a current of 200mA through a conductor converts 40 joules of electrical energy into heat in 30 seconds determine the potential drop across the conductor

Answers

Answer 1

Answer:

V = 6.65 [volt]

Explanation:

We must first find the power generated, power is defined as the amount of energy consumed or generated in a given time.

[tex]P=\frac{E}{t}[/tex]

where:

P = power [w]

E = energy = 40 [J]

t = time = 30 [s]

[tex]P =40/30\\P = 1.33[w][/tex]

Now we can calculate the voltage or potential drop by means of the power, the power is calculated by means of the product of the voltage by the current.

[tex]P =V*I[/tex]

where:

V = voltage [volts]

I = current = 200mA = 0.2 [A]

[tex]V = P/I\\V = 1.33/0.2\\V = 6.65 [Volt][/tex]


Related Questions

SI unit differ from one country to another . true or false​

Answers

Answer:

false ..........false

Answer:

FALSE                                            

Explanation:

Is a parked car potential or kinetic ?

Answers

Answer:

Potential energy is the energy that is stored in an object. ... When you park your car at the top of a hill, your car has potential energy because the gravity is pulling your car to move downward; if your car's parking brake fails, your vehicle may roll down the hill because of the force of gravity.

7. It is the art of drawing solid objects on two-dimensional surfaces.

Answers

Explanation:

Perspective- the art of drawing solid objects on a two-dimensional surface so as to give the right impression of their height, width, depth, and position in relation to each other when viewed from a particular point.

A ball is thrown vertically upward with an initial velocity of 23 m/s. What are its position and velocity after 2 s?

Answers

Answer:

The position of the ball after 2 s is 26.4 mThe velocity of the ball after 2 s is 3.4 m/s

Explanation:

Given;

initial velocity of the ball, u = 23 m/s

time of motion, t = 2 s

The position of the ball after 2 s is given by;

h = ut - ¹/₂gt²

h = (23 x 2) - ¹/₂ x 9.8 x 2²

h = 46 - 19.6

h = 26.4 m

The velocity of the ball after 2 s is given by;

v² = u² + 2(-g)h

v² = u² - 2gh

v² = 23² - (2 x 9.8 x 26.4)

v² = 529 - 517.44

v² = 11.56

v = √11.56

v = 3.4 m/s

Calculate how long
in years it will take your radiation to reach Earth from Alpha Centauri (research this).
( MY RADIATION IS INFARED )

Answers

Alpha Centauri is 4.4 lightyears away. Traveling at 1/10 speed of light it would take you 44 years.

A force of 150 N is applied on an object at 60 degrees above the positive x-axis. Determine its

horizontal and vertical components.

Answers

Answer:

horizontal component=fcostita

=150cos60

use calculator to evaluate it

for vertical=fsintita

=150sin60


3 Magnetised mineral ore was found in
A Africa
B Australia
C America
D Magnesia
E None of the above

Answers

Answer:

D Magnesia

Explanation:

Ore particles Fe3O4 (magnetite) were found in the region called Magnesia.

Magnesia was an antic city in Asia, named after the inhabitants of Greek Magnesia.

Magnetite is used as an ore, abrasive, in paint production, electrophotography, as a micronutrient fertilizer, and as a high-density concrete aggregate.

You are a guest magician in a circus. One of your tricks is to place a football on an inclined plane without the football rolling over. How can you achieve this?
Determine if the weight of the football is needed for you to prevent the football from rolling over. The coefficient of static friction of the plane is 0.067.

Answers

Answer:

You are a guest magician in a circus. One of your tricks is to place a football on an inclined plane without the football rolling over is explained below in details.

Explanation:

spinning ball halts after traveling some range due to friction energy act different direction of movement of the ball. you can observe in the figure.

Let any rolling ball of mass (m ) is traveling with velocity v ,

common effect on ball (N) = mg

because of motion, friction energy develops on the contact exterior and begins to resist the movement of the rolling ball.

hence,

fr = uN = umg act on communicating exterior, so, after any time due to friction energy rolling ball gets to rest.

