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34kurt
3 years ago
15

Thinking Mathematically: Explore the quantitative dependencies of the acceleration upon the speed and the radius of curvature. T

hen answer the following questions. a. For the same speed, the acceleration of the object varies _____________ (directly, inversely) with the radius of curvature. b. For the same radius of curvature, the acceleration of the object varies _____________ (directly, inversely) with the speed of the object. c. As the speed of an object is doubled, the acceleration is __________________ (one-fourth, one-half, two times, four times) the original value. d. As the speed of an object is tripled, the acceleration is __________________ (one-third, one-ninth, three times, nine times) the original value. e. As the radius of the circle is doubled, the acceleration is __________________ (one-fourth, one-half, two times, four times) the original value. f. As the radius of the circle is tripled, the acceleration is __________________ (one-third, one-ninth, three times, nine times) the original value.
Physics
1 answer:
solong [7]3 years ago
6 0

The expression for the centripetal acceleration allows to find the results for the questions are:

A) The acceleration varies INVERSELY with the radius of curvature.

B) The acceleration varies DIRECTLY with the speed.

C) The acceleration becomes four times greater.

D) The acceleration becomes nine times greater.

E) The acceleration is reduced to half.

F) The acceleration is reduced to a third.

In circular motion there must be an acceleration towards the center of the circle, it is called cenripetal acceleration, in this case all the energy supplied to the system is used to change the direction of the speed even when its magnitude remains constant.

        a_c = \frac{v^2}{r}  

Where a_c the centripetal acceleration, v is the speed and r the radius of curvature of the circle.

Now we can answer the questions about centripetal acceleration.

A) For the same speed, the acceleration varies INVERSELY with the radius of curvature.

B) For the same radius of curvature the acceleration varies DIRECTLY with the speed.

C) The speed is doubled

         v = 2 v₀

         a_c = \frac{(2v_o)^2 }{r}  

         a_c = 4 \ \frac{v_o^2 }{r}  

The acceleration becomes four times greater than the original value

D) The speed is tripled

         v = 3 v₀

         a_c = 9 \frac{ v_o^2}{r}  

Acceleration becomes nine times greater than the original

E) the radius of curvature is doubled

        r = 2 r₀  

        a_c = \frac{v_o^2}{2 r_o }  

        a_c = \frac{1}{2}  a_o  

Acceleration is reduced to half the original value

F) The radius of curvature is tripled

       r = 3 r₀

       a_c = \frac{v_o^2 }{3 r_o}  \\ \\a_c = \frac{1}{3} a_o  

       

The acceleration is reduced to a third of the initial one.

In conclusion using the expression for the centripetal acceleration we can find the answers for the questions are:

A) The acceleration varies INVERSELY with the radius of curvature.

B) The acceleration varies DIRECTLY with the speed.

C) The acceleration becomes four times greater.

D) The acceleration becomes nine times greater.

E) The acceleration is reduced to half.

F) The acceleration is reduced to a third.

Learn more here: brainly.com/question/6082363

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Brrunno [24]
It changes the motion as Newton's second law of motion states that a force, acting on an object, will change its velocity by changing either its speed or its direction or both. If your basketball goes rolling into the street and is hit by a bike, either the ball will change direction or its speed or both.
3 0
3 years ago
A small wooden block with mass 0.750 kg is suspended from the lower end of a light cord that is 1.72 m long. The block is initia
Ierofanga [76]

Answer:

v_{0}=319.2 m/s    

Explanation:

We need to use the momentum and energy conservation.

p_{0}}=p_{f}

mv_{0}=(m+M)V_{1}

Where:

  • m is the mass of bullet (m=0.01 kg)
  • M is the mass of the wooden (M=0.75 kg)
  • v(0) initial velocity of bullet
  • V(1) is the velocity of the combined object in the moment the bullet hist the block

Conservation of energy.

We have kinetic energy at first and kinetic and potential energy at the end.            

(1/2)(m+M)V_{1}^{2}=(1/2)(m+M)V_{2}^{2}+(m+M)gh

Here:

  • V(1) is the velocity of the combined at the initial position
  • h is the vertical height ( h = 0.800 m)

We can find V(2) using the definition of force at this point:

\Sigma F=(m+M)a_{c}=(m+M)(V_{2}^{2}/R)

T-(m+M)gcos(\theta)=(m+M)a_{c}=(m+M)(V_{2}^{2}/R)

cos(\theta) =(L-h)/L=(1.72-0.8)/1.72

\theta =57.66

Now, we can solve the equation to find V(2)

V_{2}=\sqrt{\frac{R*(T-(m+M)*g*cos(\theta))}{(m+M)}}

V_{2}=\sqrt{\frac{1.72*(4.86-(0.01+0.75)*9.81*cos(57.66))}{(0.01+0.75)}}

V_{2}=1.40 m/s        

Now we can find V(1) using the conservation of energy equation

(1/2)V_{1}^{2}=(1/2)V_{2}^{2}+gh

V_{1}=\sqrt{V_{2}^{2}+2gh}

V_{1}=\sqrt{1.40^{2}+2*9.81*0.8}          

V_{1}=4.20 m/s        

Finally, using the momentum equation we find v(0)

v_{0}=\frac{(m+M)V_{1}}{m}                

v_{0}=\frac{(0.01+0.75)*4.20}{0.01}

v_{0}=319.2 m/s        

I hope it helps you!

 

7 0
4 years ago
A steel sphere sits on top of an aluminum ring. The steel sphere (a= 1.1 x 10^-5/degrees celsius) has a diameter of 4.000 cm at
Mars2501 [29]

Answer:

C

Explanation:

To solve this question, we will need to develop an expression that relates the diameter 'd', at temperature T equals the original diameter d₀ (at 0 degrees) plus the change in diameter from the temperature increase ( ΔT = T):

d = d₀ + d₀αT

for the sphere, we were given

D₀ = 4.000 cm

α = 1.1 x 10⁻⁵/degrees celsius

we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T             EQN 1

Similarly for the Aluminium ring we have

we were given

d₀ = 3.994 cm

α = 2.4 x 10⁻⁵/degrees celsius

we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T       EQN 2

Since @ the temperature T at which the sphere fall through the ring, d=D

Eqn 1 = Eqn 2

4 + (4.4x10⁻⁵)T =3.994 + (9.58x10⁻⁵)T, collect like terms

0.006=5.18x10⁻⁵T

T=115.7K

4 0
3 years ago
A ball is kicked horizontally from a 60 meter tall cliff at 10 m/s. How far from
o-na [289]

Hi there!

We can begin by deriving the equation for how long the ball takes to reach the bottom of the cliff.

\large\boxed{\Delta d = v_it+ \frac{1}{2}at^2}}

There is NO initial vertical velocity, so:

\large\boxed{\Delta d= \frac{1}{2}at^2}}

Rearrange to solve for time:

2\Delta d = at^2\\\\t = \sqrt{\frac{2\Delta d}{g}}

Plug in the given height and acceleration due to gravity (g ≈ 9.8 m/s²)

t = \sqrt{\frac{2(60)}{(9.8)}} = 3.5 s

Now, use the following for finding the HORIZONTAL distance using its horizontal velocity:

\large\boxed{d_x = vt}\\\\d_x = 10(3.5) = \karge\boxed{35 m}

6 0
3 years ago
Can any one help plz in the question ?
kherson [118]

Yes, No, yes

mark me brainliestttt :)))

5 0
3 years ago
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