At an air show, a stunt pilot performs a vertical loop-the-loop in a circle of radius 3.63 x 103 m. During this performance the pilot whose weight is 676 N, maintains a constant speed of 2.25 x 102 m/s. At what speed, in m/s, will the pilot experience weightlessness

Answer:1 carbon atom

Explanation:AP3X

Answer:

1

Explanation:

Answer:

Explanation:

initial velocity u = 0

final velocity v = 11.3 m /s

distance covered s = 12.8 m

v² = u² + 2 a s

11.3² = 0 + 2 x a x 12.8

a = 4.99 m /s²

again ,

v = u + a t

11.3 = 0 + 4.99 t

t = 2.26 s .

Rest of the sprint will be covered with uniform velocity .

Distance covered = 100 - 12.8 = 87.2 m

speed = 11.3 m /s

time taken = 87.2 / 11.3 = 7.7 s

Total time of 100 m sprint = 7.7 + 2.26 = 9.96 m .

b )

Let the time taken to reach the top speed be t .

acceleration a = 11.3 / t

distance covered s = 1/2 a t²

= .5 x (11.3 / t) x t²

= 5.65 t

Rest of the distance = 100 - 5.65 t

time taken to cover rest of the distance = (100 - 5.65 t ) / 11.3

Total time =  (100 - 5.65 t  / 11.3 ) + t = 9.75

100 - 5.65 t + 11.3 t = 11.3 x 9.75

100 + 5.65 t = 110.175

5.65 t = 10.175

t = 1.8

acceleration a = 11.3 / t

= 11.3 / 1.8

= 6.278 m /s²

distance covered in 1.8 s

s = 1/2 a t²

= .5 x 6.278 x 1.8²

= 10.17 m .

Answer:

Answer C

Explanation:

Answer:

It is longer

Explanation:

According to the theory of special relativity, moving clocks run slower. So, the construction worker moving at a constant speed observers a time much longer than the time I observe since I am stationary. If t is the time observed by me and v is the speed of the construction worker, then, the time observed by the construction worker, t' is given by

t' = t/√[1 - (v/c)²] where c = speed of light

So, the construction worker reports a longer time interval than me since his time runs slower.

Answer:

A) v1 = -0.256 m/s

v2 = 0.128 m/s

B) v1 = -0.0642 m/s

v2 = 0 m/s

C) v1 = v2 = 0 m/s

Explanation:

We are given;

Spring constant; k = 3.85 N/m

Distance compressed; x = 8 cm = 0.08 m

Mass of left block; m1 = 0.25 kg

Mass of right block; m2 = 0.5 kg

A) From conservation of energy;

½kx² = ½m1•v1² + ½m2•(v2)²

Also from conservation of linear momentum, we know that;

m1v1 = m2v2

Now, v2 = m1•v1/m2

Plugging this for v2 in the first equation gives;

½kx² = ½m1•v1² + ½m2•(m1•v1/m2)²

Making v1 the subject, we have;

v1 = √(kx²(m2))/(m1•m2 + (m1)²))

v1 = √[(3.85 × 0.08² × 0.5)/((0.25 × 0.5) + 0.5²))

v1 = 0.256 m/s

This is the left block which is in the negative x direction and so v1 = -0.256 m/s

We saw that v2 = m1•v1/m2

Thus; v2 = (0.25 × 0.256)/0.5

v2 = 0.128 m/s

B) Force exerted by spring is;

F_s = kx = 3.85 × 0.08

F_s = 0.308 N

Normal forces will be calculated for both blocks as;

N1 = m1•g = 0.25 × 9.8 = 2.45 N

N2 = m2•g = 0.5 × 9.8 = 4.9 N

Let's calculate force of static friction for both blocks;

F_s1 = μN1 and F_s2 = μN2

We are given coefficient of friction as

μ = 0.1.

