A force of 41 N acts on an object which has a mass of 2.4 kg. What acceleration (in m/s2) is produced by the force

[tex]a = \frac{vf - vi}{t} [/tex]

here initial velocity vi=0 as ball release from rest

the final velocity is vf=4.0

time is t=6

so putting all these values in above equation

[tex]a = \frac{ 4.0- 0}{6} [/tex]

[tex]a = 0.6667m \s {}^{2} [/tex]

Answer:

Options B & E are correct

Explanation:

Looking at all the options, B & E are the correct ones.

Option B is correct because the thicker the wire per unit length, the lesser resistance it will posses and the lesser the energy that will be dissipated by the wire and in return more energy will be dissipated by the bulb.

Option E is also correct because the resistance of the copper wires is low enough to ensure that there's not much drop in voltage across the copper wires. Thus, there will not be any noticeable differences in the voltage across the bulb.

Option A is not correct because the current is not used up and thus the charge is conserved, and it will circulate just through the circuit.

Option C is not correct because although the Electric field along the wire is not zero, it is very small.

Option D is not correct because the wires and the light bulb are connected in series and as such, the current in both the wires and the light bulb will be identical.

The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is :

b. Very little energy is dissipated in the thick connecting wires.

e. The electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.

The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is very little energy is dissipated in the thick connecting wires and the electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.

The thicker the wire per unit length, the lesser resistance it'll posses and the lesser the vitality that will be scattered by the wire and in return more vitality will be disseminated by the bulb.

The resistance of the copper wires is low sufficient to guarantee that there's not much drop in voltage over the copper wires. Hence, there will not be any noticeable contrasts within the voltage over the bulb.

Thus, the correct answer is B and E.

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Answer:

hello your question is incomplete attached below is the complete question

21) The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C

22) The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)

hence the voltage in the battery will be equal to the voltage across each bulb

Explanation:

The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C

The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)

hence the voltage in the battery will be equal to the voltage across each bulb

Answer:

The charge pass through your hair dryer is 3000 C.

Explanation:

Given that,

Power = 1200 W

Voltage = 120 V

Flow time = 5 min

We need to calculate the current

Using formula of power

[tex]P=VI[/tex]

[tex]I=\dfrac{P}{V}[/tex]

Put the value into the formula

[tex]I=\dfrac{1200}{120}[/tex]

[tex]I=10\ A[/tex]

We need to calculate the charge pass through your hair dryer

Using formula of current

[tex]I=\dfrac{Q}{t}[/tex]

[tex]Q=It[/tex]

Put the value into the formula

[tex]Q=10\times5\times60[/tex]

[tex]Q=3000\ C[/tex]

Hence, The charge pass through your hair dryer is 3000 C.

Answer:

The object has to have mass and speed

Explanation:

You can increase both speed and mass to increase the kinetic energy, hope this answers your question.

Happy Halloween!

Answer:

147.27N

Explanation:

Power = workdone/time

Power = Force*distance/time

Given

Power = 90Watts

Distance = 1.1m

Time = 1.8secs

Force = ?

Substitute the given parameters into the formula:

[tex]90 = \frac{1.1d}{1.8}\\cross \ multiply\\ 90 \times 1.8 = 1.1F\\162 = 1.1F\\1.1F = 162\\F = \frac{162}{1.1} \\F = 147.27N[/tex]

Hence the magnitude of the force that you exert on the handle is 147.27N

Answer:

Just to help, periods on the periodic table are those running horizontally from left to right

Answer:

The answer is A.Boron and carbon are likely together in one period because they have very close atomic numbers, while gallium and germanium are likely together in another period because they have very close atomic numbers.

