using this free-body diagram, calculate the net force.

The apparatus that can used to measure the mass of the wooden block​ by the student is called beam balance.

A beam balance, often referred to as a double-pan balance, is a straightforward tool for determining an object's weight. Two pans or trays are hung from either end of a horizontal beam that is suspended from a pivot point in the middle.

The thing to be weighed is put on one tray, and then the second tray is filled with standard weights until it balances, showing the weight of the object. From little ones used in laboratories to larger ones used in enterprises, beam balances can be found in a variety of shapes and sizes. Because they are precise and operate without electricity or batteries, they are widely used.

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Answer:Calculate the work done using:

work done (in joules) = force (in newtons) x distance moved (in metres)

To practice calculations involving force, distance and work done.

Explanation: I hope this helps srry if I'm wrong

Answer: Thermal energy transfer involves the transfer of internal energy. The three types of thermal energy transfer are conduction, convection and radiation. Conduction involves direct contact of atoms, convection involves the movement of warm particles and radiation involves the movement of electromagnetic waves.

Explanation:

If a 5kg mass starts from rest at xo = -1 and moves under the action of a variable force F(x) = √1-x² to point xf = 1. Then the total work done by the force is equal to π/2 + 1.

To calculate the total work done by the force in this scenario, we can use the formula for work:

Work = ∫F(x) dx

where F(x) is the force as a function of position and dx represents an infinitesimal displacement.

In this case, the force is given by F(x) = √(1 - x²), and we need to find the total work done as the object moves from xo = -1 to xf = 1.

Let's break down the calculation step by step:

Write the integral for work:

Work = ∫F(x) dx

Substitute the given force:

Work = ∫√(1 - x²) dx

Integrate with respect to x:

To integrate the square root of (1 - x²), we use the trigonometric substitution. Let's substitute x = sin(θ) and dx = cos(θ) dθ.

Work = ∫√(1 - sin²(θ)) cos(θ) dθ

Simplify the integrand:

Using the trigonometric identity sin²(θ) + cos²(θ) = 1, we can rewrite the integrand as cos²(θ).

Work = ∫cos²(θ) dθ

Apply the power-reducing formula:

The power-reducing formula states that cos²(θ) = (1 + cos(2θ)) / 2. We can use this formula to simplify the integrand further.

Work = ∫(1 + cos(2θ))/2 dθ

Integrate the terms separately:

Work = (1/2) ∫dθ + (1/2) ∫cos(2θ) dθ

The first integral, ∫dθ, is simply θ, and the second integral, ∫cos(2θ) dθ, can be calculated as sin(2θ)/2.

Work = (1/2) θ + (1/2) (sin(2θ)/2) + C

Evaluate the integral limits:

To find the total work done, we need to evaluate the integral at the upper and lower limits of integration.

At xf = 1, the angle θ is π/2, and at xo = -1, the angle θ is -π/2.

Work = (1/2) (π/2) + (1/2) (sin(2(π/2))/2) - [(1/2) (-π/2) + (1/2) (sin(2(-π/2))/2)]

Simplifying further:

Work = π/4 + (1/2) - (-π/4 + (1/2))

Work = π/4 + 1/2 + π/4 + 1/2

Work = π/2 + 1

Therefore, the total work done by the force is equal to π/2 + 1.

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Answer:

Mass doubles

Explanation:

Since F = ma, and given that the acceleration stayed constant, the mass must have changed. (Assuming thatt when the problem states the force doubles, they mean the net force doubles). Note, that the mass does not have to change if the acceleration is zero.

Answer:

a

Explanation:

How long would it take a machine to do 5.000

joules of work if the power rating of the machine

is 100 watts?

The given machine will take 50 s to complete the work. the power is the rate of performing work.

It can be defined as the rate of performing work. It can also be written as the amount of work divided by the time it takes to complete the work.

\(p = \dfrac wt\)

So

\(t = \dfrac w p\)

Where,

\(p\) - power = 100 watt  = 100 J/s

\(t\) - time = ?

\(w\)- work = 5000 J

Put the values in the formula,

\(t = \dfrac{ 5000 \rm \ J} {100 \rm \ J/s}\\\\t = 50 \rm \ s\)

Therefore, the given machine will take 50 s to complete the work.

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The magnitude of the electric flux through the sheet is 4.05 N m² C⁻¹ (Option E).

The electric flux through a surface is given by the product of the electric field strength and the area of the surface projected perpendicular to the electric field.

