Please help! Which statements correctly describe the effect of distance in determining the gravitational force and the electrical force? Check all that apply.

Please Help! Which Statements Correctly Describe The Effect Of Distance In Determining The Gravitational

Answers

Answer 1

There are six statements on the list.

The first 2 are true, and the last 2 are true.

The 2 in the middle aren't true.  They are false.


Related Questions

As you drive down the road at 13 m/s , you press on the gas pedal and speed up with a uniform acceleration of 1.02 m/s2 for 0.70 s. If the tires on your car have a radius of 33 cm, what is their angular displacement during this period of acceleration?

Answers

Answer:

The angular displacement is  [tex]\theta = 28.33 \ rad[/tex]

Explanation:

From the question we are told that

     The speed of the driver is  [tex]v =13 \ m/ s[/tex]

     The acceleration of the driver is  [tex]a = 1.02 \ m/s^2[/tex]

      The time taken is [tex]t = 0.70 \ s[/tex]

      The radius of the tire is  [tex]r = 33 cm = 0.33 \ m[/tex]

The distance covered by the car during this  acceleration can be  calculated using the equation of motion as follows

        [tex]s = v*t +\frac{1}{2} * a * t^2[/tex]

Now substituting values  

       [tex]s = 13 * 0.70 +\frac{1}{2} * 1.02 * (0.700)^2[/tex]

      [tex]s = 9.35 \ m[/tex]

Now the angular displacement of the car with respect to the tire movement can be  represented mathematically as

      [tex]\theta = \frac{s}{r}[/tex]

substituting values

      [tex]\theta = \frac{9.35}{0.33}[/tex]

      [tex]\theta = 28.33 \ rad[/tex]

A 6.50-m-long iron wire is 1.50 mm in diameter and carries a uniform current density of 4.07 MA/m^2. Find the voltage between the two ends of the wire.

Answers

Answer:

V = 0.45 Volts

Explanation:

First we need to find the total current passing through the wire. That can be given by:

Total Current = I = (Current Density)(Surface Area of Wire)

I = (Current Density)(2πrL)

where,

r = radius = 1.5/2 mm = 0.75 mm = 0.75 x 10⁻³ m

L = Length of Wire = 6.5 m

Therefore,

I = (4.07 x 10⁻³ A/m²)[2π(0.75 x 10⁻³ m)(6.5 m)]

I = 1.25 x 10⁻⁴ A

Now, we need to find resistance of wire:

R = ρL/A

where,

ρ = resistivity of iron = 9.71 x 10⁻⁸ Ωm

A = Cross-sectional Area = πr² = π(0.75 x 10⁻³ m)² = 1.77 x 10⁻⁶ m²

Therefore,

R = (9.71 x 10⁻⁸ Ωm)(6.5 m)/(1.77 x 10⁻⁶ m²)

R = 0.36 Ω

From Ohm's Law:

Voltage = V = IR

V = (1.25 x 10⁻⁴ A)(0.36 Ω)

V = 0.45 Volts

If an object is determined to have a negative charge of 1.6 micro Coulomb, you can conclude that the object has an excess of

Answers

Answer:

The object has an excess of [tex]10^{13}[/tex] electrons.

Explanation:

When an object has a negative charge he has an excess of electrons in its body. We can calculate the number of excessive electrons by dividing the charge of the body by the charge of one electron. This is done below:

[tex]n = \frac{\text{object charge}}{\text{electron charge}}\\n = \frac{-1.6*10^{-6}}{-1.6*10^{-19}} = 1*10^{-6 + 19} = 10^{13}[/tex]

The object has an excess of [tex]10^{13}[/tex] electrons.

Can an object travel at the speed of
light? Why or why nbt?

Answers

Answer:

no the only things that can travel at the speed of light are waves in the electromagnetic spectrum

No because, the object shrinks as it moves forward compared to light, it never does.