Explanation:

just took test

Answer:

(a) Wavelength = 3.21 m (b) Time = [tex]1.07\times 10^{-4}\ s[/tex]

Explanation:

Given that,

The frequency of FM radio station, f = 93.4 MHz

(a) We need to find the wavelength of the radio wave associated with this signal. The relation between wavelength and frequency is given by :

[tex]c=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{93.4\times 10^6}\\\\\lambda=3.21\ m[/tex]

(b) It is given that, an FM radio station, 20 miles away. Let t is time taken for signal to reach your radio from the station. So,

[tex]t=\dfrac{d}{c}\\\\t=\dfrac{20\times 1609.34}{3\times 10^8}\\\\t=1.07\times 10^{-4}\ s[/tex]

Hence, this is the required solution.

Answer:

1.680kN/m

Explanation:

Work done by the spring is expressed as shown:

[tex]W = \frac{1}{2}ke^2[/tex] where:

k is the spring constant

e is the extension

Given

W = 525Joules

extension = 25cm = 0.25m

Substitute into the formula:

[tex]525 = \frac{1}{2}k(0.25)^{2} \\525 = \frac{0.0625k}{2}\\ 525 = 0.03125k\\k = \frac{525}{0.3125}\\k = 1680N/m\\k = 1.680kN/m[/tex]

Hence the force constant of the spring is 1.680kN/m

Answer:

its true trust me

Explanation:

Answer: true

Explanation: edge

Answer:

The wavelength is [tex]\lambda  =  1.2  * 10^8 nm[/tex]

Explanation:

From the question we are told that

   The frequency of operation of the microwave is  [tex]f =  2.50 GHz  =  2.50 *10^{9} \ Hz[/tex]

     Generally the wavelength is mathematically represented as

          [tex]\lambda  =  \frac{c}{f}[/tex]

Here c is the speed of light with value [tex]c =  3.0 *10^{8} \  m/s[/tex]

So  

         [tex]\lambda  =  \frac{3.0 *10^{8}}{  2.50 *10^{9}}[/tex]

=>       [tex]\lambda  =  0.12 \  m [/tex]

converting to nanometer

           [tex]\lambda  =  1.2  * 10^8 nm[/tex]

Answer:

The number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second

Explanation:

Given;

emf of the battery, V = 1.5 V

electron density, = 9 × 10²⁸ mobile electrons per m³

mobility of electron, u = 7 × 10⁻⁵ (m/s)/(N/C)

length of thin wire, L = 6 cm = 0.06 m

cross sectional area of the thin wire, A = 1.3 x 10⁻⁸ m²

The magnitude of the electric field in the thin wire is given by;

E = V/L

E = (1.5) / (0.06)

E = 25 N/C

the number of electrons entering the thin wire every second is given by;

[tex]e/s = mobility \ x \ Electric \ field\\\\number \ of \ electrons \ per \ second =\frac{7*10^{-5} (m/s)}{N/C} *25 (N/C)\\\\number \ of \ electrons \ per \ second = 1.75*10^{-3} \ m/s[/tex]

Therefore, the number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second

The number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second

Since

emf of the battery, V = 1.5 V

electron density, = 9 × 10²⁸ mobile electrons per m³

mobility of electron, u = 7 × 10⁻⁵ (m/s)/(N/C)

length of thin wire, L = 6 cm = 0.06 m

cross sectional area of the thin wire, A = 1.3 x 10⁻⁸ m²

So here the magnitude should be

E = V/L

E = (1.5) / (0.06)

E = 25 N/C

Now the number of electrons should be

= 7 × 10⁻⁵  *25

= 1.75 x 10⁻³ mobile

hence, The number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second

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Answer:

Explanation:

The total distance is how far you walk from the starting point.

Distance through west = 18.0m

Distance through north = 25.0m

Total distance covered = 18.0+25.0m

Total distance covered = 43.0m

This means that I am 43.0m from the starting point

Displacement is the distance covered in a specified direction. The displacement will be gotten using the Pythagoras theorem as shown:

[tex]d^2 = 25^2 + 18^2\\d^2 = 625+324\\d^2 = 949\\d = \sqrt{949}\\ d = 30.81m[/tex]

The direction of your displacement is 30.81m

Direction is gotten according to the formula;

[tex]\theta = tan ^{-1}{\frac{y}{x} }\\\theta = tan ^{-1}{\frac{25}{-18} }\\\theta = tan ^{-1}-1.3889}\\\theta = -60.27^0\\\theta = 180-60.27\\\theta = 119.7^0[/tex]

Note that the direction to the west is negative, that is why the x is -18.0m

The distance from the starting point is 43 m, the displacement vector is 30.81 m and the direction of the displacement is 119.7 degrees.

