A car accelerates uniformly from rest and reaches a speed of 9.9 m/s in 11.4 s. The diameter of a tire is 86.9 cm. Find the number of revolutions the tire makes during this motion, assuming no slipping. Answer in units of rev.

Answers

Answer 1

Answer:

Number of revolutions = 20.71 rev.

Explanation:

Given the following data;

Initial speed, u = 0m/s

Final speed, v = 9.9m/s

Time, t = 11.4secs

Diameter = 86.9cm to meters = 86.9/100 = 0.869m

To find the acceleration;

Acceleration, a = (v - u)/t

Acceleration, a = (9.9 - 0)/11.4

Acceleration, a = 9.9/11.4

Acceleration, a = 0.87m/s²

Now we would find the distance covered by the tire using the second equation of motion.

S = ut + ½at²

S = 0(11.4) + ½*0.87*11.4²

S = 0 + 0.435*129.96

S = 56.53m

The circumference of the tire is calculated using the formula;

Circumference = 3.142 * diameter

Circumference = 3.142 * 0.869

Circumference = 2.73m

Number of revolutions = distance/circumference

Number of revolutions = 56.53/2.73

Number of revolutions = 20.71 rev.

Therefore, the number of revolutions the tire makes during this motion is 20.71 revolutions.


Related Questions

When catching a baseball, a catcher's glove moves by 11 cm along the line of motion of the ball. If the baseball exerts a force of 464 N on the glove, how much work is done by the ball

Answers

Answer:

51.04 joules

Explanation:

The movement by the gloves = 11cm

Force by the baseball = 464N

First of all we are going to convert cm to metres

11cm = 11/100 meters = 0.11m

The formula for workdone is given as:

W = f x d

W = workdone

F = force

D = distance

Workdone = 464 x 0.11

= 51.04 joules

The workdone by the ball is 51.04 joules

Thank you

A projectile is launched with an initial velocity of
200 meters per second at an angle of 30° above the
horizontal. What is the magnitude of the vertical
component of the projectile's initial velocity by?

(1) 200 m/s x cos 30°
(2) 200 m/s X sin 30°
(3) (200 m/s)/(cos 30 °)
(4) (200 m/s)/(sin 30 °)

Answers

i believe the answer is (3)

The magnitude of the vertical component of the projectile's initial velocity is 200 m/s × sin 30°.

The diagrammatic representation of the velocity of the projectile can be seen in the attached image below.

From the diagram, let consider the ΔOAP where Vector OP makes an ∠θ = 30° to the horizontal x-axis.

where;

|OP| = magnitude of projectile velocity|OA| = magnitude of the horizontal component|OB|/|AP| = vertical component of the projectile

Using trigonometric approach for ΔOAP;

[tex]\mathbf{sin\theta = \dfrac{AP}{OP}}[/tex]

[tex]\mathbf{AP =OP\times sin \theta}}[/tex]

AP = 200 × sin 30°

Learn more about projectile here:

https://brainly.com/question/20689870?referrer=searchResults

A hockey puck slides across the ice and eventually comes to a stop. Why did the puck stop?

Answers

Answer: the total energy of the puck, ice surface, and surrounding air decreases to zero

Explanation:

A hockey puck slides across the ice and eventually comes to a stop because of friction between surface of the puck and ice surface.

What is Friction ?

Friction is a resistance to motion of the object. for example, when a body slides on horizontal surface in positive x direction, it has friction in negative x direction and that measure of friction is a frictional force. frictional force is directly proportional to the Normal(N). i.e. [tex]F_{fri}[/tex] ∝ N

[tex]F_{fri}[/tex] = μN where μ is called as coefficient of the friction. It is a dimensionless quantity.

When a body is kept on horizontal surface, its normal will be straight upward which is reaction of mg. i.e. N=mg.

Frictional force is equal to

[tex]F_{fri}[/tex] = μmg

When hockey puck slides across the ice, friction between surface of puck and surface of ice produces resistance to the motion of the puck, due to resistance puck slow down slowly and eventually come to a stop. Motion and frictional force are opposite to each other. we can calculated the exact value of frictional force when we know the coefficient of friction between puck and ice surface.

hence due to friction, puck come to a stop

To know more about frictional force, click :

https://brainly.com/question/30280752

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A student is transmitting sound waves through various materials. Through which metal in the table will the sound waves travel the fastest? Aluminum Copper Lead Brass (70% Cu, 30% Zn) 6,420 4,700 5,010 1,960 Speed of sound (m/s) O
A. Brass O
B. Aluminum
C. Copper O
D. Lead​

Answers

Answer:

Aluminum

Explanation:

promise

give  me brainlest plleaseee

Answer:aluminum

Explanation:

A plane drops a package for delivery. The plane is flying horizontally at a speed of 120\,\dfrac{\text m}{\text s}120 s m 120, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction, and the package travels 255\,\text m255m255, start text, m, end text horizontally during the drop. We can ignore air resistance.