Source: google

A crane lifts a 425 kg steel beam vertically a distance of 64 m. How much work does the crane do on the beam if the beam accelerates upward at 1.8 m/s2

Answers

Answer:

work done= 48.96 kJ

Explanation:

Given data

mass of  load m= 425 kg

height/distance h=64 m

acceleration a= 1.8 m/s^2

The work done can be calculated using the expression

work done= force* distance

but force= mass *acceleration

hence work done= 425*1.8*64= 48,960 J

work done= 48.96 kJ

A man stands on a merry-go-round that is rotating at 2.5 rad/s. If the coefficient of static friction between the man’s shoes and the merry-go-round is µs = 0.5, how far from the axis of rotation can he stand without sliding?

Answers

Answer:

0.8 m

Explanation:

Draw a free body diagram.  There are three forces:

Weight force mg pulling down,

Normal force N pushing up,

and friction force Nμ pushing towards the center.

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum of forces in the centripetal direction:

∑F = ma

Nμ = m v²/r

Substitute and simplify:

mgμ = m v²/r

gμ = v²/r

Write v in terms of ω and solve for r:

gμ = ω²r

r = gμ/ω²

Plug in values:

r = (10 m/s²) (0.5) / (2.5 rad/s)²

r = 0.8 m

The distance (radius) from the axis of rotation which the man can stand without sliding is 0.784 meters.

Given the following data:

Angular speed = 2.5 rad/s.Coefficient of static friction = 0.5

To determine how far (radius) from the axis of rotation can the man stand without sliding:

We would apply Newton's Second Law of Motion, to express the centripetal and force of static friction acting on the man.

[tex]\sum F = \frac{mv^2}{r} - uF_n\\\\\frac{mv^2}{r} = uF_n[/tex]....equation 1.

But, Normal force, [tex]F_n = mg[/tex]  

Substituting the normal force into eqn. 1, we have:

[tex]\frac{mv^2}{r} = umg\\\\\frac{v^2}{r} = ug[/tex]....equation 2.

Also, Linear speed, [tex]v = r\omega[/tex]

Substituting Linear speed into eqn. 2, we have:

[tex]\frac{(r\omega )^2}{r} = ug\\\\r\omega ^2 = ug\\\\r = \frac{ug}{\omega ^2}[/tex]

Substituting the given parameters into the formula, we have;

[tex]r = \frac{0.5 \times 9.8}{2.5^2} \\\\r = \frac{4.9}{6.25}[/tex]

Radius, r = 0.784 meters

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Two vehicles approach an intersection, a 2500kg pickup travels from E to W at 14.0m/s and a 1500kg car from S to N at 23.0m/s. Find P net of this system (direction and magnitude)

Answers

Answer:

The magnitude of the momentum is 49145.19 kg.m/s

The direction of the two vehicles is 44.6° North West

Explanation:

Given;

speed of first vehicle, v₁ = 14 m/s (East to west)

mass of first vehicle, m₁ = 1500 kg

speed of second vehicle, v₂ = 23 m/s (South to North)

momentum of the first vehicle in x-direction (E to W is in negative x-direction)

[tex]P_x = mv_x\\\\P_x = 2500kg(-14 \ m/s)\\\\P_x = -35000 \ kg.m/s[/tex]

momentum of the second vehicle in y-direction (S to N is in positive y-direction)

[tex]P_y = m_2v_y\\\\P_y = 1500kg(23 \ m/s)\\\\P_y = 34500 \ kg.m/s[/tex]

Magnitude of the momentum of the system;

[tex]P= \sqrt{P_x^2 + P_y^2} \\\\P = \sqrt{(-35000)^2+(34500)^2} \\\\P = 49145.19 \ kg.m/s[/tex]

Direction of the two vehicles;

[tex]tan \ \theta = \frac{P_y}{|P_x|} \\\\tan \ \theta = \frac{34500}{35000} \\\\tan \ \theta = 0.9857\\\\\theta = tan^{-1} (0.9857)\\\\\theta = 44.6^0[/tex]North West

is tantalum least reactive or more

Answers

Answer:

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

Find the magnitude of the resultant of forces 6N and 8N acting at 240° to each other

Answers

Answer:

magnitude of the resultant of forces is 11.45 N

Explanation:

given data

force F1 = 6N

force F2 = 8N

angle = 240°

solution

we get here resultant force that is express as

F(r) = [tex]\sqrt{F_1^2+F_2^2+2F_1F_2cos\ \theta}[/tex]    ..............1

put here value and we get

F(r) = [tex]\sqrt{6^2+8^2+2\times 6\times 8 \times cos240}[/tex]

F(r) =  11.45 N

so magnitude of the resultant of forces is 11.45 N

The voltage difference between the AA and AAA batteries should be quite small. What then might be the difference between them?