Thus;

F_s1 = 0.1 × 2.45 = 0.245 N

F_s2 = 0.1 × 4.9 = 0.49 N

F_s1 is lesser than F_s. Thus let's calculate the new compression;

x_1 = 0.245/3.85

x_1 = 0.06364 m

Thus, change in compression is;

Δx = x - x_1

Δx = 0.08 - 0.06364

Δx = 0.01636 m

From conservation of energy, since our coefficient of friction is not zero and we have frictional force, then we use the equation;

½k(x)² - F_s1•Δx = ½m1•v1² + ½k(x_1)²

Making v1 the subject, we have;

v_1 = √(k(x² - x_1²) - 2F_s1•Δx)/m1

v_1 = √(3.85(0.08² - 0.06364²) - 2(0.245 × 0.01636)/0.25

v1 = 0.0642 m/s

Since in negative x direction, then

v1 = -0.0642 m/s

F_s2 is greater than F_s. Thus, it means the right side block will not move and velocity is zero. v2 = 0 m/s

C) Coefficient of friction is now 0.463.

Thus;

F_s1 = 0.463 × 2.45 = 1.13435 N

F_s2 = 0.463 × 4.9 = 2.2687 N

They are both greater than F_s and thus no motion in both cases.

So v1 = v2 = 0 m/s

Answer:

The base runner

Explanation:

To know the correct answer to the question, we shall determine the time taken for the baseball and the base runner to get to the home plate. This is illustrated below:

For the baseball:

Horizontal velocity (u) = 30 m/s

Horizontal distance (s) = 60 m

Time (t) =?

s = ut

60 = 30 × t

Divide both side by 30

t = 60 / 30

t = 2 s

Thus, it will take the baseball 2 s to get to the home plate.

For the base runner:

Horizontal velocity (u) = 5 m/s

Horizontal distance (s) = 5 m

Time (t) =?

s = ut

5 = 5 × t

Divide both side by 5

t = 5 / 5

t = 1 s

Thus, it will take the base runner 1 s to get to the home plate.

SUMMARY:

Time taken for the baseball to get to the home plate = 2 s

Time taken for the base runner to get to the home plate = 1 s

From the calculations made above, we can conclude that the base runner will arrive at the home plate first because it took him 1 s to get to the home plate whereas the baseball took 2 s to get there.

Answer:

1) dependent

2) Experiment

3) Controlled variable

4) hypothesis

5) Inquiry

Explanation:

The dependent variables are those variables that change when the experimental or independent variable is being manipulated.

An experiment is a carefully designed study that tends to establish a cause and effect relationship between two or more variables.

A control variable is any variable which is held constant or unchanged throughout the course of the investigation(experiment).

Hypothesis is a precise statement which can be tested and predicts the outcome of the study.

The independent variable is the variable that is being manipulated.

Answer: Option A;  9.8 m/s^2

Explanation:

When an object is in the air, and there is no air resistance acting on the object, the only force that will act on the object is the gravitational force (on the vertical axis).

Then, if the only force acting on the object is the gravitational force, the acceleration of the object will be equal to the gravitational acceleration.

We know that the gravitational acceleration is equal to:

g = 9.8m/s^2

Then the acceleration on the vertical axis will be equal to:

a(t) = 9.8m/s^2

The correct option is the first one:

A. 9.8 m/s^2

The  velocity of the given graph will be 2m/s.

Speed is a scalar quantity that measures how fast an object is moving. It is the rate of change of distance traveled by an object over a specific period of time. Speed does not consider the direction of motion; it only reflects the magnitude of the velocity.

Mathematically, speed (v) is calculated as:

Speed (v) = Distance traveled (d) / Time taken (t)

where:

Speed (v) is measured in units like meters per second (m/s), kilometers per hour (km/h), miles per hour (mph), etc.

Distance traveled (d) is the total path length covered by the object in a given timeframe.

Time taken (t) represents the duration it took the object to cover that distance.

According to graph,

speed = 6-2/ 3- 1

= 4/ 2

= 2 m/s

Therefor, the velocity of the given graph will be 2m/s.

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Answer:

I think that this ans may help you

Answer:

I. a, c, f and h

II. e

III. b, d, g and i

IV. i

Explanation:

I. Chemical symbols are simple abbreviations used to represent various elements or compound. They consist entire of alphabet.

