motor starts. The work done in starting the engine is determined.

Work is defined as the product of force and displacement in the direction of the force. In this case, the force applied is the constant tension of 80 N in the cord, and the displacement is the distance traveled by the cord as the hub makes three revolutions.

To calculate the work done, we need to find the distance traveled by the cord. Each revolution of the hub corresponds to the circumference of a circle with a radius of 7.0 cm. Therefore, the distance traveled in one revolution is 2π(7.0 cm) = 44.0 cm.

Since the hub makes three revolutions, the total distance traveled by the cord is 3 × 44.0 cm = 132.0 cm.

To convert the distance to meters, we divide by 100: 132.0 cm ÷ 100 = 1.32 m.

Now we can calculate the work done: work = force × distance = 80 N × 1.32 m = 105.6 J.

Therefore, the work done in starting the lawn mower engine is 105.6 Joules.

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Answer:

600 Coulomb

Explanation:

Electric Current refers to the charge passing through unit cross section of a wire per second.

Current ( C / sec) = 5A = 5C / sec

So, 5C charge passes through the wire in one second, then, amount of charge passing through a cross section of the wire in 120 seconds is :

5C * 120 = 600 Coulumb

A body of a certain mass could become weightless in the following two conditions:

1. In microgravity

2. In neutral buoyancy

Cheers,

qxxi

Answer:

increasing the masses of the objects and increasing the distance between the objects

Answer:

 V₁ = 6 V ,  V₂ = V₃ = 3 V

Explanation:

To solve this circuit we must remember that there are two fundamental types of construction in series and parallel.

* a serial circuit there is only one path for current

in this circuit the constant current in the entire circuit and the voltage is the sum of the voltage of each term

* Parallel circuit in this there are two or more paths for the current

in this circuit the voltage is constant and the east is divided between each branch

with these principles let's analyze the proposed circuit

The DC battery is in parallel with resistor R1 and the equivalent of the other branch,

as in a parallel circuit the voltage is constant

               V₁ = 6 V

in the other branch (23) it forms a series construction, where the current is constant

               6 = iR₂ + iR₃

as they indicate that each resistance has the same value

              6 = 2 iR

              V = V₂ = V₃ = 3 V

The smallest number of DAC bits that would satisfy the requirements is 13. The correct option is a.

To determine the smallest number of DAC bits required to satisfy the given requirements, we can use the formula:

Resolution = (Voltage Range) / (2ⁿ - 1)

Here, the voltage range is 0 to 5V, and the required resolution is 1mV (0.001V). We need to find the smallest integer value of n that satisfies the equation.

0.001V = 5V / (2ⁿ - 1)

By solving for n, we get:

2ⁿ = 1 + (5V / 0.001V) = 1 + 5000 = 5001

Taking the logarithm base 2 of both sides, we get:

n = log2(5001)

n ≈ 12.287

Since n must be an integer, we round up to the next whole number:

n = 13

Therefore, the smallest number of DAC bits required is 13 (option A).

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The specific gravity of the other fluid is 2.0.

We know that the specific gravity of the fluid at the bottom of the u-tube is 4.0, which means its density is 4 times the density of water. Since the density of water is 62.4 lb/ft^3, the density of the fluid is:

ρ_1 = 4 * 62.4 lb/ft^3 = 249.6 lb/ft^3

We also know that the height of the fluid in the outer leg is 6 inches, or 0.5 feet. Substituting these values into the equation, we get:

P_atm + 249.6 lb/ft^3 * g * 0.5 ft = P_atm + ρ_2 * g * 0.5 ft

Simplifying and canceling out P_atm and g, we get:

124.8 lb/ft^3 = ρ_2

Therefore, the specific gravity of the other fluid is,

SG = ρ_2 / ρ_water = 124.8 lb/ft^3 / 62.4 lb/ft^3 = 2.0

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The values of GPE and KE are given in the attachment. Also when we compare the KE is always less than the GPE, which is also expected since some of the energy is lost due to friction.

Using the formula GPE = mgh (Gravitational potential energy), where m is the mass of the car, g is the acceleration due to gravity (9.81 m/s²), and h is the height of the ramp, we can calculate the GPE for each car at the top of the ramp relative to the bottom of the ramp.

