A piece of gum becomes stuck upon a skateboard's wheel. What is the centripetal acceleration of the piece of gum if the wheel's radius is 30 mm and the tangential velocity is 0.5 m/s?

Answers:
a. 8.33 m/s^2
b. 0.5 m/s^2
c. 30 mm/s^2
d. 0.33 m/s^2
e. 83.20 m/ft^2

This question can be solved using the equations of motion. There are two scenarios where the equations of motion can be used. The first scenario is the free-fall motion of the piece of bread. The second scenario is the uniformly accelerated motion of the bird.

The acceleration of the bird is  "a = 26.13 m/s²".

First, we will calculate the time taken by the bread to fall 3 m. Using the second equation of motion for this free-fall motion:

[tex]h = v_it + \frac{1}{2}gt^2[/tex]

where,

h = height fall = 3 m

vi = initial velocity = 0 m/s

g = acceleration due to gravity = 9.8 m/s²

t = time taken = ?

Therefore,

[tex]3\ m = (0\ m/s)t+\frac{1}{2}(9.8\ m/s^2)t^2\\t = \sqrt{\frac{(3\ m)(2)}{9.8\ m/s^2}}\\\\t = 0.78\ s[/tex]

The bird took the same time to catch the bread. Now applying the second equation of motion to the bird's motion:

[tex]s = v_it + \frac{1}{2}at^2[/tex]

where,

s = distance covered by the bird = 5 m + 3 m = 8 m

vi = initial velocity of the bird = 0 m/s

a = acceleration of the bird = ?

t = time taken = 0.78 s

Therefore, using these values we get:

[tex]8\ m = (0\ m/s)(0.78\ s)+\frac{1}{2}a(0.78\ s)^2\\\\a = \frac{16\ m}{(0.78\ s)^2}[/tex]

a = 26.13 m/s²

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Explanation:

The wheel and axle is a type of simple machine used to make tasks easier in terms of manipulating force by applying the concept of mechanical advantage.

Answer:

  an object sliding down hill

Explanation:

On a slope, the force applied is due to gravity. Its direction is straight down. If the object is sliding down the hill, its displacement is at an angle to the applied force. The angle of displacement will depend on the steepness of the hill.

you need the number of seconds to calculate the distance. I'm assuming that the seconds were written in the question but you forgot to write the here, you can calculate the distance then and find the exact point where the cyclist stopped since you have the circumference. then you can find the angle (after finding the distance he cycled) by subtracting it from 360 since the circular path will have an angle of 360 degrees.

Although your question lacks some data A general answer is provided :

The angle made vertically  = 360° - x°

where x = angle at which the the cyclist stopped

and circumference of the circular path = Total Distance travelled by the cyclist

First step : determine the distance travelled by the cyclist vertically

circumference = 2*π*r

343 = 2*π * r

∴ r = 343 / ( 2π ) = 54.59

therefore distance travelled by the cyclist vertically = 54.59 * 2 = 109.18 m

Given that the time travelled is missing

assuming the angle to the distance travelled by the cyclist before it stopped vertically = 360° - x°  

This is because the Total angle of a circular path = 360°

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Answer:

0.10 g/cm3

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Answer:

Lever systems are simple machines that change or increase the input force that we apply to the load. The lever provides us with some mechanical...

Answer:

● Effort arm or Effort distance (ED): The perpendicular distance from the fulcrum to the point of effort is called effort arm.

● Load arm or Load distance (LD): The perpendicular distance from the fulcrum to the point of load is called load arm.

Answer:

The distance covered is 40 m and the displacement is 31,6m.

Explanation:

The distance covered is the sum of the two distances (10+30). The displacement is equal to the distance of the hipotenusa of the triangle that the two distances (10 m to north and 30m to east) create. Using the Pythagoras theorem the displacent is equal to the Square root of (30^2 +10^2) .

