Can you name some people who have shown perseverance?​

Answers

Answer 1

Answer:

Milton Hershey

Steve Jobs

Simon Cowell

Thomas Edison

Explanation:


Related Questions

Devise and draw a circuit using a long, straight wire resistor, instead of a decade box, that would allow the study of the variation of the voltage with resistance (IS constant). According to Ohm's law, what would

Answers

Answer:

...

Explanation:

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A projectile is fired at time t = 0.0 s from point o at the edge of a cliff, with initial velocity components of Vox = 30 m/s and Voy = 100 m/s. The projectile rises, and then falls into the sea
at point P. The time of flight of the projectile is 25 s. Assume air resistance is negligible.
t
What is the height of the cliff?
560 m
450 m
780 m
400 m
640 m

Answers

Answer:

It would be 450 or 640. My final answer would be about 450

Explanation: Because it would't be to high if it was shot Voy = 100

btw i think i know what i know what i am talking about.

The answer would be about 450 m.

What peak is considered a cliff?

The top isn't the standard for a cliff to be reckoned as a cliff as such. Any steep rock face particularly at the edge of the sea can be specified as a cliff.

A 'clifftop' just refers to any pinnacle of a cliff. A 'plateau' is any flat extended geologic floor. An 'overhang' is a part of a structure or formation that protrudes from the primary frame and rests such that it is 'overhanging' the ground (striking above it).

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If a negative point charge is placed at P without moving the original charges, the net electrical force the charges ±Q will exert on it is

Answers

Answer:

The particle P moves directly upwards

Explanation:

Lets designate the negative point charge at point P as particle P

The +Q charge will exert an attractive force on the particle P.

The -Q charge will exert a repulsive force on the particle P

The +Q charge exerts an upwards and leftward force on particle P

The -Q charge exerts an upwards and rightward force on particle P

Since the charges are equidistant from the particle P, and are of equal magnitude, the rightward force and the leftward force will cancel out, leaving just the upward force on the particle P.

The effect of the upward force is that the particle P moves directly upwards

Which phrases accurately describe an elliptical galaxy? Check all that apply.
may be egg-shaped
may be spiral-shaped
has no recognizable shape
has no new stars being formed
has almost no gas or dust between stars​

Answers

Answer:

May be egg shaped

Has no new stars being formed.

Has almost no gas or dust between stars.

Explanation:

Elliptical galaxy is the collection of many stars which are bounded together gravitationally, which is smooth and ellipsoidal and shape and the appearance is featureless.

Elliptical galaxy is ovoid or spherical masses of stars.

It is found in galaxy clusters and compact galaxies.

It has no gas or dust between stars which result in low rates of star formation.

It is formed When two spirals collide, they lose their familiar shape, morphing into the less-structured elliptical galaxies.

Elliptical galaxy is made of old stars and have no gas and dust.

An example is elliptical galaxy m60 which shines brightly and is egg shaped.

Final naïve case: If the highest-pitch string on the piano is made of spring steel (density = 7800 kg/m3) with a diameter of 1/32" (= 0.794 mm), what will the linear density of such a string be (in kg/m)?

Answers

Answer:

The linear density is  [tex]K = 3.863 *10^{-3 } \ kg/m[/tex]

Explanation:

From the question we are told that

     The density of  steel is  [tex]\rho = 7800 \ kg/m^3[/tex]

      The diameter of the string is  [tex]d = 0.794 \ mm = 7.94 *10^{-4} \ m[/tex]

       The  radius of  the string is  evaluated as  [tex]r = \frac{D}{2} = \frac{7.94 *10^{-4}}{2} = 3.97*10^{-4} \ m[/tex]

The volume of the string is  mathematically evaluated as

       [tex]V = \pi * r ^2 * L[/tex]

Now assuming that the length of the string is  L = 2 m  

      So  

         [tex]V = 3.142 * (3.97 *10^{-4})^2 * (2)[/tex]

        [tex]V = 9.9041 *10^{-7} \ m^3[/tex]

Then the mass of the string would be  

       [tex]m = \rho * V[/tex]

substituting value  

       [tex]m = 7800*9.904 14 *10^{-7}[/tex]

      [tex]m = 7.73*10^{-3} \ kg[/tex]

Looking at the question we see that the unit of the linear density is  [tex]\frac{kg}{m}[/tex]

Hence the linear density is evaluated as

        [tex]K = \frac{m}{L}[/tex]

substituting value  

        [tex]K = \frac{7.73 *10^{-3}}{2}[/tex]

        [tex]K = 3.863 *10^{-3 } \ kg/m[/tex]

 

A standard 1 kilogram weight is a cylinder 48.5 mm in height and 49.0 mm in diameter. What is the density of the material? kg/m3

Answers

Answer:

Density = 10,933.93 kg/m^3

the density of the material is 10,933.93 kg/m^3

Explanation:

Density is the mass per unit volume

Density = mass/volume = m/V

Volume of a cylinder V = πr^2 h

Given;

Height h = 48.5mm = 0.0485 m

Radius r = diameter/2 = 49mm÷2 = 24.5mm = 0.0245m

Substituting the values;

Volume V = π×(0.0245^2)×0.0485

V = 0.000091458438030 m^3

V = 0.000091458 m^3

The mass is given as;

Mass = 1 kg

So, the density can be calculated as;

Density = 1/0.000091458

Density = 10933.92825785 kg/m^3

Density = 10,933.93 kg/m^3

the density of the material is 10,933.93 kg/m^3

ir temperature in a desert can reach 58.0°C (about 136°F). What is the speed of sound (in m/s) in air at that temperature?

Answers

Answer:

363m.s-1

Explanation:

A 95 N force exerted at the end of a 0.35 m long torque wrench gives rise to a torque of 15 N · m. What is the angle (assumed to be less than 90°) between the wrench handle and the direction of the applied force?

Answers

Answer:

The angle between the wrench handle and the direction of the applied force is 26.8°

Explanation:

Given;

applied force, F = 95 N

length of the wrench, r = 0.35 m

torque on the wrench due to the applied force, τ = 15 N.m

Torque is calculated as;

τ = rFsinθ

where;

r is the length of the wrench

F is the applied force

θ is the angle between the applied force and the wrench handle

Make Sin θ the subject of the formula;

Sinθ = τ / rF

Sinθ = 15 / (0.35 x 95)

Sinθ = 0.4511

θ = Sin⁻¹(0.4511)

θ = 26.8°

Therefore, the angle between the wrench handle and the direction of the applied force is 26.8°

I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block and it oscillates with a period of 0.13 s. What is the amplitude of oscillation

Answers

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

[tex]T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\[/tex]

Now, determine the amplitude of oscillation, A;

[tex]E = \frac{1}{2} kA^2[/tex]

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

[tex]E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm[/tex]

Therefore, the amplitude of the oscillation is 2.82 cm

is tantalum least reactive or more

Answers

Answer:

it is more reactive in high temperature than in low temperature.

Which compound is composed of oppositely charged ions?

Answers

Answer:

Option A. Li2O

Explanation:

To know which of the compound contains oppositely charged ions, let us determine the nature of each compound. This is illustrated below:

Li2O is an ionic compound as it contains a metal (Lithium, Li) and non metal (oxygen, O). Ionic compounds are charactized by the presence of aggregate positive and negative charge ions. This is true because they are formed by the transfer of electron(s) from the metallic atom to the non-metallic atom.

2Li —> 2Li^+ + 2e

O2 + 2e —> O^2-

2Li + O2 + 2e —> 2Li^+ + O^2- + 2e

2Li + O2 —> 2Li^+ O^2- —> Li2O

OF2 is a covalent compound as it contains non metals only (i.e oxygen, O and fluorine, F). Covalent compounds are characterised by the presence of molecules. This is true because they are formed from the sharing of electron(s) between the atoms involved.

PH3 is a covalent compound as it contains non metals only (i.e phosphorus, P and hydrogen, H).

SCl2 is a covalent compound as it contains non metals only (i.e sulphur, S and chlorine, Cl).

From the above information, we can see that only Li2O contains oppositely charged ions.

Answer:

A

Explanation:

Just took the test

A vector quantity has direction, a scalar quantity does not.

Answers

Explanation:

hope you like then comment plz

A cylindrical shell of radius 7.00 cm and length 2.21 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 15.2 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C. (a) Find the net charge on the shell.