Answers

Answer:

-22.1

Explanation:

1 / 4

Step 1. List horizontal (xxx) and vertical (yyy) variables

xxx-direction yyy-direction

t=\text?t=?t, equals, start text, question mark, end text t=\text?t=?t, equals, start text, question mark, end text

a_x=0a  

x

​  

=0a, start subscript, x, end subscript, equals, 0 a_y=-9.8\,\dfrac{\text m}{\text s^2}a  

y

​  

=−9.8  

s  

2

 

m

​  

a, start subscript, y, end subscript, equals, minus, 9, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction

\Delta x=255\,\text mΔx=255mdelta, x, equals, 255, start text, m, end text \Delta y=\text ?Δy=?delta, y, equals, start text, question mark, end text

v_x=v_{0x}v  

x

​  

=v  

0x

​  

v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript v_y=?v  

y

​  

=?v, start subscript, y, end subscript, equals, question mark

v_{0x}=120\,\dfrac{\text m}{\text s}v  

0x

​  

=120  

s

m

​  

v, start subscript, 0, x, end subscript, equals, 120, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction v_{0y}=0v  

0y

​  

=0v, start subscript, 0, y, end subscript, equals, 0

Note that there is no horizontal acceleration, so v_x=v_{0x}v  

x

​  

=v  

0x

​  

v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript. The time is the same for the xxx and yyy directions.

Also, the package has no initial vertical velocity.

Our yyy-direction variable list has too many unknowns to solve for \Delta yΔydelta, y directly. Since both the yyy- and xxx-directions have the same time ttt and horizontal acceleration is zero, we can solve for ttt from the xxx-direction motion by using equation:

\Delta x=v_xtΔx=v  

x

​  

tdelta, x, equals, v, start subscript, x, end subscript, t

Once we know ttt, we can solve for \Delta yΔydelta, y using the kinematic equation that does not include the unknown variable v_yv  

y

​  

v, start subscript, y, end subscript:

\Delta y=v_{0y}t+\dfrac {1}{2}a_yt^2Δy=v  

0y

​  

t+  

2

1

​  

a  

y

​  

t  

2

delta, y, equals, v, start subscript, 0, y, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, start subscript, y, end subscript, t, squared

Hint #22 / 4

Step 2. Find ttt from horizontal variables

\begin{aligned}\Delta x&=v_xt \\\\ t&=\dfrac{\Delta x}{v_{0x}} \\\\ t&=\dfrac{255\,\text m}{120\dfrac{\text m}{\text s}} \\\\ &=2.125\,\text s \end{aligned}  

Δx

t

t

​  

 

=v  

x

​  

t

=  

v  

0x

​  

 

Δx

​  

 

=  

120  

s

m

​  

 

255m

​  

 

=2.125s

​  

 

Hint #33 / 4

Step 3. Find \Delta yΔydelta, y using ttt

Using ttt to solve for \Delta yΔydelta, y gives:

\begin{aligned}\Delta y&=v_{0y}t+\dfrac{1}{2}a_yt^2 \\\\ &=\cancel{ (0 )t}+\dfrac{1}{2}\left (-9.8\dfrac{\text m}{\text s^2}\right )\left(2.125\,\text s\right)^2 \\\\ &=-22.1\,\text m \end{aligned}  

Δy

​  

 

=v  

0y

​  

t+  

2

1

​  

a  

y

​  

t  

2

 

=  

(0)t

​  

+  

2

1

​  

(−9.8  

s  

2

 

m

​  

)(2.125s)  

2

 

=−22.1m

​  

 

Hint #44 / 4

The correct answer is -22.1\,\text m−22.1mminus, 22, point, 1, start text, m, end text.

The volume of a solid object tends to increase as its temperature increases. Which quantities determine how large the change in volume is

Answers

Answer:

I. The change in temperature of the object.

II. The initial volume of the object.

Explanation:

Generally, the volume of a solid object tends to increase as its temperature increases and this phenomenon is known as thermal expansion.

Hence, the quantities which determine how large the change in volume of a solid object is includes;

I. The change in temperature of the object.

II. The initial volume of the object.

This ultimately implies that, when a solid object is heated, the atoms of the object vibrate rapidly about their fixed points and thus, causing an increase in the volume of the object.

In conclusion, this scientific phenomenon known as thermal expansion is valid and true for all the three (3) states of matter;

Solid. Liquid. Gas.

4. Friction is required for :
(a) Fast movement
(0) Both of the above
(b) Stopping the objec
(d) None of the above​

Answers

Both of the above! :)

Answer:

b) stopping the object

Explanation:

Friction always slows a moving object down. ... Friction can be a useful force because it prevents our shoes slipping on the pavement when we walk and stops car tyres skidding on the road. When you walk, friction is caused between the tread on shoes and the ground. This friction acts to grip the ground and prevent sliding

which of the following are not units used to measure energy?
a. joules
b. newtons
c. BTU
d. calories​

Answers

Answer:

The BTU, or British thermal unit, is actually a measure of heat.

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