For the diagram given above, the labelled parts which represent chemical symbol are: a, c, f and h

II. Coefficients are numbers written before the chemical symbol of elements or compound.

For the diagram given above, the labelled part which represent Coefficient is: e

III. Number of atoms of element present in a compound is simply obtained by taking note of the numbers written as subscript in the chemical formula of the compound.

For the diagram given above, the labelled part which represent the number of atoms of the element are: b, d, g and i

IV. When no number is written as subscript in the formula of the element in the compound, it means the element has just 1 atom in the compound.

For the diagram given above, the labelled part which indicates that only 1 atom of the element is present is: i

Answer:

i. a, c, f, h

ii. e

iii. b, d, g, i

iv. i

Answer:

The formula for speed is speed = distance ÷ time. To work out what the units are for speed, you need to know the units for distance and time. In this example, distance is in metres (m) and time is in seconds (s), so the units will be in metres per second (m/s).

Answer:

thats answer

Explanation:

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Answer:

a) 202.7 N

b) 58.1 N

c) 74.1º N of E.

d) 210.9 N

Explanation:

a)

      [tex]F_{Ay} = F_{A} * cos \theta = 155 N* cos 22 = 143.7 N (1)[/tex]

      ⇒ Fy = FH + FAy = 59 N + 143.7 N = 202.7 N (2)

b)

      [tex]F_{x} = F_{A} * sin \theta = 155 N* sin 22 = 58.1 N (3)[/tex]

c)

       [tex]tg \theta = \frac{F_{y}}{F_{x}} = \frac{202.7N}{58.1N} = 3.5 (4)[/tex]

      ⇒ θ = tg⁻¹ (3.5) = 74.1º N of E. (5)

d)

      [tex]F_{DS} = \sqrt{(F_{x} ^{2} +F_{y} ^{2})} = \sqrt{(58.1N)^{2}) + (202.7N)^{2} } = 210.9 N (6)[/tex]

Electronegativity increases from left to right in the periodic table.

A table of chemical elements arranged in atomic number order, commonly in rows, with elements having comparable atomic structures (and hence chemical characteristics) appearing in vertical columns.

Electronegativity increases as you move  from left to right across a period on the periodic table, and drops as you move down a group.

As a result, the most electronegative elements appear on the periodic table's top right, while the least electronegative elements appear on the bottom left.

Hence, the correct option is C.

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Answer:

the moment of inertia is 4.5 × 10⁻⁵ kg.m²  

Explanation:

Given that;

point mass m = 0.005 g = ( 0.005 / 1000 ) = 5 × 10⁻⁶ kg

perpendicular distance r = 3m

We know that a point mass doesn't have a moment of inertia around its own axis but, but using the parallel axis theorem, a moment of inertia around a distant axis of rotation can be determined using;

[tex]I_{}[/tex] = mr²

so we substitute

[tex]I_{}[/tex] = (5 × 10⁻⁶ kg) × (3 m)²

[tex]I_{}[/tex] = (5 × 10⁻⁶ kg) × 9 m²

[tex]I_{}[/tex]  = 4.5 × 10⁻⁵ kg.m²  

Therefore; the moment of inertia is 4.5 × 10⁻⁵ kg.m²  

The moment of inertia of given point mass is 4.5 × 10⁻⁵ kgm² at a perpendicular distance of 3 m.

The moment of inertia of given point mass can be determined by,

[tex]I = mr^2[/tex]

Where,

[tex]I[/tex]- moment of inertia

[tex]m[/tex]- mass = 0.005 g = ( 0.005 / 1000 ) = 5 × 10⁻⁶ kg

[tex]r[/tex] - perpendicular distance = 3 m

Put the values in the formula,

[tex]I = (5 \times 10^{-6}{\rm \ kg}) \times (3 {\rm \ m})^2\\\\I = 5 \times 10^{-6}{\rm \ kg} \times 9 {\rm \ m}\\\\I = 4.5 \times 10^{-5} kgm^2[/tex]

Therefore; the moment of inertia of given point mass is 4.5 × 10⁻⁵ kgm².