Answers

Answer:

The net charge on the shell is 30x10^-9C

Explanation:

Pls see attached file

Match each term to the best description. ::
1. Coherent
2. Diffraction
3. Grating
4. Interference
5. Specular dot
a. Composed of numerous narrowly spaced parallel slits or grooves
b. Having the same wavelength, frequency, and in-phase
c. Interaction of waves where they meet in space
d. The bending of waves near a boundary or as a wave passes through an opening
e. The zeroth order direct reflection fringe

Answers

These are the matched answers:

Faraday's Law states that the negative of the time rate of change of the flux of the magnetic field through a surface is equal to which of the following quantities?

a. The flux of the magnetic field through a surface which has the loop as its boundary.
b. The negative of the time rate of change of the flux of the magnetic field through a surface which has the loop as its boundary.
c. The line integral of the magnetic field around the closed loop.
d. The flux of the electric field through a surface which has the loop as its boundary.

Answers

Answer:

(C). The line integral of the magnetic field around a closed loop

Explanation:

Faraday's law states that induced emf is directly proportional to the time rate of change of magnetic flux.

This can be written mathematically as;

[tex]EMF = -\frac{\delta \phi _B}{\delta t}[/tex]

[tex](\frac{\delta \phi _B}{\delta t} )[/tex] is the rate of change of the magnetic flux through a surface bounded by the loop.

ΔФ = BA

where;

ΔФ is change in flux

B is the magnetic field

A is the area of the loop

Thus, according to Faraday's law of electric generators

∫BdL = [tex]\frac{\delta \phi _B}{\delta t}[/tex] = EMF

Therefore, the line integral of the magnetic field around a closed loop is equal to the negative of the rate of change of the magnetic flux through the area enclosed by the loop.

The correct option is "C"

(C). The line integral of the magnetic field around a closed loop

Faraday's Law states that the negative of the time rate of change of the flux of the magnetic field through a surface is equal to: D. The flux of the electric field through a surface which has the loop as its boundary.

In Physics, the surface integral with respect to the normal component of a magnetic field over a surface is the magnetic flux through that surface and it is typically denoted by the symbol [tex]\phi[/tex].

Faraday's Law states that the negative of the time rate of change ([tex]\Delta t)[/tex] of the flux of the magnetic field ([tex]\phi[/tex]) through a surface is directly proportional to the flux ([tex]\phi[/tex]) of the electric field through a surface which has the loop as its boundary.

Mathematically, Faraday's Law is given by the formula:

[tex]E.m.f = -N\frac{\Delta \phi}{\Delta t}[/tex]

Where:

N is the number of turns.

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Four identical charges particles of charge 1Uc, 2Uc,
3Uc and 4Uc
are placed at x = lm, x=2m,
x=3m and
x=5m. The electric field intensity
at origin is?

Answers

Answer:

17.94 kN/C is the electric field intensity at the origin due to the charges.

Explanation:

From the question, we are told that

The distance of 1 μC from origin = 1 m

The distance of 2 μC from origin = 2 m

The distance of 3 μC from origin = 3 m

The distance of 4 μC from origin = 5 m

Therefore, for us to find the electric field intensity, we'll solve below:

The formula for Electric field intensity = ( k * q ) / ( r * r )

where , r is distance ,

k = 9 * 10^9 ,

and , q is charge .

now ,

electric field intensity at the origin = [ k * 10^(-6) / 1 * 1 ] +[ k * 2 * 10^(-6) / 2 * 2 ] + [ k * 3 * 10^(-6) / 3 * 3 ] + [ k * 4 * 10^(-6) / 5 * 5 ]

=> electric field intensity at the origin = k * 10^(-6) [ 1 + 1/2 + 1/3 + 4/25 ] N/C

=> electric field intensity at the origin = 9 * 10^9 * 10^(-6) * 1.99 N/C

=> electric field intensity at the origin = 17.94 kN/C

Suppose the ring rotates once every 4.30 s . If a rider's mass is 53.0 kg , with how much force does the ring push on her at the top of the ride?

Answers

The complete question is;

In an amusement park ride called The Roundup, passengers stand inside a 16-m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane.

Suppose the ring rotates once every 4.30 s . If a rider's mass is 53.0 kg , with how much force does the ring push on her at the top of the ride?

Answer:

F_top = 385.36 N

Explanation:

We are given;

mass;m = 52 kg

Time;t = 4.3 s

Diameter;d = 16m

So,Radius;r = 16/2 = 8m

The formula for the centrifugal force is given as;

F_c = mω²R

Where;

R = radius

Angular velocity;ω = 2πf

f = frequency = 1/t = 1/4.3 Hz

F_c = 53 × (2π × 1/4.3)² × 8 = 905.29 N.

The force at top would be;

F_top = F_c - mg

F_top = 905.29 - (9.81 × 53) N

F_top = 385.36 N

The force at the top of ride will be "385.36 N".

Force and mass:

According to the question,

Rider's mass, m = 52 kg

Time, t = 4.3 s

Diameter, d = 16 m

Radius, r = [tex]\frac{16}{2}[/tex] = 8 m

Frequency, f = [tex]\frac{1}{t}[/tex] = [tex]\frac{1}{4.3}[/tex] Hz

We know the formula,

Centrifugal force,  [tex]F_c[/tex] = mω²R

or,

Angular velocity, ω = 2πf

By substituting the values in the above formula,

[tex]F_c = 53(2\pi \times (\frac{1}{4.3})^2\times 8 )[/tex]

    [tex]= 905.29[/tex] N

hence,

The top force will be:

→ [tex]F_{top} = F_c[/tex] - mg

By substituting the values,

          [tex]= 905.29-(9.81\times 53)[/tex]

          [tex]= 385.36[/tex] N

Thus the above response is correct.  

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(5 pt) You tie a cord to a pail of water, and your swing the pail in a vertical circular 0.700 m. What is the minimum speed must you give the pail at the highest point of the circle if no water is to spill from it

Answers

Answer:

The minimum speed required  is 2.62m/s

Explanation:

The value of  gravitational acceleration = g = 9.81 m/s^2

Radius of the vertical circle = R = 0.7 m

Given the mass of the pail of water = m

The speed at the highest point of the circle = V

The centripetal force will be needed must be more than the weight of the pail of water in order to not spill water.

Below is the calculation:

[tex]\frac{mV^{2}}{R} = mg[/tex]

[tex]V = \sqrt{gR}[/tex]

[tex]V = \sqrt{9.81 \times 0.7}[/tex]

[tex]V = 2.62 m/s[/tex]

A small cylinder made of a diamagnetic material is brought near a bar magnet. One end of the cylinder is brought near the North pole of the magnet and is repelled. What happens when the other end of the cylinder is brought near the North pole of the magnet? Group of answer choices

Answers

Answer:

it attracts

Explanation:

since in a magnetic body there are two poles

(north and south poles)if the first pole was repeled when brought near the North Pole therefore the other end is going to attarct because the first end was also a North Pole while the second end will be a south pole

A 90.0-kg ice hockey player hits a 0.150-kg puck, giving the puck a velocity of 45.0 m/s. If both are initially at rest and if the ice is frictionless, how far does the player recoil in the time it takes the puck to reach the goal 15.0 m away

Answers

Answer:

0.0241 m

Explanation:

mass of the hockey player m1 = 90 kg

mass of puck m2 = 0.150 kg

puck velocity v1= 45 m/s

distance traveled by puck to reach the goal =15.0 m.

now accoding to momentum conservation law

90×45+0.15×v2 = 0 [ since, If both are initially at rest and if the ice is frictionless,]

therefore, v2= -0.0725 m/s.

Now time taken by the puck to reach the goal

t= 15/45 = 1/3 sec.

therefore, how far does the player recoil in the time

=0.0725×1/3= 0.0241 m.

the distance travelled by the player( recoil ) in the time the puck reach the goal is 0.025m.

Given the data in the question

Mass of the player; [tex]m_1 = 90.0kg[/tex]Mass of puck; [tex]m = 0.150kg[/tex]

Since they were both at rest initially

Initial velocity of player; [tex]u_1 = 0[/tex]Initial velocity of puck; [tex]u = 0[/tex]Velocity of player after the hit; [tex]v_1 = \ ?[/tex]Velocity of puck after the hit; [tex]v = 45.0m/s[/tex]Distance to the goal; [tex]s = 15.0m[/tex]

Using conservation of liner momentum:

[tex]mu + m_1u_1 = mv+ m_1v_1[/tex]

Now, Since they were both at rest initially

[tex]0 = mv+ m_1v_1[/tex]

We substitute in our values to find the velocity of the player after the hit ( recoil velocity )

[tex]0 =[ 0.150kg * 45.0m/s ] + [ 90.0kg * v_1 ]\\\\0 = 6.75kg.m/s + [ 90.0kg * v_1 ]\\\\90.0kg * v_1 = -6.75kg.m/s \\\\v_1 = -\frac{6.75kg.m/s}{90.0kg} \\\\v_1 =- 0.075m/s[/tex]

{ The negative sign shows that the velocity of both the player and the puck are in opposite direction }

Hence, recoil velocity of the player is 0.075m/s

Now, we determine the time taken for the puck to trach the goal using the relation between distance, velocity and time .

Time = Distance / Velocity

We substitute our values into the expression

[tex]t = \frac{s}{v} \\\\t = \frac{15.0m}{45m/s} \\\\t = 0.3333s[/tex]

Hence, the time taken for the puck to reach the goal is 0.3333 seconds.

Next, we determine the distance travelled by the player( recoil ) in the time the puck reach the goal using the relation between distance, velocity and time .

Time = Distance / Velocity

We substitute in our values

[tex]t = \frac{s}{v}\\\\0.3333s = \frac{s}{0.075m/s} \\\\s = 0.3333s * 0.075m/s\\\\s = 0.025m[/tex]

Therefore, the distance travelled by the player( recoil ) in the time the puck reach the goal is 0.025m.

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Angular velocity in the z direction of a flywheel is w(t)=A + Bt2 The numerical values of the constants are A=2.75 and B=1.50. What is the angular acceleration α(t) when t=0s and t=5.00s?

Answers

Answer:

α(0) = 0 rad/s²

α(5) = 15 rad/s²

Explanation:

The angular velocity of the flywheel is given as follows:

w(t) = A + B t²

where, A and B are constants.

Now, for the angular acceleration, we must take derivative of angular velocity with respect to time:

Angular Acceleration = α (t) = dw/dt

α(t) = (d/dt)(A + B t²)

α(t) = 2 B t

where,

B = 1.5

AT t = 0 s

α(0) = 2(1.5)(0)

α(0) = 0 rad/s²

AT t = 5 s

α(5) = 2(1.5)(5)

α(5) = 15 rad/s²

If a sound with frequency fs is produced by a source traveling along a line with speed vs. If an observer is traveling with speed vo along the same line from the opposite direction toward the source, then the frequency of the sound heard by the observer is fo = c + vo c − vs fs where c is the speed of sound, about 332 m/s. (This is the Doppler effect.) Suppose that, at a particular moment, you are in a train traveling at 32 m/s and accelerating at 1.3 m/s2. A train is approaching you from the opposite direction on the other track at 48 m/s, accelerating at 1.9 m/s2, and sounds its whistle, which has a frequency of 439 Hz. At that instant, what is the perceived frequency that you hear? (Round your answer to one decimal place.) Hz

Answers

Answer: The frequency heard is 562.7 Hz.

Explanation: Doppler Effect happens when there is shift in frequency during a realtive motion between a source and the observer of that source.

It can be calculated as:

[tex]f_{o} = f_{s}(\frac{c+v_{o}}{c-v_{s}} )[/tex]

where:

c is the speed of light (c = 332m/s)

all the subscripted s is related to the Source (frequency, velocity);

all the subscripted o is related to the Observer (frequency, velocity);

As the source is moving towards the observer and the observer is moving towards the source, the velocities of each are opposite related to direction.

So, the frequency perceived by the observer:

[tex]f_{o} = 439(\frac{332+32}{332-48} )[/tex]

[tex]f_{o} = 439(\frac{364}{284} )[/tex]

[tex]f_{o} = 439(1.282 )[/tex]

[tex]f_{o}[/tex] = 562.7 Hz

At this condition, the observer hears the train's horn in a perceived frequency of 562.7 Hz

The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the bal

Answers

Complete Question

The complete question is gotten from OpenStax

A soccer player starts at rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s ? The player’s mass is 70.0 kg, and air resistance is negligible.

The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball

Answer:

The  time it will take is  [tex]t = 1.4907 \ s[/tex]  

Explanation:

From the  question we are told that

    The force experienced by the player is  [tex]F = 126 \ N[/tex]

     The distance of the ball from the player is  [tex]d = 2.00 \ m[/tex]

      The initial velocity is  u =  0 m/s because the player stopped

From the Newton law the acceleration of the player is mathematically evaluated as

             [tex]a = \frac{F}{m }[/tex]    [i,e  F =  ma  ]

substituting values

             [tex]a = \frac{126}{70}[/tex]

             [tex]a = 1.8 \ m/s^2[/tex]

Now from the equation of motion  we have that

           [tex]s = ut + \frac{1}{2} at^2[/tex]

substituting values              

             [tex]2.0 = 0 + \frac{1}{2} * 1.8 * t^2[/tex]

             [tex]t = \sqrt{ \frac{2.0}{0.9} }[/tex]

            [tex]t = 1.4907 \ s[/tex]

Find the displacement. Will give brainliest!

Answers

Answer:

1000 m upwards

Explanation:

Displacement Formula: Average Velocity = Displacement/Total Time

Simply plug in our known variables and solve:

100 m/s = x m/10 seconds

100 m/s(10 s) = x m

m = 1000

A small ferryboat is 4.70 m wide and 6.10 m long. When a loaded truck pulls onto it, the boat sinks an additional 5.00 cm into the river. What is the weight of the truck

Answers

Answer:

   M = 1433.5 kg

Explanation:

This exercise is solved using the Archimedean principle, which states that the hydrostatic thrust is equal to the weight of the desalinated liquid,

              B = ρ g V

with the weight of the truck it is in equilibrium with the push, we use Newton's equilibrium condition

           Σ F = 0

           B-W = 0

           B = W

       body weight

           W = M g

the volume is

           V = l to h

           rho_liquid g (l to h) = M g

           M = rho_liquid l a h

           

we calculate

            M = 1000 4.7 6.10 0.05

           M = 1433.5 kg

A proton moving in the positive x direction with a speed of 9.9 105 m/s experiences zero magnetic force. When it moves in the positive y direction it experiences a force of 1.6 10-13 N that points in the positive z direction. Determine the magnitude and direction of the magnetic field.

Answers

Answer:

The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction

Explanation:

In order to calculate the magnitude and direction of the magnetic field, you take into account the following equation for the magnetic force on the proton:

[tex]\vec{F_B}=q\vec{v}\ X\ \vec{B}[/tex]       (1)

v: speed of the proton = 9.9*10^5 m/s

q: charge of the proton = 1.6*10^-19C

B: magnetic field = ?

FB: magnetic force on the proton = 1.6*10^-13N

When the proton travels in the positive y direction (^j), you have that the proton experiences a force in the positive z direction (+^k). To obtain this direction of the magnetic force on the proton, it is necessary that the magnetic field points in the negative x direction, in fact, you have:

^j X (-^i) = -(-^k)=^k

To obtain the magnitude of the magnetic field you use:

[tex]F_B=qvBsin90\°=qvB\\\\B=\frac{F_B}{qv}=\frac{1.6*10^{-13}N}{(1.6*10^{-19}C)(9.9*10^5m/s)}\\\\B=1.01T[/tex]

The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction

A single-turn circular loop of radius 6 cm is to produce a field at its center that will just cancel the earth's field of magnitude 0.7 G directed at 70 below the horizontal north direction. Find the current in the loop.

Answers

Answer:

The current is  [tex]I = 6.68 \ A[/tex]

Explanation:

From the question we are told that  

     The radius of the loop is  [tex]r = 6 \ cm = 0.06 \ m[/tex]

     The  earth's magnetic field is [tex]B_e = 0.7G= 0.7 G * \frac{1*10^{-4} T}{1 G} = 0.7 *10^{-4} T[/tex]

      The  number of turns is  [tex]N =1[/tex]

Generally the magnetic field generated by the current in the loop is mathematically represented as

        [tex]B = \frac{\mu_o * N * I}{2 r }[/tex]

Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth

         [tex]B = B_e[/tex]

=>     [tex]B_e = \frac{\mu_o * N * I }{ 2 * r}[/tex]

     Where  [tex]\mu[/tex] is the permeability of free space with value  [tex]\mu _o = 4\pi * 10^{-7} N/A^2[/tex]

       [tex]0.7 *10^{-4}= \frac{ 4\pi * 10^{-7} * 1 * I}{2 * 0.06}[/tex]

=>     [tex]I = \frac{2 * 0.06 * 0.7 *10^{-4}}{ 4\pi * 10^{-7} * 1}[/tex]

       [tex]I = 6.68 \ A[/tex]

The current in the loop will be "6.68 A".

Magnetic field

According to the question,

Radius of loop, r = 6 cm or,

                           = 0.06 m

Earth's magnetic field, [tex]B_e[/tex] = 0.7 G or,

                                          = 0.7 × [tex]\frac{1\times 10^{-4}}{1 G}[/tex]

                                          = 0.7 × 10⁻⁴ T

Number of turns, N = 1

We know the relation,

→ B = [tex]\frac{\mu_0\times N\times I}{2r}[/tex]

or,

  B = [tex]B_e[/tex]

then,

→         [tex]B_e[/tex] = [tex]\frac{\mu_0\times N\times I}{2r}[/tex]

By substituting the values,

0.7 × 10⁻⁴ = [tex]\frac{4 \pi\times 10^{-7}\times 1\times I}{2\times 0.06}[/tex]  

hence,

The current will be:

               I = [tex]\frac{2\times 0.06\times 0.7\times 10^{-4}}{4 \pi\times 10^{-7}\times 1}[/tex]

                 = 6.68 A

Thus the above approach is correct.    

             

Find out more information about Magnetic field here:

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A piano tuner hears a beat every 2.20 s when listening to a 266.0 Hz tuning fork and a single piano string. What are the two possible frequencies (in Hz) of the string? (Give your answers to at least one decimal place.)

Answers

Answer:

The lower frequency is [tex]f_1 = 265.55 \ Hz[/tex]

The higher frequency is  [tex]f_2 = 266.4546 \ Hz[/tex]

Explanation:

From the question we are told that

     The period is   [tex]T = 2.20 \ s[/tex]

      The frequency of the tuning fork is  [tex]f = 266.0 \ Hz[/tex]

Generally the beat frequency is mathematically represented as

       [tex]f_b = \frac{1}{T}[/tex]

substituting values

      [tex]f_b = \frac{1}{2.20}[/tex]

      [tex]f_b = 0.4546 \ Hz[/tex]

Since the beat  frequency is gotten from the beat produced by the tuning fork and and  the string   then

The possible frequency of the string ranges from

     [tex]f_1 = f- f _b[/tex]

to

    [tex]f_2 = f + f_b[/tex]

Now  substituting values

    [tex]f_1 = 266.0 - 0.4546[/tex]

    [tex]f_1 = 265.55 \ Hz[/tex]

For  [tex]f_2[/tex]

    [tex]f_2 = 266 + 0.4546[/tex]

    [tex]f_2 = 266.4546 \ Hz[/tex]

In a high school swim competition, a student takes 1.6 s to complete 1.5 somersaults. Determine the average angular speed of the diver, in rad/s, during this time interval.

Answers

Answer:

The  angular speed is [tex]w = 5.89 \ rad/s[/tex]

Explanation:

From the question we are told that

    The time taken is  [tex]t = 1.6 s[/tex]

    The number of somersaults  is n  =  1.5

The total angular displacement during the somersault is mathematically represented as

         [tex]\theta = n * 2 * \pi[/tex]

substituting values

        [tex]\theta = 1.5 * 2 * 3.142[/tex]

       [tex]\theta = 9.426 \ rad[/tex]

 The angular speed is mathematically represented as

         [tex]w = \frac{\theta }{t}[/tex]

substituting values

         [tex]w = \frac{9.426}{1.6}[/tex]

          [tex]w = 5.89 \ rad/s[/tex]

     

Which kind of energy is found in an atom's nucleus? (A) Nuclear (B) Elastic (C) Thermal (D) Electromagnetic

Answers

Answer:

The answer is option A.

Nuclear

Hope this helps you

Answer:

Answer is A Nuclear

Explanation:

Just answered this question on my test

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