You get out of school and walk north for 20 minutes to travel 1.6 km what is your average velocity?

Answers

Answer 1

Answer: 12.5 km/h

Explanation: 20/1.6 = 12.5 km/h

Answer 2

The average velocity of an individual will be "0.08 km/min". To understand the calculation, check below.

Distance and Velocity

According to the question,

Time, t = 20 minutes

Distance, d = 1.6 km

We know the formula,

Distance = Speed × Time,

Or,

Speed, v = [tex]\frac{Distance}{Time}[/tex]

By substituting the values,

               = [tex]\frac{1.6}{20}[/tex]

               = 0.08 km/min

Thus the avg. velocity will be "0.08 km.min" is correct.

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Related Questions

You toss a ball straight up with an initial speed of 30m/s. How high does it go, and how long is it in the air (neglecting air resistance)?

Answers

Explanation:

Given that,

A ball is tossed straight up with an initial speed of 30 m/s

We need to find the height it will go and the time it takes in the air.

At its maximum height, its final speed, v = 0 and it will move under the action of gravity. Using equation of motion :

v = u +at

Here, a = -g

v = u -gt

i.e. u = gt

[tex]t=\dfrac{u}{g}\\\\t=\dfrac{30\ m/s}{9.8\ m/s^2}\\\\t=3.06\ s[/tex]

So, the time for upward motion is 3.06 seconds. It means that it will in air for 3.06×2 = 6.12 seconds

Let d is the maximum distance covered by it.

[tex]d=ut-\dfrac{1}{2}gt^2[/tex]

Putting all values

[tex]d=30(3.06)-\dfrac{1}{2}\times 9.8\times (3.06)^2\\\\d=45.91\ m[/tex]

Hence, it will go to a height of 45.91 m and it will in the air for 6.12 seconds.

Question 2
What two qualities of objects does gravity depend on?
What each object is made up of and the distance between them.
What each object is made up of and what material is between them.
Each object's mass and its shape.
Each object's mass and the distance between them.

Answers

Answer:

in pretty sure the last one

The qualities of objects do gravity depend on Each object's mass and the distance between them. Hence the correct option is (4).

Gravity is the force of attraction between two objects with mass. The strength of the gravitational force depends on the mass of each object and the distance between them.

The mass of an object refers to the amount of matter it contains. The greater the mass of an object, the stronger its gravitational pull.

The distance between two objects is also a crucial factor. The gravitational force weakens as the distance between the objects increases. In other words, the force of gravity decreases with increasing distance.

So, the two qualities that gravity depends on are each object's mass and the distance between them.

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A 50.0-g Super Ball traveling at 25.5 m/s bounces off a brick wall and rebounds at 21.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 4.45 ms, what is the magnitude of the average acceleration of the ball during this time interval

Answers

Answer:

The magnitude of the average acceleration of the ball during this time interval is 10,449.44 m/s²

Explanation:

Given;

mass of super ball, m = 50 g

initial velocity of the ball, u = 25.5 m/s

final velocity of the ball, v = -21 m/s (re-bouncing backward)

time in contact with the wall, t = 4.45 ms = 0.00445 s

The average acceleration of the ball during this time interval is given by

[tex]a = \frac{dv}{dt} = \frac{v-u}{t} \\\\a = \frac{-21-25.5}{0.00445}\\\\a = -10449.44 \ m/s^2\\\\|a| = 10,449.44 \ m/s^2[/tex]

Therefore, the magnitude of the average acceleration of the ball during this time interval is 10,449.44 m/s²

Newtons first law of motion describes

Answers

Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.