Using the formula KE = 1/2mv², where m is the car's mass, and v is the car's speed at the bottom of the ramp, we can calculate the KE for each car at the bottom the ramp. (KE = Kinetic energy)

The completed table is given in the attachment.

We are able to observe that as the height of the ramp increases, both the GPE and KE of the car at the bottom of the ramp increase. This makes sense, as the car gains more potential energy as it is raised to a greater height, and this potential energy is converted to kinetic energy as the car rolls down the ramp. Also, the energy loss also happens due to air friction.

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The magnitude of the centripetal force acting on the rider at the top of the Ferris wheel is equal to the sum of the rider's weight and the normal force exerted by the seat.

In a vertical circular motion, the rider experiences a centripetal force directed towards the center of the circle. At the top of the circle, the direction of the centripetal force is downward, towards the center of the Ferris wheel. The rider's weight, which is equal to the force of gravity acting on them (mg, where m is the mass of the rider and g is the acceleration due to gravity), also acts downward. To maintain the rider's motion in a circle, the normal force exerted by the seat must provide the upward force component to balance the rider's weight and the downward force component to provide the centripetal force.

The magnitude of the centripetal force is equal to the sum of these two forces. Therefore, at the top of the circle, the magnitude of the centripetal force (Fc) is given by Fc = mg + Fn, where Fn is the magnitude of the normal force exerted by the seat on the rider. This total force ensures that the rider moves in a vertical circle at a constant speed without falling off the seat.

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The DC resistance in ohms per kilometer for an aluminum conductor with a 3 cm diameter is 0.00402 Ω/km. The correct option is b.

The cross-sectional area of the conductor is given by:

A = πr² = π(0.015 m)² = 7.07 × 10⁻⁴ m²

The resistance R of a conductor is given by:

R = ρL/A

where ρ is the resistivity of the material, L is the length of the conductor, and A is the cross-sectional area.

To find the resistance per unit length or the DC resistance in ohms per kilometer, we need to divide both sides of the above equation by the length of the conductor and then multiply by 1000 to convert the result to ohms per kilometer. Thus:

R/1000 = ρL/(1000A)

R/1000 = (2.83 × 10⁻⁸ Ω-m) L/(1000 × 7.07 × 10⁻⁴ m²)

R/1000 = 0.00402 L

Therefore, the DC resistance in ohms per kilometer for an aluminum conductor with a 3 cm diameter is 0.00402 Ω/km. Answer choice (b) is the closest to this value, rounded to three significant figures.

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Answer:either a telescope or  benoclulers

!!

Explanation:

The persοn is mοving at a speed οf 55 miles per hοur in the Sοuth directiοn when the persοn is 50 miles West and 70 miles Sοuth οf the car, and the distance between them is increasing at a rate οf 55 miles per hοur.

Tο determine the speed at which the persοn is mοving, we can use the cοncept οf relative velοcity.

Let's cοnsider the hοrizοntal and vertical cοmpοnents separately:

Hοrizοntal Cοmpοnent:

The persοn is walking due East, which is perpendicular tο the Nοrth directiοn οf the car. Therefοre, the hοrizοntal cοmpοnent οf the persοn's velοcity dοes nοt affect the speed at which the persοn is mοving away frοm the car.

Vertical Cοmpοnent:

The persοn is 70 miles Sοuth οf the car, and the distance between them is increasing at a rate οf 55 miles per hοur. This indicates that the persοn's vertical pοsitiοn is changing with time. Since the persοn is mοving in the Sοuth directiοn and the distance is increasing, the persοn's speed can be determined by the rate οf change οf the vertical distance.

Given that the distance is increasing at a rate οf 55 miles per hοur, the persοn's speed in the Sοuth directiοn is 55 miles per hοur.

Therefοre, the persοn is mοving at a speed οf 55 miles per hοur in the Sοuth directiοn when the persοn is 50 miles West and 70 miles Sοuth οf the car, and the distance between them is increasing at a rate οf 55 miles per hοur.

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A cart is transferred in an instant line. if we double each the mass and speed of the cart, the magnitude of its momentum trade: a doubling of the momentum.

In arithmetic, the magnitude or length of a mathematical object is a property which determines whether the object is greater or smaller than different objects of the equal type. more officially, an object's magnitude is the displayed end result of an ordering —of the magnificence of gadgets to which it belongs.