Answers

Answer:

The major difference is the capacity of both batteries. The AA battery has a higher capacity (a higher current) than the AAA battery.

Explanation:

The AA batteries and the AAA batteries are very similar in their voltage; both of them have 1.5 V.

The difference between these two batteries is their size and also the current that they have. The AAA battery is smaller than the AA battery, which means that the amount of electrochemical material is lower, so the AA battery has a higher capacity (a higher current) than the AAA battery. Generally, AA battery has 2400 mAh capacity and AAA battery has a capacity of 1000mAh; this means that AA battery has almost three times the capacity of an AAA battery.      

Furthermore, the size of the AA battery makes it more common than the AAA battery and therefore has higher commercial demand.                                  

I hope it helps you!

Find the change in the force of gravity between two planets when the distance between them becomes 10 times smaller. g

Answers

Answer:

F' = F/16

So, the force of gravity has become 16 times less than initial value.

Explanation:

The force of gravity between two planets, is given by the following formula:

F = Gm₁m₂/r²   ----------- equation 1

where,

F = Force of gravity between two planets

G = Gravitational Constant

m₁ = Mass of one planet

m₂ = Mass of other plant

r = Distance between two planets

Now, if the distance between the planets (r) is 10 times smaller, then Force of gravity will become:

F' = Gm₁m₂/(4r)²

F' = (1/16) (Gm₁m₂/r²)

using equation 1:

F' = F/16

So, the force of gravity has become 16 times less than initial value.

At what temperature will silver have a resistivity that is two times the resistivity of iron at room temperature? (Assume room temperature is 20° C.)

Answers

Answer:

The temperature of silver at this given resistivity is 2971.1 ⁰C

Explanation:

The resistivity of silver is calculated as follows;

[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\[/tex]

where;

Rt is the resistivity of silver at the given temperature

Ro is the resistivity of silver at room temperature

α is the temperature coefficient of resistance

To is the room temperature

T is the temperature at which the resistivity of silver will be two times the resistivity of iron at room temperature

[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\\R_t = 1.59*10^{-8}[1 + 0.0038(T-20)][/tex]

Resistivity of iron at room temperature = 9.71 x 10⁻⁸ ohm.m

When silver's resistivity becomes 2 times the resistivity of iron, we will have the following equations;

[tex]R_t,_{silver} = 2R_o,_{iron}\\\\1.59*10^{-8}[1 + 0.0038(T-20)] =(2 *9.71*10^{-8})\\\\\ \ (divide \ through \ by \ 1.59*10^{-8})\\\\1 + 0.0038(T-20) = 12.214\\\\1 + 0.0038T - 0.076 = 12.214\\\\0.0038T +0.924 = 12.214\\\\0.0038T = 12.214 - 0.924\\\\0.0038T = 11.29\\\\T = \frac{11.29}{0.0038} \\\\T = 2971.1 \ ^0C[/tex]

Therefore, the temperature of silver at this given resistivity is 2971.1 ⁰C

A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energy that is equal to the proton rest energy, the speed of the kaon is most nearly:___________.
A. 0.25c
B. 0.40c
C. 0.55c
D. 0.70c
E. 0.85c

Answers

Answer:

0.85c

Explanation:

Rest mass of Kaon [tex]M_{0K}[/tex] = 494 MeV/c²

Rest mass of proton [tex]M_{0P}[/tex]  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy [tex]E_{0K}[/tex] = 494c² MeV

for the proton, rest energy [tex]E_{0P}[/tex] = 938c² MeV

Recall that the rest energy, and the total energy are related by..

[tex]E[/tex] = γ[tex]E_{0}[/tex]

which can be written in this case as

[tex]E_{K}[/tex] = γ[tex]E_{0K}[/tex] ...... equ 1

where [tex]E[/tex] = total energy of the kaon, and

[tex]E_{0}[/tex] = rest energy of the kaon

γ = relativistic factor = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex]

where [tex]\beta = \frac{v}{c}[/tex]

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

[tex]E_{K}[/tex] = [tex]E_{0P}[/tex] ......equ 2

where [tex]E_{K}[/tex] is the total energy of the kaon, and

[tex]E_{0P}[/tex] is the rest energy of the proton.