The CEO, ellen misk, left her martian office but accidentally left a cylindricall can of coke (3.1 inches in diameter, 5.42 inches in height) on her desk. If the can exerts a pressure of 510 Pascals, what is the specific gravity of the can?

Answers

Answer:

Specific Gravity = 0.378

Explanation:

First, we will find the force exerted by the can on the table. This force will be equal to the weight of the can:

Pressure = Force/Area = Weight/Area

Weight = Pressure*Area

where,

Area = πdiameter²/4 = π[(3.1 in)(0.0254 m/1 in)]²/4 = 4.8 x 10⁻³ m²

Weight = (510 N/m²)(4.8 x 10⁻³ m²)

Weight = 2.48 N

Now, the weight is given as:

Weight = mg

2.48 N = m(9.8 m/s²)

m = (2.48 N)/(9.8 m/s²)

m = 0.25 kg

Now, we calculate volume of can:

Volume = (Area)(Height) = (4.8 x 10⁻³ m²)(5.42 in)(0.0254 m/1 in)

Volume = 6.6 x 10⁻⁴ m³

Hence, the density of can will be:

Density of Can = m/Volume = 0.25 kg/6.6 x 10⁻⁴ m³

Density of Can = 378.32 kg/m³

So, the specific gravity of Can will be:

Specific Gravity = Density of Can/Density of Water

Specific Gravity = (378.32 kg/m³)/(1000 kg/m³)

Specific Gravity = 0.378

The x and y coordinates of a particle at any time t are x = 5t - 3t2 and y = 5t respectively, where x and y are in meter and t in second. The speed of the particle at t = 1 second is​

Answers

Answer:

[tex]v=\sqrt{26}~m/s[/tex]

Explanation:

Parametric Equation of the Velocity

Given the position of the particle at any time t as

[tex]r(t) = (x(t),y(t))[/tex]

The instantaneous velocity is the first derivative of the position:

[tex]v(t)=(v_x(t),v_y(t))=(x'(t),y'(t))[/tex]

The speed can be calculated as the magnitude of the velocity:

[tex]v=\sqrt{v_x^2+v_y^2}[/tex]

We are given the coordinates of the position of a particle as:

[tex]x=5t-3t^2[/tex]

[tex]y=5t[/tex]

The coordinates of the velocity are:

[tex]v_x(t)=(5t-3t^2)'=5-6t[/tex]

[tex]v_y(t)=(5t)'=5[/tex]

Evaluating at t=1 s:

[tex]v_x(1)=5-6(1)=-1[/tex]

[tex]v_y(1)=5[/tex]

The velocity is:

[tex]v=\sqrt{(-1)^2+5^2}[/tex]

[tex]v=\sqrt{1+25}[/tex]

[tex]\mathbf{v=\sqrt{26}~m/s}[/tex]

student measures the weight of a bag of bananas with a spring balance.
Describe what is inside a spring balance and explain how it works.

Answers

A spring balance measures the weight of an object by opposing the force of gravity acting with force of an extending spring. May be used to determine mass as well as weight by recalibrating the scale. Some spring balances are available in gram or kilogram markings and are used to measure the mass of an object. Spring balances consist of a cylindrical tube with a spring inside. One end (at the top) is fixed to an adjuster which can be used to calibrate the device. The other end is attached to a hook on which you can hang masses etc.

Convert 451 milliliters to fluid

ounces. Round your answer to 2

decimal places. **There are 29.57

milliliters in 1 fluid ounce***

Answers

Answer:

451 milliliters equals 15.25 fluid ounces

Explanation:

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three.

To solve a direct rule of three, the following formula must be followed:

a ⇒ b

c ⇒ x

So: [tex]x=\frac{c*b}{a}[/tex]

The direct rule of three is the rule applied in this case where there is a change of units.

In this case, the rule of three can be applied in the following way: if there are 29.57 milliliters in 1 fluid ounce, in 451 milliliters how many fluid ounces are there?