Given-

Distance travel through the west is 18 m.

Distance travel through the north is 25 m.

To know the total distance, add both the covered distance. Thus total distance x is,

[tex]x=18+25[/tex]

[tex]x=43[/tex]

Hence, the distance from the starting point is 43 m.

Displacement is calculated as the shortest distance between starting and final point. This shortest distance can be calculated using the Pythagoras theorem which states that in a right-angled triangle, the square of the hypotenuse [tex]d[/tex] is equal to the sum of the squares of the other two sides. Therefore,

[tex]d^2=18^2+25^2[/tex]

[tex]d^2=324+625[/tex]

[tex]d^2=949[/tex]

[tex]d=\sqrt{949}[/tex]

[tex]d=30.81[/tex]

The displacement vector is 30.81 m.

The direction of displacement [tex]\theta[/tex] with these two sides can be calculated with the formula,

[tex]\theta=tan^{-1}\dfrac{25}{-18}[/tex]

Here due to the west direction(opposite side), the sign is taken negatively.

[tex]\theta=tan^{-1}(-1.389)[/tex]

[tex]\theta=-60.27^o[/tex]

For the other quarter,

[tex]\theta=180-60.27=119.7^o[/tex]

Hence, the distance from the starting point is 43 m, the displacement vector is 30.81 m and the direction of the displacement is 119.7 degrees.

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Answer:

F = 58.35 [N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that force is equal to the product of mass by acceleration. But first we must use the following equation of kinematics.

We have to convert speeds from kilometers per hour to meters per second

[tex]\frac{26km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=\frac{7.22m}{s} \\\frac{54km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=15\frac{m}{s}[/tex]

[tex]v_{f}=v_{o}+(a*t) \\[/tex]

where:

Vf = final velocity = 15 [m/s]

Vi = initial velocity = 7.22 [m/s]

a = acceleration [m/s^2]

t = time = 4 [s]

Note: the positive sign of the above equation is because the car increases its speed

15 = 7.22 + (a*4)

a = 1.945 [m/s^2]

Now we can use the Newton's second law:

F = m*a

F = 30*1.945

F = 58.35 [N]

Answer:

1.5m/s^2

Explanation:

Answer:

1.5 m/s2. accerelation =force ÷mass

Analyzing the question:                                                                                        

We are given:

initial velocity (u) = 100 m/s

final velocity (v) = v m/s

distance (s) = 125 m

acceleration (a) = 5 m/s²

Solving for Final Velocity (v):                                                                              

from the third equation of motion:

v² - u² = 2as

v² - (100)² = 2(5)(125)

v² - 10000 = 1250

v² = 1250 + 10000

v² = 11250

v = 106.06 m/s

(a) Differentiate the position vector to get the velocity vector:

r(t) = (3.00 m/s) t i - (4.00 m/s²) t² j + (2.00 m) k

v(t) = dr/dt = (3.00 m/s) i - (8.00 m/s²) t j

(b) The velocity at t = 2.00 s is

v (2.00 s) = (3.00 m/s) i - (16.0 m/s) j

(c) Compute the electron's position at t = 2.00 s:

r (2.00 s) = (6.00 m) i - (16.0 m) j + (2.00 m) k

The electron's distance from the origin at t = 2.00 is the magnitude of this vector:

||r (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m

(d) In the x-y plane, the velocity vector at t = 2.00 s makes an angle θ with the positive x-axis such that

tan(θ) = (-16.0 m/s) / (3.00 m/s)   ==>   θ ≈ -79.4º

or an angle of about 360º + θ ≈ 281º in the counter-clockwise direction.

