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Rudik [331]
3 years ago
6

If a rock is thrown upward on the planet mars with a velocity of 11 m/s, its height (in meters) after t seconds is given by h =

11t − 1.86t2. (a) find the velocity of the rock after two seconds. 3.56 m/s (b) find the velocity of the rock when t =
a. 11−3.72a m/s (c) when will the rock hit the surface? (round your answer to one decimal place.) t = s (d) with what velocity will the rock hit the surface? m/s
Physics
1 answer:
Butoxors [25]3 years ago
3 0
(a) 3.56 m/s 
(b) 11 - 3.72a 
(c) t = 5.9 s 
(d) -11 m/s  
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule. 
y = 11t - 1.86t^2 
y' = 11 - 3.72t 
Now that you have the first derivative, it will give you the velocity as a function of t. 
(a) Velocity after 2 seconds. 
y' = 11 - 3.72t 
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56 
So the velocity is 3.56 m/s  
(b) Velocity after a seconds. 
y' = 11 - 3.72t 
y' = 11 - 3.72a  
So the answer is 11 - 3.72a  
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.  
(d) Plug in the value of t calculated for (c) into the velocity function, so: 
y' = 11 - 3.72a
 y' = 11 - 3.72*5.913978495
 y' = 11 - 22
 y' = -11 
 So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
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A billiard ball is moving in the x-direction at 30.0 cm/s and strikes another billiard ball moving in the y-direction at 40.0 cm
____ [38]

Answer:

53.13 °

Explanation:

In order to do this, we just need to apply the following:

tanα = Dy/Dx

Where:

Vy: speed of the ball in the y axis.

Vx: speed of the ball in the x axis.

At this point we do not need the speed of the first ball after the collision because in that moment is already heading in the direction that we are looking for. Therefore, we just need to use the innitial data to calculate the direction which the first ball will go.

According to this, then:

tanα = (40/30)

tanα = 1.3333

α = tan⁻¹(1.3333)

<h2>α = 53.13°</h2>

This means that the final direction of the first ball is 53.13° and in the x axis because the starting momentum of this ball in the x axis has not dissapeared.

Hope this helps

6 0
2 years ago
uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station
Alex787 [66]

Answer:

The magnitude of the tangential velocity is v= 0.868 m/s

The magnitude of the resultant acceleration at that point is  a = 4.057 m/s^2

Explanation:

From the question we are told that

      The mass of the uniform disk is m_d = 40.0kg

       The radius of the uniform disk is R_d = 0.200m

       The force applied on the disk is F_d = 30.0N

Generally the angular speed i mathematically represented as

             w = \sqrt{2 \alpha  \theta}

Where \theta is the angular displacement given from the question as

           \theta  = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}

                 =1.257\  rad

   \alpha is the angular acceleration which is mathematically represented as

                    \alpha = \frac{torque }{moment \ of  \ inertia}  = \frac{F_d * R_d}{I}

    The moment of inertial is mathematically represented as

                     I = \frac{1}{2} m_dR^2_d

Substituting values

                    I = 0.5 * 40 * 0.200^2

                        = 0.8kg \cdot m^2

Considering the equation for angular acceleration

               \alpha = \frac{torque }{moment \ of  \ inertia}  = \frac{F_d * R_d}{I}

Substituting values

               \alph\alpha = \frac{(30.0)(0.200)}{0.8}

                   = 7.5 rad/s^2

Considering the equation for angular velocity

    w = \sqrt{2 \alpha  \theta}

Substituting values

     w =\sqrt{2 * (7.5) * 1.257}

         = 4.34 \ rad/s

The tangential velocity of a given point on the rim is mathematically represented as

                 v = R_d w

Substituting values

                    = (0.200)(4.34)

                     v= 0.868 m/s

The radial acceleration at hat point  is mathematically represented as

            \alpha_r = \frac{v^2}{R}

                  = \frac{0.868^2}{0.200^2}

                 = 3.7699 \ m/s^2

The tangential acceleration at that point is mathematically represented as

               \alpha _t = R \alpha

Substituting values

           \alpha _t = (0.200) (7.5)

                 = 1.5 m/s^2

The magnitude of resultant acceleration at that point is

                 a = \sqrt{\alpha_r ^2+ \alpha_t^2 }

Substituting values

                a = \sqrt{(3.7699)^2 + (1.5)^2}

                   a = 4.057 m/s^2

         

7 0
3 years ago
A spring has a natural length of 28.0 cm. If a 23.0-N force is required to keep it stretched to a length of 36.0 cm, how much wo
krek1111 [17]

Answer:

0.23 J

Explanation:

k*(36 - 28) = 23

so k = 23/8 N/cm

W = k(32 - 28)²/2 = 23/8 * 4²/2 = 23 N-cm = 0.23 J

4 0
2 years ago
How are mixtures and pure substances related?
Elden [556K]
You are talking about make sure's and pearl substance I thought you was talking about mix in with something
4 0
3 years ago
Help pls i need this right now
pantera1 [17]

Answer:

The x-component of F_{3} is 56.148 newtons.

Explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O (1)

Where:

\vec F_{1}, \vec F_{2}, \vec F_{3} - External forces exerted on the ring, measured in newtons.

\vec O - Vector zero, measured in newtons.

If we know that \vec F_{1} = (70.711,70.711)\,[N], \vec F_{2} = (-126.859, 46.173)\,[N], F_{3} = (F_{3,x},F_{3,y}) and \vec O = (0,0)\,[N], then we construct the following system of linear equations:

\Sigma F_{x} = 70.711\,N - 126.859\,N +F_{3,x} = 0\,N (2)

\Sigma F_{y} = 70.711\,N + 46.173\,N+F_{3,y} = 0\,N (3)

The solution of this system is:

F_{3,x} = 56.148\,N, F_{3,y} = -116.884\,N

The x-component of F_{3} is 56.148 newtons.

5 0
2 years ago
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