[tex]fluid ounces=\frac{451 mL*1 fluid ounce}{29.57 mL}[/tex]

fluid ounces= 15.25

451 milliliters equals 15.25 fluid ounces

At what distance on the axis of a current loop is the magneticfield half the strength of the field at the center of the loop? Give your answer as a multiple of R. z=?R

Answers

Answer:

x = 1.26 R

Explanation:

For this exercise let's find the magnetic field using the Biot-Savart law

            B = μ₀ I/4π ∫ ds x r^ / r²

In the case of a loop or loop, the quantity ds is perpendicular to the distance r, therefore the vector product reduces to the algebraic product and the direction of the field is perpendicular to the current loop

suppose that the spiral eta in the yz plane, therefore the axis is in the x axis

         B = μ₀ I/4π ∫ ds / (R² + x²)

     

The total magnetic field has two components, one parallel to the x axis and another perpendicular, this component is annual when integrating the entire loop, so the total field is

            B = Bₓ i^

using trigonometry

            Bₓ = B cos θ

we substitute

            Bₓ = μ₀ I/4π ∫ ds cos θ / (x² + R²)

the cosine function is

            cos θ = R /√(x² + R²)

The differential is

            ds = R dθ

we substitute

             Bₓ = μ₀ I/4π ∫ (R dθ)  R /√( (x² + R²)³ )

we integrate from 0 to 2π

              Bₓ =μ₀ I/4π R² / √(x² + R²)³   2pi

therefore the final expression is

            B = μ₀ I R²/ 2√(x² + R²)³   i^

In our case the distance is requested where B is half of B in the center of the bone loop x = 0

Spire center field   x=0

              B₀ = μ₀ I/2R

Field at the desired point (x)

              B = B₀ / 2

               

we substitute

              R² /√(x² + R²)³ = ½  1 /R

              2R³ =√(x² + R²)³

              (x² + R²)³ = 4 (R²)³

              (x²/R² + 1)³ = 4

               

The exact result is the solution of this equation, but it is quite laborious, we can find an approximate result assuming that the distance x is much greater than R (x »R)

           B = μ₀ I/2x³ 

we substitute

            R² / x³ = 1/2   1 / R

            2R³ = x³

          x = ∛2  R

            x = 1.2599 R

The distance at which the magnetic field strength is half the strength of the field at the center of the loop in terms of  R is 0.766 R

Suppose we consider a magnetic field located at point z on the axis of the current loop with radius R carrying a current (I), then the magnetic field can be represented as:

[tex]\mathbf{B = \dfrac{\mu_o}{2} \dfrac{IR^2}{(z^2+R^2)^{^{\dfrac{3}{2}}}}}[/tex]

And the field situated at the center of the loop is:

[tex]\mathbf{B_{z=0} =\dfrac{\mu_o}{2} \dfrac{I}{R} }[/tex]

Let consider a distance (z) on the axis of the loop, in which the magnetic field as a result of the loop is equal to half the strength of the magnetic field at the center of the loop;

Then;

[tex]\mathbf{B(z) = \dfrac{1}{2}B_{z=0}}[/tex]

[tex]\mathbf{\dfrac{\mu_o}{2} \dfrac{IR^2}{(z^2+R^2)^{\frac{3}{2}}} =\dfrac{1}{2} \Big (\dfrac{\mu_o}{2} \dfrac{I}{R} \Big )}[/tex]

Multiply both sides by (2);

[tex]\mathbf{\dfrac{R^2}{(z^2+R^2)^{\frac{3}{2}}} = \dfrac{1}{2R}}[/tex]

Cross multiply;

[tex]\mathbf{2R^3 = (z^2 +R^2)^{\dfrac{3}{2}}}[/tex]

[tex]\mathbf{4R^6 = (z^2 +R^2)^3}[/tex]

[tex]\mathbf{z = \sqrt{4^{1/3} R^2 -R^2 }}[/tex]

[tex]\mathbf{z = R\sqrt{4^{1/3} -1 }}[/tex]

[tex]\mathbf{z = R\sqrt{0.587401052 }}[/tex]

z = 0.766 R

Therefore, we can conclude that the distance at which the magnetic field strength is half the strength of the field at the center of the loop in terms of  R is 0.766 R

Learn more about the magnetic field here:

https://brainly.com/question/23096032?referrer=searchResults

The earth's radius is 6.37×106m; it rotates once every 24 hours.What is the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator? (Hint: what is the radius of the circle in which the point moves?) Express your answer to two significant figures and include the appropriate units.

