HELP ME NOW PLZ
What is the speed of a boy moving around a circular park of
radius 6 m if he goes once around the park in 20 s?
A-1.88m/s
B-3.76m/s
C-5.65m/s
D-None of the above

Answer:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.

b) The weight of the sled is 490.35 newtons.

c) A force of 147.105 newtons is needed to start the sled moving.

d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Explanation:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:

[tex]F[/tex] - External force exerted on the sled, measured in newtons.

[tex]f[/tex] - Friction force, measured in newtons.

[tex]N[/tex] - Normal force from the ground on the mass, measured in newtons.

[tex]W[/tex] - Weight, measured in newtons.

b) The weight of the sled is determined by the following formula:

[tex]W = m\cdot g[/tex] (1)

Where:

[tex]m[/tex] - Mass, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

If we know that [tex]m = 50\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], the weight of the sled is:

[tex]W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]

[tex]W = 490.35\,N[/tex]

The weight of the sled is 490.35 newtons.

c) The minimum force needed to start the sled moving on the horizontal ground is:

[tex]F_{min,s} = \mu_{s}\cdot W[/tex] (2)

Where:

[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.

[tex]W[/tex] - Weight of the sled, measured in newtons.

If we know that [tex]\mu_{s} = 0.3[/tex] and [tex]W = 490.35\,N[/tex], then the force needed to start the sled moving is:

[tex]F_{min,s} = 0.3\cdot (490.35\,N)[/tex]

[tex]F_{min,s} = 147.105\,N[/tex]

A force of 147.105 newtons is needed to start the sled moving.

d) The minimum force needed to keep the sled moving at constant velocity is:

[tex]F_{min,k} = \mu_{k}\cdot W[/tex] (3)

Where [tex]\mu_{k}[/tex] is the kinetic coefficient of friction, dimensionless.

If we know that [tex]\mu_{k} = 0.1[/tex] and [tex]W = 490.35\,N[/tex], then the force needed to keep the sled moving at a constant velocity is:

[tex]F_{min,k} = 0.1\cdot (490.35\,N)[/tex]

[tex]F_{min,k} = 49.035\,N[/tex]

A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

[tex]k = \frac{1}{2} m {v}^{2} \\ [/tex]

m is the mass

v is the velocity

From the question we have

[tex]k = \frac{1}{2} \times 1500 \times {8}^{2} \\ = 750 \times 64[/tex]

We have the final answer as

Hope this helps you

Answer:

5.35 rad/s

Explanation:

From the question, we are toldthat an Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted so as to rotate freely about a pivot at the 0-cm mark on the meter stick.

Solving this question, the potential energy of the particles must equal to the Kinectic energy i.e

P.E=K.E

Mgh= m½Iω²-------------eqn(*)

Where M= mass of the particles

g= acceleration due to gravity= 9.81m/s^2

ω= angular speed =?

h= height of the particles in the stick on the metre stick= ( 50cm + 80cm)= (0.5m + 0.8m)= 130cm=1.3m

If we substitute the values into eqn(*) we have

m×9.81× (1.3m)= 1/2× m×[ (0.5m)² + [(0.8m)²]× ω²

m(12.74m²/s²)= 1/2× m× (0.25+0.64)× ω

m(12.74m²/s²)= 1/2× m× 0.89× ω²

We can cancel out "m"

12.74= 1/2×0.89 × ω²

12.74×2= 0.89ω²

25.48= 0.89ω²

ω²= 28.629

ω= √28.629

ω=5.35 rad/s

Hence, the angular speed of the meter stick as it swings through its lowest position is 5.35 rad/s

Answer:

the answer is B

Explanation:

Answer:

The value is  [tex]V = 2.8284 *10^{-9 } \ Volts[/tex]

Explanation:

From the question we are told that

   The measure voltage is  [tex]E_{rms} = 1.0 \ n V = 1.0 *10^{-9} \ V[/tex]