In this case, the electric field strength is 18 N C⁻¹, and the area of the sheet projected perpendicular to the electric field is 0.450 m²

(since the normal to the sheet makes an angle of 60° with the electric field). Multiplying these values gives the electric flux:

Electric flux = Electric field strength × Area

Electric flux = 18 N C⁻¹ × 0.450 m²

Electric flux = 8.1 N m² C⁻¹

In summary, the magnitude of the electric flux through the sheet is 4.05 N m² C⁻¹. This value is obtained by multiplying the given electric field strength by the projected area of the sheet perpendicular to the electric field.

The angle of 60° is taken into account to determine the effective area for calculating the flux.(Option E).

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Answer:

No of proton is 13 and nucleus is 13

Answer:

magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission

Explanation:

Given the data in the question;

Speed of carousel N = 24 rpm

From the diagram below, selected path direction defines the Axis of slip.

Hence, The Coriolis is acting along the axis of transmission

Now, we determine the angular speed ω of the carousel.

ω = 2πN / 60

we substitute in the value of N

ω = (2π × 24) / 60

ω = 2.5133 rad/s

Next, we convert the given velocity from mph to ft/s

we know that; 1 mph = 1.4667 ft/s

so

\(V_{slip\) = 6 mph = ( 6 × 1.4667 ) = 8.8002 ft/s

Now, we determine the magnitude of the Coriolis acceleration

\(a_c\) = 2( \(V_{slip\) × ω )

we substitute

\(a_c\) = 2( 8.8002 ft/s × 2.5133 rad/s )

\(a_c\) = 44.235 ft/s²

Hence, magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission

The effort distance of a lever should be increased to lift a heavier load because it provides a mechanical advantage, allowing for easier lifting of the load.

The effort distance of a lever should be increased to lift a heavier load because it allows for a mechanical advantage that compensates for the increased weight.

In a lever system, the effort distance is the distance between the point of application of the input force (effort) and the fulcrum, while the load distance is the distance between the point of application of the output force (load) and the fulcrum. The mechanical advantage of a lever is determined by the ratio of the load distance to the effort distance.

By increasing the effort distance, the mechanical advantage of the lever system is increased. This means that for the same input force (effort), a greater output force (load) can be achieved. When dealing with a heavier load, a higher mechanical advantage is required to overcome the increased resistance.

By increasing the effort distance, the lever system can effectively multiply the applied force, making it easier to lift the heavier load. This allows for the redistribution of force and facilitates the efficient use of human effort in various applications, such as in construction, engineering, and even everyday tools like scissors and pliers.

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A would land before since its heavier

Answer:

0.1667 meters/second.

Explanation:

To find Jibari's average velocity for the trip, we need to first calculate his average velocity for each leg of the trip. To do this, we need to divide the distance he traveled by the time it took him to travel that distance.

For the first leg of the trip, Jibari walked 40.0 meters east in 120.0 seconds. His average velocity for this leg of the trip is therefore 40.0 meters / 120.0 seconds = 0.3333 meters/second east.

For the second leg of the trip, Jibari walked 30.0 meters west in 60.0 seconds. His average velocity for this leg of the trip is therefore 30.0 meters / 60.0 seconds = 0.5000 meters/second west.

To find Jibari's average velocity for the entire trip, we need to add the velocities for each leg of the trip, taking into account their direction. Since the first leg of the trip was east and the second leg was west, we can simply add the two velocities together to find the overall average velocity.

Jibari's overall average velocity for the trip is therefore 0.3333 meters/second east + 0.5000 meters/second west = 0.1667 meters/second.

Therefore, Jibari's average velocity for the trip was 0.1667 meters/second.

Jibari's average velocity for the trip is 0.7 m/s east.

To solve for Jibari's average velocity, we will first need to find his total displacement. We can do this by adding the magnitudes of his eastward and westward displacements. The magnitude of Jibari's eastward displacement is 40.0 meters, and the magnitude of his westward displacement is 30.0 meters. So, his total displacement is

40.0 - 30.0 = 10.0 meters east.

We can then find Jibari's average velocity by dividing his total displacement by the total time it took him to complete the trip. The total time it took Jibari to complete the trip is

120.0 + 60.0 = 180.0 seconds.

So, his average velocity is

10.0 / 180.0 = 0.7 m/s east.

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While the penumbra of a sunspot can "grow" and become bigger.The significance is that the area around the sunspot is becoming darker and larger.

Sunspots are cooler, darker areas on the sun's surface that stand out from the sun's brighter surroundings. They are brought on by a blockage in the flow of heat and energy from the interior of the sun caused by intense magnetic activity. Sunspots can be anywhere from a few hundred to many thousands of kilometers in diameter and are most common near the sun's poles. Depending on their size and activity, sunspots can last for days, weeks, or even months.