Answers

Answer:

v = 120 m/s

Explanation:

We are given;

earth's radius; r = 6.37 × 10^(6) m

Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s

Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.

The angle will be;

θ = ¾ × 90

θ = 67.5

¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.

The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:

v = r(cos θ) × ω

v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)

v = 117.22 m/s

Approximation to 2 sig. figures gives;

v = 120 m/s

If you start at a speed of 4m/s and slow down to 2m/s in 4s what is your
acceleration?

Answers

Answer:

penis

Explanation:

A projectile is fired with an initial speed of 36.6 m/s at an angle of 42.2° above the horizontal on a long flat firing range. Determine (a) the maximum height reached by the projectile, (b) the total time in the air, (c) the total horizontal distance covered (that is, the range), and (d) the speed of the projectile 1.50 s after firing.

Answers

Answer:

Explanation:

a) Maximum height id expressed as;

H = u²sin² theta/2g

H = 36.6²(sin42.2)²/2(9.8)

H = 1,339.56(0.6717)/19.6

H = 899.81/19.6

H = 45.91m

Hence the maximum height is 45.91m

b) The Time of flight is the total time in air expressed as;

T = 2usin theta/g

T = 2(36.6)sin42.2/9.8

T = 73.2sin42.2/9.8

T = 49.17/9.8

T = 5.017secs

Hence the total time in air is 5.017scs

c) Range = U²sin2(theta)/g

Range = 36.6²sin2(42.2)/9.8

Range = 1,339.5(0.9952)/9.8

Range = 1,333.11/9.8

Range = 136.03m

Hence the range is 136.03m

d) USing the rime of flight formula;

T =2Usintheta/g

1.5 = 2Usin42.2/9.8

2Usin42.2 =1.5*9.8

2Usin42.2 = 14.7

U = 14.7/2sin42.2

U = 14.7/1.3434

U = 10.94m/s

Hence the speed of the projectile is 10.94m/s

write an email to your friend about your
preparation for the upcoming exam

Answers

Answer:

You will be very glad to know that my preparation for HSC examination is very good. I have prepared myself for the HSC Examination 2015. I am confident that I will obtain very good marks in the examination. I have revised them several times and I am hopeful of obtaining distinction marks in all the subjects.

Explanation:

This is from Brainly hope this person helped you

A sealed cubical container 28.0 cm on a side contains three times Avogadro's number of molecules at a temperature of 24.0°C. Find the force exerted by the gas on one of the walls of the container.

Answers

Answer:

3.32 atm

__________________________________________________________

We are given:

side of the cubical container = 28 cm

number of molecules in the container = 3 * Nₐ  

[where Nₐ is the Avogadro's number]

Temperature = 24°C  OR  297 K

We need to find the pressure exerted by the gas on the walls of the container

__________________________________________________________

Some Calculations:

Volume of the container

we are given the side of the cubical container = 28 cm

Volume of the cubical container = side³

Volume = 28³

Volume = 21952 cm³

We know that 1 cm³ = 1 mL

So,

Volume = 21952 mL

We also know that 1 L = 1000 mL

Volume = 21.952 L

Number of moles of Gas

We know that:

number of moles = number of molecules / Avogadro's number

number of moles = 3 * Nₐ / Nₐ                   [number of molecules = 3 * Nₐ]

number of moles = 3 moles

__________________________________________________________

Pressure Exerted by the Gas:

Using the ideal gas equation:

PV = nRT

Since the volume is in L, and Temperature is in K. R is equal to

0.082 L atm /mol K and the pressure will be in atm

P(21.952) = 3*(0.082)*(297)        

P = 3.32 atm

Hence, the gas will exert a pressure of 3.32 atm on the walls of the container

Which phrase desenbes an irregular galaxy ?
has a round shape
contains many young stars
has arms that extend from the center
Is larger than other types of galaxies

Answers

Answer:

contains many young stars

Explanation:

Irregular galaxies have no definite shape, which means that the first option is incorrect. They are definitely not round.