A hot air balloon is ascending straight up at a constant speed of 6.60 m/s. When the balloon is 11.0 m above the ground, a gun fires a pellet straight up from ground level with an initial speed of 30.0 m/s. Along the paths of the balloon and the pellet, there are two places where each of them has the same altitude at the same time. How far above ground level are these places

Answers

Answer:

[tex]H_1 =39.05 \ m[/tex] OR [tex]H_2 =14.5 \ m)[/tex]

Explanation:

From the question we are told that

The constant speed of the balloon is [tex]v = 6.60 \ m/s[/tex]

The height of the balloon is [tex]h = 11.0 \ m[/tex]

The initial speed of the pellet is [tex]u = 30 \ m/s[/tex]

Generally the height of the balloon at the point it is the same altitude with the pellet is mathematically represented as

[tex]H = h + v * (t)[/tex]

Note: vt is the distance covered by the balloon before the pellet got to it

Generally the height of the pellet when it is the same height with the balloon is mathematically represented using kinematics equation

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

So

H = s

=> [tex] ut + \frac{1}{2} gt^2 = h + v * (t)[/tex]

=> [tex] 30t + \frac{1}{2} *( -9.8)t^2 = 11 + 6.60t[/tex]

=> [tex] 4.9t^2 -23.4t + 11= 0[/tex]

using the quadratic formula to solve the above equation

From the quadratic formula calculation

[tex]t_1 = 4.25 \ s[/tex]

   OR

     [tex]t_1 =  0.529  \  s[/tex]

So the height of this two place above the ground is mathematically evaluated as

[tex]H_1 = h + v * (4.25)[/tex]

[tex]H_1 = 11 + 6.60 * (4.25)[/tex]

[tex]H_1 =39.05 \ m[/tex]

OR

[tex]H_2 = h + v * ( 0.529)[/tex]

[tex]H_2 = 11 + 6.60 * (0.529)[/tex]

       [tex]H_2  =14.5 \  m)[/tex]

Consider a monochromatic electromagnetic plane wave propagating in the x direction. At a particular point in space, the magnitude of the electric field has an instantaneous value of 941 V/m in the positive y-direction. What is the instantaneous magnitude of the Poynting vector at the same point and time? The speed of light is 2.99792 x 108 m/s, the permittivity of free space is 8.85419 x 10-12 C2/N/m2 and the permeability of free space is 47 x 10-7 T N/A.
What is the direction of the instantaneous magnetic field?
1. B = - 2.
2. B = +î.
3. B = +j.
4. B = - j.
5. B = +k.
6. B = -k.
7. The magnetic field vector does not have an instantaneous direction.
What is the direction of the instantaneous Poynting vector?
1. Ŝ= -î.
2. Ŝ= -k
3. The Poynting vector does not have an instantaneous direction.
4. Ŝ = tî.
5. Ŝ = +Î.
6. Ŝ= -Î.
7. Ŝ= +ê.

Answers

Answer:

a)   S = 2.35 10³   J/m²2 ,  

b)and the tape recorder must be in the positive Z-axis direction.

the answer is 5

c) the direction of the positive x axis

Explanation:

a) The Poynting vector or intensity of an electromagnetic wave is

          S = 1 /μ₀ E x B

if we use that the fields are in phase

          B = E / c

we substitute

         S = E² /μ₀ c

let's calculate

        s = 941 2 / (4π 10⁻⁷  3 10⁸)

        S = 2.35 10³   J/m²2

 

b) the two fields are perpendicular to each other and in the direction of propagation of the radiation

In this case, the electro field is in the y direction and the wave propagates in the ax direction, so the magnetic cap must be in the y-axis direction, and the tape recorder must be in the positive Z-axis direction.

the answer is 5

C) The poynting electrode has the direction of the electric field, by which or which should be in the direction of the positive x axis

To understand the electric force between charged and uncharged conductors and insulators. When a test charge is brought near a charged object, we know from Coulomb's law that it will experience a net force (either attractive or repulsive, depending on the nature of the object's charge). A test charge may also experience an electric force when brought near a neutral object. Any attraction of a neutral insulator or neutral conductor to a test charge must occur through induced polarization. In an insulator, the electrons are bound to their molecules. Though they cannot move freely throughout the insulator, they can shift slightly, creating a rather weak net attraction to a test charge that is brought close to the insulator's surface. In a conductor, free electrons will accumulate on the surface of the conductor nearest the positive test charge. This will create a strong attractive force if the test charge is placed very close to the conductor's surface.Consider three plastic balls (A, B, and C), each carrying a uniformly distributed charge equal to either +Q, -Q or zero, and an uncharged copper ball (D). A positive test charge (T) experiences the forces. The test charge T is strongly attracted to A, strongly repelled from B, weakly attracted to C, and strongly attracted to D. Assume throughout this problem that the balls are brought very close together. What is the nature of the force between balls A and B? a. Strongly attractive b. Strongly repulsive c. Weakly attractive d. Neither attractive nor repulsive