Generally the peak voltage  is   mathematically represented as

       [tex]E_{max} = \sqrt{2} * E_{rms}[/tex]

=>    [tex]E_{max} = \sqrt{2} * 1.0 *10^{-9}[/tex]

=>    [tex]E_{max} = 1.4142 *10^{-9} \ volts[/tex]

Generally the potential difference between the two extremes is mathematically represented as

       [tex]V = 2 * E_{max}[/tex]

=>    [tex]V = 2 * 1.4142 *10^{-9}[/tex]

=>    [tex]V = 2.8284 *10^{-9 } \ Volts[/tex]

Explanation:

The attatched figure shows the vector diagram for a force that has magnitude of 60 N and it is acting at an angle of 30° from the horizontal​.

When it is resolved, the horizontal and vertical components are given by :

[tex]F_x=F\cos\theta\\\\=60\times \cos30\\\\=51.96\ N[/tex]

And

[tex]F_y=F\sin\theta\\\\=60\times \sin30\\\\=30\ N[/tex]

Hence, this is the required solution.

Answer:

the new frequency is : 302.25Hz

using the formula F2 = [tex]\frac{F1 L1}{L2}[/tex]

Explanation:

frequency of a string is the number of vibrations of a plucked string per second. it is measured in Hertz.

the frequency of a string is inversely proportional to twice the length of the string. which means the longer the string, the smaller the freqency and the higher the string the higher the frequency.

f ∝ 1/2L.

f = k/2L

where f = frequency

                         L = length of string

                         k = constant

k = 2fL is a constant

given data

L1 = 0.62m

f1 = 234Hz

L2 = 0.48m

2f1L1 = 2f2L2

f1L1 = f2L2

f2 = f1L1/L2

f2 = [tex]\frac{234 x 0.62}{0.48}[/tex] = 302.25Hz

in conclusion, the new frequency is 302.25Hz

learn more about frequency of a vibrating string: brainly.in/question/1149252

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Answer:

OD. The wave diffracted as it passed through the opening in the  barrier.

Explanation:

A progressive wave (i.e waves in motion) has the capacity to bend around an obstacle on its path. This is one of the general properties of waves called diffraction. Others are: reflection, refraction, interference. Note that only transverse waves undergo polarization.

Diffraction of waves is the ability of waves to bend around an obstacle on its path during progression.

Thus, the bending of the part of waves as it passes through the barrier implies that the wave diffracted as it passed through the opening in the  barrier.

Answer:

The wave diffrected as it passed through the opening

Explanation:

Answer:

in pulley there are different kinds.but the most common one are fixed,moveable and compound pulley. in this question we asked about movable pulley.

Explanation:

Given request solutions

L=500N a,M.A=? a,M.A=L/E =5/3

E=300N b,V.R=? b,V.R=2

c,efficiency =? c,£=M.A/V.R=5/6

£=IS NOT THE REAL SYMBOLS OF EFFICIENCY BUT I IS LOOK LIKE THIS

I THINK I HELPED

Answer:

0.068kg

Explanation:

In compressed state, a spring 3.00 meters long hanging from the ceiling has a potential energy of 2.0 joules. If the entire potential energy is used to release a ball of mass

Potential energy is expressed using the formula

PE = mgh

m is the mass

g is the acceleration due to gravity = 9.8m/s²

h is the height =  3.0m

PE = 2.0Joules

Required

Mass of the ball

Substitute the given values into the formula as shown;

2.0 = m(9.8)(3)

2 = 29.4m

m = 2/29.4

m = 0.068kg

Hence the mass of the ball is 0.068kg

Answer:

Q = 189000 [J]

Explanation:

The internal energy or heat can be calculated by means of the following expression.