The time period value is defined as “how a good deal of a quantity”. as an example, the value can be used for explaining the contrast between the speeds of a car and a bicycle. it is able to also be used to provide an explanation for the distance traveled by means of an object or to provide an explanation for the quantity of an item in phrases of its value.

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There could be a few possible causes for feeling pushed down to the floor while waking up in a spaceship. One possibility is: that the spaceship is experiencing a sudden acceleration or change in velocity, causing the sensation of increased gravity or g-forces.

Another possibility is that the artificial gravity system on the spaceship is malfunctioning, resulting in an increase in the force of gravity felt by the occupants.

Alternatively, the sensation could be a result of waking up in a low-gravity environment after being used to Earth's higher gravity, which can cause a feeling of heaviness or difficulty moving at first.

It is probable that the spaceship is encountering a sudden acceleration or velocity change, resulting in an increased sensation of gravity or g-forces. The artificial gravity system on the spaceship might also be malfunctioning, resulting in an increase in the gravitational force felt by the occupants.

Lastly, waking up in a low-gravity environment after being used to Earth's higher gravity can cause a sensation of heaviness or difficulty moving at first.

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Answer:

Explanation:

This is a fast plane! At it's takeoff speed, it could fly around the world in 5 seconds!

Acceleration is defined as the change in speed over the change in time.

\(a = \frac{v_f - v_o}{t}\)

Where v_f is the final velocity, v_o is the original velocity, and t is the time.

We know v_f = 8820km/s.

v_o = 0 km/s. The plane "starts at rest before it's engine start." So we know it starts from zero velocity.

t = 32.8 seconds.

You can solve the equation from there.

Answer:

but for which activity donu need the answer and mention the question clearly

Explanation:

It is acted on by gravity     F = m * 9.81 m/s^2

  and acted on by an equal but opposite spring force

     it thus has NET ZERO forces and is in equilibrium.

Answer:

B

Explanation:

B. Must be the answer I. E 10kg

The solution to the problems in correct significant figures are;

a. 10.8 g - 8.264 g = 2.536 g

b. 4.75 m - 0.4168 m = 4.3332 g

c. 139 cm x 2.3 cm = 319.7 cm²

d. 13.78 g/11.3 mL = 1.22 g/ml

e. 1.6 km + 1.62 m + 1200 cm =  1,613.62 m

Significant figures are used to establish the number which is presented in the form of digits.

a. 10.8 g - 8.264 g = 2.536 g

b. 4.75 m - 0.4168 m = 4.3332 g

c. 139 cm x 2.3 cm = 319.7 cm²

d. 13.78 g/11.3 mL = 1.22 g/ml

e. 1.6 km + 1.62 m + 1200 cm = 1,600 m + 1.62 m + 12 m = 1,613.62 m

Thus, the solution to the problems in correct significant figures are;

a. 10.8 g - 8.264 g = 2.536 g

b. 4.75 m - 0.4168 m = 4.3332 g

c. 139 cm x 2.3 cm = 319.7 cm²

d. 13.78 g/11.3 mL = 1.22 g/ml

e. 1.6 km + 1.62 m + 1200 cm =  1,613.62 m

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a) The average annual evaporation from the catchment is approximately 180.29 mm/year.

b) The exact value of evapotranspiration from the irrigated area cannot be calculated due to missing information.

a) Average annual evaporation from the catchment in mm/year:

First, we calculate the total annual rainfall that is collected by the catchment area:

800,000,000 m² × 0.2 m = 160,000,000 m³/year

Since this is the only source of water for the catchment, the total amount of water available to the catchment area per year will be 160,000,000 m³/year.