Answers

Answer:

the correct  is a  Strongly ATTRACTIVE

Explanation:

For this exercise we must use that charges of the same sign repel and charges of the opposite sign attract, the attraction is strong if they are charged or weak if the charges are induced.

Let's apply this to our case.

The test load T is attracted by the sphere A, this implies that the charges are of different sign

the test charge T is repelled by the sphere B, therefore the charges are of equal sign

As the test charge cannot change the sign, this implies that the spheres A and B are of different sign, therefore attractive forces.

Now let's analyze the intensity, as in the exercise they indicate that spheres A and B are charged and are insulators, these charges cannot move, so the attraction must be Strong.

When reviewing the statements, the correct one is a  Strongly ATTRACTIVE

The current through a 10 m long wire has a current density of 4 cross times 10 to the power of 6 space open parentheses bevelled A over m squared close parentheses. The wire conductivity is 2 cross times 10 to the power of 7 space open parentheses bevelled S over m close parentheses. Find the voltage drop across the wire. (Answer with the numeric value, don't write the unit V)

Answers

Answer:

The voltage drop across the wire is 2 V

Explanation:

Given;

length of wire, L = 10 m

current density, I/A, μ = 4 x 10⁶ (A/m²)

wire conductivity, σ = 2 x 10⁷ (S/m)

The resistivity of wire is given by;

[tex]\rho = \frac{RA}{L} \\\\But \ R = V/I\\\\\rho = \frac{VA}{IL}[/tex]

Conductivity, σ = ¹/ρ

[tex]\sigma = \frac{IL}{VA}\\ \\V = \frac{IL}{ A \sigma}\\\\V = (\frac{I}{A})\frac{L}{\sigma}\\ \\V = (\mu)\frac{L}{\sigma}\\\\V = (4*10^{6} )*\frac{10}{2*10^{7} } \\\\V = 2 \ V[/tex]

Therefore, the voltage drop across the wire is 2 V

Which statement describes a controlled experiment?

A. It includes more than one control group.

B. It includes a control group and an experiment group.

C. It includes only the control group.

D. It includes only an experimental group.

Answers

Answer:

B

Explanation:

I think this is right.

If it is Good luck!

The controlled experiment includes a control group and an experiment group, this statement describes a controlled experiment, therefore the correct answer is option B.

What is the scientific investigation?

Scientific discovery is the method of finding solutions through in-depth analysis and discovering them through the outcomes of experiments.

True experimental outcomes that can be backed up by data are incredibly important to scientific inquiry.

A regulated investigation where all sensors to measure have been held constant and the scientific element is utilized as a factor associated.

Since the statement depicts a controlled experiment with a control group and an experiment group, option B is the appropriate response.

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Convert a speed of 4.50 km/h to units of ft/min. (1.00 m = 3.28 ft).a. 165 ft/min.b. 82.3 ft/min.c. 886 ft/min.d. 246 ft/min.e. 0.246 ft/min.

Answers

Answer:

246ft/min

Explanation:

In this problem we are faced with unit conversion.

We are expected to convert from km/h to ft/min

Given that

1.00 m = 3.28 ft

1km(1000m)= 3280 ft

But we need to convert 4.5km to ft

Hence in ft 4.5km= 3280*4.5

= 14760ft

We now convert hour to minutes

60min make 1 hour

Therefore the convert form km/h to ft/min is

= 14760/60= 246ft/min

The answer is 246ft/min

For a point charge, how does the potential vary with distance from the point charge, r?
b. r.c. 1/r.d. 1/r2.e. r2.