Complete Question

The complete question is shown on the first uploaded image

Answer:

The equation is   [tex]c = \frac{1}{\sqrt{ \mu_o *  \epsilon_o} }[/tex]

The value of  c is [tex]c = 2.998 *10^{8} \  m/s  [/tex]

Explanation:

From the question we are told that

   Generally the equation that  relates the speed of light to other fundamental electromagnetic constants is

     [tex]c = \frac{1}{\sqrt{ \mu_o *  \epsilon_o} }[/tex]

Here  c is the speed of light

[tex]\mu_o[/tex] is the permeability of free space with value

    [tex]\mu_o = 4\pi *10^{-7} \ N/A^2[/tex]

and  [tex]\epsilon_o[/tex] is the permittivity of free space  with value  

      [tex]\epsilon_o  =  8.85*10^{-12} \ C/V \cdot m[/tex]

So

      [tex]c = \frac{1}{\sqrt{ 4\pi *10^{-7}  *  8.85*10^{-12}} }[/tex]

=>   [tex]c = 2.998 *10^{8} \  m/s  [/tex]

Answer:

We must apply a force to keep the object moving at a constant velocity due to gravitational force or weight (in case of lifting), and due to frictional force (in case of pushing).

Explanation:

LIFTING:

When an object is lifted, we first need to overcome the force exerted on it by the field of gravity. Due to this force, which is also called the weight of object, we must apply a force on the object to keep it moving at constant speed, otherwise the gravity force will cause the object to slow down and eventually fall back on ground.

PUSHING:

When pushing an object the person must apply the force to first overcome the frictional force. The frictional force acts in opposite direction of motion. Thus, to move the object at constant speed we must apply force to it.

Hence, we must apply a force to keep the object moving at a constant velocity due to gravitational force or weight (in case of lifting), and due to frictional force (in case of pushing).

Answer:

The correct answer is (a)

Explanation:

A spring scale measures the weight of an object not the mass because according to hooke's law the extension of a spring is directly proportional to the load or force attached/applied to it. The force of gravity acting on the mass of any substance as it goes up actually reduces and increases as it comes down.

If F = ma, as a increases, F will also increase and vice versa

Where F = force

m = mass

a = acceleration (due to gravity in this case)

From the above explanation, it can be deduced that the scale will read more than mg as it gets to the ground because of an increase in the force of gravity (which also increases a) as it approaches the ground.

Answer:

Explanation:

The needle is showing north south direction . when current starts flowing in the wire which is held vertical to the ground , it deflects towards east .

a )

Therefore a magnetic field towards east has been created . It is possible only if current flows towards the surface in the vertical wire .

b )

magnetic field created at the magnetic needle B = 10⁻⁷ x  2I / d where I is current and d is distance .

B = 10⁻⁷ x  2 x 26.3  / .27

= 194.81 x 10⁻⁷ T

angle of deflection of solenoid = 22.9°

Tan 22.9 = B /H

.422 = 194.81 x 10⁻⁷ / H

H = 461.63 x 10⁻⁷ T

= .46 x 10⁻⁴ T .

A) The current in the wire flows towards the Earth's surface

B) The magnitude of the horizontal component of the Earth's magnetic field is :   0.46 x 10⁻⁴ T

A) The compass needle held horizontally points in a North-south direction of the earth and also deflects eastwards when current is allowed to flow through it. The deflection of the needle indicates the presence/generation of a magnetic field on the earth surface. which is facilitated by the flow of the current in the wire towards the Earth's surface

B) Determine The magnitude of the horizontal component of the Earth's magnetic field

B ( magnetic field ) = 10⁻⁷ * 2I / d ---- ( 1 )

where : l = 26.3 A,   d = 0.27 m

Back to equation ( 1 )

B = 10⁻⁷ * 2 * 26.3 / 0.27

  = 194.81 * 10⁻⁷ T

Final step : Calculate the magnitude of horizontal component  ( H )

Tan ∅ = B / H ---- ( 2 )

where : ∅ ( angle of deflection ) = 22.9°

∴ H = B / Tan ( 22.9° )

      = (  194.81 * 10⁻⁷ ) / 0.422

      = 0.46 x 10⁻⁴ T

Hence we can conclude that The current in the wire flows towards the Earth's surface and  The magnitude of the horizontal component of the Earth's magnetic field is :   0.46 x 10⁻⁴ T

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Answer:

he peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.