From [tex]E_{K}[/tex] = [tex]E_{0P}[/tex] = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex] = 1.89

1.89[tex]\sqrt{1 - \beta ^{2} }[/tex] = 1

squaring both sides, we get

3.57( 1 - [tex]\beta^{2}[/tex]) = 1

3.57 - 3.57[tex]\beta^{2}[/tex] = 1

2.57 = 3.57[tex]\beta^{2}[/tex]

[tex]\beta^{2}[/tex] = 2.57/3.57 = 0.72

[tex]\beta = \sqrt{0.72}[/tex] = 0.85

but, [tex]\beta = \frac{v}{c}[/tex]

v/c = 0.85

v = 0.85c

You have a circuit of three resistors in series connected to a battery. You add a fourth resistor, also in series, to the combination. As a result:_______.
A. Power delivered from the battery to combination increases.
B. It is impossible to give the answer without knowing the actual resistances and voltage of the battery.
C. Power delivered from the battery to combination is unchanged.
D. Power delivered from the battery to combination decreases.

Answers

Answer:

D

Explanation:

The power equation is P= V^2/R

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As a result of the given scenario, power delivered from the battery to combination decreases. The correct option is D.

What is a resistors?

A resistor is a two-terminal passive electrical component that uses electrical resistance as a circuit element.

Resistors are used in electronic circuits to reduce current flow, adjust signal levels, divide voltages, and bias active elements.

A resistor is a component of an electronic circuit that limits or regulates the flow of electrical current. Resistors can also be used to supply a fixed voltage to an active device such as a transistor.

The current through resistors is the same when they are connected in series. The battery voltage is divided among resistors.

Adding more resistors to a series circuit increases total resistance and thus lowers current. However, in a parallel circuit, adding more resistors in parallel creates more options while decreasing total resistance.

Thus, the correct option is D.

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An electron, moving west, enters a magnetic field of a certain strength. Because of this field the electron curves upward. What is the direction of the magnetic field?

Answers

Answer:

Towards the west.

Explanation:

The direction of a magnetic field lines is the direction north end of a compass needle points. The magnetic field exert force on positive charge.

Using the magnetic rule,which indicate that in order to find the direction of magnetic force on a moving charge, the thumb of the right hand point in the direction of force, the index finger in the direction of velocity charge and the middle finger in the direction of magnetic field.

According to the right hand rule, the electron moving moving west which is the thumb, the direction of the electron is west which is the middle finger and it is upward

a jet fighter accelerates at 17.7 m/s^2 , increasing its velocity from 119 m/s to 233 m/s. how much time does that take?

Answers

Answer:

6.44 s

Explanation:

Given:

v₀ = 119 m/s

v = 233 m/s

a = 17.7 m/s²

Find: t

v = at + v₀

(233 m/s) = (17.7 m/s²) t + (119 m/s)

t = 6.44 s

supose at 20 degree celsius the resistance of Tungsten thermometer is 154.9. WHen placed in a particular solution , the resistance is 207.4 What is the temerature (in degree Celsius of this solution

Answers

Answer:

T₂ = 95.56°C

Explanation:

The final resistance of a material after being heated is given by the relation:

R' = R(1 + αΔT)

where,

R' = Final Resistance = 207.4 Ω

R = Initial Resistance = 154.9 Ω

α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹

ΔT = Change in Temperature = ?

Therefore,

207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]

207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT

1.34 - 1 = (0.0045°C⁻¹)ΔT

ΔT = 0.34/0.0045°C⁻¹

ΔT = 75.56°C

but,

ΔT = Final Temperature - Initial Temperature

ΔT = T₂ - T₁ = T₂ - 20°C

T₂ - 20°C = 75.56°C

T₂ = 75.56°C + 20°C

T₂ = 95.56°C

A friend throws a heavy ball toward you while you are standing on smooth ice. You can either catch the ball or deflect it back toward your friend. What should you do in order to maximize your speed right after your interaction with the ball?
A. You should catch the ball.
B. You should let the ball go past you without touching it.
C. You should deflect the ball back toward your friend.
D. More information is required to determine how to maximize your speed.
E. It doesn't matter. Your speed is the same regardless of what you do.

Answers

Answer:

C You should deflect the ball back toward your friend.

Explanation:

This is because it would result in a completely inelastic collision, and the final velocity of me would be found using,

with m= mass, V=velocity, i=initial, f=final:

mV(me,i) +mV(ball,i) = [m(me)+m(b)]V(f)

So V(f) would be just the momentum of the ball divided by just MV mass of the ball and it will be higher resulting in inelastic collision

Answer:

A. You should catch the ball.

Explanation:

Catching the ball maximizes your speed by converting most of the momentum of the flying ball into the momentum of you and the ball. Since the ice is smooth, the friction between your feet and the ice is almost negligible, meaning less energy is needed to set your body in motion. Catching the ball means that you and the ball undergoes an inelastic collision, and part of the kinetic energy of the ball is transferred to you, setting you in motion. Deflecting the ball will only give you a relatively small speed compared to catching the ball.

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]

Determine the maximum height and range of a projectile fired at a height of 6 feet above the ground with an initial velocity of 100 feet per second at an angle of 40 degrees above the horizontal.Maximum heightRange Question 20 options:a) 70.56 feet183.38 feet b) 92.75 feet310.59 feet c) 92.75 feet183.38 feet d) 70.56 feet314.74 feet e)

Answers

Answer:

C is the correct answer

Explanation:

A 2.0-m wire carrying a current of 0.60 A is oriented parallel to a uniform magnetic field of 0.50 T. What is the magnitude of the force it experiences

Answers

Answer:

The force experienced  is 0.6 N

Explanation:

Given data

length of wire L= 2 m

current in wire I= 0.6 A

magnetic field B= 0.5

The force experienced can be represented as

[tex]F= BIL[/tex]

[tex]F= 0.5*0.6*2\\\F= 0.6 N[/tex]

A kinesin that is transporting a secretory vesicle uses approximately 80 ATP molecules/s. Each ATP provides a kinesin molecule with an energy of about 0.8 × 10-19 J. If the velocity of the kinesin is 800 nm/s, can you determine the force the kinesin is exerting, if you assume that all the ATP energy is used (100% efficiency)? If you can, find it and give your answer in newtons. If not, answer with 0.

Answers

Answer:

The force is  [tex]F = 8*10^{-12} \ N[/tex]

Explanation:

From the question we are told that

     The rate at which ATP molecules are used is [tex]R = 80 ATP/ s[/tex]

       The energy provided by a single ATP is  [tex]E_{ATP} = 0.8 * 10^{-19} J[/tex]

       The velocity of the kinesin is  [tex]v = 800 nm/s = 800*10^{-9} m/s[/tex]

The power provided by the ATP in one second is  mathematically represented as

       [tex]P = E_{ATP} * R[/tex]

substituting values

       [tex]P = 80 * 0.8*10^{-19 }[/tex]

       [tex]P = 6.4 *10^{-18}J/s[/tex]

Now  this power is mathematically represented as

       [tex]P = F * v[/tex]

Where  F  is  the force the kinesin is exerting

  Thus  

          [tex]F = \frac{P}{v}[/tex]

substituting values

            [tex]F = \frac{6.4*0^{-18}}{800 *10^{-9}}[/tex]

           [tex]F = 8*10^{-12} \ N[/tex]

The force exerted by the kinesin  is 8  × 10-12 N.

Let us recall that power is defined as the rate of doing work. Hence, power = Energy/Time.