[tex]Q=m*c_{p}*DT[/tex]

where:

Q = internal energy or heat [J]

m = mass = 15 [kg]

Cp = 840 [J/kg*°C]

DT = temperature change = 15 [°C]

[tex]Q = 15*840*15\\Q = 189000 [J][/tex]

Answer: the length of the pendulum should be 1.4% longer

Explanation:

Given that;

when its noon, the clock reads 11:50 am,

i.e we have 10 minutes delay ⇒ 10min × 60 = 600secs

we know that in simple pendulum

T = 2π√(l/g)  

d means delay and c means correct;

24hrs = 86400 secs

Now

Td/Tc = (86400-600) / 86400  = 0.993 = [2π√(ld/gc)] / [2π√(ld/gc)] = √(ld/lc)

so ld/lc = 0.98616  

lc = 1.014 ld

lc/l.d = -1 + 1.014 = 0.014 × 100 = 1.4% longer

therefore the length of the pendulum should be 1.4% longer

This question is incomplete, the complete question is;

An l-c cirucit with a 70 mh inductor and a 0.54 μF capacitor oscillates. The maximum charge on the capacitor is 11.5 μC. What are the oscillation frequency and the maximum current in this circuit ;

Options

a) 1.07 kHz, 63.4 mA

b) 4.38 kHz, 101.3 mA

c) 6.74 kHz, 55.7 mA

d) 2.31 kHz, 93.5 mA

e) 0.82 kHz, 59.1 mA

Answer:

the oscillation frequency and the maximum current in this circuit are; 0.82 kHz and 59.1 mA respectively.

so Option e) 0.82 kHz, 59.1 mA is the correct answer

Explanation:

Given that;

inductor L = 70 mH = 70 × 10⁻³ H

Capacitor C = 0.54 μf = 0.54 × 10⁻⁶ f

Qmax on capacitor = 11.5 μf = 11.5 × 10⁻⁶ c

oscillation frequency in L-C circuit;

f = 1/2π√(LC)

we substitute our values;

f = 1/2π√(70 × 10⁻³ × 0.54 × 10⁻⁶  )

f = 0.0818 × 10⁴ Hz

f = 0.082 × 10³ Hz ≈ 0.82 kHz

Maximum circuit in L-C circuit is given by

I_max = Qmax/√(LC)

we substitute

I_max = 11.5 × 10⁻⁶ / √(70 × 10⁻³ × 0.54 × 10⁻⁶  )

= 59.1 × 10³ A ≈  59.1 mA

Therefore the oscillation frequency and the maximum current in this circuit are; 0.82 kHz and 59.1 mA respectively.

so Option e) 0.82 kHz, 59.1 mA is the correct answer

Answer:

The value is  [tex]\beta_f = 84.95 \ dB[/tex]  

Explanation:

From the question we are told that  

    The intensity level  of the shout of a single person is [tex]\beta = 48.1 \ dB[/tex]

     The number of fans  is  [tex]n = 4841[/tex]

Gnerally intensity level is mathematically represented as

                     [tex]\beta = 10 log * \frac{I}{I_o }[/tex]

Here [tex]I_o[/tex] is the minimum intensity of sound human ear can pick and the value is  

          [tex]I_o = 1 * 10^{-12} \ W/m ^2[/tex]

when  [tex]\beta = 48.1 \ dB[/tex]

          [tex]48.1 = 10 log * \frac{I}{ 1 * 10^{-12}}[/tex]

    =>   [tex]4.81 = log ( \frac{ I}{ 1 * 10^{-12}} )[/tex]

taking antilog of  both sides

        [tex]64565.42 = \frac{I}{ 1 *10^{-12}}[/tex]

=>     [tex]I = 6.457 *10^{-8} \ W/m^2[/tex]

Generally  the intensity for the whole fans is mathematically represented as  

         [tex]I_f = n * I[/tex]

=>      [tex]I_f = 4841 * 6.457 *10^{-8 }[/tex]

=>      [tex]I_f = 0.0003126 \ W/m^2[/tex]

Gnerally the intensity level for the whole fans is mathematically represented as    

               [tex]\beta_f = 10 log [ \frac{I_f }{I_o } ][/tex]