Explanation:

In a resonance experiment, the amplitude of the system is plotted as a function of the frequency, finding maximums for the values ​​where some natural frequency of the system coincides with the excitation frequency.

In a Fourier transform spectrum, the amplitude of the frequencies present is the signal, whereby each peak corresponds to a natural frequency of the system.

From this explanation we can see that in the first case the peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.

Explanation:

According to newton's second law of motion.

[tex]\sum Fx = ma\\\\\sum Fx = 1061 - 915\\\\\sum Fx = 146N[/tex]

m is the mas of the sky diver = 93.4kg

a is the acceleration of the skydiver

From the formula above;

[tex]a = \frac{\sum Fx}{m}\\ \\a = \frac{146}{93.4}\\\\a = 1.563m/s^2[/tex]

Hence the acceleration of the sky diver is 1.563m/s²

Answer:

The voltage is   [tex]V =   0.993V_b[/tex]

Explanation:

From the question we are told that

   The time that has passed is  [tex]t = \frac{\tau}{2}[/tex]

 Here [tex]\tau[/tex] is know as the time constant

    The voltage of the  power source is   [tex]V_b[/tex]

Generally the voltage equation for charging a capacitor is mathematically represented as

       [tex]V =  V_b  [1 - e^{- \frac{t}{\tau} }][/tex]

=>   [tex]V =  V_b  [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }][/tex]

=>   [tex]V =  V_b  [1 - e^{- \frac{\tau}{2\tau} }][/tex]

=>   [tex]V =  V_b  [1 - e^{- \frac{1}{2} }][/tex]

=>   [tex]V =   0.993V_b[/tex]    

Answer:

8.93*10^13 N.

Explanation:

       [tex]F_{g}= \frac{G*m_{1}*m_{s}}{r_{s}^{2} }[/tex]

       [tex]F_{g}= \frac{6.67e-11Nm^2/kg^2*1.99e30 kg*67.3 kg}{10e4m^2} = 8.93e13 N[/tex]

Answer:

Explanation:

We are expected to solve for the velocity with no slip condition

we know that the expression that relate coefficient of friction and velocity is given as

μs = v^2/rg

Given

coefficient of friction μs = 0.3

radius r= 0.15

assume g=9.81m/s^2

substituting into the expression we have

0.3= v^2/0.15*9.81

v^2=0.3*0.15*9.81

v^2=0.44145

v=√0.44145

v=0.66

therefore the velocity is 0.66m/s

Analysing the question:

Since the stone was dropped, there was no initial velocity applied on it and hence it's initial velocity of the stone is 0 m/s

We are given:

height of the tower (h) = 66 m

mass of the stone (m) = 0.5 kg

initial velocity of the stone (u) = 0 m/s

time taken by the stone to reach the ground (t) = t seconds

acceleration due to gravity = 10 m/s²

** Neglecting air resistance**

Finding the time taken by the stone to reach the ground:

from the second equation of motion

h = ut + 1/2at²

replacing the variables

66 = (0)(t) + 1/2 (10)(t)²

66 = 5t²

t² = 13.2

t = 3.6 seconds

I initially wanted to subtract the height of the student from the height of the tower since the time i calculated is the time taken by the stone to reach the ground and that means that the stone has already hit the student before 3.6 seconds

but since we were NOT given the height of a student, the person who posed this question wants the time taken by the stone to reach the ground and that is what we solved