We know that the average annual discharge at the outlet of a catchment is 0.5 m³/s, and since there are 31,536,000 seconds in a year, we can calculate the total volume of water that is discharged per year:

0.5 m³/s × 31,536,000 s = 15,768,000 m³/year

So, the total volume of water that is lost through evaporation per year will be:

160,000,000 m³/year - 15,768,000 m³/year = 144,232,000 m³/year

To convert this into millimeters, we need to divide this value by the area of the catchment in square meters, and then multiply by 1000 (since 1 m = 1000 mm):

144,232,000 m³/year ÷ 800,000,000 m² × 1000 mm/m = 180.29 mm/year

Therefore, the average annual evaporation from the catchment is approximately 180.29 mm/year.

b) Evapotranspiration from the irrigated area in mm/year:

Since we know that the size of the irrigated area is 10 km² = 10,000,000 m², we can calculate the total volume of water that is used for irrigation each year by multiplying this area by the amount of discharge that is lost as a result of the irrigation project:

10,000,000 m² × (0.5 m³/s - 0.175 m³/s) × 31,536,000 s/year = 4,422,480,000 m³/year

To calculate the amount of water that is lost through evapotranspiration from the irrigated area, we need to know the crop coefficient and the reference evapotranspiration (ET0) for the area. However, since this information is not provided in the question, we cannot calculate the exact value of evapotranspiration from the irrigated area.

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Answer:

The energy that can be stored in the capacitor is 239 nJ

Explanation:

We first calculate the capacitance of each material. Let C₁ be the capacitance of pyrex glass and C₂ be the capacitance of polystyrene.

C₁ = κ₁ε₀A/d where κ₁ = dielectric constant of pyrex glass = 5, A = area of plates = L² where L = length of square plate = 7.40 cm = 7.40 × 10⁻² m and d = thickness of pyrex slab = 1.60 mm = 1.60 × 10⁻³ m and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

C₁ = κ₁ε₀A/d = κ₁ε₀L²/d = 5 × 8.854 × 10⁻¹² F/m × (7.40 × 10⁻² m)²/1.60 × 10⁻³ m = 2424.2252/1.60 × 10⁻¹¹ F = 1515.14 × 10⁻¹¹ F = 15.2 × 10⁻⁹ F = 15.2 nF

C₂ = κ₂ε₀A/d where κ₂ = dielectric constant of polystyrene = 3, A = area of plates = L² where L = length of square plate = 7.40 cm = 7.40 × 10⁻² m and d = thickness of polystyrene slab = 1.60 mm = 1.60 × 10⁻³ m and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

C₁ = κ₁ε₀A/d = κ₁ε₀L²/d = 3 × 8.854 × 10⁻¹² F/m × (7.40 × 10⁻² m)²/1.60 × 10⁻³ m = 1454.5351/1.60 × 10⁻¹¹ F = 909.08 × 10⁻¹¹ F = 9.09 × 10⁻⁹ F = 9.09 nF

Since the capacitors are in series, we find their effective capacitance C from

1/C = 1/C₁ + 1/C₂

C = C₁C₂/(C₁ + C₂)

= 15.2 × 10⁻⁹ F × 9.09 × 10⁻⁹ F/(15.2 × 10⁻⁹ F + 9.09 × 10⁻⁹ F)

= 138.168 × 10⁻¹⁸/24.29 × 10⁻⁹ F

= 5.69 × 10⁻⁹ F

The amount of electrical energy stored in a capacitor is given by W = 1/2CV² where C = capacitance and v = voltage applied. Now C = 5.69 × 10⁻⁹ F and V = 84.0 V for this capacitor

So W = 1/2 × 5.69 × 10⁻⁹ F × 84.0 V

= 238.98 × 10⁻⁹ J

≅ 239 × 10⁻⁹ J

= 239 nJ

So the energy that can be stored in the capacitor is 239 nJ

The second-order minimum thickness of the film required is 1.41 μm.

The minimum thickness required for a thin film to reflect a given color is half the wavelength of the light in the film material. For a second-order minimum thickness, the formula is given by;

t2=2nλwhere t2 represents the second-order minimum thickness of the film, n is the refractive index of the film material, and λ is the wavelength of the light in air.

If the wavelength of the light in air is 470 nm, then the second-order minimum thickness of the film required is given by;t2=2nλ= 2 × 1.5 × 470 nm = 1410 nm = 1.41 μm.

The second-order minimum thickness of the film required is 1.41 μm.

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Explanation:

here's the answer to your question

The density (ρh) of hot air inside the balloon, assuming it is uniform throughout, can be expressed in terms of Th, Tc  and ρc (density of surrounding air) using the ideal gas law as follows: ρh = (ρc * Tc) / Th

The ideal gas law states that for a given amount of gas, the product of its pressure (P) and volume (V) is proportional to the product of its temperature (T) and the number of moles (n). Mathematically, it can be expressed as:

PV = nRT

Where R is the ideal gas constant.

