if the instantaneous current in the circuit is giveen by I=3 sin theta amperes, the rms value of the current will be

Answers

Answer 1

Answer:

[tex]I_{rms}=2.12\ A[/tex]

Explanation:

Given that,

The instantaneous current in the circuit is giveen by :

[tex]I=3\sin\theta\ A[/tex]

We need to find the rms value of the current.

The general equation of current is given by :

[tex]I=I_o\sin\theta[/tex]

It means, [tex]I_o=3\ A[/tex]

We know that,

[tex]I_{rms}=\dfrac{I_o}{\sqrt2}\\\\=\dfrac{3}{\sqrt2}\\\\=2.12\ A[/tex]

So, the rms value of current is 2.12 A.


Related Questions

An object that is initially traveling at 17.6 m/s has
1,743 J of work performed on it. If the mass of the
object is 283 kg, its final kinetic energy will be
J.

Answers

CnnsmamfmdmmaldlMnvnnakwkkfkfosksmnfnfnek

In which reaction are the atoms of elements rearranged?
A. A neutron fuses with hydrogen-1 to form hydrogen-2.
B. Uranium-235 breaks apart into cesium, rubidium, and two
neutrons.
C. A neutron fuses with uranium-238 to form uranium-239.
D. Hydrogen combines with oxygen to form water.

Answers

Answer:

D. Hydrogen combines with oxygen to form water.

Explanation:

Hydrogen combining with oxygen to form water is a typical example of chemical reaction.

During a chemical reaction, atoms of elements are rearranged. Most chemical reactions obey the law of conservation of mass which states that "matter is neither created nor destroyed in a chemical reaction but atoms are simply rearranged".

The other choices given are nuclear reactions. In such reactions, atoms are not rearranged but are simply destroyed and made in the process.

Answer:

D.) Hydrogen combines

Explanation:

BE HAPPY

g You want to make simultaneous measurements of the position and momentum of an electron and a proton that are moving along a straight line. An alternate statement of the uncertainty principle involves relationship between position (Δx) and momentum (Δp) uncertainties in the form . If both of them are located with an uncertainty of 1 × 10-10 m, what is the ratio of uncertainty in the velocity of the electron to that of the proton?

Answers

Answer:

1832

Explanation:

From;

Δp Δx = h/4π

Δp = uncertainty in momentum

Δx = uncertainty in position

h= Plank's constant

But p =mv hence, Δp= Δmv

m= mass, v= velocity

mass of electron = 9.11 * 10^-31 Kg

Mass of proton = 1.67 * 10^-27 Kg

since m is a constant,

Δv = h/Δxm4π

For proton;

Δv = 6.6 * 10^-34/4 * 3.14 * 1.67 * 10^-27 * 1 * 10^-10

Δv = 315 ms-1

For electron;

Δv = 6.6 * 10^-34/4 * 3.14 * 9.11 * 10^-31 * 1 * 10^-10

Δv = 577000 ms-1

Ratio of uncertainty of electron to that of proton = 577000 ms-1/315 ms-1= 1832

Which type of bond has elements with the greatest difference in electronegativity?

Covalent
Ionic
Metallic
Polar

Plz Help

Answers

Answer:

inonic bond has greatest difference in electronegativiru

The type of bond that has elements with the greatest difference in electronegativity ionic bond. The correct option is is B.

What is electronegativity?

Electronegativity is a chemical property that describes an atom's or functional group's tendency to attract electrons toward itself.

An atom's electronegativity is affected by both its atomic number and the distance between its valence electrons and the charged nuclei.

Electronegativity is a measurement of an atom's proclivity to attract electrons (or electron density) towards itself.

It governs the distribution of shared electrons between two atoms in a bond. The greater an atom's electronegativity, the more strongly it attracts electrons in its bonds.

Ionic bonds are formed by elements with large differences in electronegativity. Covalent bonds are formed between atoms of elements with similar electronegativity.

Thus, the correct option is B.

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A car slows down uniformly from a speed of 21.0m/s to rest in 6.00s. How far did it travel in that time?

Answers

(sorry something wrong w my keyboard so write each line for the explnation!)

63.0 m

Explanation:

Acceleration of car

=

v

u

t

=

0 ms

1

21.0 ms

1

6.00 s

=

3.50 ms

2

S

=

v

2

u

2

2a

S

=

(

0 ms

1

)

2

(

21.0 ms

1

)

2

2

×

3.50 ms

2

S

=

63.0

m

How long is a simple pendulum with a period of 1 second?​

Answers

Answer:

the second pendulum ( also called the royal pendulum) o.994m (39.1 in) long, in which each swing takes one second, became widely used in quality clocks.