Sunspots are linked to increased solar activity, which can alter Earth's magnetic fields, so this is significant for us on Earth. Geomagnetic storms and additional radiation that reaches Earth's surface as a result of this could disrupt electronic systems like power grids and radio transmissions.

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The pressure of the tank, when filled with water at a depth of 6 m, is approximately 580.124 atmospheres (atm). To calculate the pressure of the tank, one can use the equation: Pressure (P) = Density (ρ) × g × Depth (h)

Pressure (P) = Density (ρ) × g × Depth (h)

Given: Density of water (ρ) = 1000 kg/m³

Acceleration due to gravity (g) = 9.8 m/s²

Depth (h) = 6 m

Using the given values, one can calculate the pressure:

Pressure = 1000 kg/m³ × 9.8 m/s² × 6 m Pressure

= 58800 kg·m⁻¹·s⁻²

Now, let's convert the units to pascals (Pa) using the conversion 1 atm = 1.013 x \(10^5\) Pa:

Pressure = 58800 kg·m⁻¹·s⁻² × (1 atm / 1.013 x\(10^5\) Pa)

Pressure = 580.124 atm

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The number of people you will stack to reach the top of the CN Tower (553 m) is 316 people

We'll begin by converting 175 cm to m. This can be obtained as illustrated below:

100 cm = 1 m

Therefore,

175 cm = (175 cm × 1 m) / 100 cm

175 cm = 1.75 m

Thus, 175 cm is equivalent to 1.75 m

The number of people needed to be stacked to get to the top of the CN tower can be o btained asfollow:

Number of people needed = Height of tower / height of a person

Number of people needed = 553 / 1.75

Number of people needed = 316 people

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The mass of each object is 0.004 kg, to the nearest kg.

What is mass?

The mass is a measure of the amount of matter an object contains. It is typically measured in units of kilograms (kg) or grams (g). The mass of an object is distinct from its weight, which is the force exerted on an object by gravity and is measured in units of newtons (N) or pounds (lb).

The mass of an object will remain the same, regardless of its location in the universe, while its weight will vary depending on the strength of the gravitational force acting on it.

To find the mass of each object, we can use the formula for gravitational force: F = G * (m1 * m2) / r^2.

By substituting in the given values and solving for m1 or m2, we get:

F = G * (m1 * m2) / r^2

2.67 x 10^-10 N = (667 x 10^-11 N-m^2/kg^2) * (m1 * m2) / (10 m)^2

m1 = m2 = (2.67 x 10^-10 N * (10 m)^2) / (667 x 10^-11 N-m^2/kg^2)

m1 = m2 = 0.004 kg

The mass of each object is 0.004 kg, to the nearest kg.

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In the diagram tha shows snapshots of an oscillator at different times, the frequency of the oscillation is 0.555 Hz.

The period of the oscillation is the time taken for the for the object to return to its original position. (ie. Displacement = 0). From the above snapshot,

Period of oscillation = 1.80s.

From here, finding the frequency is simple as, Frequency = 1/Period. Hence,

Frequency = 1/1.80

= 0.555 Hz (3 sf).

The frequency of the oscillation is indeed 0.555 Hz. The frequency represents the number of oscillations or cycles per second. In this case, the object completes approximately 0.555 oscillations per second.

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Answer: False

Explanation:

2) john's speed is  5 m/s

3) Lauren's average speed is 3.2 m/s

4) the speed of  the mouse is 0.125 m/s

5) the average speed of the rocket is 96m/s

6) the speed of the cyclist is   5 km/h

7) the average speed  of the  earth is  1083 mph

8) the average  speed to travel Through the city is 48 km/h

9) the distance traveled by the whale is 136 m

10) the  distance the girl travels is 120 m

11) the distance jin traveled is 36 m

12) the  time taken for the journey is 400 s

13) the time taken by the train is 200 s

14)the time taken to cover the distance is 7s

15) the time taken for the beetle to run is 12 s

16) the time taken for the journey is 0.299 h

The above problem we are dealing with is related to average speed or we can average speed, where Average speed is calculated by dividing the whole distance that something has traveled by the entire sum of time it took it to travel that distance. Speed is how quickly something is going at a specific minute. Average speed measures the normal rate of speed over the degree of a trip

The formula referring to calculating the speed is

speed = distance travelled /time taken

From we can derive the formula for both the distance traveled and time is taken which  is

time  taken = distance travelled /  speed

distance traveled =  speed x time taken.

For problem 1 we are provided we distance of 100 m

and time is taken 20.s m , So the speed = 100m /20s =  5 m/s

The same procedure applies to problems 2- 8

Next, in problem 9 we are provided with  speed =8 m/s and time = 17s

So the distance traveled = 8m/s x 17  =136 m

The same procedure applies to problems 10-11

For problem 12 , we are provided with distance =26000m and speed 65m/s. So the time is taken = 26000/65 =400 s

The same procedure applies to problems 13-16

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Answer:

A)    E / B = 2.99 10⁸  V/ mT,  B)     B = 2.50 10⁻⁶ T

Explanation:

A) the expressions for the energy densities are:

         u_E = ½ ε₀ E²

         u_B = ½ B² /μ₀

indicate that the two densities are equal

         ½ ε₀ E² = ½ B² /μ₀  

         E / B =  1 /\(\sqrt{\epsilon_o \ \mu_o }\)

we calculate

         E / B = 1 / √( 8.85 10⁻¹² 4π 10⁻⁷

         E / B = 1 /√( 11.1212 10⁻¹⁸)

         E / B = 0.29986 10⁹9

         E / B = 2.99 10⁸  V/ mT

B) for this case E = 750 V / m, ask the magnetic field

         E / B =  1 /\(\sqrt{\epsilon_o \ \mu_o}\)

         B = E \(\sqrt{\epsilon_o \ \mu_o}\)

we calculate

        B = 750 √(8.85 10⁻¹² 4π 10⁻⁷

        B = 750 3.3348 10⁻⁹

        B = 2.50 10⁻⁶ T

Energy densities can be expressed as

        u_E = ½ ε₀ E²

        u_B = ½ B² /μ₀

When the two densities are equal we use the formula

        ½ ε₀ E² = ½ B² /μ₀  

             E / B =  1 / μ₀  

             E / B = 1 / √( 8.85 10⁻¹² 4π 10⁻⁷)

             E / B = 1 /√( 11.1212 10⁻¹⁸)

             E / B = 0.29986 10⁹

             E / B = 2.99 10⁸  V/ mT

      E / B =  1 / √ ξ₀ μ₀

       B = E √ ξ₀ μ₀

We substitute the values into the equation

        B = 750 √(8.85 10⁻¹² 4π 10⁻⁷)

        B = 750 3.3348 10⁻⁹

        B = 2.50 10⁻⁶ T

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Answer:

your neighbors biycyle inner tubes

Explanation:

hope this helps :)

Explanation:

c because there is element

Answer:

C. H2 and N2 would react to produce more NH3

Explanation:

A.P.E.X

The thing that happens to the speed of light when it passes from a medium of low index of refraction  into a medium of high index of refraction is that it tend to speed up.

Light moves more quickly when it enters a medium with a lower refractive index, as air from water. The light deviates from the straight line. Light will move more slowly and obliquely as it penetrates a substance with a higher refractive index.

The speed of light increases with decreasing refractive index. The lesser refractive index belongs to medium A. When moving at a speed equal to the speed of light divided by the refractive index, light will move more quickly through medium A.

Hence, Light bends in the direction of the normal as it passes through an interface into a medium with a greater index of refraction. On the other hand, light bending away from an interface from higher n to lower n.

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Answer:

(a) 159.84 m

(b) 1.89 m/s²

Explanation:

Applying,

(a)

s = (v+u)t/2.................. Equation 1

Where s = distance traveled by the car, u = initial velocity, v = final velocity, t = time.

From the question,

Given: u = 0 m/s ( from rest), v = 55 mi/h = (55/2.237) m/s = 24.59 m/s, t = 13 s

Substitute these values into equation 1

s = (24.59+0)13/2

s = 159.84 m

(b)

Also applying,

a = (v+u)/t................. Equation 2

Where a = acceleration of the car.

substituting into equation 2,

a = (24.59+0)/13

a = 1.89 m/s²

Answer:

The quartering act of 1765 required the colonies to house British soldiers in barracks provided by the colonies.

Answer:

true! : )

(i underlined the place where the answer is the other information is just as important but if you do not want to read it you do not have to)

Explanation:

Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases. the greater the mass, the greater the gravitational pull. gravitational pull decreases with an increase in the distance between two objects. Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases.

The evidence that Wegener did NOT use to support his idea of continental drift is "the thickness of layers of ice in the Antarctic.

option C

Wegener's primary evidence for continental drift included the fit of the coastlines of different continents, the distribution of fossils across different continents, and the alignment of rock strata on different continents.

So the thickness of layers of ice in the Antarctic, was not used by Wegener to support his idea of continental drift. While this evidence is important for supporting the theory of glaciation, it is not relevant to the theory of continental drift.

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