However, they contain many young stars because the degree of star formation is fast. They also contain old stars. Thus, the second choice is correct.

The "spiral galaxy" is the type of galaxy that has arms that extend from the center. These arms look "spiral," which influenced its name. This makes the last choice incorrect.

They are actually smaller than the other types of galaxies. This makes them prone to collisions. This makes the last choice incorrect.

Answer:

Contains many young stars

Explanation:

A cheetah can maintain a maximum constant velocity of 34.2 m/s for 8.70 s. What is

the displacement the cheetah covered at that velocity?

Answers

Answer:

297.54m

Explanation:

step one:

given data

velocity v=34.2m/s

time t= 8.7s

Step two

Required is the distance the cheetah has covered on the condition

we know that speed= distance/time

make distance subject of formula we have

distance= velocity *time

distance= 34.2*8.7

distance = 297.54m

Therefore the displacement the cheetah covered at that velocity

is 297.54m



5.List the four goals of Psychology. Give your own example for each one using a behavior

Answers

Answer:

describe, explain, predict, and change/control behavior.

Explanation:

describe: What are they doing? -Pavlov noticed that dogs were salivating when they would see his lab assistant before food was presented to them. This observation acted as a description of what was happening to them.

explain: Why are they doing that?- Pavlov started to look into why they were doing it. There was a stimulus, the assistant giving them food in the past to where they started to salivate at the sight of the lab assistant

predict: What would happen if I responded in this way?- Pavlov predicted that he could get the same reaction if he used a bell as a stimulus. Using this he was able to condition dogs to salivate at the ring of the bell.

change/control: What can I do to get them to stop doing that? Because of this discovery we can use conditioning today. For example, in the classroom teachers can use conditioning with their students to make it easier, parents to teach their children right from wrong and to have good behavior. (you do this bad thing you get time out, do a good thing and I will praise you, etc) It can be used when training employees and many other places.

A small block of mass m1 = 0.4 kg is placed on a long slab of mass m2 = 2.8 kg. Initially, the slab is stationary and the block moves at a speed of vo = 3 m/s. The coefficient of kinetic friction between the block and the slab is 0.15 and there is no friction between the slab and the surface on which it moves.

Determine the speed v1.

Determine the distance traveled by the slab before it reaches the speed v1.

Answers

Answer:

v₁ = 0.375 m / s ,   x = 0.335 m

Explanation:

Let's analyze this interesting exercise, the block moves and has a friction force with the tile, we assume that the speed of the block is constant, so the friction force opposes the block movement. For the only force that acts (action and reaction) this friction force exerted by the block that is in the direction of movement of the tile.

We can also see that the isolated system formed by the block and the tile will reach a stable speed where friction cannot give the system more energy, this speed can be found by treating the system with the conservation of linear momentum.

initial moment. Right at the start of the movement

       p₀ = m v₀ + 0

final moment. Just when it comes to equilibrium

      [tex]p_{f}[/tex] = (m + M) v₁

how the forces are internal

       p₀ =p_{f}

       m v₀ = (m + M) v₁

       v₁ = m /m+M    v₀

let's calculate

       v₁ = 0.4 /(0.4 + 2.8)  3

       v₁ = 0.375 m / s

 

Let's apply Newton's second law to the Block, to find the friction force

Y axis

       N - W = 0

       N = W

       N = m g

where m is the mass of the block

the friction force has the formula

      fr = μ N

      fr = μ m g

We apply Newton's second law to slab    

X axis

       fr = M a

where M is the mass of the slab

       μ m g = M a

       a = μ g m / M

let's calculate

       a = 0.15  9.8  0.4 / 2.8

       a = 0.21 m / s²

With kinematics we can find the position

       v²= v₀²+2 a x

as the slab is initially at rest, its initial velocity is zero

       v² = 2 a x

       x = v2 / 2a

let's calculate

        x = 0.375²/2 0.21

        x = 0.335 m

What is the voltage drop across the 10.0 2 resistor?
10.00
120.0 V
20.00
30.00

Answer: 20.0V

Answers

Answer:

20.0V

SANA MAKATULONG

The voltage drop across the 10.0-ohm resistor would be 20.0 volt.

