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velikii [3]
3 years ago
14

g The tires of a car make 75 revolutions as the car reduces its speed uniformly from 91 km/h to 48 km/h . The tires have a diame

ter of 0.78 m . Part A What was the angular acceleration of the tires
Physics
1 answer:
aleksandrvk [35]3 years ago
5 0

Answer:

   α = -0.01625 rad / s²

Explanation:

This is an exercise in angular kinematics, we can use the relation

         w = w₀ + 2 α θ

linear and angular variables are related

        v = w r

         w = v / r

Let's reduce the magnitudes to the SI system

         v₀ = 91 km / h (1000m / 1km) (1h / 3600s) = 25.278 m / s

         v = 48 km / h = 13,333 m / s

         θ  = 75 rev (2π rad / 1 rev) = 471.24 rad

Let's find the angular velocities

         w₀ = v₀ / r

         w₀ = 25.278 / 0.78

         w₀ = 32,408 rad / s

         w = v / r

         w = 13.333 / 0.78

         w = 17.09 rad / s

we calculate the angular acceleration

         α = (w- w₀) / 2θ  

         α = (17.09 - 32.408) / (2 471.24)

         α = -0.01625 rad / s²

the negative sign indicates that the wheel is stopping

You might be interested in
which statements about velocity are true? check all that apply. a. for velocity, you must have a number, a unit, and a direction
satela [25.4K]
a). for velocity, you must have a number, a unit, and a direction.
Yes.  This one isn't bad.  The 'number' and the 'unit' are the speed.

b). the si units for velocity are miles per hour.
No.  That's silly. 
'miles' is not an SI unit, and 'miles per hour'
is only a speed, not a velocity. 

c). the symbol for velocity is .
You can use any symbol you want for velocity, as long as
you make its meaning very clear, so that everybody knows
what symbol you're using for velocity.
But this choice-c is still wrong, because either it's incomplete,
or else it's using 'space' for velocity, which is a very poor symbol.

d). to calculate velocity, divide the displacement by time.
Yes, that's OK, but you have to remember that the displacement
has a direction, and so does the velocity.
3 0
3 years ago
Positive charge Q is distributed uniformly along the x-axis fromx=0 to x=a. A positive point charge q is located on the positive
dybincka [34]

Answer:

a. b- x= y

dx = -dy

b. F = \frac{-kQqi}{r (a+r)}

c.  F = \frac{-kQqi}{r^{2} }

Explanation:

a. x components:

dE = \frac{kdq}{(a+r-x^{2}) } \\

     = \frac{kQdx}{(a(a+r-x)^2}

Integrating and solving gives:

b- x= y

dx = -dy

b. the force is given by the equation derived from (a.):

F = \frac{-kQqi}{r (a+r)}

c. Given that r>>a, the expression becomes:

F = \frac{-kQqi}{r^{2} }

Explanation:

When the size of the charge distribution is less than the distance to the deviation point of the charge then the charge distribution would produce the same effect such as a linear charge.

6 0
3 years ago
PLEASE HELP FAST The object distance for a convex lens is 15.0 cm, and the image distance is 5.0 cm. The height of the object is
IgorLugansk [536]

Answer:

The image height is 3.0 cm

Explanation:

Given;

object distance, d_o = 15.0 cm

image distance, d_i = 5.0 cm

height of the object, h_o = 9.0 cm

height of the image, h_i = ?

Apply lens equation;

\frac{h_i}{h_o} = -\frac{d_i}{d_o}\\\\ h_i = h_o(-\frac{d_i}{d_o})\\\\h_i = -9(\frac{5}{15} )\\\\h_i = -3 \ cm

Therefore, the image height is 3.0 cm. The negative values for image height indicate that the image is an inverted image.

4 0
3 years ago
already did the work, i just need someone to see if i did the tangent line right? the line has to touch the point 0.6! thank you
Dmitrij [34]

The tangent looks good.

The curve is a bit crooked, at the 0.9 and 1.

But overall, cool graph.

5 0
2 years ago
1 2 3 4 5 6 7 8 9 10
7nadin3 [17]

Answer:

<h2>C. <u>0.55 m/s towards the right</u></h2>

Explanation:

Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Momentum = Mass (M) * Velocity(V)

BEFORE COLLISION

Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s

Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s

AFTER COLLISION

Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s

<u>x is the final velocity of the 0.25kg ball</u>

Momentum of 0.15kg body moving at 0.75m/s(body at rest) =

0.15 * 0.75kgm/s = 0.1125 kgm/s

Using the law of conservation of momentum;

0.25+0 = 0.25x + 0.1125

0.25x = 0.25-0.1125

0.25x = 0.1375

x = 0.1375/0.25

x = 0.55m/s

Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at <u>0.55 m/s towards the right</u>

<u></u>

4 0
2 years ago
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