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Answer:

a) Remember the relationship:

v = f*λ

where:

v = velocity of the wave, remember that for an electromagnetic wave, this always is equal to the speed of light, then:

v = c = 3*10^8 m/s

f is the frequency of the wave

λ is the wavelength.

So if we want to compute the frequency, we have the equation:

f = c/λ

then:

We know that UVA has the range from 320 nm to 400nm, converting these to meters we hav:

1nm = 1m*10^(-9)

then:

320mm = (320*10^(-9)) m = 0.00000032 m

400nm = (400*10^(-9)) m =  0.00000040 m

Replacing these in our equation we can find that the range of frequencies is:

f = (3*10^8 m/s)/( 0.00000032 m) = 9.375*10^(14) Hz

f =  (3*10^8 m/s)/( 0.00000040 m) = 7.5*10^(14) Hz

Then the range of frequencies is:

( 7.5*10^(14) Hz to 9.375*10^(14) Hz)

Similar for the case of UVB:

the wavelengths are:

280 nm to  320 nm

Rewriting these in meters we get:

280nm = 280*10^(-9) m = 0.00000028m

320nm = 320*10^(-9) m = 0.00000032 m

Then the frequencies are:

f = (3*10^8 m/s)/( 0.00000032 m) = 9.375*10^(14) Hz

f = (3*10^8 m/s)/( 0.00000028 m) = 1.07*10^(15) Hz

The range of frequencies for UVB is:

(9.375*10^(14) Hz to 1.07*10^(15) Hz)

B) We want to know the range of wavenumbers for both types of waves.

The wave number is calculated as:

1/λ

Which represents the number of waves in a given unit of length.

Then the range for UVA will be:

1/320nm = 0.0031 nm^-1

1/400nm = 0.0025 nm^-1

Then the range is:

( 0.0025 nm^-1, 0.0031 nm^-1)

And for the case of UVB we will have:

1/320 nm = 0.0031 nm^-1

1/280 nm = 0.0036 nm^-1

Then the range is:

(0.0031 nm^-1, 0.0036 nm^-1)

Answer:

 t ’= [tex]\frac{1450}{0.6499 + 2 v_r}[/tex],  v_r = 1 m/s       t ’= 547.19 s

Explanation:

This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.

By the time the boat goes up the river

        v_b - v_r = d / t

By the time the boat goes down the river

        v_b + v_r = d '/ t'

let's subtract the equations

       2 v_r = d ’/ t’ - d / t

       d ’/ t’ = 2v_r + d / t

       [tex]t' = \frac{d'}{ \frac{d}{t}+ 2 v_r }[/tex]

In the exercise they tell us

         d = 1.22 +1.45 = 2.67 km= 2.67 10³ m

         d ’= 1.45 km= 1.45 1.³ m

at time t = 69.1 min (60 s / 1min) = 4146 s

the speed of river is v_r

      t ’= [tex]\frac{1.45 \ 10^3}{ \frac{ 2670}{4146} \ + 2 \ v_r}[/tex]

      t ’= [tex]\frac{1450}{0.6499 + 2 v_r}[/tex]

In order to complete the calculation, we must assume a river speed

          v_r = 1 m / s

let's calculate

      t ’= [tex]\frac{ 1450}{ 0.6499 + 2 \ 1}[/tex]

      t ’= 547.19 s

Answer:

15.7

Explanation:

Since N1 sin(angle1)= N2 sin(angle2), you can set up the equation to find the missing angle using inverse sin( 1.00 sin(21.2)/ 1.33). You would then get that the other angle is equal to 15.7 or 15.8 if you round up. I hope this helped! :)

Answer:

15.7

Explanation:

Acellus

Answer: 0.24 m/s

Explanation:

Answer:

53.2

Explanation:

You can use the kinematic equation: displacement of x = (initial velocity + final velocity)*t/2

Subsititing: 17+21 = 38 * 2.8/2 = 53.2

Note: Displacement = distance between the 2 points

on smooth surface like mirror.

Answer:

Honestly, I don't know ¯\_(ツ)_/¯

Explanation: