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Kaylis [27]
3 years ago
14

A tourist stands at the top of the Grand Canyon, holding a rock, overlooking the valley

Physics
1 answer:
kkurt [141]3 years ago
8 0

Answer:

a. Vf = 39.24 [m/s]

b. Vf = 31.24 [m/s]

c. Vf = 47.24 [m/s]

Explanation:

To solve this problem we can use the following equation of kinematics. We have to keep in mind that the gravitational acceleration acts downwards, therefore when the rock falls towards the abyss it has the same direction of the acceleration and that is why the gravitational acceleration has a positive sign in the equation.

v_{f}=v_{o}+g*t

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0 (when the rock is dropped)

g = gravitational acceleration = 9.81 [m/s²]

t = time = 4 [s]

a.

v_{f}=0+9.81*4\\v_{f}= 39.24 [m/s]

b.

In this particular situation, the acceleration will be taken as negative because the gravity is pointing in the opposite direction of the movement of the rock.

v_{f}=8-(9.81*4)\\v_{f}=-31.24[m/s]

The negative sign in the answer tells us that the rock no longer moves up instead it does downwards when 4 seconds have passed.

c.

v_{f}=8+(9.81*4)\\v_{f}=47.24[m/s]

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v = omega . r

which means it's directly proportional to radius...

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4 years ago
A motor produces less mechanical energy than the energy it uses because the motor?
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<span>A motor produces less mechanical energy than the energy it uses because the motor looses some energy to heat.</span>
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If a car is traveling at a velocity of +10m/s suddenly turns around and continues at the same speed, what is its new velocity?
Lady_Fox [76]

Answer:

-10m/s

Explanation:

only the magnitude of the velocity will change due to the change in direction of the car.

6 0
3 years ago
Consider the following distribution of objects: a 2.00-kg object with its center of gravity at (0, 0) m, a 2.20-kg object at (0,
adelina 88 [10]

Answer:

body position 4 is (-1,133, -1.83)

Explanation:

The concept of center of gravity is of great importance since in this all external forces are considered applied, it is defined by

               x_cm = 1 /M   ∑ x_{i} m_{i}

               y_cm = 1 /M   ∑ y_{i} mi

Where M is the total mass of the body, mi is the mass of each element

give us the mass and position of this masses

body 1

m1 = 2.00 ka

x1 = 0 me

y1 = 0 me

body 2

m2 = 2.20 kg

x2 = 0m

y2 = 5 m

body 3

m3 = 3.4 kg

x3 = 2.00 m

y3 = 0

body 4

m4 = 6 kg

    x4=?

   y4=?

mass center position

x_cm = 0

y_cm = 0

let's apply to the equations of the initial part

X axis

    M = 2.00 + 2.20 + 3.40

    M = 7.6 kg

    0 = 1 / 7.6 (2 0 + 2.2 0 + 3.4 2 + 6 x4)

     x4 = -6.8 / 6

     x4 = -1,133 m

Axis y

    0 = 1 / 7.6 (2 0 + 2.20 5 +3.4 0 + 6 y4)

    y4 = -11/6

    y4 = -1.83 m

body position 4 is (-1,133, -1.83)

7 0
3 years ago
where σ(t) and σ(0) represents the time-dependent and initial (i.e., time =0) stresses, respectively, and t and τ denote elapsed
lesya [120]

Answer:

E_r(6)=4.35614\ MPa

Explanation:

\epsilon = Strain = 0.49

\sigma _0 = 3.1 MPa

At t = Time = 32 s \sigma = 0.41 MPa

\tau = Time-independent constant

Stress relation with time

\sigma=\sigma _0exp\left(-\frac{t}{\tau}\right)

at t = 32 s

0.41=3.1exp\left(-\frac{32}{\tau}\right)\\\Rightarrow exp\left(-\frac{32}{\tau}\right)=\frac{0.41}{3}\\\Rightarrow -\frac{32}{\tau}=ln\frac{0.41}{3}\\\Rightarrow \tau=-\frac{32}{ln\frac{0.41}{3}}\\\Rightarrow \tau=16.0787\ s

The time independent constant is 16.0787 s

E_{r}(t)=\frac{\sigma(t)}{\epsilon_0}

At t = 6

\\\Rightarrow E_{r}(6)=\frac{\sigma(6)}{\epsilon_0}

From the first equation

\sigma(t)=\sigma _0exp\left(-\frac{t}{\tau}\right)\\\Rightarrow \sigma(6)=3.1exp\left(-\frac{6}{16.0787}\right)\\\Rightarrow \sigma(6)=2.13451

E_r(6)=\frac{2.13451}{0.49}\\\Rightarrow E_r(6)=4.35614\ MPa

E_r(6)=4.35614\ MPa

6 0
3 years ago
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