Answer: i)A to B : (ice) freezing

ii) B to C (water) boiling

C to D  (steam) evaporating

explanation:  0° is the freezing point of water  when temperature increases from 0° the water starts melting. As 100° is the boiling point of water so at 100° the water  completely melts and it starts boiling during boiling water changes into steam(water vapour) and it evaporates

Answer:

It's a pretty simple suvat linear projectile motion question, using the following equation and plugging in your values it's a pretty trivial calculation.

V^2=U^2+2*a*x

V=0 (as it is at max height)

U=30ms^-1 (initial speed)

a=-g /-9.8ms^-2 (as it is moving against gravity)

x is the variable you want to calculate (height)

0=30^2+2*(-9.8)*x

x=-30^2/2*-9.8

x=45.92m

Answer:

35

Explanation:

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Answer:

a) v = 0.4799 m / s,  b)  K₀ = 1600.92 J,    K_f = 5.46 J

Explanation:

a) How the two players collide this is a momentum conservation exercise. Let's define a system formed by the two players, so that the forces during the collision are internal and also the system is isolated, so the moment is conserved.

Initial instant. Before the crash

        p₀ = m v₁ + M v₂

where m = 95 kg and his velocity is v₁ = -3.75 m / s, the other player's data is M = 111 kg with velocity v₂ = 4.10 m / s, we have selected the direction of this player as positive

Final moment. After the crash

       p_f = (m + M) v

as the system is isolated, the moment is preserved

       p₀ = p_f

       m v₁ + M v₂ = (m + M) v

       v =[tex]\frac{m v_1 + M v_2}{m+M}[/tex]

let's calculate

        v = [tex]\frac{ -95 \ 3.75 \ + 111 \ 4.10}{95+111}[/tex]

        v = 0.4799 m / s

b) let's find the initial kinetic energy of the system

         K₀ = ½ m v1 ^ 2 + ½ M v2 ^ 2

         K₀ = ½ 95 3.75 ^ 2 + ½ 111 4.10 ^ 2

         K₀ = 1600.92 J

the final kinetic energy

         K_f = ½ (m + M) v ^ 2

         k_f = ½ (95 + 111) 0.4799 ^ 2

         K_f = 5.46 J

average speed of the car is 23.9 km/h

Answer:

hlo

Explanation:

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Answer:

D

Explanation:

Cá xương, chim, thú, cá sấu không có sự pha trộn máu giàu O2 và máu giàu CO2 ở tim vì tim cá có 2 ngăn, tim các loài chim, thú, cá sấu có 4 ngăn

Answer:

an element with a small number of valence electrons and a large atomic radius

Answer:

1) Motion of air mass moving from equator northward (closer to earth axis)

2) Motion of object in orbit

3) Collision of 2 objects

4) Skater changing rotation by extension of arms

5) Motion of rocket due to velocity of expelled gas

Answer:

4m/s4

Explanation:

Answer:

divide the density of solution by density of water

EXPLANATION:

LIKE:

1.25÷1000kgm-3

Answer:

Alkali metals

Explanation:

Elements in this group are highly reactive, soft, lustrous and highly conductive.

An element which is highly conductive, highly reactive, soft, and lustrous is most likely an alkali metal.

Alkali metals are in group 1 of the Periodic table which means that they have only a single valence electron.

This causes them to be soft and highly reactive because:

Examples of alkali electrons include:

In conclusion therefore, alkali metals are highly reactive and soft and so the element described above is most likely an alkali metal.

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Answer:

Distance = 6.667 kilometres

Explanation:

Given the following data;

Speed = 20 km/h

Departure time = 7:00

Arrival time = 7:20

Time taken = 20 minutes

To calculate the distance travelled from home to school;

First of all, we would have to convert the value of time in minutes to hours.

Conversion:

60 minutes = 1 hour

20 minutes = X hours

Cross-multiplying, we have;

X = 20/60 = 1/3 hours

Mathematically, the distance travelled by an object is calculated by using the formula;

Distance = speed * time

Distance = 20 * 1/3

Distance = 20/3 =

Distance = 6.667 kilometres

Chlorine has the smallest atomic radius since the atomic radius decreases as you travel to the right and up

Answer:

The tendency of undisturbed objects to stay at rest or to keep moving with the same velocity is called inertia. This is why, the first law of motion is also known as the law of inertia.