A large pendulum with a 200-lb gold-plated bob 12 inches in diameter is on display in the lobby of the United Nations building. The pendulum has a length of 75 ft. It is used to show the rotation of the Earth-for this reason it is referred to as a Foucault pendulum. What is the least amount of time it takes for the bob to swing from a position of maximum displacement to the equilibrium position of the pendulum?

Answers

Answer:

2.4s

Explanation:

The length of the pendulum = 75ft

Diameter d = 12 inches

The time period of the pendulum is given as

T = 2pi(L/g)^1/2

Then the time it takes to move from displacement to equilibrium is given as:

t = T/4

= (Pi/2)*(L/g)^1/2

= pi/2 x [(75x0.3048)/9.81]^0.5

= 1.57x[22.86/9.81)^0.5

= 2.4s

2.4 seconds is the least amount of time that it would take.

The inductor in a radio receiver carries a current of amplitude 200 mA when a voltage of amplitude 2.4 V is across it at a frequency of 1400 Hz. What is the value of the inductance

Answers

Answer:

The value of the inductance is 1.364 mH.

Explanation:

Given;

amplitude current, I₀ = 200 mA = 0.2 A

amplitude voltage, V₀ = 2.4 V

frequency of the wave, f = 1400 Hz

The inductive reactance is calculated;

[tex]X_l = \frac{V_o}{I_o} \\\\X_l = \frac{2.4}{0.2} \\\\X_l =12 \ ohms[/tex]

The inductive reactance is calculated as;

[tex]X_l = \omega L\\\\X_l = 2\pi fL\\\\L = \frac{X_l}{2 \pi f}[/tex]

where;

L is the inductance

[tex]L = \frac{12}{2 \pi \times \ 1400} \\\\L = 1.364 \times \ 10^{-3} \ H\\\\L = 1.364 \ mH[/tex]

Therefore, the value of the inductance is 1.364 mH.

The value of inductance will be "1.36 mH".

Given values are:

Current, [tex]I_0 = 200 \ mA, or \ 0.2 \ A[/tex]Voltage, [tex]V_0 = 2.4 \ V[/tex]Frequency, [tex]f = 1400 \ Hz[/tex]

As we know the formula,

→ [tex]X_l = \frac{V_0}{I_0}[/tex]

By substituting the values, we get

       [tex]= \frac{2.4}{0.2}[/tex]

       [tex]= 12 \ ohms[/tex]

Now,

The inductive reactance will be:

→ [tex]X_l = \omega L[/tex]

or,

→ [tex]X_l = 2 \pi f L[/tex]

Hence,

The Inductance will be:

→ [tex]L = \frac{X_i}{2 \pi f}[/tex]

By putting the values, we get

      [tex]= \frac{12}{2 \pi\times 1400}[/tex]

      [tex]= 1.364\times 10^{-3} \ H[/tex]

      [tex]= 1.36 \ mH[/tex]

Thus the above answer is right.

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How much is a ball that is tethered to a post accelerating if its linear speed is 3.5 m/s and the ball is 0.91 meters away from the pole? a = v^2 / r

Answers

Answer:

[tex]a_{c} = 13.46\ m/s^2[/tex]

Explanation:

The motion of the tethered ball around the pole can be modeled as uniform circular motion. The acceleration of the objects executing uniform circular motion is due to the change in direction of their velocity. This is called centripetal acceleration. It is given by the following formula:

[tex]a_{c} = \frac{v^2}{r}[/tex]

where,

ac = centripetal acceleration = ?

v = speed of ball = 3.5 m/s

r = distance of ball from center of rotation = 0.91 m

Using these values in equation, we get:

[tex]a_{c} = \frac{(3.5\ m/s)^2}{0.91\ m}\\\\a_{c} = 13.46\ m/s^2[/tex]

To understand the meaning of the variables that appear in the equations for rotational kinematics with constant angular acceleration. Rotational motion with a constant nonzero acceleration is not uncommon in the world around us. For instance, many machines have spinning parts. When the machine is turned on or off, the spinning parts tend to change the rate of their rotation with virtually constant angular acceleration. Many introductory problems in rotational kinematics involve motion of a particle with constant, nonzero angular acceleration. The kinematic equations for such motion can be written as