=>             [tex]\beta_f = 10 log [ \frac{ 0.0003126 }{1*10^{-12}}[/tex]

=>             [tex]\beta_f = 84.95 \ dB[/tex]  

The question is incomplete. The complete question is :

First Law of Thermodynamics - Sign Convention The first law of thermodynamics applies the conservation of energy principle to systems where heat transfer and doing work are the methods of transferring energy into and out of the system. The first law of thermodynamics states that the change in internal energy of a system equals the net heat transfer into the system minus the net work done by the system. In equation form, the first law of thermodynamics is AU = Q-W. Here AU is the change in internal energy U of the system. Q is the net heat transferred into the system that is, Q is the sum of all heat transfer into (positive) and out of (negative) the system. W is the net work done by the system—that is, W is the sum of all work done by (positive) and on (negative) the system. We use the following sign conventions: if Q is positive, then there is a net heat transfer into the system; if W is positive, then there is net work done by the system. So positive Q adds energy to the system and positive W takes energy from the system. Thus AU = Q-W. Note also that if more heat transfer into the system occurs than work done, the difference is stored as internal energy. The first law of thermodynamics AU = 9 - W Ur-U, Heat Work System AU--W system w Qin: talu Qout:

The first law of thermodynamics AU = Q - W U-U Heat Work System AUQ-W Qin: ta Qout: - Wout: + WK w Win: - - volume expands t volume decreases o All answers can be positive or negative. (a) Suppose there is heat transfer of 42 ) into a system, while the system does 6 ) of work. Later, there is heat transfer of 22 J out of the system while 6 ) of work is done on the system. What is the net heat transfer? 20 Correct (100.0%) Submit What is the total work? Enter a number Submit (5 attempts remaining) What is the net change in internal energy of the system? Enter a number

What is the net change in internal energy of the system? Enter a number Submit (5 attempts remaining) (b) What is the change in internal energy of a system when a total of 140 J of heat transfer occurs out of (from) the system and 165 ) of work is done on the system? Enter a number Submit (5 attempts remaining) (c) An athlete doing push-ups performs 645 kJ of work and loses 440 kJ of heat. What is the change in the internal energy (in kJ) of the athlete? Enter a number Submit (5 attempts remaining) kJ (d) An athlete doing push-ups performs 690 kJ of work and loses 450 kJ of heat. Then he takes in 830 kJ of energy from eating food, What is the total change in the internal energy (in kJ) of the athlete? Enter a number kJ.

Solution :

a). Given :

[tex]$Q_1 = 42 \ J$[/tex] , [tex]$Q_2 = -22 \ J , \ W_1 = 6 \ J, \ W_2 = -6 \ J $[/tex]

Net heat transfer

[tex]$Q= Q_1+Q_2$[/tex]

   = 42 + (-22)

   = 20 J

Total work

[tex]$W= W_1+W_2$[/tex]

   = 6 + (-6)

   = 0 J    

∵ ΔU = Q - W

       = 20 - 0

        = 20 J

This is the net change in the internal energy of the system.

b). ΔU = Q + W

           = (-140) + (-165)

           = -305 J

c). ΔU = Q + W

           = (-440) + (645)

           = 205 J

d). ΔU = Q + W

           = (-450) + (690)

           = 240 J

Answer:

[tex]v_=-8.5\times 10^{-4}\ m/s[/tex]

Explanation:

Given that,

Mass of a wrench, m₁ = 4 g = 0.004 kg

Speed of wrench, v₁ = -15 m/s

Mass of the Astronaut, m₂ = 70 kg

We need to find Astronaut's velocity. Let it is v₂. Using the conservation of linear momentum to find it.

[tex]m_1v_1=m_2v_2\\\\v_2=\dfrac{m_1v_1}{m_2}\\\\v_2=\dfrac{0.004\times (-15)}{70}\\\\=-8.5\times 10^{-4}\ m/s[/tex]

So, the speed of Astronaut is [tex]8.5\times 10^{-4}\ m/s[/tex].