Example.: A ball at rest on the ground continues to be at rest unless someone kicks it or any external force acts on it.

Explanation:

I hope this will help you buddy

Answer:

A lunar eclipse is when the Earth passes between the moon and the sun, casting a shadow on the moon. This can only occur when the sun, Earth and moon are aligned exactly, or very closely so, with the Earth in the middle.

Answer:

I. Period = 1.3 seconds

II. Frequency = 0.769 Hertz

Explanation:

Given the following data;

Number of oscillation = 240 cycles

Time = 5.2 minutes.

Conversion:

1 minute = 60 seconds

5.2 minutes = X seconds

X = 60 * 5.2

X = 312 seconds

To find the following;

I. Period

Mathematically, the number of oscillation of a pendulum is given by the formula;

[tex] Number \; of \; oscillation = \frac {Time}{Period} [/tex]

Making period the subject of formula, we have;

[tex] Period = \frac {Time}{Number \; of \; oscillation} [/tex]

Substituting into the formula, we have;

[tex] Period = \frac {312}{240} [/tex]

Period = 1.3 seconds

II. Frequency

[tex] Frequency = \frac {1}{Period} [/tex]

Substituting the values into the formula, we have;

[tex] Frequency = \frac {1}{1.3} [/tex]

Frequency = 0.769 Hertz

Answer:

a) The velocity of the object as a function of time, v(t) is 2·b·t

b) The acceleration of the function of time, a(t) is 2·b

c) The time at which the object is at the flagpole is t = √(c/b)

Explanation:

The function that gives the position of the object north of the flagpole, x(t) is presented as follows;

x(t) = b·t² - c (b and c are constants)

a) The velocity of the object as a function of time, v(t), is derived as follows

v(t) = x'(t) = d(b·t² - c)/dt = 2·b·t

The velocity of the object as a function of time, v(t) = 2·b·t

b) The acceleration of the function of time, a(t) = v'(t) = d(2·b·t)/dt = 2·b

c) The time at which the object is at the flagpole is given by the x-intercept of the function, where x(t) = 0, as follows;

At the x-intercept, we have, x(t) = 0 and x(t) = b·t² - c

∴ 0 = b·t² - c, which gives

b·t² = c

t² = c/b

t = ±√(c/b), we reject the negative value to get;

The time at which the object is at the flagpole, t = √(c/b).

Answer:

không có điện

Explanation:

Answer:

a. Acceleration, a = 1.47 * 10^{-9} m/s²

b. Mass = 4.57 * 10^{10} kilograms

c. Force = 67.12 Newton

Explanation:

Given the following data;

Distance = 93 cm to meters = 93/100 = 0.93 meters

Weight = 67 N

Time = 9.89 hours to seconds = 35604 seconds

Initial velocity = 0 m/s (since it's starting from rest)

Acceleration due to gravity, g = 9.8 m/s

a. To find the acceleration, we would use the second equation of motion;

[tex] S = ut + \frac{1}{2} at^{2} [/tex]

Where;

S is the distance covered or displacement of an object.

u is the initial velocity.

a is the acceleration.

t is the time.

Substituting the values into the equation, we have;

[tex] 0.93 = 0*35604 + \frac{1}{2} * a*35604^{2} [/tex]

[tex] 0.93 = 0 + \frac{1}{2} * 1267644816a [/tex]

[tex] 0.93 = 633822408a [/tex]

[tex] Acceleration, a = \frac{0.93}{633822408} [/tex]

Acceleration, a = 1.47 * 10^{-9} m/s²

b. To find the mass

Weight = mass * acceleration due to gravity

67 = mass * 1.47 * 10^{-9}

[tex] Acceleration, a = \frac{67}{1.47 * 10^{-9}} [/tex]

Mass = 4.57 * 10^{10} kilograms

c. To find the force;

Force = mass * acceleration

Force = 4.57 * 10^{10} * 1.47 * 10^{-9}

Force = 67.12 Newton