Answers

Answer:

   ω = ω₀ + α t

      ω² = ω₀² + 2 α  θ

      θ = θ₀ + ω₀ t + ½ α t²

Explanation:

Rotational kinematics can be treated as equivalent to linear kinematics, for this change the displacement will change to the angular displacement, the velocity to the angular velocity and the acceleration to the angular relation, that is

     x → θ

     v → ω

     a → α

with these changes the three linear kinematics relations change to

      ω = ω₀ + α t

      ω² = ω₀² + 2 α  θ

      θ = θ₀ + ω₀ t + ½ α t²

where it should be clarified that to use these equations the angles must be measured in radians

The aorta carries blood away from the heart at a speed of about 42 cm/s and has a diameter of approximately 1.1 cm. The aorta branches eventually into a large number of tiny capillaries that distribute the blood to the various body organs. In a capillary, the blood speed is approximately 0.08 cm/s, and the diameter is about 0.0005 cm. Treat the blood as an incompressible fluid, and determine N the approximate number of capillaries in the human body.

Answers

Answer:

ok fine I'll answer u on comment

An Airplane moves 70 m/s as it travels around a vertical circular loop which has a 3 km radius. What is the magnitude of the net force on the 95 kg pilot at the bottom of this loop?

Answers

Answer:

155.17N.

Explanation:

The magnitude of the net force is expressed as;

F = mv²/r where;

m is the mass

v is the velocity of the airplane

r is the radius of the loop

Given

m = 95kg

v = 70m/s

r = 3km = 3000m

Required

Magnitude of the net force

F = 95*70²/3000

F = 95*4900/3000

F = 95*49/30

F = 4655/30

F = 155.17N

Hence the magnitude of the net force on the 95 kg pilot at the bottom of this loop is 155.17N.

The net force on the pilot is 1551.67 N.

What is force?

Force can be defined as the product of mass and acceleration.

To calculate the magnitude of the net force at the bottom of the loop, we use the formula below.

Formula:

F = mv²/r............. Equation 1

Where:

m = mass of the pilotv = velocity of the Airplaner = radius of the loopF = net force on the pilot.

From the question,

Given:

m = 95 kgr = 3 km = 3000 mv = 70 m/s

Substitute these values into equation 1

F = 95(70²)/300F = 465500/300F = 1551.67 N

Hence, the net force on the pilot is 1551.67 N.

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A 1500 kg car sits on a 3.5° inclined hill. Find the force of friction required to keep it from
sliding down the hill. The coefficient of static friction is μ=0.45

Answers

Answer:

The force of friction required to keep the car from sliding down the hill is 898.054 newtons and since [tex]f<f_{max}[/tex], the car will not slide down the hill.

Explanation:

The free body diagram is included below and now we prepare each equation of equilibrium for respective orthogonal axis:

[tex]\Sigma F_{x'} = f-m\cdot g \cdot \sin \theta = 0[/tex] (1)

[tex]\Sigma F_{y'} = N-m\cdot g \cdot \cos \theta = 0[/tex] (2)

Where:

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

[tex]N[/tex] - Normal force from the inclined hill to the car, measured in newtons.

[tex]m[/tex] - Mass of the car, measured in newtons.

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

[tex]\theta[/tex] - Angle of inclination of the hill, measured in sexagesimal degrees.

If we know that [tex]m = 1500\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\theta = 3.5^{\circ}[/tex], then the friction force required to keep the car from sliding down the hill is:

[tex]f = m\cdot g \cdot \sin \theta[/tex]

[tex]f = (1500\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \sin 3.5^{\circ}[/tex]

[tex]f = 898.054\,N[/tex]

The force of friction required to keep the car from sliding down the hill is 898.054 newtons.

If the car does not slide down the hill, then the following condition must be observed:

[tex]f\le \mu_{s}\cdot m\cdot g \cdot \cos \theta[/tex] (2)

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

If we know that [tex]\mu_{s} = 0.45[/tex], [tex]m = 1500\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\theta = 3.5^{\circ}[/tex], then the maximum allowable static friction force is:

[tex]f_{max} = \mu_{s}\cdot m \cdot g \cdot \cos \theta[/tex]

[tex]f_{max} = (0.45)\cdot (1500\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \cos 3.5^{\circ}[/tex]

[tex]f_{max} = 6607.378\,N[/tex]

Since [tex]f<f_{max}[/tex], the car will not slide down the hill.