Answer:

i am!

Explanation:

Answer:

0.00129rad/s

Explanation:

The angular velocity is expressed as;

v = wr

w is the angular velocity

r is the radius

Given

v =  20,000 mph

r = 4300mi

Get w;

w = v/r

w = 20000* 0.44704/4300*1609.34

w = 8940.8/6,920,162

w = 0.00129rad/s

Hence the angular velocity generated is 0.00129rad/s

Answer:

Test bullets fired from the suspected gun

Explanation:

This is the correct answer!! I took the test!!

Answer:

8 seconds

Explanation:

as per 1st equation of motion

v=u+at,

u= initial velocity, v= final velocity , t=time, a=acceleration

since it starts from rest , u=0

v=16m/s

a=2m/s^2

16=0+2t

16=2t

t=8

The matching of each method of transferring electric charge with the correct description should be explained below.

The friction means the transfer of electric charge via rubbing. The conduction means the transfer of electric charge via the direct contact.

Also, the induction means the transfer of the electric charge without the direct contact.

In this way it should be matched.

Learn more about electric here: https://brainly.com/question/23056096

Answer:

this is the answer that is correct

Answer:

The correct answer is "24 V".

Explanation:

The given values are:

Current,

I = 0.50 A

Resistance,

R = 12 W

As we know,

⇒ [tex]I = 0.5\times (\frac{E}{2R})[/tex]

On substituting the given values, we get

⇒ [tex]0.5= (\frac{E}{4\times 12} )[/tex]

⇒ [tex]0.5= (\frac{E}{48} )[/tex]

⇒   [tex]E=24 \ V[/tex]  

Answer:

it comprises of the DNA/RNA bipolymer molecules

Answer:

a. 6602.7 N b. 64.44 km/h

Explanation:

a. Find the force of friction required to keep it from  sliding down the hill.

The frictional force f equals the component of the weight of the car, W perpendicular to the inclined hill = Wcosθ times the coefficient of static friction, μ = 0.45.

Since f = μN = μWcosθ = μmgcosθ where m = mass of car = 1500 kg, g = acceleration due to gravity = 9.8 m/s² and θ = angle of incline of hill = 3.5°

So, f = μmgcosθ

= 0.45 × 1500 kg × 9.8 m/s²cos3.5°

= 6615cos3.5°

= 6602.7 N

b. Determine the speed of the car after it emerges from the mud (in km/h)

Since the car drops a vertical height of 14 m, its potential energy decreases by mgh and its kinetic energy increases by mgh where m =mass of car and h = height drop = 14 m. So its kinetic energy increase is ΔK = mgh = 1500 kg × 9.8 m/s² × 14 m = 205800 J

Since it has an initial velocity of u = 50 km/h = 50 km/h 1000m/3600 s = 13.89 m/s, its initial kinetic energy is K = 1/2mu² = 1/2 × 1500 kg × (13.89 m/s)² = 144699.08 J.

Its new kinetic energy after the drop is thus K' = K + ΔK = 144699.08 J + 205800 J = 350499.08 J

Let v be its velocity after the drop, since K' = 1/2mv²,

v = √(2K'/m) = √(2 × 350499.08 J/1500 kg) = √(700998.16 J/1500 kg) = √(467.332 J/kg) = 21.62 m/s

Now, from work kinetic energy principles, the kinetic energy change in the car is the work done on car by friction

So, ΔK" = -fd = -μmgd

Let v' be the velocity of the car after emerging from the mud and moving a distance d = 30.0 m.

So, 1/2m(v'² - v²) = -μmgd

v'² - v² = -2μgd

v'² = v² - 2μgd

Substituting the values of the variables, we have

v'² = (21.62 m/s)² - 2 × 0.25 × 9.8 m/s² × 30.0 m

v'² = 467.42 m²/s² - 147 m²/s²

v'² = 320.42 m²/s²

taking square root of both sides, we have

v' = √(320.42 m²/s²)

= 17.9 m/s

Converting v to km/h we have v' = 17.9 m/s × 3600 s/h × 1 km/1000 m = 64.44 km/h.