Answers

For a point charge, how does the potential vary with distance from the point charge, r?

a constant

b. r.

c. 1/r.

d. [tex]1/r^2[/tex].

e. [tex]r^2[/tex].

Answer:

The  correct option is  C

Explanation:

Generally for a point charge the electric potential is mathematically represented as

    [tex]V  =  \frac{k  Q  }{r }[/tex]

Here we can deduce that the electric potential varies inversely with the distance i.e

      [tex]V  \  \alpha \  \frac{1}{r}[/tex]

So

   

The potential should vary with distance from the point charge r should be option c. 1/r.

What is a point charge?

The point charge with respect to the electric potential should be expressed in

V = KQ/r

Here we can say that the electric potential should be varied oppositely with the distance

Here is the electric file with respect to the point charge should be obtained from the Coulomb law. It should be radially outward from the point charge in all types of directions.

Therefore, the option c is correct.

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An ideal monatomic gas initially has a temperature of T and a pressure of p. It is to expand from volume V1 to volume V2. If the expansion is isothermal, what are the final pressure pfi and the work Wi done by the gas? If, instead, the expansion is adiabatic, what are the final pressure pfa and the work Wa done by the gas? State your answers in terms of the given variables.

Answers

Answer:

Isothermal :   P2 = ( P1V1 / V2 ) ,  work-done [tex]pdv = nRT * In( \frac{V2}{v1} )[/tex]

Adiabatic : : P2 = [tex]\frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }[/tex]  , work-done =

W = [tex](3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)[/tex]

Explanation:

initial temperature : T

Pressure : P

initial volume : V1

Final volume : V2

A) If expansion was isothermal calculate final pressure and work-done

we use the gas laws

= PIVI = P2V2

Hence : P2 = ( P1V1 / V2 )

work-done :

[tex]pdv = nRT * In( \frac{V2}{v1} )[/tex]

B) If the expansion was Adiabatic show the Final pressure and work-done

final pressure

[tex]P1V1^y = P2V2^y[/tex]

where y = 5/3

hence : P2 = [tex]\frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }[/tex]

Work-done

W = [tex](3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)[/tex]

Where    [tex]T2 = T1V1^(2/3)/V2^(2/3)[/tex]



A basketball is shot. After the ball leaves the player's hand, in which direction does the ball accelerate?
A)It always accelerates in the opposite direction that the object is moving.
B) It always accelerates in a upward direction.
C) It always accelerates in a downward direction.
D) It accelerates upward when the ball is rising and downward when it is falling.

Answers

Answer: Always accelerates in a downward direction

Explanation: Gravity is pulling the ball downward

What statements are true regarding the skydivers jump and fall to Earth?
Regardless of their weight, before opening their parachutes, the guys felt at the same velocity.
According to the data, the acceleration due to gravity is 9.8m/s/s.
Once the parachutes were opened, all five skydivers continued to fall at exactly the same velocity
Opening the parachutes provided air resistance and changed their velocity
By free-falling for a longer period of time, the skydivers would have changed the acceleration due to
gravity

Answers

Answer: Regardless of their weight, before opening their parachutes, the guys fell at the same velocity.

According to the data, the acceleration due to gravity is 9.8 m/s/s

Opening the parachutes provided air resistance and changed their velocity.

Explanation:

When in free fall, mass does not affect the speed of moving objects (only gravitational force is acting upon it).

If you take a look at the data, you can see that the velocity is increasing at a constant rate in intervals of 9.8. This is the acceleration which makes sense because acceleration is the change of velocity divided by time.

Think about it: parachutes slow down the object that is falling because it creates a barrier between the object falling and the gravitational force it encounters. The change in the effects of the gravitational force will alter the velocity.

I hope this was helpful. Have a nice day!

The speed of sound in humid air than in dry air. Why ​

Answers

Answer:

When moisture is removed from air, its density increases. The speed of sound in a humid air is inversely proportional to the square root of its density. Therefore, the speed of sound in moist air is humidity increases, the velocity of sound increase and vice-versa.