What is resistance?

Resistance is the obstruction of electrons in an electrically conducting material.

The mathematical relation for resistance can be understood with the help of the empirical relation provided by Ohm's law.

V=IR

For calculating equivalent resistance in series combination.

Re = R1  + R2 + R3

For the given problem the total resistance of the circuit would be as all the three resistors are connected in the series combination.

Re= 10+20+30

Re=60 ohm

As given in the figure all the resistance are connected in the series combination therefore the current flowing through them would be the same.

For the given problem we have to design and construct a circuit that has two resistors connected in series.

By using Ohms law

V=IR

120 = 60×I

I = 2 ampere

Given that we have a 120 V battery, that will produce a current of 2 Ampere

By using Ohm's law we can calculate the voltage drop across a 10-ohm resistor

V=IR

  =2×10

  = 20 volt

Thus, the voltage drop across the 10.0-ohm resistor comes out to be 20.0 volt.

Learn more about resistance from here

brainly.com/question/14547003

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An ideal gas expands quasi-statically and isothermally from a state with pressurepand volumeVto a state with volume 4V. How much heat is added to the expanding gas?

Answers

Answer:

Q = PV(In 4)

Explanation:

We are told that the volume expands from V to a state with volume 4V.

Thus, initial volume is V and Final volume is 4V.

We want to find How much heat is added to the expanding gas.

For an isothermal process, the work done is calculated from;

W = nRT(In(V_f/V_i))

Where;

V_f is final volume

V_i is initial volume

Thus;

W = nRT(In(4V/V))

W = nRT(In 4)

Now, from ideal gas equation, we know that;

PV = nRT

Thus;

W = PV(In 4)

Now from first law of thermodynamics, we know that internal energy is zero and thus; Q = W

Where Q is quantity of heat

Thus;

Q = PV(In 4)

A circular conducting loop with a radius of 1.00 m and a small gap filled with a 10.0 Ω resistor is oriented in the xy-plane. If a magnetic field of 2.0 T, making an angle of 30º with the z-axis, increases to 11.0 T, in 2.5 s, what is the magnitude of the current that will be caused to flow in the conductor?

Answers

Answer:

ill get back to this question once i find the answer to it

An airtight box, having a lid of area 80.0 cm^2, is partially evacuated. Atmospheric pressure is 1.01 Times 10^5 Pa. A force of 108 lb is required to pull the lid off the box. The pressure in the box was:_________.

Answers

Answer:

5×10^4Pa

Explanation:

Given force of 108 lb is required to pull the lid off the box,

To convert "Ib"to Newton ,we use conversation rate below

1 pounds = 4.4482216282509 newtons

Then 108 lb=x Newton

Cross multiply we have

X= 480.41Newton

The force that is needed to open the lid is F and pressure P.

We know that Pressure= Force/Area

Area is given as 80.0 cm^2, we can convert to m^2 for unit consistency since 1cm^2= 0.001m^2 then

80.0 cm^2 = 80×10^-4m^2

Substitute to the equation of the pressure we have

P= 480.41Newton/(80×10^4m^2)

P=6×10^4 Pa

The pressure in the box will be difference between the initial pressure and final pressure

=( 1.01 ×10^5 Pa)-(6×10^4 Pa)

= 50100Pa

= 5×10^4Pa

Therefore, The pressure in the box was

5×10^4Pa

A student kicks a soccer ball upward at a 30º angle with an initial speed of 20 m∕s. What expression should the student use to calculate the magnitude of the ball’s initial velocity in the horizontal direction?

Answers

Answer:

[tex]\displaystyle x=10\sqrt{3}\ m/s[/tex]

[tex]y=10\ m/s[/tex]

Explanation:

Rectangular coordinates of vectors in 2D

Given a vector with a magnitude v and angle θ with respect to the positive horizontal direction, the x and y components of the vector are given by:

[tex]x=v\cos\theta[/tex]

[tex]y=v\sin\theta[/tex]

The soccer ball is kicked upward at an angle θ = 30° and at a speed v=20 m/s.