So, the car emerges from the mud with a speed of 64.44 km/h

Answer:

v = 35 km/s

Explanation:

Given that,

Orbital radius of Venus, [tex]r=10.8\times 10^{10}\ m=10.8\times 10^{10}\times 10^{-3}=10.8\times 10^{7}\ km[/tex]

The rotation period is, [tex]T = 224.7\ \text{days}=19414080\ s[/tex]

We need to find the orbital speed of the Venus. The formula for the orbital speed is given by :

[tex]v=\dfrac{2\pi r}{T}\\\\v=\dfrac{2\pi \times 10.8\times 10^{7}}{19414080}\\\\=34.95\ km/s[/tex]

or

v = 35 km/s

So, the orbital speed of Venus is 35 km/s.

Answer:

See the answers and the explanation below.

Explanation:

To solve these questions we must understand that speed is the relationship between the displacement and the time that the displacement lasts.

1. 300 [km], in 5 [hr]

d = displacement [km]

[tex]v = d/t[/tex]

where:

v = velocity [km/h]

t = time [hr]

[tex]v = 300/5\\v = 60 [km/h][/tex]

In one hour : [tex]d = 60*1\\d = 60 [km][/tex]

b. Car B 100 [km] in 2 [hr]

[tex]v =100/2\\v = 50 [km/h]\\[/tex]

C. The car A has the greatest average speed.

Answer:

Explanation:

The impulse of an object can be found by using the formula

impulse = force × time

From the question we have

impulse= 23 × 3.1

We have the final answer as

Hope this helps you

Answer:

75kg

Explanation:

[tex]F=mg[/tex]

[tex]m=\frac{F}{g}[/tex]

[tex]m=\frac{735}{9.8}[/tex]

[tex]m=75kg[/tex]

Therefore, the answer is the third option 75 kg

s = 40m - 4m = 36m

W = F × s

= 10N × 36m = 360J

W = Work (J)

F = Force / weight (N)

s = distance (m)

Work done in physics is the product of force and displacement. The displacement for the object is 36 m and force acts on it is 10 N. Then the work done is 360 J.

Work done is the dot product of force acting on a body and the resultant displacement. When a force applied on an object results in a displacement from its position, the force is said to be work done.

Work done is a vector quantity thus, characterised by a magnitude and direction.  The common unit of work done is joule.

Given that force applied on the weight = 10 N

displacement occurred = 40 m - 4 m = 36 m

Work done = F . ds

ds = 36 m and f = 10 N

Then W = 10 N × 36m

             = 360 J.

Therefore, the work is required to lift a 10-newton weight from 4.0 meters to 40 meters above the surface of Earth is 360 J.

To find more on work done, refer here:

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Answer:

The strength of the source charge's electric field could be measured by any other charge placed somewhere in its surroundings. The charge that is used to measure the electric field strength is referred to as a test charge since it is used to test the field strength. The test charge has a quantity of charge denoted by the symbol q.

Explanation:

Electric field strength is a vector quantity; it has both magnitude and direction. The magnitude of the electric field strength is defined in terms of how it is measured. Let's suppose that an electric charge can be denoted by the symbol Q. This electric charge creates an electric field; since Q is the source of the electric field, we will refer to it as the source charge. The strength of the source charge's electric field could be measured by any other charge placed somewhere in its surroundings. The charge that is used to measure the electric field strength is referred to as a test charge since it is used to test the field strength. The test charge has a quantity of charge denoted by the symbol q. When placed within the electric field, the test charge will experience an electric force - either attractive or repulsive. As is usually the case, this force will be denoted by the symbol F. The magnitude of the electric field is simply defined as the force per charge on the test charge.