Hope it helps :)

An object travels due North and covers 500 miles in 5 hours. Which one of the following is the average velocity of the object?
a) 100 mph
b) 2500 mph
c) 2500 mph due North
d) 100 mph due North

Answers

Answer:

100 mph.

Explanation:

Given that,

Distance covered by an object is 500 miles in 5 hours.

We need to find the average velocity of the object. It can be given by :

[tex]v=\dfrac{d}{t}\\\\v=\dfrac{500\ \text{miles}}{5\ \text{hour}}\\\\v=100\ \text{mph}[/tex]

Hence, the average velocity of the object is 100 mph.

Which of these statements describes a species? Group of answer choices
A. A species is a group of organisms that are closely related and produce fertile offspring
B. A species is a group of organisms that are closely related.
C. A species is a group of organisms that live in the same habitat
D. A species is a group of organisms that live in an area and interact.

Answers

I believe the answer would be A .
The answer haves to be A or it could be also D but most likely I am positive that is A



What positively charged subatomic particle is found in the nucleus?

Answers

Answer:

Protons

Explanation:

The Proton is one of the subatomic particles that are found in the nucleus of an atom. The protons are positively charged. They have an electric charge of 1.6 * 10^-19 C. Protons have an approximately 1 atomic mass unit, or otherwise called, 1 amu.

One or more protons, at least, are found in every nucleus if an atom, as they are an essential and very necessary part of the nucleus. Ernest Rutherford, in 1920, named protons, proton

A computer monitor uses 200 W of power. How much energy does it use in
10 seconds?
A. 20 J
B. 200 J
C. 20,000 J
D. 2000 J
SUBMIT

Answers

Answer: D (2000 J)

Explanation: p= 200 W

t= 10 s

E=pt

E=200 W * 10s

E= 2000 J

Answer:

it is D.

Explanation:

Calculate the average speed of a complete round trip in which the outgoing 220 kmkm is covered at 92 km/hkm/h , followed by a 1.0-hh lunch break, and the return 220 kmkm is covered at 55 km/hkm/h . Express your answer to two significant figures and include the appropriate units.

Answers

Answer:

60 km/hr

Explanation:

220 km is covered at 92km/hr in the round trip

There is 1 hour lunch break

During the return of the round trip 220 km is covered at 55km/hr

The first step is to calculate the total distance

= 220km + 220km

= 440km

The time spent during the entire round trip can be calculated as follows

= (220/92) + 1 + (220/55)

= 2.3913 + 1 + 4

= 7.3913

Therefore the average speed can be calculated as follows

= distance/ time

= 440/7.3913

= 59.52

= 60km/hr (approximated to 2 significant figures)

Hence the average speed is 60 km/hr

Why does wine go sour faster
it the cork is removed from the
bottle

Answers

When the cork is removed the gases from the wine are reacting to the oxygen and it creates a chemical reaction while when the cork is still on the chemicals never interact and stay there in the bottle

A mountain lion jumps to a height of 3.25 m when leaving the ground at an angle of 43.2°. What is its initial speed (in m/s) as it leaves the ground?

Answers

Recall that

[tex]{v_f}^2={v_i}^2+2a\Delta y[/tex]

where [tex]v_i[/tex] and [tex]v_f[/tex] are the lion's initial and final vertical velocities, [tex]a[/tex] is its acceleration, and [tex]\Delta y[/tex] is the vertical displacement.

At its maximum height, the lion has 0 vertical velocity, so we have

[tex]0={v_i}^2-2gy_{\rm max}[/tex]

where g is the acceleration due to gravity, 9.80 m/s², and we take the starting position of the lion on the ground to be the origin so that [tex]\Delta y=y_{\rm max}-0=y_{\rm max}[/tex].

Let v denote the initial speed of the jump. Then

[tex]v_i=v\sin(43.2^\circ)=\sqrt{2\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(3.25\,\mathrm m)}\implies\boxed{v\approx11.7\dfrac{\rm m}{\rm s}}[/tex]

The initial speed will be "11.7 m/s".