A tissue is a group of similar cells that perform a specific function in an organism true or false

Answers

Answer:

true

Explanation:

Answer:

This is true

Given a force of 100N and an acceleration of 5 m/s2 , what is the mass?

Answers

Answer:

20kg

Explanation:

Given parameters:

Force  = 100N

Acceleration  = 5m/s²

Unknown:

Mass  = ?

Solution:

According to Newton's second law of motion:

 Force  = mass x acceleration;

 So;

          Mass  = [tex]\frac{force }{acceleration}[/tex]  

          Mass  = [tex]\frac{100}{5}[/tex]   = 20kg

If a paratroopers legs have enough strength to support five times her weight, and they can do so over a travel distance of 0.5 m as they flex, from what height can she free fall to a solid surface without injury

Answers

Answer: the required height is 2.5 m

Explanation:

given the given data in question

Lime to come in rest;

using S = ((U1 + U2)/2) t

⇒ 0.5 = ((√(2gh) + 0)/2) t

t = (0.5 × 2) / √(2gh)

now fat = impulse = change in momentum dp

F = 5mg = dp/dt = [(m√(2gh))/(0.5 × 2)] × √(2gh)

5mg = 2mgh/(0.5 × 2)

5 = h / 0.5

h = 5 × 0.5

h = 2.5 m

Therefore the required height is 2.5 m

Question 1 of 25
Two asteroids with masses 3.71 x 10 kg and 1.88 x 104 kg are separated by
a distance of 1,300 m. What is the gravitational force between the asteroids?
Newton's law of gravitation is F gravity
Gm, 2 The gravitational
constant Gis 6.67 x 10-11 Nm²/kg?
A. 275 x 10"N
B. 4.13 x 10°N
C. 2.04 x 10°N
O D. 3.58 x 10-N
SUBMIT

Answers

Answer:

2.753*10^-11N

Explanation:

According to Newton's law of gravitation, the force between the masses is expressed as;

F = GMm/d²

M and m are the distances

d is the distance between the masses

Given

M = 3.71 x 10 kg

m = 1.88 x 10^4 kg

d = 1300m

G = 6.67 x 10-11 Nm²/kg

Substitute into the formula

F = 6.67 x 10-11* (3.71 x 10)*(1.88 x 10^4)/1300²

F = 46.52*10^(-6)/1.69 * 10^6

F = 27.53 * 10^{-6-6}

F = 27.53*10^{-12}

F = 2.753*10^-11

Hence the gravitational force between the asteroid is 2.753*10^-11N

How are clouds formed? What are the three major
Types of clouds and what do they look like?
O

Answers

Answer:

clouds are created when water vapor and invisible gas turns into water droplets and these water droplets form tiny particles like dust and float in the air.

you have cumulus

cirrus

cirrostraus  

Explanation:

A car accidently roll of a cliff. As it leaves the cliff it has horzontal velocity of 13 m/s it hits the ground 60m from the shoreline. Calculate the hight of the cliff

Answers

Answer: The height of the cliff is 104.59 m

Explanation:

The horizontal speed of the car when it leaves the cliff is 13 m/s, and it hits the ground 60m from the shoreline.

Here we can use the relationship:

Time*Speed = Distance.

To find the time that the car is in the air, we know that:

speed = 13m/s

distance = 60m

time = T

13m/s*T = 60m

T = (60m)/13m/s = 4.62 s

This means that the car is falling for 4.62 seconds.

Now let's analyze the vertical problem.

As the car leaves the cliff, it only has horizontal velocity, this means that the vertical initial velocity will be zero

The only force acting in the vertical axis is the gravitational force, this means that the acceleration will be equal to the gravitational acceleration, which is:

g = 9.8m/s^2

then:

a = -9.8m/s^2

Where the negative sign is because the acceleration is pulling the car downwards.

To get the vertical velocity, we could integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the constant of integration and the initial vertical velocity, that we already know that is equal to zero, then the vertical velocity as a function of time can be written as:

v(t) = (-9.8m/s^2)*t

To get the vertical position equation, we need to integrate again over the time:

P(t) = (1/2)*(-9.8m/s^2)*t^2 + H

Where H is the constant of integration and the initial vertical position, then H will be the height of the cliff.