The rectangular components of the vector are:

[tex]x=20\cos 30^\circ[/tex]

[tex]\displaystyle x=20\cdot \frac{\sqrt{3}}{2}[/tex]

Operating:

[tex]\mathbf{\displaystyle x=10\sqrt{3}\ m/s}[/tex]

[tex]y=20\sin 30^\circ[/tex]

[tex]\displaystyle y=20\cdot \frac{1}{2}[/tex]

Operating:

[tex]\mathbf{y=10\ m/s}[/tex]

7. Answer each of the following questions if the student applied a 550N force to a 100kg box on the same surface. What would be the magnitude of the box's acceleration?

Answers

F=ma
550=100a
a=550/100
a=5.5 m/s²

Acceleration of an object is the force divided by its mass. Hence, acceleration of the 100 Kg box with an applied force of 550 N is 5.5 m/s².

What is acceleration?

Acceleration is a physical quantity having both magnitude and direction. This vector measures the rate of change of velocity of a moving body. Thus, acceleration has the unit of m/s².

According to Newton's second law of motion, force acting on a body  is the product of its mass and acceleration. Thus, force is directly proportional to the mass and acceleration of the body.

Given the mass of the box = 100 Kg

 Force applied on it = 550 N.

Acceleration = 550 N / 100 Kg

                      = 5.5 m/s² ( 1 N  = 1 Kg m/s²)

Therefore, the acceleration of the box is 5.5 m/s².

To find more on acceleration, refer here:

https://brainly.com/question/2303856

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You are hanging on to the edge of a merry-go-round, and must exert a force of 100 N to hang on. If the speed of the merry-go-round doubles, how much force will you need to exert to hang on?

Answers

Answer:

If the speed of the merry-go-round doubles, the force you will need to exert to hang on is 400 N.

Explanation:

Given;

initial force exerted to hang on, F₁ = 100 N

The force exerted on the merry-go-round in order to hang on must be an inward force known as centripetal force.

Centripetal force is given by;

[tex]F_c = \frac{mv^2}{r} \\\\keeping \ "m" \ and \ "r" \ constant, we \ will \ have \ the \ following \ equation;\\\\\frac{F_c_1}{v_1^2} = \frac{F_c_2}{v_2^2} \\\\F_c_2 = \frac{F_c_1*v_2^2}{v_1^2}\\\\when \ the \ speed\ doubles \ i.e, v_2 = 2v_1\\\\ F_c_2 = \frac{F_c_1*(2v_1)^2}{v_1^2}\\\\ F_c_2 = \frac{F_c_1*4v_1^2}{v_1^2}\\\\F_c_2 = F_c_1 *4\\\\F_c_2 = 4(F_c_1)\\\\F_c_2 = 4 (100 \ N)\\\\F_c_2 = 400 \ N[/tex]

Therefore, If the speed of the merry-go-round doubles, the force you will need to exert to hang on is 400 N.

You serve a tennis ball of mass 60g at a speed of 50
m/s, what is the impulse exerted on the ball? ( ball starts from rest )

Answers

Answer:

[tex]J = 3~Kg.m/s[/tex]

Explanation:

Impulse and Momentum

The impulse-momentum theorem states that the change in momentum of an object equals the impulse applied to it.

The equation can be written as follows:

[tex]J =\Delta p = p_2-p_1[/tex]

Where:

J    = Impulse

p2 = Final Momentum

p2 = Initial Momentum

The momentum can be calculated as:

p = m.v

Where m is the mass of the object and v is the velocity.

The tennis ball with mass m=60 g = 0.06 Kg was served from rest (v1=0) to v2=50 m/s. The change in momentum is:

[tex]\Delta p = 0.06Kg~50~m/s-0[/tex]

[tex]\Delta p = 3~Kg.m/s[/tex]

Thus the impulse is:

[tex]\marhbf{J = 3~Kg.m/s}[/tex]

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