Speed and Displacement:

Given:

Height = 3.25 mAngle = 43.2°Acceleration due to gravity = 9.8 m/s²

We know,

→ [tex]vf^2=v_i^2+2a\Delta y[/tex]

At max. height, vertical height be zero, then

→ [tex]0 = v_i^2-2gy_{max}[/tex]

or,

→ [tex]\Delta y = y_{max} -0 = y_{max}[/tex]

now,

The initial speed,

→ [tex]v_i = v \ Sin(43.2^{\circ})[/tex]

By substituting the values,

      [tex]= \sqrt{1\times 9.8\times 3.25}[/tex]

      [tex]= 11.7 \ m/s[/tex]

Thus the solution above is correct.

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The amount of kinetic energy an object has depends on which feature of the object?

Answers

Answer:

Mass and speed.

Explanation:

The amount of kinetic energy of an object depends on the object's mass and speed.

Answer:

its a motion hope this helps

Explanation:

A person running down the hallway at a velocity of 4.2 m/s comes to a stop in a time of 1.8 seconds to avoid hitting the wall. What was the acceleration of the person?
A.) 7.6 m/s/s
B.) -7.6 m/s/s
C.) 2.3 m/s/s
D.) -2.3 m/s/s

Answers

Hshshshhshdhdhdhdhddhdhdhdhdhdh

Answer:

It's not -7.6

Explanation:

All I can tell you

Please HELP I NEED TO WRITE A PARAGRAPH BUT DON'T KNOW HOW TO EXPLAIN IT!

How does protein, atoms, vitamin d, magnesium, and zinc work together?

Answers

Explain how they interact with each other. I'm not a chem genius but taht would make the most sense.

Hope it helps! Comment if you have any questions and have an amazing day!

Problem: You and your friend are discussing last week’s recitation problem, in which an electron falling down a tunnel through a uniformly charged earth exhibits simple harmonic motion. Instead of being uniformly charged, imagine that the Earth is given a spherically symmetric charge density ⍴(r)=b/r, where b is a constant. If an electron is released from rest inside this tunnel at some initial distance from the center, find an expression for the acceleration of the electron inside the tunnel (ignore gravity). If the electron is released from rest at the surface, what value of b will ensure that the electron reaches the center of the Earth in 12 seconds? ………………………………………………………………………………………

Answers

Answer:

a. a = eb/2ε₀m b. b = 8.91 × 10⁻¹⁸ kg/V

Explanation:

a. Using Gauss' law

ε₀∫E.dA = q where E = electric field and q = charge enclosed.

Now , the charge has  a spherical symmetric charge density ⍴(r) = b/r, the total charge enclosed at distance r is given by

q = ∫∫∫ρ(r)r²sinθdrdθdΦ

q = ∫∫∫(b/r)r²sinθdrdθdΦ

q = ∫∫∫brsinθdrdθdΦ

we integrate r from 0 to r, θ from 0 to π and Φ from 0 to 2π

q = ∫∫br[∫sinθdθ]drdθdΦ

q = ∫∫br[-cosθ]drdΦ

q = ∫∫br-[cosπ - cos0]drdΦ

q = ∫∫br-[-1 - 1]drdΦ

q = ∫2brdr∫dΦ

q = ∫2brdr[Φ]

q = ∫2brdr[2π - 0]

q = 4πb∫rdr

q = 4πb[r²/2]

q = 4πb[r²/2 - 0]

q = 2πbr²

The let side of the equation is

ε₀∫E.dA = ε₀E∫dA = ε₀E4πr² since there is spherical symmetry

Equating both sides of the equation

ε₀E4πr² =  2πbr²

E =  2πbr²/ε₀4πr²

E = b/2ε₀

Now, the force acting on the electron , F = ma where m = mass of electron and a = acceleration of electron.