Since we are considering a balloon filled with hot air, the pressure and volume inside the balloon remain constant. Therefore, we can modify the ideal gas law to relate the densities of hot air (ρh) and surrounding air (ρc) to their respective temperatures:

(ρh * V) / Th = (ρc * V) / Tc

Here, V represents the volume of the balloon, which cancels out on both sides of the equation.

Simplifying the equation:

ρh / Th = ρc / Tc

Rearranging the equation to solve for ρh:

ρh = (ρc * Tc) / Th

The density (ρh) of hot air inside the balloon, assuming it is uniform throughout, can be calculated using the ideal gas law. The expression for ρh in terms of Th (temperature of hot air), Tc (temperature of surrounding air), and ρc (density of surrounding air) is given as: ρh = (ρc * Tc) / Th.

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When a beam of protons traveling at 1.70 km/s enters a uniform magnetic field perpendicularly and exits the field, the beam changes its direction by 90°. The beam travels a distance of 1.20 cm while inside the magnetic field.

When charged particles such as protons move through a magnetic field, they experience a force known as the magnetic Lorentz force. The magnitude of this force is given by the equation:

F = qvB

where F is the force, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field strength.

In this scenario, the protons are traveling perpendicularly to the magnetic field, which means the angle between their velocity vector and the magnetic field is 90°. Therefore, the force acting on the protons is perpendicular to their velocity and causes them to change direction.

The protons move in a circular path while inside the magnetic field, with the radius of the path given by:

r = mv / (qB)

where r is the radius, m is the mass of the particle, and v and B are the velocity and magnetic field strength, respectively.

The distance traveled by the beam inside the magnetic field is given as 1.20 cm. This distance corresponds to the circumference of the circular path, so we can write:

2πr = 1.20 cm

From here, we can solve for the radius of the circular path.

Once the protons exit the magnetic field, they continue moving in a straight line due to the absence of the magnetic force. The change in direction of the beam by 90° is a result of the magnetic field's effect on the protons' motion.

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ETo determine the acceleration of the crate, we need to resolve the applied force into its horizontal and vertical components. The horizontal component of the force will contribute to the acceleration, while the vertical component will not affect the motion of the crate on a frictionless surface.
Given:
Mass of the crate (m) = 40 kg
Magnitude of the applied force (F) = 140 N
Angle of the force with the horizontal (θ) = 30°

To find the horizontal component of the force (F_horizontal), we can use trigonometry:
F_horizontal = F * cos(θ)
F_horizontal = 140 N * cos(30°)
F_horizontal = 140 N * √3/2
F_horizontal = 140 N * 0.866
F_horizontal ≈ 121.24 N
Since there is no friction or vertical forces acting on the crate, the horizontal component of the applied force will be responsible for the acceleration.
Using Newton's second law of motion, which states that the force applied to an object is equal to the mass of the object multiplied by its acceleration (F = m * a), we can calculate the acceleration (a).
a = F_horizontal / m
a = 121.24 N / 40 kg
a ≈ 3.03 m/s²
Therefore, the acceleration of the crate is approximately 3.03 m/s².

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Answer:

The frequency is =2⋅10 18 H z

Explanation:

Apply the equation

Frequency

×

Wavelength

=

Speed

f

×

λ

=

c

The wavelength is

λ

=

1.5

⋅

10

−

10

m

The speed is

c

=

3

⋅

10

8

m

s

−

1

The frequency is

f

=

c

λ

=

3

⋅

10

8

1.5

⋅

10

−

10

=

2

⋅

10

18

H

zApply the equation

Frequency

×

Wavelength

=

Speed

f

×

λ

=

c

The wavelength is

λ

=

1.5

⋅

10

−

10

m

The speed is

c

=

3

⋅

10

8

m

s

−

1

The frequency is

f

=

c

λ

=

3

⋅

10

8

1.5

⋅

10

−

10

=

2

⋅

10

18

H

z

Amplitude...........

Answer:

Amplitude

Explanation:

we took a test today and this same question showed up and it said it was Anplitude