We know that the car needs 4.62 seconds to hit the ground, this means that:

P(4.6s) = 0m

Then:

P(t) = (1/2)*(-9.8m/s^2)*(4.62s)^2 + H = 0

            (-4.9m/s^2)*(4.62s)^2 + H = 0

          H =  (4.9m/s^2)*(4.62s)^2 = 104.59 m

This means that the cliff is 104.59 meters high

X- rays with a wavelgnth of 0.0700 nm diffract from a crystal. Two adjacne angels of x-ray diffraciton are 45.6 and 21.0 degrees. What is the distance in nm between the atomic planes responsible for the diffraction? (Use 2dcos(θ)=m(λ))

Answers

Answer:

0.15 nm

Explanation:

d = Distance between the atomic planes

m = Order

[tex]\theta_{m}[/tex] = First angle = [tex]45.6^{\circ}[/tex]

[tex]\theta_{m+1}[/tex] = Adjacent angle = [tex]21^{\circ}[/tex]

[tex]\lambda[/tex] = Wavelength = 0.07 nm

From Bragg's relation we know

[tex]2d\cos\theta_{m}=m\lambda[/tex]

[tex]2d\cos45.6^{\circ}=m0.07[/tex]

[tex]2d\cos\theta_{m+1}=(m+1)\lambda[/tex]

[tex]2d\cos21^{\circ}=(m+1)0.07\\\Rightarrow 2d\cos21^{\circ}=m(0.07)+0.07[/tex]

So

[tex]2d\cos21^{\circ}=2d\cos45.6^{\circ}+0.07\\\Rightarrow 2d(\cos21^{\circ}-\cos45.6^{\circ})=0.07\\\Rightarrow d=\dfrac{0.07}{2(\cos21^{\circ}-\cos45.6^{\circ})}\\\Rightarrow d=0.14962\ \text{nm}[/tex]

The distance between the atomic planes is 0.15 nm.

Do dwarf planets has several other bodies in their path orbiting the Sun just as they do?​

Answers

Answer:

No they just orbit the sun

Explanation:

A lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffraction standpoint is A) The same for all wavelengths B) In the near-ultraviolet C) In the visible D) In the near-infrared E) Indeterminate

Answers

Answer:

B) In the near-ultraviolet

Explanation:

The best option to this problem is : In the near-ultraviolet

What do your observations show you about the forces exerted on objects in a collision?
Scroll for more

Answers

I need more information

A(n) 70.1 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 41.4 m away from the shuttle and moving with zero speed relative to the shuttle. She has a(n) 0.916 kg camera in her hand and decides to get back to the shuttle by throwing the camera at a speed of 12 m/s in the direction away from the shuttle. How long will it take for her to reach the shuttle

Answers

Answer:

4.4 min

Explanation:

To solve this, we use the law of conservation of momentum

0 = -m(a).v(a) + m(c).v(c), where

m(a) = mass of the astronaut

m(c) = mass of the camera

v(a) = velocity of the astronaut

v(c) = velocity of the camera

Making v(a) subject of the formula, we have

v(a) = m(c).v(c)/m(a)

Now, we make one last assumption, that the astronaut is moving at constant speed and time, thus the time needed will be

t = s/v -> since s = vt

Now, we already have our v from above, so we substitute

t = s/[m(c).v(c)/m(a)]

t = s.m(a)/m(c).v(c)

Applying the values, we have

t = (41.4 * 70.1) / (0.916 * 12)

t = 2902.14 / 10.992

t = 264 seconds or 4.4 minutes

The pendulum consists of two slender rods AB and OC which each have a mass of 3 kg/m. The thin plate has a mass of 10 kg/m2. Determine the location of the center of mass G of the pendulum, then calculate the moment of inertia of the pendulum about an axis perpendicular to the page and passing through G.

Answers

Answer:

the answer is below

Explanation:

The diagram of the problem is given in the image attached.

Mass of rod AB = (0.4 m + 0.4 m) * 3 kg/m = 2.4 kg

Mass of rod OC = (1.5 m) * 3 kg/m = 4.5 kg

Mass of plate = 10 kg/m²[ (π* 0.3²) - (π* 0.1²)] = 2.513 kg

The center of mass is:

[tex]\hat{y}[2.4+2.513+4.5]=(0.75*4.5)+(2.513*0.5)\\\\\hat{y}=0.839\\\\I_{AB}=\frac{1}{12}*2.4*(0.4+0.4)^2= 0.128\ kg/m^2\\\\I_{OC}=\frac{1}{12}*4.5*(1.5)^2= 3.375\ kg/m^2\\\\I_{Gplate}=\frac{1}{2}*(\pi*0.3^2*10)*(0.3)^2-\frac{1}{2}*(\pi*0.1^2*10)*(0.1)^2= 0.126\ kg/m^2\\\\I_{plate}=I_{Gplate}+md^2=0.126+2.513(1.8^2)=8.27\ kg/m^2\\\\I_o=I_{plate}+I_{AB}+I_{OC}\\\\I_{o}=0.128+3.375+8.27=11.773\ kg/m^2 \\\\I_G=I_o-m_{tot}\hat{y}^2\\\\I_G=11.773-(9.413*0.839^2)\\\\I_G=5.147\ kg/m^2[/tex]

Which of the following best illustrates Lewin's interactionist perspective?

Answers

Answer:

The options are

a. Sally is a very creative kind of person who likes to build things.

b. Jerry only works because he receives a very large income.

c. Rikki is usually shy, but at work she appears to be quite outgoing.

d. Maury gives money to charities because he wants other people to think he is very generous.

The answer is c. Rikki is usually shy, but at work she appears to be quite outgoing.

Lewin's interactionist perspective explains that an individual’s behavior is usually dependent on his personal behavior/ trait and the environment. The best option is that Rikki is usually shy which is her personal behavior but at work she appears to be quite outgoing due to her environment.

The energy released in the fission of one 235U92 is 206.6 MeV. A Nuclear reactor that uses this element has an output of 28.7 Mega Watts. What is the mass of 235U92 that is consumed in one day in this reactor

Answers

Answer: the required mass is 1.7628 × 10²⁵ μ

Explanation:

Given that;

energy released in the fission of one ²³⁵U₉₂ is 206.6 MeV

power p = 28.7 Mega Watts = 28.7 × 10⁶ W = 28.7 × 10⁶/ 1.6× 10⁻¹³ = 17.9375 × 10¹⁹ MeV/s

now fission need per second will be;

⇒ power / energy released i  fission

= 17.9375 × 10¹⁹ MeV /  206.6 MeV = 8.68 × 10¹⁷ per second

now fission need per day will be;

⇒ ( 8.68 × 10¹⁷ × 360 × 24 ) = 7.5 × 10²² per day

hence mass of ²³⁵U₉₂ that is consumed in one day in this reactor will be;

⇒ (235 × 7.5 × 10²²)μ

= 1.7628 × 10²⁵ μ

Therefore the required mass is 1.7628 × 10²⁵ μ

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

Answers

Answer: 1.88 m/s

Explanation:

The speed of the boy moving around a circular park of radius 6 m and taking time 20 seconds is 1.88 m/s.

What is speed?

Speed is a measurement of how quickly an object's distance traveled changes. Speed is a scalar, which implies it has magnitude but no direction as a unit of measurement.

A thing that moves quickly and with great speed, covering a lot of ground in a short time. On the other hand, a slow-moving object traveling at a low speed covers a comparatively small distance in the same amount of time.

Given information:

The radius of the circular park, r = 6 m,

The time is taken to cover one round, t = 20s,

The total distance covered by the boy is equal to the circumference of the park, so

d = 2[tex]\pi[/tex]r,

[tex]d = 2*3.14*6\\d = 37.68 m[/tex]

Now,

the speed of the boy, s = [tex]37.68/20[/tex] ,

s = 1.88 m/s

Therefore, the speed of the boy moving around a circular park of radius 6 m and taking time 20 seconds is 1.88 m/s.

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Two objects of same material are travelling near you. Object A is a 1.1 kg mass traveling 10.2 m/s; object B is a 2 kg mass traveling 5 m/s. Which object would make you feel worse if you are hit by it

Answers

Answer:

Object A

Explanation:

The object that would make you feel worse if you're hit by it is the object possessing the highest momentum. Thus, we need to find the momentum of the two objects.

Momentum of an object is the product of its mass and that of it's velocity. Momentum is given by the formula

P = M * V, where

P = momentum

M = mass of the object

V = velocity of the object

Now, solving for object A, we have

P(a) = 1.1 * 10.2

P(a) = 11.22 kgm/s

And then, solving for object B, we have

P(b) = 2 * 5

P(b) = 10 kgm/s

The object when the highest momentum is object A, and thus would make you feel worse when hit by it

What are various names of tropical storms
based upon the ocean they occur in?

Answers

Answer:

hurricanes,Typhoons and cyclones

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