F also equal F = eE where e = electron charge and E = electric filed acting on electron

So, eE = ma

a = eE/m

a = eb/2ε₀m

b. I(f the electron is released from rest and is to reach the center of the earth, it covers a distance of the radius of the earth. Using s = ut + 1/2at² where s = radius of earth = R = 6400 km = 6.4 × 10⁶ m  , u = initial velocity = 0 m/s. and a = eb/2ε₀m. Substituting these values into s we have

R = 0t + 1/2(eb/2ε₀m)t²

R = ebt²/4ε₀m

making b subject of the formula, we have

b = 4ε₀mR/et²

when t = 12 s and e = 1.609 × 10⁻¹⁹ C, ε₀ = 8.854 × 10⁻¹² F/m and m = 9.109 × 10⁻³¹ kg.

Substituting these values into b we have

b = (4 × 8.854 × 10⁻¹² F/m × 9.109 × 10⁻³¹ kg × 6.4 × 10⁶ m)/(1.609 × 10⁻¹⁹ C × 12²)

b = 2064.67/ 231.696 × 10⁻¹⁸

b = 8.91 × 10⁻¹⁸ Fkg/C

b = 8.91 × 10⁻¹⁸ kg/V

1 gram is 0.035 ounces.How many ounces is 200 grams

Answers

Answer:

7.05479

Explanation:

Answer:

200 grams:     ounces:7.05479239 that's the answer

Explanation:

how was the atomic bomb different from conventional bombs? ​

Answers

Answer:

A conventional bomb releases most of its energy in the form of blast. Atomic bombs on the other hand, release 50 per cent energy as blast, 35 per cent as heat and 15 per cent as nuclear radiation.

Explanation:

A conventional bomb releases most of its energy in the form of blast. Atomic bombs on the other hand, release 50 per cent energy as blast, 35 per cent as heat and 15 per cent as nuclear radiation.

Which laboratory activity involves a chemical change?

Answers

Answer:

A: leaving a copper penny in vinegar until it turns green

Explanation:

The corrosion is from oxidation -- a chemical reaction between the metal and oxygen, water, and carbon dioxide in the air. Rust is the term used to describe this process when it happens to iron instead of other metals. I also took it on E2020. If you need further explanation, please comment below.

The laboratory activity which involves a chemical change is leaving a copper penny in vinegar until it turns green.

What are chemicals?

The chemicals are the artificially prepared or purified and a distinct compound or substance.

The corrosion is the result of oxidation that is a chemical reaction between the metal and oxygen, water, and carbon dioxide in the air. Rust is the term used to describe the corrosion process.

Corrosion also takes place in copper due to which the copper turns green when oxidized.

Thus, The laboratory activity which involves a chemical change is leaving a copper penny in vinegar until it turns green.

Learn more about chemicals.

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You collect some data on horse racing along a straight track. You are able to fit the motion of the horse to a function during this interval, where you’ve chosen a particular spot on the track to be your origin and started your clock (t = 0) when you started collecting this new data.Required:a. What is the horse’s velocity as a function of time? Does the horse ever turn around during this time?b. What is its acceleration as a function of time?

Answers

Answer:

The equation is missing in the question. The equation is [tex]$10 m + 5(m/s^2)t^2+3(m/s^3)t^3$[/tex]

a). [tex]$v=10 t +9t^2$[/tex] , the horse will not turn.

b). a(t) = 10 + 18t

Explanation:

Given :

[tex]$x(t)=10 m + 5(m/s^2)t^2+3(m/s^3)t^3$[/tex]

∴ At t =0, x = 10 m

a). Velocity as a function of time

[tex]$v = \frac{dx}{dt} $[/tex]

  = [tex]$10 t +9t^2$[/tex]

Turning velocity must be zero.

v(t) = 0

[tex]$10 t +9t^2=0$[/tex]

[tex]$\therefore t = 0 \text{ or}\ t =-\frac{10}{9}$[/tex]

Taking the positive value of time.

The horse will not turn.

b). Acceleration as a function of time.

[tex]$a(t)=\frac{dv}{dt}$[/tex]

     = 10 + 18t

∴ a(t) = 10 + 18t

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