Since;

Energy  =  0.8 × 10-19 J/molecule

Number ATP molecules transported per second = 80 ATP molecules/s

Power =  0.8 × 10-19 J/molecule × 80 ATP molecules/s

Power = 6.4  × 10-18 J

Again, we know that;

Power = Force × Velocity

Velocity of the ATP molecules =  800 nm/s or 8 × 10-7 m/s

Force = Power/velocity

Force =  6.4  × 10-18 J/ 8 × 10-7 m/s

Force = 8  × 10-12 N

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A uniform disk of 10 kg and radius 4.0 m can rotate in a horizontal plane about a vertical axis through its center. The disk is rotating at an angular velocity of 15 rad/s when a 5-kg package is dropped vertically on a point that is 2.0 m from the center of the disk. What is the angular velocity of the disk/package system

Answers

Answer:

18.75 rad/s

Explanation:

Moment of inertia of the disk;

I_d = ½ × m_disk × r²

I_d = ½ × 10 × 4²

I_d = 80 kg.m²

I_package = m_pack × r²

Now,it's at 2m from the centre, thus;

I_package = 5 × 2²

I_package = 20 Kg.m²

From conservation of momentum;

(I_disk + I_package)ω1 = I_disk × ω2

Where ω1 = 15 rad/s and ω2 is the unknown angular velocity of the disk/package system.

Thus;

Plugging in the relevant values, we obtain;

(80 + 20)15 = 80 × ω2

1500 = 80ω2

ω2 = 1500/80

ω2 = 18.75 rad/s

symbol of science hhshsjsiwtwwisjzhJava​

Answers

Answer:

is this a company name.? java is a computer software right..

A medieval city has the shape of a square and is protected by walls with length 500 m and height 15 m. You are the commander of an attacking army and the closest you can get to the wall is 100 m. Your plan is to set fire to the city by catapulting heated rocks over the wall (with an initial speed of 80 m/s). At what range of angles should you tell your men to set the catapult? (Assume the path of the rocks is perpendicular to the wall. Round your answers to one decimal place. Use g ≈ 9.8 m/s2. Enter your answer using interval notation. Enter your answer in terms of degrees without using a degree symbol.)

Answers

Answer:

  θ₁ = 85.5º       θ₂ = 12.98º

Explanation:

Let's analyze this projectile launch problem, the catapults are 100 m from the wall 15 m high, the objective is for the walls, let's look for the angles for which the rock stops touching the wall.

Let's write the equations for motion for this point

X axis

          x = v₀ₓ t

          x = v₀ cos θ t

Y axis

         y = [tex]v_{oy}[/tex] t - ½ g t2

         y = v_{o} sin θ t - ½ g t²

let's substitute the values

         100 = 80 cos θ t

           15 = 80 sin θ t - ½ 9.8 t²

we have two equations with two unknowns, so the system can be solved

let's clear the time in the first equation

           t = 100/80 cos θ

         15 = 80 sin θ (10/8 cos θ) - 4.9 (10/8 cos θ)²

         15 = 100  tan θ - 7.656 sec² θ

we can use the trigonometric relationship

         sec² θ = 1- tan² θ

we substitute

       15 = 100 tan θ - 7,656 (1- tan² θ)

       15 = 100 tan θ - 7,656 + 7,656 tan² θ

        7,656 tan² θ + 100 tan θ -22,656=0

let's change variables

       tan θ = u

         

        u² + 13.06 u + 2,959 = 0

let's solve the quadratic equation

       u = [-13.06 ±√(13.06² - 4  2,959)] / 2

       u = [13.06 ± 12.599] / 2

        u₁ = 12.8295

        u₂ = 0.2305

now we can find the angles

         u = tan θ

         θ = tan⁻¹ u

        θ₁ = 85.5º

         θ₂ = 12.98º

Given that the velocity of blood pumping through the aorta is about 30 cm/s, what is the total current of the blood passing through the aorta (in grams of blood per second)?

Answers

Answer:

94.248 g/sec

Explanation:

For solving the total current of the blood passing first we have to solve the cross sectional area which is given below:

[tex]A_1 = \pi R^2\\\\A_1 = \pi (1)^2\\\\A_1 = 3.1416 cm^2[/tex]

And, the velocity of blood pumping is 30 cm^2

Now apply the following formula to solve the total current

[tex]Q = \rho A_1V_1\\\\Q = (1)(3.1416)(30)\\\\[/tex]

Q =  94.248 g/sec

Basically we applied the above formula So, that the total current could come

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]

A guitar string 0.65 m long has a tension of 61 N and a mass per unit length of 3.0 g/m. (i) What is the speed of waves on the string when it is plucked? (ii) What is the string's fundamental frequency of vibration when plucked? (iii) At what other frequencies will this string vibrate?

