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3 years ago

13

What is the change in its velocity, v, during this 0.80-s interval?

1 answer:

lyudmila [28]3 years ago

7 0

This link should help you out https://quizlet.com/40330134/biomechanics-questions-flash-cards/

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Which explanation describes the forces involved for a person to walk down the sidewalk?

Svet_ta [14]

The person walking down the sidewalk follows the newton's third law of motion.

Explanation:

A person is able to walk down the sidewalk by using the reaction forces from the ground.
In simple term, feet pushes the ground and the reaction forces makes the feet able to walk.
Another important force included in the walking mechanism is friction. With out friction one cannot walk down the sidewalks.
Hence the forces involved in the walking of a person down the sidewalk are:

Friction force
Action and reaction force between ground and person's feet.

4 0

3 years ago

A 2.43ug particle moves at 1.97 x 108 m/s. What is its momentum?

Ira Lisetskai [31]

Answer:

0.48 kgm/s

Explanation:

$m$ = mass of the particle = 2.43 μg = 2.43 x 10⁻⁶ x 10⁻³ kg = 2.43 x 10⁻⁹ kg

$v$ = velocity of the particle = 1.97 x 10⁸ m/s

$p$ = momentum of the particle

momentum of the particle is given as

$p = m v$

inserting the values

$p = (2.43\times 10^{-9})(1.97\times 10^{8})$

$p = 0.48$ kgm/s

4 0

4 years ago

A motor keep a Ferris wheel (with moment of inertia 6.97 × 107 kg · m2 ) rotating at 8.5 rev/hr. When the motor is turned off, t

Talja [164]

Answer:

$P = 133.13 Watt$

Explanation:

Initial angular speed of the ferris wheel is given as

$\omega_i = 2\pi f$

$\omega_i = 2\pi(8.5/3600)$

$\omega_i = 0.015 rad/s$

final angular speed after friction is given as

$\omega_f = 2\pi f$

$\omega_f = 2\pi(7.5/3600)$

$\omega_f = 0.013 rad/s$

now angular acceleration is given as

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

$\alpha = \frac{0.015 - 0.013}{15}$

$\alpha = 1.27 \times 10^{-4} rad/s^2$

now torque due to friction on the wheel is given as

$\tau = I \alpha$

$\tau = (6.97 \times 10^7)(1.27 \times 10^{-4})$

$\tau = 8875.3 N m$

Now the power required to rotate it with initial given speed is

$P = \tau \omega$

$P = 8875.3 \times 0.015$

$P = 133.13 Watt$

8 0

3 years ago

Two cars are facing each other. Car A is at rest while car B is moving toward car A with a constant velocity of 20 m/s. When car

lapo4ka [179]

Answer:

Let's define t = 0s (the initial time) as the moment when Car A starts moving.

Let's find the movement equations of each car.

A:

We know that Car A accelerations with a constant acceleration of 5m/s^2

Then the acceleration equation is:

$A_a(t) = 5m/s^2$

To get the velocity, we integrate over time:

$V_a(t) = (5m/s^2)*t + V_0$

Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:

$V_a(t) = (5m/s^2)*t$

To get the position equation we integrate again over time:

$P_a(t) = 0.5*(5m/s^2)*t^2 + P_0$

Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:

$P_a(t) = 0.5*(5m/s^2)*t^2$

Now let's find the equations for car B.

We know that Car B does not accelerate, then it has a constant velocity given by:

$V_b(t) =20m/s$

To get the position equation, we can integrate:

$P_b(t) = (20m/s)*t + P_0$

This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:

$P_b(t) = (20m/s)*t + 100m$

Now we can answer this:

1) The two cars will meet when their position equations are equal, so we must have:

$P_a(t) = P_b(t)$

We can solve this for t.

$0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0$

This is a quadratic equation, the solutions are given by the Bhaskara's formula:

$t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)} }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}$

We only care for the positive solution, which is:

$t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s$

Car A reaches Car B after 11.48 seconds.

2) How far does car A travel before the two cars meet?

Here we only need to evaluate the position equation for Car A in t = 11.48s:

$P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m$

3) What is the velocity of car B when the two cars meet?

Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s

4) What is the velocity of car A when the two cars meet?

Here we need to evaluate the velocity equation for Car A at t = 11.48s

$V_a(t) = (5m/s^2)*11.48s = 57.4 m/s$

7 0

3 years ago

A shopper walks eastward 3.2 meters and then westward

Musya8 [376]

11.347 yards or 34.1 ft

4 0

3 years ago