Answers

Answer:

i

  [tex]v = 142.595 \ m/s[/tex]

ii

  [tex]f = 109.69 \ Hz[/tex]

iii1 )

  [tex]f_2 =219.4 Hz[/tex]

iii2)

   [tex]f_3 =329.1 Hz[/tex]

iii3)

    [tex]f_4 =438.8 Hz[/tex]

Explanation:

From the question we are told that

    The length of the string is  [tex]l = 0.65 \ m[/tex]

     The tension on the string is  [tex]T = 61 \ N[/tex]

     The mass per unit length is  [tex]m = 3.0 \ g/m = 3.0 * \frac{1}{1000} = 3 *10^{-3 } \ kg /m[/tex]

     

The speed of wave on the string is mathematically represented as

       [tex]v = \sqrt{\frac{T}{m} }[/tex]

substituting values

      [tex]v = \sqrt{\frac{61}{3*10^{-3}} }[/tex]

     [tex]v = 142.595 \ m/s[/tex]

generally the  string's  frequency is mathematically represented as

         [tex]f = \frac{nv}{2l}[/tex]

n = 1  given that the frequency we are to find is the fundamental frequency

So

      substituting values

       [tex]f = \frac{142.595 * 1 }{2 * 0.65}[/tex]

       [tex]f = 109.69 \ Hz[/tex]

The  frequencies at which the string would vibrate include

1       [tex]f_2 = 2 * f[/tex]

Here [tex]f_2[/tex] is  know as the second harmonic and the value is  

      [tex]f_2 = 2 * 109.69[/tex]

      [tex]f_2 =219.4 Hz[/tex]

2

[tex]f_3 = 3 * f[/tex]

Here [tex]f_3[/tex] is  know as the third harmonic and the value is  

      [tex]f_3 = 3 * 109.69[/tex]

     [tex]f_3 =329.1 Hz[/tex]

3

     [tex]f_3 = 4 * f[/tex]

Here [tex]f_4[/tex] is  know as the fourth harmonic and the value is  

      [tex]f_3 = 4 * 109.69[/tex]

     [tex]f_4 =438.8 Hz[/tex]

A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magnitude 950 \text{ N/C}950 N/C. What is the change in the electric potential energy of the proton-field system when the proton travels to x

Answers

Answer:

The change in potential energy is  [tex]\Delta PE = - 3.8*10^{-16} \ J[/tex]

Explanation:

From the question we are told that

     The  magnitude of the uniform electric field  is  [tex]E = 950 \ N/C[/tex]

      The  distance traveled by the electron is  [tex]x = 2.50 \ m[/tex]

Generally the force on this electron is  mathematically represented as

     [tex]F = qE[/tex]

Where F is the force and  q is the charge on the electron which is  a constant value of  [tex]q = 1.60*10^{-19} \ C[/tex]

    Thus  

      [tex]F = 950 * 1.60 **10^{-19}[/tex]

      [tex]F = 1.52 *10^{-16} \ N[/tex]

Generally the work energy theorem can be mathematically represented as

          [tex]W = \Delta KE[/tex]

Where W is the workdone on the electron by the  Electric field and  [tex]\Delta KE[/tex]  is the change in kinetic energy

Also  workdone on the electron can also  be represented as

        [tex]W = F* x *cos( \theta )[/tex]

Where  [tex]\theta = 0 ^o[/tex] considering that the movement of the electron is along the x-axis  

        So

             [tex]\Delta KE = F * x cos (0)[/tex]

substituting values

         [tex]\Delta KE = 1.52 *10^{-16} * 2.50 cos (0)[/tex]

          [tex]\Delta KE = 3.8*10^{-16} J[/tex]

Now From the law of energy conservation

       [tex]\Delta PE = - \Delta KE[/tex]

Where [tex]\Delta PE[/tex] is the change  in  potential energy  

Thus  

        [tex]\Delta PE = - 3.8*10^{-16} \ J[/tex]

               

a point charge q is located at the center of a cube with edge length d. whatis the value of the flux over one face of the cube

Answers

Answer:

q/6Eo

Explanation:

See attached file pls

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